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STOICHIOMETRIC CALCULATIONS

The term stoichiometry is derived from two Greek words, 'stoicheon' which means element and 'metron' whichmeans measure. Stoichiometry essentially deals with the study of quantitative relationship among elements in acompound or in a chemical reaction. Stoichiometry hence is divided into two types.

(i) Composition stoichiometry:This deals with the mass relationship among elements in a particular compound. For example, water(H

2O) contains

the elements H and O in the mass ratio 2:16 = 1:8 and methane(CH4) contains C and H in the ratio 12:4 =3:1 and

so on. The law of definite or fixed proportion is the basis of this stoichiometry i.e there is a fixed ratio of massesamong elements in a compound.PERCENT COMPOSITION, EMPIRICAL AND MOLECULAR FORMULA

PERCENTAGE COMPOSITIONThe percentage by mass of each element present in a compound can be found out and compared. Take thesimplest case of H

2O. What is the percentage of H and O present in H

2O? If in every 18gm(M.M) of H

2O,

there are 16gms of O and 2gm of H, For 100gms of H2O, there is (16/18)X100=88.89gms of O and

(2/18)X100=11.11gms of H. So % of O = 88.89% and that of H=11.11%. Mass of an element present in100gms of the compound gives the percentage composition of that element.SAQ 1: Calculate the percent by mass of O in Ca(ClO

3)

2.

SAQ 2: Find the percent of nitrogen in (i)NH4NO

3 and (ii)(NH

4)

2SO

4 and indicate which is a better

nitrogenous fertilizer?SAQ 3: Calculate the percent composition of K

2Cr

2O

7. (K=39, Cr=52, O=16)

SAQ 4: What is the percent by mass of Cu in CuO? From this, find mass of CuO that will be required toproduce 100kg of Cu? (Cu=63.5, O=16)SAQ 5: A sample impure Cu

2O contain 66.6% Copper. What is percentage of pure Cu

2O in the sample?

EMPIRICAL FORMULA AND MOLECULAR FORMULA:Empirical formula of a compound is the simplest whole number ratio of atoms which are present in amolecule. For example glucose has the molecular formula C

6H

12O

6 but its empirical formula will be obtained

by dividing all the coefficients by the highest common factor(HCF) 6. Hence its empirical formula is CH2O.

Molecular Formula, on the other hand, represents actual number of atoms of each kind present in a molecule.Empirical formula is related to the molecular formula of a compound as follows.

Molecular Formula = (Empirical Formula)n

(Where n is a whole number)

SAQ 6: The molecular formula of a compound is C7H

14, find its empirical formula.

SAQ 7: The molecular formula of a compound is C4H

8O

2. What is its empirical formula?

Empirical formula Mass:The mass(relative)obtained from the empirical formula is called empirical formula mass. For example, theempirical formula mass of CH

2O is (12+2+16)=30, but the molecular mass is mass obtained from the

molecular formula. The molecular mass of C6H

12O

6= 72+12+96=180. So if you know the molecular mass of

a compound and its empirical formula mass, you can easily find the molecular formula. Look to this example.Example: The empirical formula of a compound is CH

2O and its molecular mass is 180, find the

molecular formula.Solution: The empirical formula mass = 12+2+16=30

We know that Molecular Formula = (Empirical Formula)n:(Where n is a whole number)

So molecular mass = n × empirical formula massn = molecular mass/empirical formula mass = 180/30 =6So molecular formula = (CH

2O)

6 = C

6H

12O

6

Determination of Empirical Formula:The empirical formula of a compound can be calculated from the percent composition of the compound. Thisis just the opposite of finding percentage composition from the molecular formula of a compound discussedbefore. The difference is that here we cannot find the molecular formula. In stead we shall get the empiricalformula which may be same or different from the actual molecular formula. This is because both empiricalformula and molecular formula have the same percent composition of elements. For example, the empiricalformula CH

2 and the molecular formula C

3H

6 have the same % of C and H by mass.

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Example: Find the empirical formula of a compound which contains 60% O and 40% S by mass.If its molecular mass is 80, what is its molecular formula.Solution: Let us take 100gms of the compound. This amount contains 60gms of O and 40gms of S.Let us calculate how many gm. atoms(mole of atoms) of each element present in these amounts and whatis their ratio. This is obtained by dividing the mass of the element by the atomic mass of the element. Theratio of gm atoms is same as ratio of the atoms present in the formula , hence it will give the empiricalformula. In this case,

the no. of gm. atoms of O= 60/16=3.75the no. of gm. atoms of S= 40/32= 1.25

So the ratio of gm. atoms and hence atoms of O and S present in the formula =3.75:1.25But you know that in the formula(empirical or molecular) there are whole number of atoms of the elements.So we have to convert it into simplest whole number ratio. This is done by dividing each value by the lowestof the values. In this case the lowest is 1.25.

So whole number ratio = 3.75/1.25 : 1.25/1.25 = 3 : 1Hence the empirical formula = O

3S or SO

3 : So its empirical formula mass=32+48=80

n= Molecular mass/empirical formula mass = 80/80=1So Molecular Formula = (Empirical Formula)

n,

So Molecular Formula =(SO3)

1=SO

3(same as its empirical formula).

STEPS FOR FINDING EMPIRICAL FORMULA:(i) First the percent composition data of the elements are divided by the respective atomic masses toget ratio of atoms present in the formula(ii) Make this ratio the simplest whole number ratio by dividing all the values by the lowest of the values.If this does not give the whole number ratio, multiply 2, 3 etc.to get the simplest whole number ratio.For example, if it comes 1:1.5:2.5 , you multiply throughout by 2 to make the ratio 2:3:5. If the ratio is1.33 : 2, then multiply by 3 to get the whole number ratio i.e 4:6 and similarly if the ratio is 1.25 :1, then multipyby 4 to get the whole number ratio i.e 5:4.(iii) Then empirical formula is written by placing these whole numbers as coefficients of the respectiveelements.(iv) Empirical formula mass is then found out. Then divide the molecular mass by the empirical formulamass to find the value of n. Usually the division does not give a whole number value to 'n'. Round it off toits nearest whole number to get the value of 'n'.(v) Multiply 'n' in all the coefficients of the empirical formula to get the molecular formula.To determine the empirical formula, it is better to calculate in tabular form. The previous example of SO

3

is given in the following table for better understanding. SO3 contains 60% S and 40% O by mass.

___________________________________________________________Element % At. Relative number Simple whole Empirical

Mass of atoms no. of atoms Formula___________________________________________________________

O 60 16 60/16=3.75 3.75/1.25=3SO

3

S 40 32 40/32=1.25 1.25/1.25=1___________________________________________________________

Example: A compound gave on analysis the following percent composition: K=26.57%,Cr=35.36%, O=38.07%.Derive the empirical formula.Solution: Let us take the help of the table for easy calculation.

___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________K 26.57 39 26.57/39=0.68 0.68/0.68=1X2= 2Cr 35.36 52 35.36/52=0.68 0.68/0.68=1X2= 2 K

2Cr

2O

7

O 38.07 16 38.07/16=2.379 2.379/0.68=3.5X2=7___________________________________________________________

Remember that fractional values like 3.95 or 3.99 which is close to a whole number is rounded off to thenearest whole number(4). But if it is 1.5, 1.33 or 2.25 etc. you cannot convert it to the nearest whole number.In that case you have to multiply a suitable minimum factor to make it a whole number.

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SAQ 8: Find the empirical formula of a hydrocarbon that on analysis gave the following percent composition:C=85.63%, H=14.37%. If the molecular mass of the compound is 56, what is its molecular formula.SAQ 9: An oxide of nitrogen contains 30.4% nitrogen. What is its empirical formula?SAQ 10:Determine the simplest formula of a compound that has Cr=26.52%, S=24.52% and O=48.96%.Could you suggested the formula in proper order i.e in terms of acid and basic radicals with proper valences?

PRACTICE QUESTIONS1. A compound contains 21.6% sodium, 33.3% chlorine and 45.1% oxygen. Derive its empirical formula.2. Calculate the empirical formula of a compound formed when 7.3g of iron powder reacts completlywith 6.30gm of powder sulphur.3. A compound has the following percent composition: C=40%, H=6.66%, O=53.34%. Its molecularmass is 60. Derive its molecular formula.4. A compound consisting of 82.66% carbon and 17.34% hydrogen and has a vapour density 29.05.Determine its molecular formula.

(ii) Reaction StoichiometryThis deals with the quantitative relationship between the reactants and products in a chemical reaction. The lawof conservation of mass is the basis of this stoichiometry. Matter can neither be created nor destroyed i.e thetotal mass of the reactants should be equal to the total mass of the products in a chemical reaction.We know that chemical equations are used to describe chemical reactions.Balancing of an equation is doneonly to satisfy the law of conservation of mass.

2H2 + O

2 2 H

2O

2 ×2 32 2 × 18In the above reaction represented by a balanced equation, the total mass of the reactants is 36gm(2 g moles ofH

2 and one g. mole of O

2) and the total mass of the product is also 36g(2 g moles of H

2O)

Example: Calculate the mass of oxygen required to completely react with 24g of hydrogen gas.Solution: 4g of H

2 completely reacts with 32g of O

2.

So 24g of H2 will completely react with (32/4) × 24 = 192g of O

2

Example: Calculate the mass of water formed when 8g of oxygen completely reacts with excess of hydrogen.How much of H

2 is required for the reaction?

Solution: 32g of O2 on complete reaction produces 36g of water

So, 8g of O2 will produce (36/32) × 8 = 9 g of water(answer)

Again, 32g of oxygen requires 4g of hydrogen gasSo, 8g of oxygen will require (4/32) × 8 = 1 g of hydrogen gas(answer)That is why we have taken excess of hydrogen gas out of which 1g will be used to completely react with 8g ofoxygen gas and the rest of the hydrogen gas will remain unreacted. If we would have taken limited quantity ofhydrogen(say less than 1 g), then all the other reactant(oxygen in this case) would not have reacted completely.CONCLUSION: If the mass of any species i.e either any one of the reactants or any one of the products isgiven, the mass of any other species in the reactant or product side can be calculated from the balancedchemical equation.This is called stoichiometric calculations.So to study stoichiometric calculations two things are essential.(i)To predict the products correctly.(ii)To balance the equation correctly.If any one of the above two becomes wrong, the stoichiometric calculation also goes wrong.We shall study reaction stoichiometry in two parts.

(1)Stoichiometric calculations involving masses or gaseous volumes of the reactants and products

(2)Stoichiometric calculations involving the concentrations of solutions like molarity, normality etc. forthe reactants

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STOICHIOMETRIC CALCULATIONS INVOLVING THE MASSES ANDGASEOUS VOLUMES OF THE REACTANTS AND PRODUCTSFor simplifying this study it is divided into three types.

(i)Mass-mass relationship(ii)Mass-volume relationship(iii)Gas analysis(Volume-volume relationship)

Mass-mass relationship is the original basis of stoichiometry. The other two are applicable for reactions involvinggases. Since mass of a gas is related to its volume at a particular temperature and pressure, we can directlystudy the relationship between mass of one species with the volume of the other species(gaseous) without usingthe mass of the latter. This is called mass-volume relatioship. Similary we can relate the volume of one gaseousspecies with the volume of another gaseous species instead of using their masses. This is called volume-volumerelationship.Mass-mass relationshipWorking steps:(i)Predict products and balance the equation correctly. For balancing redox reaction, follow ON method orother methods of your choice.(ii)Write the molecular masses for molecular substances and atomic masses for atomic substances below therespective reactants and products which are involved in stoichiometric calculation. Multiply the coefficientsfrom the balanced equation with the respective molecular masses or atomic masses.(iii)Then proceed with chemical arithmetic. Look to this example.Example: Find out the mass of KClO

3 that would produce 8 g of oxygen gas.

Solution: 2KClO3

2KCl + 3O2

2(39+35.5+48) 3 × 32g=2 × 122.5g

In this case, the species involved in the calculation are KClO3 and O

2. The mass of the latter is given while the

mass of the former is to be calclated . Since the mass of the product is given, it is to be written in the LHS whiledoing calculation by unitary method.3 ×32g of oxygen is produced by 2 ×122.5g of KClO

3

So, 8 g of oxygen is produced by (2×122.5)/(3×32) ×8 = 20.416g (answer)(Note that we had to use passive voice while constructing the sentence, since the product is written in the LHS).Example: Calculate the mass of chlorine gas obtained by the reaction of 8.7g of pure MnO

2 with excess of

dilute hydrochloric acid. Also calculate the mass of HCl consumed in the reaction.MnO

2 + 4HCl MnCl

2 + Cl

2 + 2H

2O

(55+32) 4×(1+35.5) 2×35.5

From the balanced equation, we know that one mole of MnO2 reacts with 4 moles of HCl to give one mole of

Cl2, one mole of MnCl

2 and 2 moles of H

2O.

(i)87g of MnO2(1 mole) produces 71 gm of chlorine

So, 8.7 g of MnO2 produces (71/87) × 8.7 = 7.1 gm of chlorine

(ii)Again, 87 gm of MnO2 reacts wit 4 × 36.5 gm of HCl

So 8.7 gm of MnO2 reacts with [(4×36.5)/87] × 8.7 = 14.6 gm of HCl

Alternative Method:We can use the mass of Cl

2 produced obtained in (i) to get mass of HCl required.

2 × 35.5 gm of Cl2 is produced by 4 × 36.5 gm of HCl

7.1 gm of Cl2 is produced by 14.6 gm of HCl

Hence we can use any data(either reactant or product) to find out the mass of any other reactant or product.SAQ 1: Calculate the mass of CO

2 obtained by completely burning 30gms of ethane with excess oxygen?

Also calculate the mass of H2O produced during the reaction.

SAQ 2: Calculate the number of moles and mass in gm of CaCl2 needed to react with excess of silver nitrate

to produce 6.6g of AgCl?(Ag=108, Ca=40)SAQ 3: Calculate the mass of CaO which will completely react with 6.92g of HCl?

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SAQ 4: Calculate the mass of BaCO3 produced when excess CO

2 is bubbled through a solution containing

0.205mole of Ba(OH)2. (Ba = 137)

SAQ 5: What mass of sodium bicarbonate on strong heating will produce 1.5 mole of CO2?

Mass-Volume RelationshipIn this case we shall have to establish a relationship between the mass of one species(reactant or product)with the volume of another species(reactant or product). Here we shall make use of gram molar volumeof gases at NTP equal to 22.4 litres i.e one mole of any gas at NTP will have a volume equal to 22.4 litres.

Example: What volume of O2 gas be evolved at NTP when 2gms of KClO

3 is strongly heated?

3X22.42(39+35.5+48)KClO3 KCl O2+ 322

From the balanced equation, we know that 2 moles of KClO3 produces 3 moles of O

2 gas i.e 3×22.4 litres

of O2 gas at NTP. So can we not find the volume of O

2 that will be produced by 2gms of KClO

3 from these?

So for mass-volume relationship, you have to write the molecular mass with the coefficient multiplied withit below the species whose mass data is given or asked for and the molar volume (22.4litres) multiplied withits coefficient, below the species whose volume data is given or asked for. The rest of the procedure is sameas mass-mass relationship explained earlier. In the above example,

2(39+35.5+48)=2 ×122.5 gms of KClO3 produce 3×22.4 litres of O

2 gas at NTP

So 2 gms of KClO3 must produce 0.5485l =548.5ml of O

2at NTP.

Example : What mass of sodium nitrate on heating will produce 12 litres of O2 gas at NTP?

Solution: We know that NaNO3 on heating produces NaNO

2 and O

2. Let us first write the balanced equation

and the molecular mass and molar volume of the involved species i.e NaNO3 in LHS and O

2 in RHS.

22.4 l2(23+14+48)

22 +NaNO2 O2NaNO3

22.4 litres of O2 is produced by 2 × 85gms of NaNO

3

So 12 litres of O2 will be produced by = 91.07gms of NaNO

3.

SAQ 6: What volume of chlorine gas at NTP will be produced when 8.7gms of MnO2 reacts with excess

conc. HCl? (Mn=55, O=16, Cl=35.5)SAQ 7: What mass of sodium bicarbonate on reacting with excess dil. HCl will give 5.6 litres of CO

2 gas

at NTP?SAQ 8: What volume of NO

2 and what volume of O

2 be evolved at NTP when 10 gms of lead nitrate is

strongly heated?(Pb=207, N=14)SAQ 9: What mass of ammonium dichromate on strong heating will produce 11.2 litres of nitrogen gas atNTP?(N=14, Cr=52, O=16)

Procedure when volume of gas is given/asked at temperature and pressure other than NTP:(i) When volume data is given at temperature and pressure other than NTP, first you convert the givenvolume to NTP conditions by applying the combined gas equation(P

1V

1)/T

1 = (P

2V

2)/T

2.Then establish the

mass-volume relationship in the same way as done previously.(ii) When volume data is asked in the temperature and pressure other than NTP, then the conversion ofvolume from NTP conditions to the required conditions is done at the end.

Example : What mass of calcium carbonate on heating will produce 250ml of CO2 gas at 270C and

800mm of Hg pressure?Solution: Since volume of the gas has been given at a particular temperature and pressure, our first job isto convert it to NTP conditions.

273 A760mm X V2

(273+27)A =800mm X 250ml ⇒ V

2 = 239.47ml

Then we write the balanced chemical reaction. CaCO

3 CaO + CO

2

(40+12+48) 22400ml22400ml of CO

2 is evolved at NTP from 100gm of CaCO

3

So 239.47ml of CO2 is evolved at NTP from (100/22400) × 239.47 = 1.069gm(answer)

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Example: What volume of hydrogen gas will be evolved at 4500C and 700mm pressure by treating5gms of superheated iron with sufficient steam.Solution: Since in this case the volume of hydrogen is asked, we shall carry out the volume conversion atthe end. First write down the balanced equation.

3Fe + 4H2O Fe

3O

4 + 4H

2

3 ×56 gm 4 × 22400 ml

3×56gms of iron produces 4×22400ml of hydrogen gas at NPT.So 5gms of iron will produce (4×22400)/(3×56)×5= 2666.67ml of H

2 at NTP.

But we want to know the volume at some other conditions. So we shall convert now from the NTP conditionsto the given condition by using the combined gas equation.

760mm X 2666.67ml=273A

700mm X V 2(273+ 450) A ⇒ V

2 = 7667.6ml

So the volume of hydrogen collected at 4500C and 700mm pressure is 7667.7ml.

SUCCESSIVE REACTIONSIf the product of one reaction is used as a reactant in the second reaction and you are asked to establishthe relationship between the reactant of first reaction with product of second reaction, you can do so veryeasily by writing the balanced chemical reactions for both. The following example will make it more clear.Example: What mass of potassium chlorate on heating gives just sufficient oxygen gas to completelyburn 8gms of sulphur?Solution: Let us write the first reaction

2KClO3 2KCl + 3O

2

2(39+35.5+48) 3×32The oxygen evolved in the first reaction is used to burn sulphur in the second reaction.

S + O2 SO

2

32 32Since the reactant of first reaction has been asked for, we start to solve this problem from the second reactionand know how much of oxygen is necessary to completely burn 8gms of sulphur.2nd reaction:

32gms of sulphur needs 32gms of oxygen gas for complete burning.So 8gms of sulphur will need 8gms of oxygen gas

Now let us make use of the amount of oxygen (8gms) in the first reaction and know how much KClO3 on

heating will give 8gms of oxgyen.1st reaction:

3×32gms of O2 gas is produced by 2(39+35.5+48)=2×122.5gms of KClO

3

So 8gms of O2 gas will be produced by (2×122.5)/(3×32)×8=20.416gms of KClO

3.

Hence 20.416gms of KClO3 on heating will give enough oxygen gas(8gms) to burn 8gms of sulphur completely

to SO2.

SAQ 10: What mass of ammonium sulphate is required which on reacting with excess of alkali will produceenough ammonia which can reduce 2.5 gm of cupric oxide completely to metallic copper.(Cu=63.5)

(Hint: (NH4)

2SO

4 + NaOH → Na

2SO

4 + NH

3 + H

2O

CuO + NH3 → N

2 + Cu + H

2O

SAQ 11: 4 gm of CaCO3 was strongly heated and the residue obtained was mixed with excess of

ammonium chloride and heated. Calculate the volume of ammonia produced at 270C and 800mm pressure.SAQ 12: Calculate the mass of KMnO

4 required to produce enough chlorine(by reacting with conc. HCl)

which will completely react with 10gm of pure NaOH(hot and concentrated) to produce NaCl and NaClO3.

LIMITING REACTANT CONCEPTWhen the amounts of two reactants used in a reaction are known there can be two possibilities(i) The two reactants might be consumed wholly and there would be no excess reactant left after thereaction is over. This is the case when exact stoichiometric proportions of reactants are taken.(ii) If one reactant is completely exhausted and the other reactant remains in excess after the end thereaction.If in such a case, you are asked to calculate the amount of product formed, then how you proceed? Firstyou have to check on trial and error basis if both reactants are exhausted (type i) or one reactant is exhausted

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(type ii). If it belongs to type (i), then you can take any one of the reactant data and calculate the amountof the product. However if the reaction belongs to type (ii), you have to use the limiting reactant i.e thereactant which has been exhausted completely to find the amount of the product. If you take the otherreactant which has been taken in excess for calculating the product, then you will get a wrong answer,because the excess amount of the reactant won't react as there is no equivalent amount of the other reactant.Example: 6 g of carbon was burned with 20 g of oxgyen gas, how much of CO

2 gas will be

formed? What volume of the gas will be formed at NTP.Solution:

C + O2 CO

2

12g 32g 44g ( or 22.4 l)Let us take the carbon amount:

12gm of carbon requires 32 gms of oxygen for complete burning.So 6gms of carbon must require 16gms of oxgyen for complete burning.

But 20gms of oxygen gas has been taken. So oxygen gas will remain in excess(4gms) after the end of thereaction. So carbon is the limiting reactant and we have to use the mass of carbon to find the amount ofproduct(not the mass of oxygen).

12gms of of carbon produce 44gms of CO2 gas,

So 6gms of carbon must produce (44/12)×6 =22gms of CO2 gas.

Again 12gms of carbon produces 22.4 litres of CO2 gas at NTP.

So 6gms of carbon must produce 11.2 litres of CO2 at NTP .(You can also convert mass of CO

2

directly into volume)Example : A mixture of 100 g H

2 and 100 g O

2 is ignited so that water is formed according to

the reaction, 2H2+O

2 2H

2O. How much water is formed?

Solution:Let us first take hydrogen data to check whether it is limiting reactant or excess reactant.

2H2 + O

2 2H

2O

2×2 32 2X184 gms of H

2 needs 32 gms of O

2 for complete reaction.

So 100gms of H2 must need (32/4)×100 =800gms of O

2.

But we have only 100gms of O2 in the reaction. So all H

2 cannot be used and therefore O

2 is the limiting

reactant and H2 will remains in excess after the reaction. Note that you can know which one is the limiting

reactant and which remains in excess by selecting any reactant data and then by finding the amount of theother reactant needed for it from the balanced equation. So now let us take the oxgyen data(limiting reactant)to find the amount of water formed.

32gms of O2 produce 36gms of H

2O

So 100gms of O2 must produce (36/32)×100= 112.5gms.

SAQ 13: Calculate the mass of NaCl produced by the reaction of 5.3 gm of Na2CO

3 with 5.3 gm of pure

HCl.SAQ 14: Find the mass of barium sulphate formed when 0.1 mole of barium chloride reacts with 0.05 moleof sodium phosphate. (Ba=137, P=31, O=16, Cl=35.5)

PROBLEMS BASED ON PURITY OF SAMPLE:In this case the mass of an impure sample which reacts with another reactant is given and the product amountis also given, you are asked to find the percentage of purity of the original impure sample. What you willdo in this case is to find the mass of the reactant that would produce the given quantity of the product fromthe balanced equation. In such case you will not use the mass data of the impure sample because thatamount contains impurity which does not react. After finding the mass of pure sample from stoichiometricanalysis, then find the percentage of purity.

percentage of purity = (mass of pure sample)/(mass of impure sample)×100.

Example: 4gms of an impure sample of calcium carbonate (containing sand as impurity) is treatedwith an excess of hydrochloric acid. 0.88g of CO

2 is produced. What is the percentage of pure CaCO

3

in the original sample?Solution:The impure sample taken may be lime stone, marble, chalk etc. which are the minerals containing CaCO

3.

The impure sample contains sand as impurity. Our interest is to find the mass of pure CaCO3 present in the

sample. Note that when the sample will react with acid, only CaCO3 reacts but not the impurity, according

8

to the equation.CaCO

3 + 2HCl CaCl

2 + CO

2 + H

2O

(40+12+48) 4444gms of CO

2 is produced by (40+12+48)100gms of CaCO

3

0.88gm of CO2 must be produced by (100/44)×0.88= 2gm of CaCO

3

So the amount of pure CaCO3 present in 4gm of impure sample is 2gms.

So percentage of purity = (2/4) ×100= 50SAQ 15: Calculate the amount of lime(CaO) that can be prepared by heating 200kg of limestone that is 95%pure CaCO

3?

SAQ 16: 1.0 gm sample of impure zinc on treatment with excess dil. H2SO

4 produced 250 mL of hydrogen

ver water at 200C and 780 mm pressure. Calculate the percentage of purity in the sample. (aqueous tensionat 200C = 15 mm of Hg) Zn = 65.5 (68.5)SAQ 17: A silver coin weighing 7.0 gm was completely dissolvoed in conc. HNO

3 and then excess of HCl

solution was added. The white precipitate formed was dried and weighed to be 9.0 gm. Calculate thepercentage of silver present in the coin.(Ag=108) (96.7)SAQ 18: 35.0 gm of an impure sample of potassium dichromate on complete reaction with excess of H

2S

deposited 9.6 gm of sulphur in the acidic medium. Calculate the % of purity in the sample. (Cr=52, K=39,S=32, O=16) (84)

BASED ON COMPOSITION OF A MIXTURE CONTAINING ONE REACTIVE SPECIES:This is similar to the problems based on purity discussed before. One of the components of the mixture isreactive while the other remains inert.Example: 10 gm of a mixture of KCl and KNO

3 was dissolved in water and treated with excess of AgNO

3

to produce a white precipitate weighing 14.35 g. Determine the composition of the mixture.(Ag=108, K=39)Solution:

Out of the two components present in the mixture, KCl takes part in the reaction with AgNO3 while

KNO3 remains as such without any change. The mass of KCl can be calculated by the help of the massof AgCl formed.

KCl + AgNO3 KNO3 + AgCl

(39 + 35.5) (108+ 35.5) 74.5 143.5143.5 gm of AgCl is produced by 74.5 gm of KClSo, 14.35 gm of AgCl is produced by (74.5/143.5) × 14.35 = 7.45 gm

Since the mass of the mixture is 10gm, mass of KNO3 = 10-7.45 = 2.55 gm

So percentage of KNO3 = (2.55/10) × 100 = 25.5% and percentage of KCl = 74.5%

SAQ 19: A 3.0 gm of sample of NaHCO3 and Na

2CO

3 lose 0.45 gm when heated strongly until constant

mas is attained. What is the percentage composition of the original mixture. (NaHCO3= 40.64%)

COMPOSITION OF A MIXTURE CONTAINING TWO REACTIVE SPECIES:In this case both the components in the mixture are reactive. In such case we have to solve by algebraicmethod taking the mass of one as x gm and the other as total mass minus x gm.Example: A mixture of NaCl and KCl weighing 3.0 gm was dissolved in water and treated with excess ofAgNO

3 solution. The white precipitate obtained was filtered and dried to give a constant mass of 6.3 gm.

Find the % of KCl in the mixture.Solution:In this case both NaCl and KCl react with AgNO

3 solution.

Let the mass of NaCl = x gm and hence mass of KCl = (3-x) gmNaCl + AgNO

3 NaNO

3 + AgCl

(23 + 35.5) (108+35.5)=58.5 =143.5

58.5 gm of NaCl produces 143.5 gm of AgClSo, x gm of NaCl will produce (143.5/58.5) × x gm of AgCl

KCl + AgNO3 KNO

3 + AgCl

(39+35.5) 143.5= 74.5

74.5 gm of KCl proudces 143.5 gm of AgCl

9

So, (3-x) gm of KCl will produce [143.5/74.5) × (3-x) gm of AgClTotal mass of AgCl = 6.3 = (143.5/58.5) x + [143.5/74.5) × (3-x)On solving we get x = 1 gm(mass of NaCl)So, mass of KCl = 3-1 = 2 gm, hence its percentage= (2/3) X 100 = 66.67%

SAQ 20: 1.0 gm of an alloy of Al and Mg when heated with excess of HCl forms aluminium chloride andmagnesium chloride and hydrogen. The evolved hydrogen collected over mercury at 00C has a volume of 1.2litres at 0.92 atm. pressure. Calculate the composition of the alloy. (Al = 55%) (IIT 1978)

SAQ 21: A solid mixture(5.0 gm) consisting of lead nitrate and sodium nitrate was heated strongly to constantmass. If the loss in mass is 28%, find the amount of lead nitrate and sodium nitrate in the mixture.

[Pb(NO3)

2 = 3.33 gm] (IIT 1990)

GAS ANALYSIS: (GAY LUSSAC'S LAW)1. Volume-Volume Relationship:We know from Gay Lussac's Law of combining gaseous volumes that when gaseous reactants form gaseousproducts they do so in simple ratio with respect to their volumes, provided the reaction is carried out atconstant temperature and pressure. The ratio of their volumes is equal to the ratio of the stoichiometriccoefficients in the balanced equation. Look to this example.Example : 5 litre of hydrogen gas was allowed to react completely with 5 litres of chlorine gas atthe same temperature and pressure, how many litres of hydrogen chloride gas will be obtained at thattemperature and pressure?Solution: Write the balanced equation

H2(g) + Cl

2(g) 2 HCl(g)

1 mole of H2 reacts with 1 mole of Cl

2 to give 2 moles of HCl gas. Therefore

22.4litres of H2 gas reacts with 22.4 litres of Cl

2 to give 2×22.4 litres of HCl at NTP

1 litre of H2 reacts with 1 litre of Cl

2 to give 2 litres of HCl at the same temperature and pressure.

In general,1 volume of H

2 reacts with 1 volume of Cl

2 to produce 2 volumes of HCl gas at the same temperature

and pressure. The ratio of the volumes of reactants and products is simple i.e 1:1:2. This is Gay Lussac'sLaw. One can find this ratio by looking to the coefficients in the balanced equation. But one thing you mustnot forget that this law holds good for the gaseous reactants and gaseous products and not the product whichliquifies on cooling like H

2O. This will be made more clear later.

In the above example, 5 litres of H2 reacts with 5 litres of Cl

2 to give 10 litres of HCl gas.

Example : What maximum volume of HCl gas will be obtained if 5 litres of H2 is allowed to react

with 10litres of Cl2 gas at the same temperature and pressure? Which reactant gas remains unreacted

at the end and by what volume? What is the total volume of the mixture after the reaction?Solution: H

2 + Cl

2 2 HCl

From Gay Lussac's law we know that1 volume of H

2 will reacts with 1 volume of chlorine to produce 2 volumes of HCl gas (at the same

temperature and pressure conditions)Therefore 5 litres of H

2 needs only 5 litres of Cl

2. But you are given 10litres of Cl

2 out of which only 5 litres

will be used and remaining 5 litres of Cl2 will be left excess. So H

2 is the limiting reactant and we shall use

that to find the volume of HCl gas formed.5 litres of H

2 will react with 5 litres of Cl

2 to produce 10 litres of HCl gas.

The total volume of gaseous product along with excess gaseous reactant after the end of the reaction =volume of HCl gas + volume of excess Cl

2 not used in the reaction.

Total volume = 10 litre of HCl gas + 5 litres of unreacted Cl2 gas=15 litres.

Example : 10 litres of CO were allowed to burn with excess of oxygen gas to produce CO2 gas.

What volume of CO2 gas will be obtained at the same temperature and pressure? What volume of O

2

gas needed for the purpose at the same conditions?Solution: The balanced equation for the reaction is

2CO + O2

2CO2

2 litres 1 litre 2 litres (at same temperature and pressure)2 litres of CO will produce 2 litres of CO

2

So 10 litres of CO must produce 10 litres of CO2.

Again 2 litres of CO require 1 litre of O2

So 10 litres of CO must require 5 litres of O2.

10

SAQ 22: 24 litres of N2 gas was allowed to react with excess of H

2 gas at a particular temperature

and pressure. What volume of NH3 gas at the same conditions would be formed if all the N

2 is made to

react? What volume of H2 gas is required for the purpose.(48L, 72L)

SAQ 23: 500ml of O2 gas was converted to O

3 gas at the same temperature and pressure and a 50ml

reduction in volume was observed at the end of the reaction. What is the volume of O3 gas formed? Find

the volume of the unreacted O2 gas and the volume of the mixture (ozonised oxygen). (100, 350, 450 mL)

SAQ 24: 20 litres of H2 gas was allowed to mix with 15 litres of O

2 gas in a vessel operated at

constant pressure (1atm) and temperature(1000C) conditions. An electric spark was formed inside the mixturewhere H

2O vapour is produced with a pop sound. Find out the volume of water vapour that would be formed

and which gas would remain in excess? What is the total volume of the resulting mixture? (20, 25 L)SAQ 25: 10 litres of O

2 gas was allowed to react with excess of sulphur. What volume of SO

2 gas

will be formed at the same temperature and pressure?SAQ 26: Calculate the volume of CO

2 gas and water vapour evolved by completely burning 3 litres

of CH4 gas at a particular temperature and pressure. (10 L)

PERCENTAGE OF GASEOUS MIXTURE:Example: 1 litre of a mixture of CO and CO

2 is passed through a tube containing red hot

charcoal. The volume now becomes 1.6 litres. The volumes are measured under the same conditionsof temperature and pressure. Find the composition of the original mixture.Solution: Out of the two gases present in the mixture, it is only CO

2 which will react with charcoal(C)

to be reduced to CO and the original CO present in the mixture will remain as such.CO

2(g) + C(s) 2CO(g)

1 litre 2 litresLet us assume that out of 1 litre of the mixture, the volume of CO is x litre and CO

2 is (1-x) litre.

From the balanced equation we know that1 litre of CO

2 produces 2 litres of CO.

So (1-x) litres of CO2 must produce 2(1-x) litres of CO.

So the total volume after the reaction = volume of original CO + the volume of CO formed from CO2 by

its reaction with C = x + 2(1-x)The initial volume before the reaction was 1 litre and the volume after the reaction is x + 2(1-x).In the question, it is given that the final volume is 1.6 litres.

x + 2(1-x) = 1.6 ⇒ x + 2 -2x =1.6, ⇒ x = 0.4litre(volume of CO).So the volume of CO

2 in the original mixture was 1-0.4= 0.6 litre.

So the percentage of CO = (0.4/1.0)X100= 40%So the percentage of CO

2 = 100-40= 60%

EUDIOMETRYThe composition of a mixture of gases is determined by this method. The mixture of gases along with O

2

gas is taken in a eudiometer tube which is a long and narrow glass tube closed at one end. The tubecontaining the mixture of gases is inverted over a mercury trough so as to keep always at same pressurecondition(1 atm. pressure). The reaction among the gases is allowed to take place by creating electric sparkinside the eudiometer tube. Reaction among gases takes place with an explosive sound (pop sound). Supposewe have a mixture of CH

4, CO and O

2 gases. CH

4 reacts with O

2 to give CO

2 and H

2O while CO reacts

with O2 to give CO

2. After the reaction the gaseous mixture is allowed to cool so that any H

2O formed in

the reaction is condensed to liquid. Thus the volume of the mixture is reduced because H2O vapour present

in the mixture is removed and liquid water formed has negligible volume compared to volume of gas and doesnot contribute to gaseous volume any more. Then the eudiometer tube is transferred carefully to a jarcontaining an alkali like caustic soda or caustic potash or lime water [Ca(OH)

2]. The tube is shaken well to

absorb all CO2 gas present in the mixture. The volume is further reduced. The volume reduced at this stage

is equal to the volume of CO2 present in the mixture. The mixture is then treated with alkaline pyrogallol

to absorb any excess (unreacted) O2 gas that might be present in the mixture. Again the volume decreases

and this decrease is equal to the volume of O2 present in the mixture. In this way the gases are selectively

removed with specific reagents one after the other and volume reduction is measured at each step. Thenusing the balanced equation and Gay Lussac's law, we can find the composition of the original mixture. Readthe following example mindfully.

11

(a) Composition of a gas mixture:Example: 20ml of a mixture of CO and C

2H

2(acetylene) are mixed with 30ml of O

2 and exploded

in a eudiometer tube. After cooling the residual gas occupied 34ml. After treatment with caustic potash,the residual oxygen gas occupied 8ml. Calculate the percentage composition of the original mixture.Solution: Let us write two separate combustion reactions, one for CO and the other for C

2H

2 reacting with

O2.

2CO(g) + O2(g) 2CO

2(g) (i)

2ml 1ml 2 ml2C

2H

2(g) + 5O

2(g) 4CO

2(g) + 2H

2O(l) (ii)

2 ml 5 ml 4ml (zero)Let us assume that out of 20ml of mixture, the volume of CO = x ml, and so the volume of C

2H

2=(20-x)ml

Equation (i)2ml of CO produces 2 ml of CO

2 gas

So x mol of CO will produce x ml of CO2 gas.

Again 2 ml of CO requires 1 ml of O2

So x ml of CO will require x/2 ml of O2

Equation (ii)2 ml of C

2H

2 produces 4 ml of CO

2 gas

(20-x)ml of C2H

2 will produce 2(20-x) ml of CO

2.

Again 2ml of C2H

2 requires 5 ml of O

2 gas

So (20-x) of C2H

2 will require (5/2) X (20-x)ml of O

2 gas.

Total CO2 formed = x + 2(20 -x)

The volume of oxygen consumed by both reactions=x/2 + (5/2) × (20-x)=(100-4x)/2

So the volume of oxygen left after the reaction = 30 - (100-4x)/2According to the question, the volume of unreacted oxgyen is 8ml.So 30-(100-4x)/2 = 8, ⇒ x = 14ml(volume of CO in the original mixture)So the volume of C

2H

2 = 20-14=6ml.

Alternative method: We can also solve this problem by taking the volume of CO2. Note that after the reaction,

the total volume was 34ml (which contain CO2 and unreacted O

2, not water which is in the liquid state)

and after absorbing the mixture by caustic postash the volume reduced to 8ml(for O2 only). So the

volume of CO2= 34-8 = 26ml

The total volume of CO2 is x +2(20-x)

So x + 2(20-x) = 26, ⇒ x =14ml(volume of CO) and hence volume of C2H

2 = 20-14 = 6ml.

Alternative method: We can also solve this problem by finding the total volume of the mixture after thereaction which is given to be 34ml.Total volume after the reaction = volume of CO

2 formed from both reactions + the volume of unreacted O

2

= x + 2(20-x) + [30-(100-4x)/2](since H

2O is converted to liquid state, its volume is negligible and hence has been taken as zero)

But it is given that the volume after the end of the reaction is 34ml.So x + 2(20-x) + [30-(100-4x)/2] = 34, ⇒ x = 14ml(volume of CO)

So the volume of C2H

2= 20-14= 6ml.

So you found that you can solve this problem in many ways and you can use the most economical way toget the result quickly. In the above example the first two ways are easier than the last method. Is'nt it?Example : 10ml of a mixture of CH

4 ,C

2H

4 and CO

2 were exploded with excess of air. After

explosion and cooling, there was a contraction of 17ml and after treatment with aq. KOH there wasfurther reduction of 14ml. Find the composition of the original mixture.Solution: CH

4 and C

2H

4 only react with O

2 to produce CO

2 and H

2O. The CO

2 present in the original mixture

remains as such.Let the volume of CH

4= x ml, volume of C

2H

4= y ml, So volume of CO

2 = 10-x-y

Now let us write two combustion reactions.CH

4(g) + 2O

2(g) CO

2(g) + 2H

2O(l) (i)

1 ml 2ml 1 ml (zero)C

2H

4(g) + 3O

2 2CO

2(g) + 2H

2O(l) (ii)

1ml 3ml 2ml (zero)Equation (i):

1 ml of CH4 produces 1 ml of CO

2

x ml of CH4 will produce x ml of CO

2

12

1 ml of CH4 requires 2ml of O

2 gas,

x ml of CH4 will require 2x ml of O

2 gas.

Equation (ii):1 ml of C

2H

4 produces 2ml of CO

2

y ml of C2H

4 will produce 2y ml CO

2.

1 ml of C2H

4 requires 3ml of O

2

y ml of C2H

4 will require 3y ml of O

2.

So total CO2 formed during the reaction= (x + 2y)

The total CO2 present in the mixture = (x+2y) + (10-x-y)

Since the volume contraction by treatment with KOH is 14ml, the total volume of CO2 present after the

reaction is 14ml.So (x+2y) + (10-x-y) = 14ml, ⇒ y = 4ml(volume of C

2H

4)

Now let us find the total volume of mixture after the reaction= x+2y +(10-x-y)This is same as the volume of total CO

2 present since the volume of water formed in the liquid state is taken

to be zero.The volume of the mixture before the reaction = volume of mixture + O

2 consumed

= 10 + 2x + 3yNote that if there is any unreacted O

2 in the mixture, that would be cancelled while finding the difference.

Hence the difference between volumes of the reactant and product mixtures=volume contraction = 17ml(10+2x+3y) - (x+2y +10-x-y) = 17, ⇒ 2x+2y =17

Substituting the value of y, we get, 2x + 2×4=17, ⇒ x = 4.5ml(volume of CH4)

So the volume of CO2 present in the original mixture =10 - 4- 4.5=1.5ml

So volume of CH4

= 4.5ml, C2H

4 = 4ml and CO

2 = 1.5ml in the original mixture.

SAQ 27: What will be volume of the products in each case of the following.(i)When 40ml of H

2 gas is allowed to react with equal volume of Cl

2 gas (80 mL)

(ii)When 200ml of O2 is allowed to react with equal volume of CO gas. (300 mL)

SAQ 28: 10ml of a mixture of ethylene(C2H

4) and acetylene(C

2H

2) require 29ml of O

2 gas for complete reaction at

the same temperature and pressure. Determine the composition of the mixture. (8, 2 mL)(b)Determination of molecular formula of an unknown gasWe can find the molecular mass of an unknown gas by eudiometry. Go through the following example.Example: 100 mL of an unknown gas is mixed with 140 mL of oxygen in a gas burette and subjected toelectric spark. The residual volume was 190 mL and after absorbing the mixture by caustic potash, theresidual volume was 90 mL. Find the molecular mass of the unknown gas.Soultion: The voume of CO

2 formed = 190 - 90 = 100 mL

The volume of oxygen consumed = 140 - 90 = 50 mLHence 100 mL of the gas has reacted with 50 mL of O

2 gas to give 100 mL of CO

2

Applying Avogadro's law we have2 molecules of the unknown gas reacted with 1 molecule of O

2 to give 2 moclecules of CO

2

So, 1 molecule of the unknown gas reacted with ½ molecule of O2 to give 1 molecule of CO

2

Since ½ molecule of O2 contains 1 O atom and 1 molecule of CO

2 contains 1 C atom, the unknown gas

is CO.Alternative method:Let the formula of the unknown gas is C

xO

y (here we presumed that it is compound of C and O only)

CxO

y+ (2x - y)/2 O

2→ x CO

2

1 mL (2x-y)/2 mL x mL100 mL (2x-y)/2 X 100 mL 100x mL

Since the volume of CO2 is 100 mL (determined before)

100x = 100 ⇒ x =1Since the volume of O

2 consumed is 50 mL(determined before)

(2x-y)/2 × 100 = 50 ⇒ y = 1Hence the formula of the unknwon gas is CO.(Note that if you presume the gas to be a hydrocarbon having formula C

xH

y, then while solving for x you

shall get -ve value, which is absurd. Only presuming it be CxO

y you will get +ve value for both x and y.

13

Example 2: 10 mL of a gaseous hydrocarbon is mixed with 100 mL of O2 in an eudiometer tube at 250C

and 1 atmosphere pressure. The mixture is exploded by an electric spark. The residual gas occupied 95 mL.On passing the mixture through KOH solution contraction of volume occurred by 20 mL. The residual gaswas completely absorbed by pyrogallol. Determine the molecular formula of the hydrocarbon.Solution: C

xH

y+ ½ ( 2x + y/2) O

2 x CO

2 + y/2 H

2O (l)

1 mL ½(2x + y/2) mL x mL zeroVolume of CO

2 formed = volume contraction by KOH = 20 mL

Volume of residual gas after explosion at room temp. = volume of CO2 + volume of excess O

2 = 95 mL

Hence volume of excess O2 = 95 - 20 = 75 mL (totally absorbed by pyrogallol)

Total volume of O2 used = 100 mL. Hence volume of O

2 consumed = 100 - 75 = 25 mL

From the above equation:1 mL of C

xH

y produces x mL of CO

2 gas

10 mL of CxH

y produces 10x mL of CO

2

Hence 10x = 20, ⇒ x = 2Again

1 mL of CxH

y requires 1/2(2x +y/2) mL of O

2 gas

So 10 mL of the gas requires 10 × 1/2(2x + y/2) mL of O2

Hence 10 × 1/2(2x +y/2) = 25Substituting the value of x in the above equation we get, y= 2Hence the molecular formula of the hydrocarbon is C

2H

2(acetylene)

SAQ 29: 0.9 g of a solid organic compound having molecular mass 90 containing C, H and O was heatedwith oxygen corresponding to volume of 224 mL at NTP. After combustion the total volume of the gas was560 mL at NTP. On treatment with KOH, the volume decreased to 112 mL. Determine the molecular formulaof the compound.

RESPONSE TO SAQs(Percent Composition, Empirical and Molecular Formula)

SAQ 1: M.M of Ca(ClO3)

2 = 40+ 2(35.5 + 48)= 207

207gms of Ca(ClO3)

2 contains 6×16=96gms of Oxygen

So 100 gms of Ca(ClO3)

2 contains (96/207) × 100=46.37gms

Hence the percent by mass of O = 46.37%SAQ 2: (i) M.M of NH

4NO

3 = 28 + 4+48= 80

80gms of NH4NO

3 contains 28gms of Nitrogen

100gms of NH4NO

3 contains (28/80) × 100 = 35gms.

So the percent of N in NH4NO

3 = 35%

(ii) M.M of (NH4)

2SO

4 = 2×18 + 32+ 64= 132

132gms of (NH4)

2SO

4 contains 28gms of Nitrogen

100gms of (NH4)

2SO

4 contains (28/132) × 100 = 21.21gms

So the percent of N in (NH4)

2SO

4 = 21.21%.

Comparing the % of N in both the cases, we find that NH4NO

3 is a better grade nitrogenous

fertilizer as it has greater percentage of nitrogen.SAQ 3: M.M of K

2Cr

2O

7 = 2×39+ 2×52 + 7×16= 294

If only percent composition is asked in the question, you are required to find the percent by mass of eachelement present in the compound.K: 294gms of K

2Cr

2O

7 contains 2×39gms of K

100gms of K2Cr

2O

7 contains (2×39)/294 × 100 = 26.53gms

Cr: 294gms of K2Cr

2O

7 contains 2×52gms of Cr

100gms of K2Cr

2O

7 contains (2×52)/294 × 100 = 35.37gms

O: 294gms of K2Cr

2O

7 contains 7×16=112 gms of O

100gms of K2Cr

2O

7 contains (112/294) X 100 = 38.1gms.

So percent of K=26.53%, Cr=35.37% and O=38.1%SAQ 4: The M.M of CuO= 63.5 + 16=79.5

79.5gms of CuO contains 63.5 gms of Cu100gms of CuO contains (63.5/79.5) × 100 = 79.87gms

14

So the percent of Cu = 79.87%79.87gms of Cu can be produced by 100gms of CuO100kg i.e100000gms of Cu can be produced by 125203.455gm=125.203kg.

SAQ 5: Let us take 100gms of impure Cu2O. This contains 66.6 gms of Cu.

But let us see how much of Cu2O can contain 66.6gms of Cu from its molecular mass.

The M.M of Cu2O= 2×63.5 + 16 = 143;

127(=2×63.5)gms of Cu is present in 143gms of Cu2O

So 66.6gms of Cu must be present (143/127)×66.6=74.99gms of Cu2O.

So 100gm of impure Cu2O contains only 74.99gms of pure Cu

2O and the rest impurity.

So the percent of purity = 74.99%.SAQ 6: Dividing the coefficients by 7, we get the empirical formula CH

2.

SAQ 7: Dividing the coefficients by 2, we get the empirical formula C2H

4O.

SAQ 8:___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________C 85.63 12 85.63/12=7.135 7.135/7.135=1 CH

2

H 14.37 1 14.37/1=14.37 14.37/7.135=2

___________________________________________________________Empirical formula= CH

2 and empirical formula mass= 12+2=14

Molecular Mass = 56; But we know that Molecular Formula = (Empirical Formula)n

n= Molecular Mass/Empirical Formula mass = 56/14=4So Molecular Formula =(CH

2)

4= C

4H

8.

SAQ 9:___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________N 30.4 14 30.4/14= 2.17 2.17/2.17=1 NO

2

O 69.6 16 69.6/16=4.35 4.35/2.17=2___________________________________________________________

SAQ 10:___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula__________________________________________________________Cr 26.52 52 26.52/52=0.51 0.51/0.51=1×2=2S 24.52 32 24.52/32=0.766 0.766/0.51=1.5×2=3 Cr

2S

3O

12

O 48.96 16 48.96/16=3.06 3.06/0.51=6×2=12__________________________________________________________

It seems that the compound is a sulphate of chromium as 3(SO4) give S

3O

12. So arranging them in proper

order it is Cr2(SO

4)

3. From this formula it is evident that it is also the molecular formula. Anyway, we cannot

officially find the molecular formula without the molecular mass data.ANSWERS TO PRACTICE QUESTIONS1.

__________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula__________________________________________________________Na 21.6 23 21.6/23=0.939 0.939/0.938=1Cl 33.3 35.5 33.3/35.5=0.938 0.938/0.938=1 NaClO

3

O 45.1 16 45.1/16=2.819 2.819/0.938=3

15

__________________________________________________________It appears that the empirical formula is also the molecular formula as it is the familiar compound sodiumchlorate.2. Total mass of the compound = 7.3+6.3 = 13.6gm

So % of Fe=(7.3/13.6) ×100 =53.68 , % of S= (6.3/13.6) ×100 =46.32After finding the % composition, we shall do in the same way as before.___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________Fe 53.68 56 53.68/56=0.958 0.958/0.958=1×2=2 Fe

2S

3

S 46.32 32 46.32/32=1.447 1.447/0.958=1.5×2=3___________________________________________________________Hence the empirical formula is Fe

2S

3.

3.___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________C 40 12 40/12=3.33 3.33/3.33=1H 6.66 1 6.66/1=6.66 6.66/3.33=2 CH

2O

O 53.34 16 53.34/16=3.33 3.33/3.33=1___________________________________________________________Empirical formula mass = 12+2+16=30; Molecular Mass(given)=60So n=60/30=2 and therefore Molecular Formula=(CH

2O)

2= C

2H

4O

2.

4.______________________________________________________El. % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula______________________________________________________C 82.66 12 82.66/12=6.88 6.88/6.88=1×2=2 C

2H

5

H 17.34 1 17.34/1=17.34 17.34/6.88=2.5×2=5______________________________________________________Since its vapour density(V.D)=29.05, its Molecular Mass=2×29.05=58.1Empirical formula mass= 24+5=29, So n = 58.1/29=2.0034 = 2So molecular formula = (C

2H

5)

2= C

4H

10.

RESPONSE TO SAQs(Stoichiometric Calculations)SAQ 1: The involved species are C

2H

6, O

2 and H

2O.

C2H6 O2 CO2 H2O+ +2 7 4 62(24+6) 4(12+32) 6(2+16)

. 2X30gms of C2H

6 on complete burning produces 4×44gms of CO

2 gas

So 30gms of C2H

6 produces (4×44)/(2×30) × 30= 88gms of CO

2

Again 2×30gms of ethane produces 6 × 18 gms of H2O

So 30gms of ethane will produce 54gms of H2O.

SAQ 2: Here the mass of AgCl is given and we are to find the mass of CaCl2.

CaCl2 AgNO3 AgCl Ca(NO3)2+ +2X(108+35.5)(40+ 2X35.5)

2 2

2×(108+35.5) gm of AgCl is produced by (40 + 2×35.5)gms of CaCl2.

6.6gms of AgCl must be produced by 2.55gms of CaCl2(do the calculation for yourself).

The number of moles of CaCl2 = 2.55/ 111 = 0.0229 mole.

SAQ 3: Here the mass of HCl is given and we have to find the mass of CaO.

CaO HCl CaCl2 H2O+ +2(40+16) 2X(1+35.5)

16

2×(1+35.5)gms of HCl requires (40+16) gms of CaOSo 6.92gms of HCl will require 5. 308gms of CaO.

SAQ 4: Here number of mole of Ba(OH)2 reacted is given.

Ba(OH)2 CO2 BaCO3 H2O+ +(137+12+48)1 mole

1 mole of Ba(OH)2 produces (137+12+48) gms of BaCO

3

0.205mole of Ba(OH)2 will produce 40.385 gms of BaCO

3.

SAQ 5: Here the moles of product formed is given.

NaHCO3 Na2CO3 CO2 H2O+ +22X(23+1+12+48) 1 mole

1 mole of CO2 is produced by 2X(23+1+12+48)gms of NaHCO

3

So 1.5moles of CO2 will be produced by 252gms of NaHCO

3.

SAQ 6: Here the mass of MnO2 is given and we are to find the volume of Cl

2.

2MnO2 HCl MnCl2 Cl2 H2O+ + +4(55+32) 22.4 l

(55+32)gms of MnO2 produces 22.4 litres of Cl

2 at NTP

So 8.7gms of MnO2 will produce 2.24 litres of Cl

2 at NTP.

SAQ 7: The volume of CO2 formed is given and we are to find the mass of NaHCO

3

NaHCO3 HCl NaCl CO2 H2O+ + +(23+1+12+48) 22.4 l

22.4 litres of CO2 at NTP is produced by (23+1+12+48)gms of NaHCO

3

5.6 litres of CO2 at NTP will be produced by 21 gms of NaHCO

3.

SAQ 8: Here you have to find both the volumes of NO2 and O

2.

Pb(NO3)2 PbO NO2 O2+ +2 2 42X[207 + 2X(14+48)] 4 X 22.4 l 22.4 l

2 X 331 gms of Pb(NO3)

2 produces 4 X 22.4 litres of NO

2 at NTP

10gms of Pb(NO3)

2 will produce 1.353 litres of NO

2 gas.

2 X 331gms of Pb(NO3)

2 produces 22.4 litres of O

2 gas at NTP.

So 10gms of Pb(NO3)

2 will produce 0.3383litre = 338.3 ml of O

2 gas.

SAQ 9: Here you have to find the mass of ammonium dichromate.

(NH4)2Cr2O7 N2 Cr2O3 H2O+ + 42X(14+4) + 2X52+112 22.4 l

22.4 litres of N2 at NTP are produced by 252 gms of (NH

4)

2Cr

2O

7

11.2 litres of N2 at NTP are produced by 126 gms of (NH

4)

2Cr

2O

7

SAQ 10: 2nd reaction:3 CuO + 2 NH

3 → 3 Cu + N

2 + 3 H

2O

3 ×(63.5+16)g 2 × 17 g

3× 79.5 g of CuO reacts with 2 × 17 g of NH3

So 2.5 g of CuO reacts with 0.356 g of NH3

1st reaction:(NH

4)

2SO

4+ 2NaOH → 2NH

3 + Na

2SO

4 + 2H

2O

[2(14+4) +32+64]g 2X17g=132 g2X 17g of NH

3 is produced by 132 g of NH

4)

2SO

4

So, 0.356 g of NH3 is produced by 1.38 g of ammonium sulphate

SAQ 11: CaCO3

→ CaO + CO2

(40+12+48)g (20+16)g100g of CaCO

3 produces 56g of CaO

4 g of CaCO3 produces 2.24 g of CaO

CaO + 2 NH4Cl → CaCl

2 + 2 NH

3+ H

2O

56g 2 X 22.4 L(NTP)

17

56 g of CaO produces 2 X 22.4 L of NH3 gas at NTP

2.24 g of CaO produces 1.792 L of NH3 at NTP

Using the combined gas equation, we can find the volume at the given conditions:The requied volume = 1.87 L

SAQ 12: 2nd reaction3 Cl

2+ 6 NaOH → NaClO

3 + 5 NaCl + 3 H

2O

3 × 71 g 6 × 40 g6×40 g of NaOH requires 3 × 71 g of Cl

2

10 g of NaOH requires 8.875 g of Cl2

1st reaction: Balance by partial equation method or NO method2KMnO

4 + 16 HCl → 2 KCl + 2 MnCl

2 + 5 Cl

2 + 8 H

2O

2(39+55+64)g 5×71g5× 71 g of Cl

2 is produced by 2 × 158 g of KMnO

4

So, 8.875g of Cl2 is produced by 7.9 g of KMnO

4

SAQ 13: Na2CO

3 + 2 HCl → 2 NaCl + CO

2+ H

2O

106g 2 × 36.5g 2 × 58.5 gFirst we have to find the limiting reactant.106 g of Na

2CO

3 reacts with 2 × 36.5 g of HCl

So, 5.3 g of Na2CO

3 reacts with 3.65 g of HCl

Since we have 5.3 g of HCl, the limting reactant is Na2CO

3

106g of Na2CO

3 produces 2 × 58.5 g of NaCl

So, 5.3 g of Na2CO

3 produces 5.85g of NaCl

SAQ 14: 3 BaCl2 + 2 Na

3PO

4 → Ba

3(PO

4)

2 + 6 NaCl

3 moles 2 moles 3×137+2(31+64)gLet us first find the limiting reactant.

3 moles of BaCl2 reacts with 2 moles of Na

3PO

4

so, 0.1 mole of BaCl2 reacts with 0.0667 mole of Na

3PO

4

But we have 0.05 mole of Na3PO

4, hence all BaCl

2 cannot be exhausted. So the limiting reactant is Na

3PO

4.

2 moles of Na3PO

4 produces 601 g of Ba

3(PO

4)

2

So, 0.05 mole of Na3PO

4 produces 15.025 g of Ba

3(PO

4)

2

SAQ 15: Mass of pure CaCO3 = 95% of 200kg = (95/100) × 200 = 190kg.

CaCO3

→ CaO + CO2

(40+12+48) (40+16)100gm of CaCO

3 produces 56gms of CaO(lime).

190,000gms of CaCO3 will produce (56/100) × 190,000 =106,400gm =106.4kg of CaO.

SAQ 16: Zn + H2SO

4→ ZnSO

4+ H

2

65.5g 22400mL at NTPBy using combined gas equation, the volume at the given conditions is converted to NTP.The NTP volume = 234.46 mL

22400 mL of H2 is produced by 65.5g Zn at NTP

So, 234.46 mL of H2 is produced by 0.685 g of ZnSo % of purity = (0.685/1)× 100 = 68.5%SAQ 17: Ag + 2 HNO

3→ AgNO

3 + NO

2 + H

2O

1 mole=108g 1 moleAgNO

3 + HCl → AgCl + HNO

3

1 mole 1mole =(108+35.5)g

In problems involving successive reactions, we can solve easily by mole method in stead of by mass methodby establishing mole relationship between the two involved molecules, one from the first reaction and the otherfrom the second reaction.1 mole of Ag(one gm. atomic mass) must produce 1 mole of AgCl as in one mole of AgCl there is one gm.atomic mass of Ag. So we can bypass the the product of the first reaction(AgNO

3) i,.e ignore the reaction

sequence through which AgCl has been prepared.143.5 g of AgCl is produced by 108 g of AgSo, 9 g of AgCl is produced by 6.77 g of AgSo % of purity = (6.77/7) × 100 = 96.7%

18

SAQ 18: Balance the equation by partial equation or ON method.K

2Cr

2O

7 + 4 H

2SO

4 + 3 H

2S → K

2SO

4 + Cr

2(SO

4)

3 + 3S + 7 H

2O

(2×39+2×52+112)g 3×32g96g of S is deposited by 294 g of K

2Cr

2O

7

9.6 g of S is deposited by 29.4 g of K2Cr

2O

7

So % of purity = (29.4/35) × 100 = 84%SAQ 19: 2 NaHCO

3 → Na

2CO

3 + H

2O + CO

2

2×(23+1+12+48)g 18g 44g(18+44)g of volatiles are lost by 2 × 84 g of NaHCO

3

0.45 g of volatiles are lost by 1.219 g of NaHCO3 (Note that Na

2CO

3 is stable to heat)

So, % of NaHCO3 = (1.219/3)×100 = 40.64%, hence % of Na

2CO

3= 59.36%

SAQ 20: 2Al + 6 HCl → 2 AlCl3 + 3 H

2

2×27g 3×22.4 L(at NTP)Mg + 2 HCl → MgCl

2 + H

2

24g 22.4 L (at NTP)First the volume at the given conditions is converted to volume at NTP by using combined gas equation.

NTP volume = 1.1 LLet the mass of Al = x g and hence mass of Mg= (1-x)g

54g of Al produces 3× 22.4 L of H2 at NTP

So, x g of Al produces 1.244 x L of H2 at NTP

24g of Mg produces 22.4 L of H2 at NTP

So, (1-x) g of Mg produces 0.933(1-x) L of H2 at NTP

Hence, 1.244x + 0.933(1-x) = 1.1 (total volume of H2 at NTP)

Solving the above equation, we get, x=0.55 g. So % of Al= 55% hence % of Mg= 45%SAQ 21: Pb(NO

3)

2 → PbO + 2 NO

2 + ½ O

2

[207+2(14+48)]g 2×46g 16gNaNO

3→ NaNO

2 + ½ O

2

(23+14+48)g 16gLet the mass of Pb(NO

3)

2 = x g and NaNO

3 = (5-x) g

331 g of Pb(NO3)

2 produces (92+16)g of volatiles(gases)

x g of Pb(NO3)

2 produces 0.326 x g of volatiles

85 g of NaNO3 produces 16g of volatiles

(5-x)g of NaNO3 produces 0.188(5-x) g of volatiles

Hence, 0.326x + 0.188(5-x) = (28/100) × 5On solving the above equation we get, x=3.33 g and NaNO

3 = 1.67g

SAQ 22: N2

+ 3H2

→ 2NH3

1vol. 3vol. 2vol.1litre of N

2 produces 2 litres of NH

3( at the same temperature and pressure)

24litres of N2 must produce 48litres of NH

3.

Again 1 litre of N2 requires 3 litres of H

2 gas

So 24 litres of N2 will require 72 litres of H

2 gas.

SAQ 23: 3O2

→ 2O3

3 vol. 2 vol.3 ml of O

2 reacts to form 2 ml of O

3 at the same temperature and pressure. So there

is a decrease of volume of 1 ml.For the decrease in volume by 1 ml, the volume of ozone(O

3) formed = 2ml

Hence for the decrease in volume by 50ml, the volume of O3 formed = 100ml.

If 2 volumes of O3 = 100ml, then what is 3 volumes? It is 150ml.

So 150ml of O2 reacted to form 100ml of O

3 in the volume ratio 3:2.

The volume of O2(unreacted) remaining= 500 - 150 = 350ml

So the total volume after the reaction = 350(O2) + 100ml(O

3) = 450 ml.

Alternative Method If 3 mL of O2 produces by 2 mL of O

3, then x mL of O

2 will produce 2x/3 mL of O

3.

So volume after reaction = 500 - x +2x/3; Then volume reduction = [500 - (500 -x + 2x/3)] =50, ⇒ x=150 mL(volume O

2 used up). So volume of unreacted O

2 = 500-150=350 and volume of O

3=2x/3=100mL,

So the total volume = 350 + 100 = 450 mL

19

SAQ 24: 2H2(g) + O

2(g) → 2H

2O(g)

2 vol. 1 vol. 2 vol.2 litres of H

2 reacts with 1 litre of O

2 to form 2 litres of water vapour

So 20 litres of H2 will react with 10 litre of O

2 to form 20 litres of water vapour

Since 15 litres of O2 gas has been taken, 5 litres of it will remain in excess.

Total volume =20 litres(water vapour) + 5 litres (excess O2) = 25 litres.

SAQ 25: S(s) + O2(g) → SO

2(g)

1 vol. 1 vol.10 litres of O

2 will produce 10 litres of SO

2. Note that S is in the solid state and therefore its volume

has not been written.SAQ 26: CH

4(g) + 2O

2(g) → CO

2(g) + 2H

2O(g)

1 vol. 2 vol. 1 vol. 2 vol.3 litre of CH

4 produces 3 litre of CO

2 and 6 litres of water vapour

SAQ 27:(i) H2(g) + Cl

2(g) → 2HCl(ga)

40 ml of H2 will react with 40ml of Cl

2 to form 80ml of HCl gas

(ii) 2CO(g) + O2(g) → 2CO

2(g)

200ml of CO will react with 100ml of O2 to produce 200ml of CO

2 gas. Here 100ml(200-

100) of O2 gas will remain unreacted.

SAQ 28: C2H

4 + 3O

2 → 2CO

2 + 2H

2O (i)

2C2H

2 + 5O

2 → 4CO

2 + 2H

2O (ii)

Let the volume of C2H

4 = x ml; Hence the volume of C

2H

2 = (10-x) ml

Eqn.(i) 1 ml of C2H

4 requires 3mls of O

2.

So x ml of C2H

4 must require 3x ml of O

2.

Eqn.(ii) 2ml of C2H

2 requires 5 ml of O

2.

So (10-x) ml of C2H

2 must require (5/2)×(10-x) ml of O

2.

Total volume of O2 needed = 3x + (5/2) × (10-x) = 29 (according to question)

⇒ x = 8 ml (volume of C2H

4); Hence the volume of C

2H

2 = 10-8 = 2ml.

SAQ 29 : CxH

yO

z + 1/2(2x+y/2-z) O

2 → x CO

2 + y/2H

2O(l)

1 mole 1/2(2x+y/2-z)×22400mL x × 22400mL zero

After combustion, the mixture contains CO2 + unreacted O

2 = 560 mL

Volume after absorption by KOH = 112 mL = volume of unreacted O2

So voume of CO2 = 560 - 112 = 448 mL

So volume of O2 consumed = 224 - 112 = 112 mL

90g of the hydrocarbon produces 22400 x mL of CO2 at NTP

So, 0.9 g of the hydrocarbon produces 224x mL of CO2 at NTP

Hence 224x = 448 ⇒ x = 2 (1)Again 90g of the hydrocarbon consumes 22400/2 ( 2x+y/2-z) mL of O

2

So, 0.9g of hydrocarbon consumes 112(2x+y/2-z) mL of O2

112(2x+y/2-z) = 112 (2)From the molecular mass of C

xH

yO

z, we get the equation

12x + y + 16z = 90 (3)Solving simultaneously equations 1,2 and 3, we get, z=4 and y =2Hence the molecular formula of the compound is C

2H

2O

4 (oxalic acid)

20

PRACTICE QUESTION

1. 200Kg of rock salt(NaCl) which contains 95% NaCl is allowed to react with excess of H2SO

4. How

much sodium sulphate will be formed? (230.5Kg)2. What volume of H

2 gas be evolved at NTP when 2g of superheated iron is allowed to completely

react with excess of superheated steam? Also find the the mass of the product residue? (Fe = 55.8)(1.070L, 2.764g of

Fe3O

4)

3. What volume of oxygen be evolved at NTP when 2.94g of K2Cr

2O

7 completely reacts with with

excess of hydrogen peroxide in presence of dil. H2SO

4?(K=39, Cr=52) (672mL)

4. How many grams and litres of CO2 formed at NTP when 1g of sucrose(C

12H

22O

11)is completely

burnt in excess of air? (1.54g and 785.96mL)5. Calculate the mass of CO

2 and water that will be produced by completely burning 1.5g of an organic

compound having molecular formula C3H

6O? (3.4g and 1.3965g)

6. When a mixture of ammonia and oxygen is passed over platinum maintained at 8000C, how muchof water will be formed from 0.5g of ammonia? Assume complete reaction.

(Hint: NH3 + O

2 ----Pt----> NO + H

2O) (0.794g)

7. 50mg of an impure sample of aluminium gave on treatment with excess dil. HCl, 50 mL of moisthydrogen at 270C and 760mm presure. Calculate the percentage of purity in the sample. (73.3)8. 1.1g of a mixture of KClO

3 and KCl gave on ignition 0.85g of residue. Find the composition of the

mixture. (KClO3=58%, KCl= 42%)

9. 1.5g of a mixture of BaCO3 and CaCO

3 was heated in a platinum crucible to a constant mass. The

mass of the residue was 1.05g. Calclulate the percentage of each metal carbonate present in themixture.(Ba=137, Ca=40) (BaCO

3=64.6%, CaCO

3=35.4%)

10. What mass of potassium chloride will be formed by passing 10L of Cl2 gas at 200C and 780mm

pressure into hot conc. KOH solution. (53g)(Hint: Cl

2 + KOH KCl + KClO

3 + H

2O)

11. 2.4g of a sample of sodium bicarbonate when strongly heated gives 300 mL of CO2 measured at 230C

and 780 mm pressure. Calculate percentage of purity of the sample. (88.7)12. What volume of oxygen gas at NTP is necessary for complete combustion of 10 litres of butane(C

4H

10)

measured at 270C and 760mm pressure? (59.15L)13. Calculate the volume of air containing 21% by volume of oxygen at NTP required to convert 400mLof SO

2 into SO

3 under same conditions. (952.38mL)

14. A portable hydrogen generator utilises the reaction of hydrolysis of CaH2. How many litres of H

2 gas

at NTP can be produced by 50g of CaH2? (53.33L)

15. 1.2g of a mixture of CaCO3 and MgCO

3 on complete ignition produced 42mL of CO

2 measured at

270C and 780mm pressure. Calculate the percentage composition of the mixture.(CaCO3=92%, MgCO

3=8%)

16. 50mL of solution containing 0.25g of an impure sample of H2O

2 reacts completely with 0.3g of

KMnO4 in presence of sulphuric acid. (i)Calculate the volume of dry oxygen gas evolved at 270C and 810mm

pressure. (ii)Find the percentage of purity of H2O

2 solution. (109.6mL, 64.4%)

17. How many grams of CaCO3 should be heated to get enough CO

2 which can convert 0.2 mole of

Na2CO

3 into NaHCO

3? (20g)

18. What mass each of magnesium and dilute H2SO

4 are required to give hydrogen just sufficient to

reduce 6.5g of zinc oxide?(Zn=65, Mg=24) (Mg=1.92g, H2SO

4=7.84g)

19. How much KCl is produced by the reaction of 1.5Kg of potassium and 2Kg of chlorine gas?(Hint: First find out which is the limiting reactant) (2.86Kg)

20. 32g of sulphur was burnt in the presence of 4g of oxygen in a closed vessel at certain temperatureand pressure. What mass of SO

2 will be formed? What is the volume of the resulting gas at NTP?

(Hint: First find out which is the limiting reactant) (8g, 2.8L)21. 5g of a natural gas consisting of methane(CH

4) and ethylene(C

2H

4) was burnt in excess of O

2producing 14.5g of CO

2 and some water as products. Find the mass percentage of ethylene in the natural

gas. (38.4%)22. A sample of gaseous hydrocarbon occupying 1.12L at NTP when completely burnt in oxygen produced1120mL of CO

2 and 1.8g of water. Calculate the mass of the hydrocarbon and the volume of O

2(at NTP)

required for its burning. Also find the molecular formula of the hydrocarbon. (0.8g, 2.24L, CH4)

23. A gaseous hydrocarbon is exploded with oxygen. The volume of O2 needed for complete combustion

and CO2 formed is in the ratio 7:4. Find the molecular formula of the hydrocarbon. (C

2H

6)

21

24. 50mL of a mixture of nitrous oxide(N2O) and nitric oxide(NO) was exploded with excess of hydrogen.

If 30mL of N2 was formed, calculate the percentage composition(mole %) of the mixture.

(N2O=20%, NO=80%)

25. When 500mL of hydrocarbon were exploded with excess of oxygen, 2500 mL of CO2 and 3L of

water vapour formed at NTP. Determine the molecular formula of the hydrocarbon. (IIT 74) (C5H

12)

LEVEL-II1. How many Kg of H

2SO

4 can by prepared from 3 Kg of cuprite(Cu

2S)? (Cu=63.5, S=32)(1.84Kg)

2. A certain mass of iron pyrite(FeS2) was roasted with excess of oxygen to produce SO

2 gas. SO

2

reacted with excess of O2 and presume that all SO

2 is completely converted to SO

3 gas at 5000C and 200

atm. pressure in presence of V2O

5 catalyst. From SO

3 4.4 g of pure H

2SO

4 was produced by an indirect

method. Find the mass of iron pyrite originally taken.(Fe=56, S=32) (2.69g)3. 3.4g of pure H

2O

2 on decomposition gives enough oxygen at 1000C and 760mm pressure which

oxidise excess of NH3 at that temperature. to N

2 and H

2O. Calculate the volume of N

2 produced.(1.02L)

4. 20 mL of a solution of KCl containing some NaCl as impurity on evaporation gave 2g of the residue.It was again made soluble in water and treated with excess AgNO

3 solution. The white precipitate of AgCl

was filtered and dried to give 4.27g. Find how much of NaCl was present in the 20 mL of the above solution.(Ag=108, K=39, Na=23, Cl=35.5) (0.8g)

5. 250 mg of a commercial sample of mercuric oxide on decomposition liberates enough oxygen forcomplete combstion of 40 mL of CO at 2000C and 760mm pressure. Caculate the percentage of purity ofthe mercuric oxide sample (Hg=200, O=16) (88.4%)6. An alloy of zinc and copper weighing 1g when treated with excess of dil. H

2SO

4 gave 224 mL of

dry hydrogen at 270C and 720 mm pressure. What is the % by mass of copper in the alloy? (44%)7. 10g of impure sample of Na

2CO

3 is dissolved in water and reacted with a solution of excess CaCl

2.

The resulting precipipite after filtration and drying was found to react completely with 5.6g of pure HCl.Calculate % of purity of Na

2CO

3. Assume that CaCl

2 bring about precipitation of Na

2CO

3 only. (81.3%)

8. Calculate the volume of CO2 evolved at 1000C and 700 mm pressure that can be obtained by heating

2 kg of limestone which is 90% pure. (598.1 L)9. 2.1 g of miixture of KNO

3 and NaNO

3 was heated to constant weight and found to have lost 0.373g.

What is the % of KNO3 in the mixture. (35.71)

10. What would be the volume at STP of each of the products formed by burning 10g of carbon disulfide.The gaseous mixture produces a burning sulfur smell. What amount caustic soda would be necessary toproduce the acidic salt of the whole of the above products?)

(CO2=2.947L, SO

2=5.894 L, acidic salt = 15.78g)

11. 1.1g of solder(a lead-tin alloy) was heated in a current of dry chlorine till the metals were completelyconverted to their chlorides PbCl

2 nd SnCl

4. If the mass increased by 52%, find the percentage composition

of solder. (Pb=207, Sn=119) (Pb=79%, Sn=20.9%) (13)12. 98 L of air at 270C and 760mm pressure were bubbled through baryta solution and yield 0.35 g ofBaCO

3 precipitate. What is the partial pressure of CO

2 in the air.(baryta=Ba(OH)

2) Ba=137 (0.339 mm)

13. A mixture of NaHCO3 and Na

2CO

3 weighing 1g was treated with excess of baryta solution to form

2.1 g of BaCO3. Find the % of NaHCO

3 in the original mixture. (50%)

14. Ignition of MnO2 converts quantitatively to Mn

3O

4. A sample of pyrolusite is of the following composition.

MnO2 80%, SiO

2 and other inert constituents 15% and rest being water. The sample is ignited in air to

constant weight. What is the percentage of Mn in the ignited sample?(Mn=55) (59.36)15. One litre of an acidified solution of KMnO

4 containing 15.8 g of KMnO

4 is decolorized by passing

sufficient amount of SO2. If the SO

2 is produced by roasting of iron pyrite(FeS

2) what will the amount of

pyrite required to produce necessary amount of SO2? (K=39, Mn=55, Fe=56, S=32) (15g)

16. 2.5 g of a mixture of CaCO3, MgCO

3 and NaHCO

3 suffered a loss of 1.16g on heating. The residue

on treatment with excess HCl gave 117.3 mL of CO2 at NTP. Calculate teh mass of each component.(CaCO

3=0.12g, MgCO

3=1.5g, NaHCO

3=0.88g)

17. In the analysis of a 0.5g of feldspar, a mixture of the chlorides of sodium and potassium is obtainedwhich is 0.118g. Subsequent treatment of the mixed chloride with AgNO

3 gives 0.2451g of AgCl. What the

% of Na2O and K

2O in feldspar? (Ag=108, Cl=35.5, K=39) (Na

2O =3.6%, K

2O=10.59%)

18. A flash bulb used for taking photograph in bar light contains 30mL of O2 at pressure of 780mm and

270C. Suppose the metal wire flashed is pure aluminium which is oxidised to aluminium oxide in the processof flashing. Calculate the minimum mass of aluminium wire that is to be used for the maximum efficiency.

(0.045g)

22

19. 500mL of O2 at NTP were passed through an ozoniser when the resulting volume was 444 mL at

NTP. This quantity of ozonised oxygen is passed through excess of KI solution. Calculate the mass of I2

liberated. (I=127) (1.27g)20. Caclulate the mass of phosphorous obtained by strongly heating 1g of calcium phosphate with excessof silica and coke in an electric furnace in the absence of air. Also find out the volume of the gas evolvedat NTP? (Ca=40, P=31) (0.2g, 0.3612L)21. What mass of P

4O

6 and P

4O

10 will be produced by the combustion of 2g of P

4 in 2g of oxygen gas

leaving no unfinished reactant(neither P4 nor O

2)? (P

4O

6=1.995g, P

4O

10 = 2.005g)

22. 0.5 g of a mixture of K2CO

3 and Li

2CO

3 produced 32.5 mL of CO

2 at NTP. Find out the composition

of the mixture. (Li=7) (Li2CO

3=0.107g, K

2CO

3=0.393g)

23. A mixture contains NaCl and an unknown chloride MCl. (i)1g of this is dissolved in water. Excessof acidified AgNO

3 solution is added to it. 2.567g of a white dry precipitate is obtained. (ii) 1g of the original

mixture is heated to 3000C. Some vapours came out which are absorbed in acidifed AgNO3, 1.341g of a white

precipitate is obtained. Find the molecular mass of the unknown chloride. (IIT 1980) (53.5)24. 10mL of a gaseous organic compound containing C, H and O only was mixed with 100 mL of O

2

and exploded under conditions which allowed the water formed to condense. The volume of the gas afterexplosion was 90 mL. On treatment with potash solution a further contraction in volume of 20mL wasobserved. Given that the vapour density of the compound is 23, deduce the molecular formula. All volumemeasurements were done under same conditions) (IIT 1977) (C

2H

6O)

Solution to Practice Questions(Level II)1. Cu

2S → → → → H

2SO

4

1 mole 1 moleCu

2S is converted to H

2SO

4 in a series of reactions. But for establishing a relationship between the starting

material and final product, we should look to the number of atoms(or moles)of one element undergoingchange. In this case, Cu

2S contains one S atom and so also H

2SO

4 contains one S atom.

So 1 mole of Cu2S must produce 1 mole of H

2SO

4

2. FeS2

→ → → → 2H2SO

4

1mole 2 molesIn FeS

2, there are two S atoms and in H

2SO

4 there is one S atom. On balancing for S atoms, we find that

one mole of FeS2 will produce 2 moles of H

2SO

4. We are least concerned about the sequence of reactions

by which the final product is obtained.3. Repeated calculation method:

H2O

2 → H

2O + ½ O

2

34g 11.2L(at NTP)3.4g 1.12L4NH

3 + 3O

2 → 2 N

2+ 6 H

2O

4×22.4L 2×22.4L(NTP)1.12L 0.746L

Converting this volume from the NTP conditions to the given condition, we get V=1.02LAlternative method(One step calculation)In stead of calculating for each step, we can add up all the step equations and establish a relationship betweenthe starting reactants with the final products. Although the overall reaction resulted from the addition of thesteps, may not really occur, it is used for stoichiometric calculation.

H2O2 H2O O 22 2

NH3 O 2 N2 H 2O4 63 2

+

+ +

X 3

H2O2 N H3 N2 H 2O6 4 2 12+ +

H2O2 N H3 N2 H 2O3 2 6+ +

3×34g 22.4L 3.4g 0.746L

Converting this volume at NTP to the given conditions we get the volume=1.02L4. We have 2g of (KCl+NaCl) mixture. Let KCl = xg and NaCl = (2-x)g

KCl + AgNO3

→ AgCl + KNO3

23

(39+35.5)g (108+35.5)gNaCl + AgNO

3→ AgCl + NaNO

3

58.5g (108+35.5)gFind the mass of AgCl produced by x g of KCl and (2-x)g of NaCl by using the above two equations. Thenadd the two to be equal to 4.27g. Then solve for x.5.

HgO Hg O 2

CO O 2 CO2

+ 12

2

1+

HgO CO Hg CO2+ +(200+16) 22400mL

First convert the volume from the given conditions to NTP conditions.NTP volume= 23 mL22400mL of CO reacts with 216g of HgO at NTP23 mL of CO reacts with 0.221g of HgO

So percentage of purity = (0.221/0.25)×100 = 88.4%Note that we solved this problem by additon method.You could have as well solved by repeated calculationmethod, i.e first finding the volume of O

2 required from the 2nd rection and then the mass of HgO required

from the 1st reaction.6. Zn + H

2SO

4(dil) → ZnSO

4 + H

2

Cu cannot displace H2 from dil. H

2SO

4.First find volume at NTP and solve from the above equation.what

mass of zince is required.7.

Na2CO3 CaCl2 CaCO3 N aCl

CaCO3 HCl CaCl2 CO2 H2O

+ +

+ + +

2

2

Na2CO3 HCl NaCl CO2 H2O2 2+ + +106g 2X36.5g

1×36.5g of HCl reacts with 106g of Na2CO

3

5.6g of HCl reacts with 8.13g of Na2CO

3.

% of purity = 81.3%8. CaCO3 → CaO + CO2

100g 22.4L(NTP)90% of 2Kg = 1.8Kg = 1800g100g of CaCO3 gives 22.4L at NTP1800g of CaCO3 gives 403.2L at NTP

Then covert this volume from NTP to the given conditions(598.1L).9. KNO

3 → KNO

2 + ½ O

2

NaNO3 → NaNO

2 + ½ O

2

Let KNO3 = x g, NaNO

3 = (2.1 - x)g

Then find the total mass of O2 lost in two separate equations. Add them to 0.373g and solve for x.

10. CS2

+ 3 O2 → CO

2 + 2 SO

2

(12+64)g 22.4L 2×22.4L(NTP)NaOH + CO

2 → NaHCO

3(acidic salt)

40g 22.4LNaOH + SO

2 → NaHSO

3 (acidic salt)

40g 22.4LFind the total mass of NaOH required.11. Pb + Cl

2 → PbCl

2

207g (207+71)gSn + 2Cl

2 → SnCl

4

119g (119+4×35.5)gTotal mass after the reaction = 1.1 + 0.52X1.1 = 1.672gLet Pb = x g and Sn = (1.1 - x)gFind the mass of PbCl

2 and SnCl

4 from the above reactions, and add them to 1.672. Then solve for x.

24

12. Ba(OH)2 + CO

2 → BaCO

3 + H

2O

22.4L (137+12+48)g197g of BaCO

3 is produced by 22.4L of CO

2 at NTP

0.35g of BaCO3 is produced by 0.03979L of CO

2 at NTP

Then convert this volume from NTP conditions to the required conditions.(V=0.0437L)So volume fraction = mole fraction = (0.0437/98) = 0.00044So partial pressure of CO

2 = 0.000446 × 760 = 0.339 mm of Hg

13. NaHCO3 + Ba(OH)

2 → BaCO

3 + NaOH + H

2O

84g 197gNa

2CO

3 + Ba(OH)

2 → BaCO

3 + 2NaOH

106g 197gLet NaHCO3 = x g, Na

2CO

3 = (1-x)g. Find the mass of BaCO

3 formed from both the reactions separately

and add them to 2.1. Then solve for x.14. 3MnO

2

→ Mn

3O

4 + O

2

3×87g 229gLet us take 100g of pyrolusite. MnO

2=80g, SiO

2=15g and H

2O = 5g

On ignition MnO2 changes to Mn

3O

4 and H

2O evaporates.

3×87g of MnO2 produces 229g of Mn

3O

4

80g of MnO2 produces 70.19g of Mn

3O

4

229g of Mn3O

4 contains 3×55g of Mn

70.19g of Mn3O

4 contains 50.57g of Mn

Total residue after ignition = 70.19(Mn3O

4) + 15(SiO

2) = 85.19g

So % of Mn in the ignited sample = (50.57/85.19) × 100 = 59.36%15. Balance the two equations independently by partial equation method or ON method and then add thetwo to get a composite equation.

FeS2 O 2 Fe2O3 S O24 11 2 8+ +

KMnO4 H2O SO2 K2SO4 MnSO4 H2SO42 2 5 2 2 + + + +

X 5

FeS2 O 2 K MnO4 H2O Fe2O3 K2SO4 MnSO4 H2SO420 55 16 16 10 8 16 16+ + + + + +

20×(56+64)g 16×158g16×158g of KMnO

4 requires 20×120g of FeS

2

15.8g of KMnO4 requires 15g of FeS

2

16. CaCO3

→ CaO + CO2

100g 56g 44gMgCO

3 → MgO + CO

2

84g 40g 44g

NaHCO3 Na2CO3 CO2 H2O22 X 84g 106g 44g 18g

+ +

Let CaCO3 = x g, MgCO

3= yg and NaHCO

3 = (2.5-x-y)g

(44/100)x + (40/84)y + (62/168)(2.5-x-y) = 1.16 (1)In the second reaction, only Na

2CO

3 from the residue reacts with HCl to produce CO

2 gas.

Na2CO

3 + 2HCl → 2NaCl + CO

2 + H

2O

106g 22400mL(NTP)(22400/106)×(2.5-x-y) =117.3 (2)Solving equations (1) and (2) simultaneously we get the results.

17. NaCl + AgNO3 → AgCl + NaNO

3

58.5g 143.5gKCl + AgNO

3 → AgCl + KNO

3

74.5g 143.5gLet NaCl = x g and KCl = (0.118 -x)g

(143.5/58.5) x + (143.5/74.5)× (0.118-x) = 0.2451,

25

Hence x= 0.034g(NaCl) and KCl = 0.084gNa

2O → 2NaCl

62g 2×58.5g2×58.5g of NaCl is produced by 62g of Na

2O

0.034g of NaCl is produced by 0.018g of Na2O

K2O → 2KCl

94g 2×74.52×74.5g of KCl is produced by 94g of K

2O

0.084g of KCl is produced by 0.053g of K2O

So % of Na2O in the original feldspar = (0.018/0.5)×100 = 3.6%

% of K2O = (0.053/0.5)×100 = 10.59%

18. 2Al + 3/2 O2 → Al

2O

3

2×27g 3/2×22400mLFirst the given volume is converted to NTP condition. NTP volume=28.018mL

3/2 × 22400 mL of O2 is produced by 54g of Al

28.018mL of O2 is produced by 0.045g of Al

19. 3 O2 → 2 O

3

3mL 2mLThe volume reduction = 500 - 444 = 56 mL

For a volume reduction of 1 mL the volume of O3 formed = 2mL

Hence for a volume reduction of 56mL the volume of O3 formed = 112mL

2KI + O3 + H

2O → 2KOH + O

2 + I

2

22400mL 2×127g22400mL of ozone produces 2×127g of iodine112mL of ozone produces 1.27g of iodine

Note that O2 does not react with KI.

Alternative method:3 O

2 → 2 O

3

3mL 2mLLet x mL of O

2 is converted to O

3

3mL of O2 forms 2 mL of O

3,

So x mL of O2 will form (2/3)x mL of O

3

Total volume after the reaction = (2/3)x + (500-x) = 444 ⇒ x=168mLSo the volume of O

3 = (2/3)x = 112mL

The subsequent calculation is same as the previous method.20. 2 Ca

3(PO

4)

2 + 6 SiO

2 + 10C → P

4 + 6 CaSiO

3 + 10CO

2×[3×40 +2(31+64)]g 4×31g 10×22.4LFrom this we can find out the mass of P

4 and volume of CO formed at NTP.

21. In this case no reactant remains in excess after the reaction. So this is not a problem involving limitingreactant that we have solved before.

P4

+ 3O2

→ P4O

6

4×31g 3×32g (3×31+96)gP

4+ 5O

2→ P

4O

10

4×31g 5×32g (4×31 + 160)gLet P

4 used in the first reaction = x g and so P

4 used in the 2nd reaction = (2-x)g

1st reaction: 124g of P4 consumes 96g of O

2

x g of P4 consumes (96/124) x g of O

2

2nd reaction: 124g of P4 consumes 160g of O

2

(2-x)g of P4 consumes (160/124)(2-x)g of O

2

According to the data(96/124) x + (160/124)(2-x) = 2

⇒ x = 1.125g (P4 used in 1st reaction), (2-x) = 0.875g(P

4 used in the 2nd reaction)

1st eaction: 124g of P4 produces 220g of P

4O

6

So 1.125g of P4 produces 1.995g of P

4O

6

2nd reaction: 124g of P4 produces 284g of P

4O

10

0.875g of P4 produces 2.005g of P

4O

10

26

Note that the total mass of products = 1.995 + 2.005 = 4g (law of conservation of mass)Alternative method:Let all 2g of P

4 is converted to P

4O

6. Let us calculate the amount of O

2 needed and amount of P

4O

6 formed

in the first step.124g of P

4 requires 96g of O

2

2g of P4 requires 1.548g of O

2

Hence the amount of O2 left out in this reaction = 2-1.548 = 0.452g

This amount will be used to convert a part of P4O

6 to P

4O

10

124g of P4 forms 220g of P

4O

6

2g of P4 forms 3.548g of P

4O

6

P4O

6 + 2O

2 → P

4O

10

2×32 g of O2 requires 220g of P

4O

6

0.452g of O2 requires 1.553g of P

4O

6

So the amount of P4O

6 left = 3.548-1.553=1.995g

64g of O2 forms 284g of P

4O

10

0.452g of O2 will form 2.005g of P

4O

10

22. Li2CO

3 → Li

2O + CO

2

74g 22400mL(NTP)Note that K

2CO

3 in the mixture does not decompose.So from the above reaction we can find the mass of

Li2CO

3 present in the mixture. The mass of K

2CO

3 can be found out.

23. NaCl + AgNO3 → AgCl + NaNO

3

58.5g 143.5gMCl + AgNO

3 → AgCl + MNO3

yg 143.5g1st experiment: Let NaCl = x g, and so MCl = (1-x)g

(143.5/58.5) x + (143.5/y) (1-x) = 2.567 (1)

2nd experiment: MCl gasAgNO3

AgCl1mole(yg) 1mole

y g of MCl produces 143.5g of AgCl(1-x)g of MCl produces (143.5/y)(1-x) g of AgClAccording to the data; (143.5/y)(1-x) = 1.341 (2)Solving equations (1) and (2), we get x and y. y= 53.5. So the MCl is NH

4Cl

which on heating gives NH3 and HCl.

24. CxH

yO

z + ½ (2x+y/2 -z) O

2 → x CO

2 + y/2 H

2O(l)

1mL 1/2(2x+y/2-z)mL x mL zeroVolume of CO

2 = 20mL

1mL of hydrocarbon gives x mL of CO2

10mL of hydrocarbon gives 10x mL of CO2.

10x =20 ⇒ x = 2 (1)1mL of hydrocarbon comsumes (x + y/4 -z/2) mL of O

2

10mL of hydrocarbon consumes 10(x+y/4-z/2)mL of O2

Volume after explosion = CO2 + excess O

2

10x + [ 100 - 10(x +y/4 -z/2)] = 90 (2)Molecular mass = 2×23 =4612x + y + 16z = 46 (3)

Solving these three equations simultaneously we get, y=6 and z =1So the molecular formula = C

2H

6O

27

SUPPLIMENTARY PRACTICE QUESTIONS(UNSOLVED)1. Chlorophyl, the green colouring matter of plants responsible for photosynthesis, contains 2.68% ofmagnesium by mass. Calculate the number of magnesium atoms in 2.0 g of chlorophyll.(1.33 X 1021)2. When 10.0g of marble chips(CaCO

3) are treated with 50 mL of HCl(d = 1.069 g/mL). the marble

dissolves giving a solution and releasing CO2 gas. The solution weighs 60.4g. How many litres of CO

2 gas

are released ? The density of the gas is 1.798 g/L. (2.4 L)3. 105 mL water at 40C is saturated with NH

3 gas, producing a solution of d = 0.9 g/mL. If the solution

contains 30% NH3 by mass, calculate its volume. (166.6 mL)

4. The atomic mass of A and B are 20 and 40. If x g of A contains Y atoms, how many atoms arepresent in 2x g of B ? ( y atoms)5. Density of a gas relative to air is 1.17. Find the molecualr mass of the gas. [ M

air = 29 g/mol]. (33.9)

6. Equal masses of mercury and iodine were allowed to react completely to form a mixture of mercurousand mercuric iodide leaving none of the reactants. Calculate the ratio of masses of Hg

2I

2 and HgI

2 formed.

( 0.532 : 1)7. Titanium which is used to make air plane engines and frames, can be obtained from titanium tetrachloridewhich in turn is obtained from titanium oxide by the following process :

3 TiO2(s) + 4C(s) + 6Cl

2(g) --------> 3 TiCl

4(g) + 2CO

2(g) + 2CO(g)

A vessel contains 4.15 g of TiO2, 5.67 g of C and 6.78 g of Cl

2, suppose the reaction goes to completion

as written, how many grams of TiCl4 can be formed ? (Ti = 48) (9.063g)

8. When you see the tip of a match fire, the chemical reaction is likely to beP

4S

3 + 8O

2 -----> P

4O

10 + 3SO

2What is the minimum amount of P

4S

3 that would have to be burned to produce at least 1.0 g of P

4O

10 and

at least 1.0 g of SO2. (1.14 g)

9. The action of bacteria on meat and fish produces a poisonous compound called cadaverine. It is58.77% C, 13.81%H and 27.42% N. Its molar mass is 102 g/mol. Determine the molecular formula ofcadaverine. (C

5H

14N

2)

10. On combustion analysis, a 0.45 g sample of caproic acid ( contained only C, H and O) gives 0.418g of H

2O and 1.023 g of CO

2. What is the empirical formula of caproic acid? If the MM of caproic acid

is 116 amu, what is its MF ? ( C3H

6O, C

6H

12O

2)

11. A sample of pure metal M weighing 1.35g was quantitatively converted to 1.88 g of pure MO.Calculate atomic mass of M. (41)12. Cu

2S and M

2S are isomorphous. The % of S in former is 20.14 and in the latter 12.94. Atomic mass

of Cu = 63.57. Calculate the atomic mass of M. (108.16)13. A compound which contains one atom of X and two atoms of Y for each three atoms of Z is madeby mixing 5.00 g of X, 1.15 X 1023 atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 g of compoundresults. Calculate atomic mass of Y if the atomic mass of X and Z are 60 and 80 respectively. (70)14. A 0.6025 g sample of a chloride salt was dissolved in water and the chloride precipitated by addingexcess of silver nitrate. the precipitate of AgCl was filtered, washed, dried and found to weigh 0.7134 g.Calcualte the pecentage of chlorine in the sample. (29.29%)15. A 0.4852 g sample of iron ore is dissolved in acid, the iron oxidised to +3 state, and then precipitatedas hydrated oxide, Fe

2O

3.xH

2O. The precipitate is filtered, washed and ignited to Fe

2O

3 which is found to

weigh 0.2481 g. Calculate the percentage of iron in the sample. (35.77%)16. A mineral consists of an equimolar mixture of the carbonate of two bivalent metals. One metal ispresent to the extent of 13.2% by mass. 2.58 g of the mineral on heating lost 1.233 g of CO

2. Calculate the

% by mass of the other metal. (21.68%)17. A 10 g sample of a mixture of CaCl

2 and NaCl is treated with Na

2CO

3 to precipitate calcium as

calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is

1.62 g. Calculate the % by mass of NaCl in the original mixture. (67.9%)18. 10 mL of a mixture of CO, CH

4 and N

2 exploded with excess of oxygen gave a contraction of 6.5

mL. There was a further contraction of 7 mL, when the residual gas treated with KOH. What is thecompostion of the original mixture. (CO = 5 mL, CH

4 = 2 mL)

19. When 100 mL of a O2 - O

3 mixture was passed through turpentine, there as reduction of volume by

20 mL. If 100 mL of such a mixture is heated , what will be the increase in volume ? (10 mL)20. 40 mL of a mixture of H

2, CH

4 and N

2 was exploded with 10 mL of O

2. On cooling, the gases

occupied 36.5 mL. After treatment with KOH the volume reduced by 3 mL and again on treatment withalkaline pyrogallol, the volume further decreased by 1.5 mL. Determine the composition of the mixture.

( H2 = 12.5%, CH

4 = 7.5%)

21. 5 mL of a gaseous hydrocarbon was exploded to 30 mL of O2. The resultant gas, on cooling is found

to measure 25 mL of which 10 mL are absorbed by KOH and the remainder by pyrogallol. Determine themolecular formula of the hydrocarbon. (C

2H

4)

28

SOLUTION STOICHIOMETRY(VOLUMETRIC ANALYSIS)

MOLARITY AND PERCENT STRENGTH

Concentration or strength of a solution:When a solution is prepared by dissolving a solute in a solvent, it is always necessary to know how much

of solute is present in how much of the solvent i.e their relative proportions in the mixture(solution). If moreamount of solute is present in a given solution, it is called a concentrated solution and if less amount of solute ispresent, it is called the dilute solution. In order to know the concentration or strength of the solution i.e to knowhow concentrated or how diluted a solution is, different terms are used. These terms are Molarity, Molality,Normality and % strength. Note that the volume of a solution should not be confused with gaseous volume. Boththe types of volumes are measured in the same unit i.e mL or cc and L. Gas laws are applicable to gaseousvolume but not to volume of solutions.Preparation of a solution of desired concentration

An accurately weighed amount of solute(solid or liquid) is taken in a measuring flask of a fixedvolume(capacity). Measuring flasks of various capacities such as 2 litres, 1 litre, 500 ml, 250 ml, 100 ml etc. areavailabe in the laboratory. Each flask has a mark at the upper part of the stem of the flask. The solute is taken ina measuring flask first and slowly distilled water is added from a wash bottle and the solute is allowed to mix withthe solvent by gentle shaking of the flask. Water is added in small amounts slowly with constant shaking upto themark located in the upper part of the flask. Finally the flask is stopperd and is shaken thoroughly. The volume ofsolution becomes exactly equal to the capcity of the flask which is written on the flask. Don't add more water soas to cross the mark on the glass. If you do so then the total volume will be more and cannot be properly recorded.Note that the volume of the solution is the total volume of solute and solvent. In case of solid solutes, particularlyionic solutes like NaCl, Na

2CO

3 etc. the dissolution involves the formation of free ions which occupy the intermolecular

space of the solvent molecules. Hence the volume change that occurs when such a solute dissolves in a solventis very small. Hence in such case the volume of the solution is approximately equal to the volume of the solvent.Of course this is true for dilute solutions prepared from solid solutes. However when a liquid solute such asH

2SO

4 is used, the addition of water brings about change in the volume of solution compared to the volume of

solvent. This is because H2SO

4 exists in the liquid state and itself possesses a lot of free intermolecular volume

like the solvent molecules. So the volumes of solute and solvent approximately add to give the volume of thesolution.

MOLARITY:Molar Solution:(1M or M solution)

If one mole of solute is present in 1 litre(1000 mL) of solution, it is called a Molar(1M or M)solution.One mole of H

2SO

4(98 gms) present in 1 litre solution is called 1M H

2SO

4.Similarly when 40gms(1mole) of

NaOH is used to prepare 1000 cc of solution, it is Molar(M) or 1M solution of NaOH.____________________________________________________________________________________SAQ 1: How much of solute in gm. will be required per litre of solution to prepare 1M solution in the followingcases.

(i)Na2CO

3(ii)HCl (iii)KMnO

4(iv)FeSO

4

SAQ 2: The solutions have the following amounts of the solutes per litre of the solution. What are their molarities.(i)56gms of KOH (ii)106gms of Na

2CO

3(iii)98gms of H

2SO

4

____________________________________________________________________________________Molarity:

The number of moles of solute present in 1000 mL(1 litre) of the solution is called the molarity of asolution.

4gms(0.1 mole) of NaOH present in 1000 cc of the solution is called 0.1M solution or the molarity of thesolution is 0.1M.

29

2gms(0.05mole) of NaOH present in 1000cc of the solution is called 0.05M solution.80gms(2 moles) of NaOH present in 1000cc of the solution is called 2M solution and so on.

SAQ 3: Find out the molarity of the solutions if the following amounts are dissolved to make 1 litre solution ineach case.

(i)49gm of H2SO

4(ii)3.65gms of HCl (iii)0.7kg of H

2SO

4

SAQ 4: A certain bottle of H2SO

4 is marked as 2M H

2SO

4 and another bottle of H

2SO

4 is marked 0.5M H

2SO

4.

What do they mean? Which is a diluted solution and which concentrated? How many times the one is moreconcentrated than the other.

Let us consider another situation. Supposing you do not have a 1 litre measuring flask with you, in steadyou have a 500 mL flask with you. How can you prepare a 1M solution of H

2SO

4? This is very simple. If 1

litre(1000cc) of solution requires 98gms of H2SO

4 to make 1M solution, 500ml of solution will need half of it i.e 49

gms. So 49gms of H2SO

4 taken to make 500 cc solution has a molarity 1M.

SAQ 5: (i)How much of NaOH will be needed to produce a 1M solution by using 250ml measuringflask?

(ii)How much of H2SO

4 will be needed to produce a 1M solution by using 100ml measuring

flask?

Now let us consider still a different situation. Supposing you want to prepare 0.2M solution of NaOH and youhave 500ml measuring flask, how shall you do? This is again very simple.

If 1000ml solution needs 0.2 mole i.e 0.2×40=8gms of NaOH for making 0.2M solution.500ml solution will need 8/2 = 4gms of NaOH to make the same 0.2M solution.

So in all these cases you have to find out in an unitary method how much of the solute is required for the givenvolume of the solution.SAQ 6: Calculate the amounts of solute present in each case of the following solutions.

(i)250ml of 0.5M H2SO

4(ii)2litres of 0.1M NaOH

(iii)100ml of 4M Na2CO

3(iv)500ml of 12M HCl

FORMULA METHOD:

Molarity =number of moles of solute

volume of solution in litre

⇒ Number of moles of solute = Volume of solution in litres × Molarity

Example: How many moles and what mass of each of the following solutes are required to prepare thefollowing solutions.

(i)500 ml of 0.8M H2SO

4(ii)100ml of 2M Na

2CO

3

(iii)2 litres of 0.1 M HCl (iv)250ml of 10M NaOHSolution: Unitary Method

(i) 1litre of 0.8M solution of H2SO

4 contains 0.8 mole of H

2SO

4

So 500ml or 0.8l of this solution must contain 0.5 ×0.8= 0.4 moleFormula method:

volume of the solution =0.5 L and molarity is 0.8M.So the number of moles of H

2SO

4= 0.5×0.8=0.4

Once you find the number of moles of the solute, it is an easy task to find the mass in gm by multiplying thenumber of moles with molecular mass.

mass of 0.4 mole of H2SO

4 = 0.4×98= 39.2gms.

(ii) Moles of Na2CO

3= volume in litre × molarity= 0.1 litre× 2= 0.2.

Mass= 0.2×106 (M.M)= 21.2gms.(iii) Moles of HCl = 2× 0.1 =0.2, so the mass of HCl= 0.2×36.5=7.3gms.(iv) Volume in litre= 0.25 l , Molarity=10, So number of moles of NaOH= 0.25×10=2.5

So the mass of NaOH=2.5×40=100gmsSAQ 7: Find out the mass of the following solutes required to prepare 2 litres of M/20 solution of

(i)H2SO

4(ii)KBr

30

____________________________________________________________________________________DETERMINATION OF MOLARITY OF A SOLUTION:Example: Find the molarity of a H

2SO

4 solution, 7gms of which is present in 250 cc solution.

Solution: Unitary method:7gms of H

2SO

4 = 7/98=1/14 mole

250cc of the solution contains 1/14 mole1000cc of the solution contain [(1/14)1000]/250 = 4/14= 0.286 mole

So the molarity of the solution= 0.286MFormula MethodWe know the formula that

volume in litre= Number of moles Molarity

In the above example , we first have to find the number of moles of solute.7gms of H

2SO

4 = 7/98=1/14 mole, volume of solution in litre= 0.25 l

So Molarity = 1/14/0.25= 0.286M.(same as obtained by unitary method)SAQ 8: Find out the molarity of the following solutions.

(i)2gms of NaOH present in 100ml solution(ii)5.3gms of Na

2CO

3 present in 2liters of solution

(iii)0.365gm of HCl present in 250ml of solution

MILLIMOLE(MMOL) METHOD:Note that in stead of solving these problems in terms of moles which often appears as a fraction, it is

more convenient to work in terms of millimoles(mmols). You know that1 mole=103 mmoles=1000 mmoles

volume in litre= Number of moles Molarity

Molarity Number of mmoles = volume in ml or ccVolume in litre × molarity = number of molesVolume in ml(cc) × molarity = number of mmoles

If we express volume of the solution in ml(cc) in stead of litre, we get millimoles(mmoles) in stead of moles.

Example: Find out the molarity of 200 ml solution containing 0.7 gm of H2SO

4

Solution: Unitary Method: 200 ml of solution contains 0.7/98 mole of H2SO

4

1000ml of solution contains 0.7/98 X1000/200 =5/140=1/28moleSo the molartiy of the solution is 1/28M

Formula method(1): volume in litre= Number of moles Molarity

The number of moles= 0.7/98= 1/140, volume of the solution=200ml=0.2 litreHence Molarity= 1/140/0.2= 1/28M

Formula method(2): Molarity Number of mmoles = volume in ml or ccThe number of moles= 0.7/98, Hence number of mmoles = 103 × no. of moles= 1000 ×0.7/98=100/14;

volume in ml= 200ml molarity =100

14 X200= 1/28 M

(You can adopt any method you like)SAQ 9: Find out the number of mmoles of the solute present in the following solutions.

(i)100ml of 0.5M NaOH solution (ii)3litres of 0.04M H2SO

4 solution

(iii)500ml of 3M Na2CO

3 solution

SAQ 10: Find out the masses of solute in each case of SAQ 9.SAQ 11: Find out the number of Na+ ions present in 100ml of M/10 Na

2CO

3 solution.

SAQ 12: You are given 500ml of 2M solution of H2SO

4. You are asked to divide this solution in two bottles, one

31

containing 200ml and the other containing 300ml. What will be molarity of H2SO

4 in the first and second bottles?

PERCENT STRENGTHLike molarity, there is another way to express the concentration of a solution. This is called percent

strength. There are two types of percentages used to find the concentration of a solution. These are(i)Volume percent(w/v) and (ii) weight percent(w/w)

(i) Volume Percent(w/v):The mass of the solute( in gram) present in 100ml of the solution is called percent strength of the

solution by volume. This pecentage is denoted by the symbol w/v i.e weight of the solute(w) and volume of thesolution(v) are considered. Say for example, 2gms of sugar used to prepare 100ml solution is a 2% solution byvolume(w/v). In molarity, we have to find out the number of moles in stead of the mass in grams and moreover thevolume of the solution is taken to be 1000ml to find molarity whereas it is 100ml in percent strength by volume.Example: Find the percent strength of a solution containing 49gms of H

2SO

4 in 10 litres of solution.

Solution: In 10 litres i.e 10,000 ml of solution, the mass of solute=49gmsSo in 100ml of solution, the mass of solute would be (49/10,000) ×100=0.49So percent strength by volume is 0.49% (w/v)

SAQ 13:Find the percent strength of the following solutions by volume.(i)NaOH solution containing 25gms of NaOH in 500ml solution(ii)Na

2CO

3 solution containing 0.005moles of Na

2CO

3 in 1 litre solution

(iv)H2SO

4 solution which has a molarity equal to 0.05M

FINDING PERCENT STRENGTH FROM MOLARITY OF SOLUTION:Example: Find the percent strength by volume of a NaOH solution whose molarity is 0.01M.

A 0.01M solution means, 1000ml of the solution contains 0.01mole of NaOH i.e 0.01×40=0.4gm.If 1000ml of the solution contains 0.4gm of NaOHThen 100 ml solution must contain (0.4/1000)×100= 0.04gm of NaOH.So the percent strength of the solution by volume= 0.04%

SAQ 14: Find the percent strength by volume in respect of the following solutions.(i)3M solution of Na

2CO

3(ii)0.5M solution H

2SO

4(iii)500ml of 2M KOH

solution.

(ii) Weight(Mass) Percent (w/w)The mass of the solute in grams that is present in 100gms of the solution is called the weight(mass)

percent of the solution. Note that in this case, for both solute and solvent we consider the mass. Therefore it isdesignated as w/w i.e weight/weight method. Let us take an examle. Supposing we took 2gms of sugar andadded with it 98gms of water, so as the make the total mass of the solution 100gms. Note that here we will notuse any measuring flask which measures exact volume of the solution. In this case the percent strength byweight is 2%(w/w).Example: 500gms of H

2SO

4 is mixed with 1kg of water to prepare a solution of H

2SO

4. Find the percent

strength by weight.Solution: Mass of the solute=500gms, Mass of the solvent= 1kg=1000gm.

So the mass of the solution= 500+1000=1500gms.If 1500gms of the solution contains 500gms of H

2SO

4

Then 100gms of the solution must contain (500/1500)×100=33.33gmsSo the percent strength by weight is 33.33%.

SAQ 15: 15gms of pure HNO3 is mixed with 150gms of water to prepare a solution. What is its percent strength

by weight?

FINDING THE WEIGHT PERCENTAGE IF VOLUME OF THE SOLUTION IS KNOWN INSTEAD OF MASS BY USIND DENISTY OF THE SOLUTION

For finding the weight percentage from the volume of the solution, we need to know the density of thesolution.

32

Example: Find out the percent strength by weight of a H2SO

4 which contains 49gms of H

2SO

4 in

500ml solution. The density of the solution is 1.32gm/ccSolution: We know that density of the solution= (mass of the solution)/(volume of the solution)d=m/v; Volume of the solution=500ml(data);

so mass= d ×V =1.32gm/ml × 500ml= 660gm.Mass of the solution is 660gms and out of which 49gms is the mass of solute.660gms of the solution contains 49gms of H

2SO

4

So 100gms of the solution must contain (49/660) ×100= 7.42gmsSo the percent strength by weight is 7.42%(w/w).

SAQ 16: Find the percent strength by volume and by weight of 18M molar solution of H2SO

4. The specific

gravity of the solution is 1.87.SAQ 17: A bottle of commercial sulphuric acid has a density 1.787gm/ml. It is levlled as 86% by weight.Find the molarity of the acid and also the percent strength by volume.

STOICHIOMETRIC CALCULATION BASED ON MOLARITY AND STRENGTH OFSOLUTION

In the previous chapter we have discussed molarity and percentage strength of the solutions. Nowwe shall use it in solving stoichiometric problems. Read this example.Example 1: How much zinc will completely react with 200ml of 2M H

2SO

4 solution? Also calculate

the volume of hydrogen gas evolved at 270C and 900mm pressure.(Zn=65)Solution: M.M of H

2SO

4 is 98.

1000ml of a 2M solution contains 2 moles i.e 2×98gms of H2SO

4

So 200ml of such solution must contain (2×98)/(1000)×200= 39.2gms.Alternative (mmole) method:

Number of mmole of H2SO

4 = volume in ml × molarity= 200×2=400

Mass of H2SO

4 = 400 × 10-3 × 98 = 39.2gms.

( Note that when mmole is converted to mol, a factor of 10-3 is multiplied)Now let us write the balanced equation.Zn + H

2SO

4 ZnSO

4 + H

2

65 g 98 g 22.4L98gms of H

2SO

4 completely reacts with 65gms of zinc

39.2gms of H2SO

4 must react with (65/98)×39.2= 26gms of Zn.

98 gms of H2SO

4 produces 22.4 litres of H

2 gas at NTP

So 39.2 gms of H2SO

4 must produce (22.4/98)×39.2= 8.96 litres of O

2 at NTP.

(Note that you can also take the data of zinc found above to calculate the volume of H2)

Now we have to convert the volume from NTP to the given conditions.

760 mm X 8.96 l273A

=900mm X V2

(273+27)A ⇒ V2 = 8.31 litres.

Example 2: What volume of 0.3M H2SO

4 is required to exactly neutralize 200ml of 0.5M NaOH

solution?Solution:

Number of mmoles of NaOH= 200 × 0.5 = 100Let us write the balanced chemical equation.2NaOH + H

2SO

4 Na

2SO

4+ H

2O

2 moles 1 mole2 mmoles 1 mmole2 mmoles of NaOH reacts with 1 mmole of H

2SO

4.

So 100 mmoles of NaOH must react with 50 mmoles of H2SO

4.

mmoles = volume in ml × Molarity ⇒ volume = 50/0.3= 166.67 ml.(You are advised also to solve this problem in terms of mass and check your answer. For that you have toconvert the mmoles of NaOH to its mass and then find out the mass of H

2SO

4 required from the equation.

Finally the volume of the 0.3M H2SO

4 solution required is found out as usual by the unitary method).

SAQ 18: How many ml of 3M HCl solution should be needed to react completely with 16.8gms of NaHCO3?

SAQ 19: What volume of 0.025M HBr solution is required to neutralize 25.0 ml of 0.02M Ba(OH)2 solution?

SAQ 20: Calculate the molarity of a HCl solution if 2.5ml of the solution took 4.5ml of 3M NaOH solution

33

for complete neutralization.SAQ 21: What volume of M/10 solution FeSO

4 solution will be required to completely react with

1.58gms of KMnO4 in dilute H

2SO

4 medium?

PROBLEMS BASED ON VOLUME STRENGTH OF HYDROGEN PEROXIDE SOLUTIONWhen we buy hydrogen peroxide from the market, we get them usually in dark coloured bottles which

are labelled as “10 Volumes” or “20 Volumes” or in general “x volumes”. In fact these are not pure H2O

2,

rather are aqueous solutions.“x Volumes” H

2O

2 solution means if 1 volume of such H

2O

2 solution(liquid) at NTP is decomposed,

‘x’ volumes of O2 gas are evolved. Note that here the two volumes are of different types. One is volume

of the solution(1 volume i.e 1L or 1 mL) and the other is gaseous volumes(i.e ‘x’ L or ‘x’ mL).“10 Volumes” of H

2O

2 means, 1 mL of such a solution on decomposition will produce 10 mL of O

2

gas at NTP. This is called volume strength of H2O

2. Volume strength can be converted to percent strength

by volume or vice versa. Look to the following examples.Example 1: Calculate the percent strength by volume of a H

2O

2 solution if its volume strength is “22.4

Volumes”.Solution: 2 H

2O

2 → 2 H

2O + O

2

2(2+32)g 22.4L(NTP)“22.4 Volumes” solution means 1 mL of such a solution liberates 22.4mL of O

2 at NTP.

1mL of the solution liberates 22.4mL of O2 at NTP

So, 100mL of the solution will liberated 2240mL of O2 at NTP

From the balanced equation,22400mL of O

2 is liberated by 2×34g of H

2O

2

So, 2240mL of O2 is liberated by 6.8g of H

2O

2

Hence 100mL of the solution contains 6.8g of H2O

2. So the percent strength is 6.8%(w/v).

Example 2: 20mL of a solution of H2O

2 labelled “15 volumes” just decolorized 100mL of KMnO

4 acidified

with dil. H2SO

4. Calculate the mass of KMnO

4 in the given solution.(K=39, Mn=55)

Solution: “15 volumes” means 1mL of the solution liberates 15 mL of O2 gas at NTP.

So 20mL of this solution will liberate 20×15 = 300mL of O2 at NTP.

2KMnO4 + 3H

2SO

4 + 5 H

2O

2 → K

2SO

4 + 2MnSO

4 + 5O

2 + 8H

2O

2×158 5×22400mL(NTP)5×22400mL of O

2 is produced by 2×158g of KMnO

4

So, 300mL of O2 is produced by 0.846g of KMnO

4

Hence the mass of KMnO4 required is 0.846g.

SAQ 22: Which is more concentrated H2O

2 solution, “10 volumes” or “20 volumes” and why?

SAQ 23: Calculate the % strength by volume of a H2O

2 solution labelled as “20 volumes”

SAQ 24: Find the volume strength on the basis of available oxygen of a H2O

2 solution which is 12% by

volume.SAQ 25: 200 mL of 0.1M BaCl

2 was mixed with 100 mL of 0.05M Na

3PO

4. What mass of barium phosphate

will be formed ? (Ba = 137)

PRACTICE QUESTIONS1. Find the number of moles of solute present in each case

(i)1 litre of 2M H2SO

4(ii)200ml of 0.02M HNO

3(iii)3.5litres of 17M HCl

(iv)500ml of 0.1M Na2CO

3

2. Find the mass of the solute required to prepare the following solutions.(i)750ml of 5M KOH (ii)2 litres of 1M H

2SO

4(iii)200ml of 0.002M HCl

(iv)20ml of 18M H2SO

4

3. Find the Molarity of the following solutions.(i)1gm of NaOH present in 50ml solution. (ii)2.12gms of Na

2CO

3 present in 250ml

solution (iii)196gms of H2SO

4 present in 2.5litres solution

4. What is the molarity of NaOH in a solution which contains 24.0gms of NaOH dissolved in 300ml solution?5. What volume of 3M NaOH solution can be prepared with 84.0g NaOH?6. What volume of 1.71M NaCl solution contains 0.2 mol NaCl?7. 0.585gm of NaCl is present in 50ml solution. What is the strength of the solution in percent by volume?8. What is the molarity of a 4% NaOH solution by volume?

34

9. 9.8gms of H2SO

4 is present in 25ml solution. The density of the solution is 1.7gm/ml. What is the percent

strength by weight.10. How many grams of a 5% NaCl solution by weight are necessary to yield 3.2g of NaCl?11. To prepare 100gm of 19.7% by weight solution of NaOH, how many g of each of NaOH and H

2O are

needed?12. What volume of dilute nitric acid of density 1.11g/mL which is 19% HNO

3 by weight contains 10g

HNO3?

13. How many gm. of conc. hydrochloric acid which is 37.9 % by weight will contain 5.0gms of HCl?14. An aqueous solution of HCl contains 28% HCl by weight and has a density of 1.2gm/ml .Find the molarityof the solution.15. What volume of 96% H

2SO

4 solution(density 1.83gm/ml)is required to prepare 2.0L of 3M H

2SO

4 solution?

16. Calculate the mass of 80% sulphuric acid by weight required to completely react with 25g of CaCO3.(30.62g)

17. Calculate the (i)mass of MnO2 and (ii)the volume of hydrochloric acid having density 1.2g/mL and

containing 50% HCl by weight needed to produce 2L of chlorine gas at NTP? ( 7 . 7 6 g ,21.7mL)18. 3.5g of MgCO

3 were added to double its mass of H

2SO

4 solution. After complete reaction, 0.5g of

MgCO3

remained unreacted. Calculate percent strength of H2SO

4 by weight and volume of CO

2 formed

at 270C and 760mm pressuire.(50%)

19. Calculate the mass of NaCl formed when 50g of Na2CO

3 reacted with 100g of 50% HCl by weight. Also

calculate the mass of excess reagent which remained unreacted. (55.18g, 65.57g ofHCl)20. A solution of nitric acid contains 65% HNO

3 by weight. What mass of this acid will be necesary to

dissolve 2g of zinc oxide. (Zn =65)(4.78g)21. Most commercial HCl is prepared by heating NaCl with conc. H

2SO

4. How much of sulphuric acid

containing 90% H2SO

4 by weight is needed for the production of 100Kg of conc. HCl containing

42% HCl by weight.(62.64Kg)

22. 10mL of a solution of H2O

2 labelled “20 volumes” just decolorizes 50mL of KMnO

4 solution acidified

with dil. H2SO

4. Calculate the amount of KMnO

4 present in the solution.

(0.564g)23. How many mL of water must be added to 250mL of 0.85M HCl to dilute the solution to 0.25M?(600mL)24. Calculate the molarity of H

3PO

4 solution if 25mL of this solution is completely neutralised by 45mL of

0.85 M/10 Ba(OH)2 solution. (0.022M)

25. Calculate the molar concentration of each of the ionic species in solution.(a)250mL of 2M BaCl

2 is diluted to 750mL

(b)200mL of 2M H2SO

4 is mixed with 100mL of 0.5M NaOH and the mixture diluted to 500mL

(c)50mL of 0.12M Fe(NO3)

2 + 100mL of 0.1M FeCl

2 + 50mL of 0.26M Mg(NO

3)

2(BITS-1990)

26. What volume of 0.5M K2Cr

2O

7 solution will react completely with 100mL of M/10 (NH

4)

2C

2O

4 solution

in acidic medium. (Cr =52, K=39)(6.66mL)27. Calculate the pecent of BaO in 35g of a mixutre of CaO and BaO which just reacts with 150mL of 6M

HCl.(Ba=137) (43.91%)28. A solution of density 1.6g/mL is 67% by weight. What will be % strength by weight of solution if it isdiluted to have density 1.1 g/mL?(16.24%)

RESPONSE TO SAQs(Solutions: Molarity)

SAQ 1: In each case we require 1 mole or gm molecular mass of the solute(i) M.M of Na

2CO

3 =106, so we need 106gms per litre

(ii) M.M of HCl = 36.5, and so we need 36.5 gms per litre(iii) M.M of FeSO

4 = 56+32+64=152, so we need 152gms per litre

SAQ 2: (i) 1M(one Molar or Molar) since 1 mole of KOH=39+16+1=56(M.M)

35

(ii) 1M since M.M of Na2CO

3 = 46+12+48=106(1 mole)

(iii) 1M since M.M of H2SO

4 =98(1 mole)

SAQ 3: (i) 98 gms of H2SO

4=1 mole, so 49 gms of H

2SO

4 = 0.5 mole.

Since 0.5mole of it is present in 1 litre solution, the molarity = 0.5M(ii) 36.5gms of HCl= 1 mole, so 3.65gms of HCl= 1/10=0.1 mole

Since 0.1 mole of HCl is present in 1 litre solution, the molarity=0.1M(iii) 98gms of H

2SO

4=1 mole, so 0.7kg i.e 700gms of H

2SO

4=700/98=7.143moles

Since 7.143moles of H2SO4 is present in 1 litre solution, the molarity=7.143MSAQ 4: The bottle marked 2M H

2SO

4 means, there would be 2 moles(2X98=196gms) of H

2SO

4 per litre of the

solution. However in the bottle marked 0.5M, there would be 0.5mole(0.5X98=49gms)of H2SO

4 per litre of

solution. So the second bottle of H2SO

4 is more diluted and the first bottle is more concentrated. The first solution

is 4 time more concentrated(49X4=196) than the second solution.SAQ 5:(i) 1000ml of solution needs 40gms(1mole) of NaOH to prepare 1M solution

So 250ml of solution must need (40/1000)X250=10gms to prepare the same 1M solution(ii) 1000ml of solution need 98gms(1mole) of H2SO4 to prepare 1M solutionSo 100ml of solution will need (98/1000)X100=9.8gms to prepare the same 1M solution.

SAQ 6: (i) 1000ml of 0.5M solution contains 0.5mole i.e 0.5X98=49gms of H2SO

4

So 250ml of 0.5M solution must contain (49/1000) X 250= 12.5gms.(ii) 1000ml of 0.1M NaOH solution contains 0.1mole i.e 0.1X40=4gms of NaOH

So 2litres i.e 2000ml of 0.1M solution must contain (4/1000)X2000= 8gms.(iii) 1000ml of 4M solution of Na

2CO

3 contains 4 moles i.e 4X106=424gms

So 100ml of 4M solution must contain (424/1000)X100=42.4gms.(iv) 1000ml of 12M solution of HCl contains 12moles i.e 12X36.5=438gms of HCl

So 500ml of 12M solution must contain 438/2=219gms.SAQ 7: Here we shall have to find the mass of the solute. Let us first find the number of moles and then mass byusing the formula method discussed in the examples in the text.

(i) Number of moles= volume in litre X molarity=2X(1/20)=1/10=0.1Mass of H

2SO

4= 0.1X98=9.8gms.

(ii) Number of moles of KBr= 2X (1/20) = 1/10=0.1 [same as (i)]Mass of KBr = 0.1X(39+80)=0.1X 119 = 11.9gms.

SAQ 8:(i) M.M of NaOH=40, 2gms of NaOH=2/40=0.05 mole,volume = 100ml=0.1 l, So Molarity= 0.05/0.1= 0.5M.

(ii) Number of moles of Na2CO

3 = 5.3/106=0.05; volume=2litres

So Molarity = 0.05/2=0.025M.(iii) Number of mole of HCl=0.365/36.5= 0.01; volume=250ml=0.25 litre

So Molarity= 0.01/0.25=0.04M.SAQ 9: We know that No. of mmoles = Molarity X Volume in ml(cc)

(i)Number of mmoles= 100X0.5=50(ii)3000X0.04= 120 (iii)500X3=1500

You found that the number of mmoles is not a fractional value(more than 1).SAQ 10: (i)50mmoles of NaOH= 5X 10-3 mole = 0.005mole= 0.005X40=0.2gm.

(ii)120mmole of H2SO

4 = 120X10-3mole=0.12 mole = 0.12X98=11.76 gms.

(iii)1500mmoles of Na2CO

3= 1500X10-3=1.5mole = 1.5X106= 159gms.

SAQ 11: The no. of mmoles of Na2CO

3 present in the solution=100 X1/10 =10

10 mmoles= 10 X10-3 moles = 0.01mole1 mole of Na

2CO

3 contains 2 moles of Na+ ions

So the number Na+ ions present in 2 moles = 2X6.023X1023 =1.2046X1024 ions0.01 mole of Na

2CO

3 will contain 0.01 X1.2046X1024 = 1.2046X1022 ions of Na+.

SAQ 12: Since the stock(original) solution is same, the molarity will remain same. Suppose you collect 500ml ofsea water and distribute the same in several containers. Do you think that the salinity of the sea water will bedifferent in these different vessels? That means the vessel containing less solution will taste less salty and thevessel containing more solution will taste more salty? The answer is NO. They will all taste same, because theconcentration of the original(stock) solution is same. Remember that the concentration of a solution gives thequality of the solution i.e how much of solute present per a fixed volume of solution(say 1 litre). Once a solutionof a particular concentration is prepared, its concentration remains the same even if this stock solution is divided

36

into any number of parts.SAQ 13: (i) 500ml of the solution contains 25gms of NaOH

So 100ml of the solution must contain (25/500)X100=5gms of NaOHSo pecent strength by volume = 5%.

(ii) M.M of Na2CO

3=106, 1 mole= 106gms, so 0.005 moles = 106X0.005= 0.53gm

1 litre i.e 1000ml of solution contains 0.53gm of Na2CO3So 100ml of the solution must contain (0.53/1000)X100= 0.053gmSo percent strength by volume=0.053%

SAQ 14: (i) M.M of Na2CO

3=106; 3M solution means

1000ml of solution contains 3moles i.e 3X106=318gms of Na2CO

3

So100ml of solution must contain (318/1000)X100=31.8gms.So percent strength by volume= 31.8%

(ii) 0.5M solution means1000ml of the solution contains 0.5mole i.e 0.5X98= 49gms of H

2SO

4

So 100 ml of the solution must contain 49/10=4.9gmsSo the percent strength by volume is 4.9%

(iii) Here 500ml solution carries no relevance and it will not be used. We have to consideronly the molarity of the solution. 2M solution means

1000ml of KBr solution contains 2 moles i.e 2X(39+80)= 2X56=238gmsSo 100ml of the solution must contain 238/10= 23.8gms of KBrSo the percent strength is 23.8% by volume.

SAQ 15: The mass of the solution = 15 + 150 = 165gms.165gms of solution contains 15 gms of HNO3,100 gms of solution must contain (15/165) X 100 = 9.09 gmsSo its percent strength is 9.09%(w/w)

SAQ 16: Volume %:18M solution means 1000ml of solution contains 18moles=18X98=1764gms of H

2SO

4.

1000ml of the solution contains 1764gms of H2SO

4.

So 100ml of the solution must contain 1764/10=176.4gmsSo percent strength by volume=176.4%

Weight %:density= m/v; 1.87gm/ml= m/1000ml,⇒ mass= 1000 X 1.87=1870 gms, So mass of 1000ml of solution is 1870gms.1870gms of the solution contains 18mole i.e 18X98=1764gms of H

2SO

4

100gms of the solution must contain (1764/1870)X100=94.33gmsSo percent strength by weight is 94.33%.

SAQ 17: Let us consider 100gm of solution. d=mass/volume,So volume of 100gm solution = 100/1.787= 55.959ml .As it is 86%solution by weight.So the mass of solute(H2SO4)in present in 100gms of solution = 86gms.

Molarity:55.959ml of solution contains 86gms or 86/98moles i.e 0.8775 mole of H

2SO

4

1000ml of the solution must contain 0.8775X (1000/55.959) = 15.68molesHence the molarity of the solution= 15.68M

Percent strength by Volume:55.959ml of solution contains 86gms of H2SO4So 100ml of solution must contain (86/55.959)X100= 153.68%(w/v)

SAQ 18:NaHCO

3 + HCl ---------> NaCl + CO

2 + H

2O

8484 gms of NaHCO

3 = 1 mole ⇒ 16.8gms = 0.2mole = 0.2 X 1000=200mmole

According the balanced equation, 1 mmole of NaHCO3 will require 1mmole of HCl

So 200mmole of NaHCO3 will require 200 mmole of HCl according to the balanced

equation.We know that mmoles = volume in ml X Molarity ⇒ Volume = 200/3=66.67ml.

37

SAQ 19:Ba(OH)

2 + 2HBr ---------> BaBr

2 + 2H

2O

No. of mmoles of Ba(OH)2 = 25 X 0.02= 0.5

According the balanced equation, 1 mmole of Ba(OH)2 will need 2mmoles of HBr.

So 0.5mmole of Ba(OH)2 will need 2 X 0.5= 1mmole of HBr

So the volume of HBr solution = mmole/Molarity = 1/0.025 = 40ml.SAQ 20: NaOH + HCl ----------> NaCl + H

2O

No. of mmoles of NaOH = 4.5 X 3= 13.5; According to the balanced equation, the number ofmmoles of HCl required = 13.5; So the molarity = mmoles/ml = 13.5/2.5= 5.4MSAQ 21:

2KMnO4 + 8H

2SO

4 + 10FeSO

4 --------> K

2SO

4 + 2MnSO

4 + 5Fe

2(SO

4)

3 + 8H

2O

158(39+55+64) gms of KMnO4 = 1 mole; So 1.58gms= 0.01mole = 10 mmoles

From the balanced equation, 2 mmoles of KMnO4 requires 10 mmoles of FeSO

4

So 10 mmoles of KMnO4 will need 50 mmoles of FeSO

4

The molarity of FeSO4 = 1/10 M; So the volume = mmoles/M = 50/1/10=500ml.

SAQ 22: “20 volume” solution is more concentrated because 1mL of this solution will evolve 20mL of O2at NTP while 1mL of “10 volumes” solution will evolve 10mL of O2 at NTP.SAQ 23: H2O2 → H2O + 1/2 O2

34g 11.2L(NTP)11.2L of O2 is produced by 34g of H2O2So, 20L of O2 is produced by 60.7g of H2O2Hence 1L or 1000mL of solution contains 60.7g of H2O2So, 100mL solution contains 6.07g. So % strength by volume =6.06%(w/v)

SAQ 24: H2O2 → H2O + 1/2 O2100mL solution contains 12g of H2O21000mL(1L) solution contains 120g of H2O234g of H2O2 liberates 11.2 L of O2 at NTP120g of H2O2 liberates 39.5 L of O2 at NTPHence 1L solution liberates 39.5L of O2 at NTP. Hence its volume strength is “39.5 volumes”.

ANSWER TO PRACTICE QUESTIONSPRACTICE QUESTIONS

1. (i) No. of Moles = Molarity X Volume in litres = 2X1=2moles(ii)0.02X0.2= 0.004mole (iii)17X 3.5= 59.5 moles (iv) 0.1X0.5= 0.05 mole

2. (i)No. of moles= 5 X 0.75= 3.75moles, Mass = 3.75X(29+16+1) = 172.5gms.(ii)No. of moles= 1X2=2moles; Mass = 2X98=196gms.(iii)No. of moles = 0.002 X 0.2= 0.0004mole; Mass= 0.0004 X 36.5= 0.0146gm(iv)No. of moles= 18 X 0.02= 0.36mole, Mass = 0.36 X 98= 35.28gm.

3. (i) Number of mmoles = 1/40 X1000 = 25mmol; volume in ml= 50mlWe know the formula; Molarity = No. of mmoles/volume in ml = 25/50= 0.5M.(ii)No. of mmoles= (2.12/106) X 1000 = 20; volume in ml = 250So Molarity = 20/250 = 0.08M(iii)No. of mmoles= (196/98)X1000= 2000; volume=2500ml;Molarity=2000/2500=0.8MNote that you can solve these problems by unitary method also.

4. mmoles = (24/40) X 1000 = 600; volume =300ml, Molarity=600/300=2M5. mmoles= (84/40)X1000 = 2100; Molarity =3M, Hence Volume =mmoles/Molarity

Volume = 2100/3=700ml = 0.7litre.6. No. of mmole = 0.2 X1000= 200; Molarity=1.71M, volume= 200/1.71=116.95ml7. 50ml of solution contains 0.585gms, So 100ml must contain 1.17gms. So percent strength b yvolume is 1.17%8. 100ml of solution contains 4gms i.e 4/40=0.1 mole of NaOH

So 1000ml of solution must contain 0.1X10=1mole of NaOH, So the molarity=1M

38

9. Let us find the mass of 25ml of solution by using the density data.m/v=1.7gm/ml ⇒ m/25 = 1.7 ⇒ mass=25 X 1.7= 42.5gms42.5gms of the solution contains 9.8gms of H

2SO

4

So 100gms of the solution must contain (9.8/42.5)X100= 23.05gmsSo the percent strength is 23.05%(w/w)

10. 5gms of pure NaCl is present in 100gms solution.3gms of pure NaCl must be present in (100/5)X3= 60gms of solution.

11. The mass of NaOH needed = 19.7gms and mass of water needed=100 - 19.7= 80.3gms.12. Let the required volume = x ml, The mass of x ml = 1.11 X x gm

Since it is 19% by weigth, 100gms of solution contains 19gms of HNO3

So 1.11x gms of solution must contain (19/100) X 1.11x gms of HNO3

According to the question; this mass is 10gm. ⇒ 0.2109x = 10x = 47.4ml.

13. 37.9gms of pure HCl is present in 100gms of solutionSo 5gms of HCl must be present in (100/37.9)X5= 13.19gms of solution

14. Let us find the mass of 100gms of solution. 100/v=1.2 ⇒ v = 83.33ml83.33ml of the solution contains 28gms i.e 28/36.5 i.e 0.767 mole of HCl.1000ml of the solution must contain (0.767/83.33)X 1000 = 9.2 moles.Hence the molarity = 9.2M

15. Let us first find the moles of H2SO

4 from the second data.

Moles = 2 X3= 6moles = 6 X 98gms; (i)Let the required volume = x mls, The mass of x mls of solution = 1.83x gms;If 100gms of solution contains 96gms,So 1.83x gms of solution must contain (96/100) X 1.83x gm (ii)Equating the relations (i) and (ii), we get x= 334.7ml.

16. CaCO3 + H2SO4 → CaSO4 + CO2 + H2O100g 98g25g 24.5g Hence (80/100)x =24.5 ⇒ x = 30.62g

17. MnO2 + 4HCl `→ MnCl2 + Cl2 + 2H2O(55+32)g 4X36.5g 22.4L22.4L of Cl2 is produced by 87g of MnO2So, 2L of Cl2 is produced by 7.7g of MnO222.4L of Cl2 is produced by 4X36.5g of HCl2L of Cl2 is produced by 13.03g of HCl50g of pure acid is present in 100g solution13.03g of pure acid is present in 2606g solutiond=(26.06/V) = 1.2 ⇒ V = 21.7mL

18. MgCO3 + H2SO4 → MgSO4 + CO2 + H2O84g 98g 22.4LMass of H2SO4 solution= 7g; Mass of MgCO3 reacted = 3.5-0.5= 3g84g of MgCO3 reacts with 98g of H2SO4So, 3g of MgCO3 will react with 3.5g of H2SO47g of H2SO4 solution contains 3.5g of pure H2SO4100g of H2SO4 solution contains 50g of pure H2SO4So % strength = 50% by weight(w/w)Find the volume of CO2 formed from 3g of MgCO3 or 3.5g o H2SO4 from the above equation.

19. Na2CO3 + 2 HCl → 2NaCl + CO2 + H2O106g 2X36.5g 2X58.5g100g of Na2CO3 reacts with 73g of HClSo, 50g of Na2CO3 reacts with 34.43g of HClSince there is 50g of HCl, the limiting reactant is Na2CO3106g of Na2CO3 produces 2X58.5g of NaCl

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50g of Na2CO3 produces 55.18g of NaClMass of solvent in the beginning = 100-50=50g; Mass of pure HCl left = 50-34.43 = 15.57gSo the mass of HCl solution left = 50+15.57 = 65.18g

20. ZnO + 2HNO3 → Zn(NO3)2 + H2O(65+16)g 2X63g81g of ZnO reacts with 126g of HNO3So, 2g of ZnO reacts with 3.11g of HNO365g of pure HNO3 is present in 100g solution3.11g of pure HNO3 is present in 4.78g of solution

21. 2NaCl + H2SO4 → Na2SO4 + 2HCl98g 2X36.5g

Mass of pure HCl = 42f%of 100Kg = 42Kg = 4200g2X36.5g of HCl is produced by 98g of H2SO44200g of HCl is produced by 56383.56g = 56.383Kg of H2SO490 Kg of pure H2SO4 is present in 100Kg solution56.383Kg of pure H2SO4 is present in 62.64Kg solution.

22. 2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 5O2 + 8H2O1mL of H2O2 solution liberates 20mL; Hence 20mL of the solution will liberate 200mL5X22400mL of O2 is liberated by 2X158g of KMnO4So, 200mL of O2 is liberated by 0.564g of KMnO4So, 50mL of KMnO4 solution contains 0.564g of KMnO4.

23. mmoles of HCl = 250 X 0.85 = 212.5V X 0.25 = 212.5 ⇒ V= 850mL (note that the mmoles will remain the same after dilution)Hence volume of water added = 850-250 = 600mL

24. 2H3PO4 + 3Ba(OH)2 → Ba3(PO4)2 + 6 H2Ommoles of Ba(OH)2 = 45 X (0.85/10) = 3.8253 mmoles of Ba(OH)2 reacts with 2 mmoles of H3PO43.825mmoles of Ba(OH)2 reacts with 2.55mmoles of H3PO4So, molarity of H3PO4 solution = (2.55/25) = 0.022M

25. (a)mmoles of BaCl2 = 250X2 = 500So, mmoles of Ba2+ = 500; mmoles of Cl- = 1000[Ba2+] = (500/750) = 0.66M; [Cl-] = (1000/750) = 1.33M(b) 2NaOH + H2SO4 → Na2SO4 + H2Ommoles of H2SO4 = 400; mmoles of NaOH = 5050mmoles of NaOH will react with 25 mmoles of H2SO4Also 50mmoles of NaOH will produce 25 mmoles of Na2SO4Excess mmoles of H2SO4 = 400-25 = 375Hence mmoles of H+ = 2X375; Molarity = 750/500 = 1.5Mmmoles of SO42- = 375+25 Molarity= 400/500 = 0.8Mmmoles of Na+ = 2X25 =50 Molarity = 50/500 = 0.1M(c) mmoles of Fe(NO3)2 = 50 X 0.12 = 6; mmoles of FeCl2 = 100X0.1= 10mmoles of Mg(NO3)2 = 50X0.26 = 13mmoles of Fe2+ = 16; Hence [Fe2+] = 16/200 = 0.08Mmmoles of Cl- = 30; [Cl-] = 30/200 = 0.15Mmmoles of NO3- = 12 + 26 = 38; [NO3-] = 38/200 = 0.19M

26. First balance the equation by partial equation or ON method.K2Cr2O7 + 7H2SO4 + 3 (NH4)2C2O4 → K2SO4 + Cr2(SO4)3 + 3 (NH4)2SO4 + 6CO2 + 7H2Ommoles of (NH4)2C2O4 = 100 X 1/10 = 103 mmoles of (NH

4)

2C

2O

4 reacts with 1 mmole of K2Cr2O7

So, 10mmoles of (NH4)

2C

2O

4 reacts with 3.333 mmoles K2Cr2O7

The molarity of solution = 0.5M, Hence the volume of the solution = 3.333/0.5 = 6.666 mL27. BaO + 2HCl → BaCl2 + H2O

(137+16)g 2X36.5gCaO + 2HCl → CaCl2 + H2O

40

(40+16)g 2X36.5gLet the mass of BaO = x g and hence CaO = (35-x)g153g of BaO reacts with 73g of HClx g of BaO reacts with 0.477x g of HCl56g of CaO reacts with 73g of HCl(35-x) g of CaO reacts with 1.3(35-x)g of HClmmoles of HCl = 150X6 = 900; Hence mass of HCl = 900 X 10-3 X 36.5 = 32.85g0.477 x + 1.3(35-x) = 32.85 ⇒ x = 15.37g(BaO), % of BaO = 43.91%

28. 1mL contains 1.6g solution; Hence 1000mL contains 1600g100g solution contains 67g pure solute; Hence 1600g solution contains 1072g of pure soluteLet us add x mL of water. Hence volume after dilution = (1000 +x)mLMass of the solution = (1600+x) g; Hence the density = (1600+x)/(1000+x) = 1.1⇒ x = 5000mL = 5000g of waterSo the total mass of solution = 1600 + 5000 = 6600 g6600g of solution contains 1072g of solute100g of solution contains 16.24g of solute. Hence percent strength by weight = 16.24%(w/w)