Quantitative Chemistry

Quantitative Chemistry

Topic 1SL + HL

1.1 Mole concept and Avogadro’s Constant

• 1 mol = • Avogadro’s Constant = • NA =• Number of atoms in 12 gram 12C (def.) =• 6.02*1023 units

• Mole = n = Amount of substance, with the unit mol

• 1 mol equals:– 6.02*1023 Hydrogen atoms, H– 6.02*1023 Hydrogen molecules, H2

– 6.02*1023 Water molecules, H20– 6.02*1023 formula units of Sodium Chloride, NaCl

If you have 1 mol of something, then you have 6.02*1023 units of that.

• In 1 mol of H2-molecules there is 2 mol Hydrogen atoms 1 = 2

• In 1 mol of H2O molecules there is 3 mol of atoms; 2 mol of H-atoms and 1 mol of O-atoms 1 = 3 =2 =1

1.2 Formulas

A 12C-atom have by definition the mass =12 All other atoms or molecules masses relates to that

mass. How much heavier or lighter they are.ÞMasses of single atoms and single molecules,

single formula units are therefore called:• Relative atomic mass, Ar

• Relative molecular mass, Mr

• Relative formula mass, Mr

Relative masses have no unit in IB (in other system the unit can be u or amu)

Relative mass => Mass of 1 mol• Mass of 1 H-atom: Ar = 1.01

Mass of 1 mol H-atoms = 1.01g Atomic mass

• Mass of 1 H2-molecule: Ar = 2.016 Mass of 1 mol H2-molecule = 2.016 g Molecular mass

• Mass of NaCl = Ar = 58.44 Mass of 1 mol NaCl = 58.44 g Formula mass

Different masses, same figures

Calculate the mass of one mole of a species from its formula.

The term molar mass, M, will be used.Unit: g/mol

If you know the formula of a compound, then you know the molar mass of the

compound.

You find the masses of the atoms in the Periodic table

The Molar mass of water , H2OM = 2*1+16 = 18g/mol

The Molar mass of (NH4)2SO4

M= (14+4*1)2 + 32 + 16*4 = 132 g/mol

The Molar mass of CuSO4*5H2OM= 63.5 + 32 + 16*4 + 5(2*1 + 16) = 249.5 g/mol

Relationship between the amount of substance in moles, mass and molar mass => The table

Quantity UnitMass, m gMolar mass, M g/mol (usually known)

Mole, n mol

Downwards: divideUpwards: multiply

You can always do a unit analysis, to check your calculation

Examples

You have 34 g of Ammonia. How many mole is that?

1. Write the Formula of the compound.2. Write the table3. Fill in what you know and ?4. Calculate

1. Write the Formula of the compound.

NH3

2. Write the table

NH3

m g

M g/mol

n mol

3. Fill in what you know and ?

NH3

m 34 g

M 17 g/mol (from periodic table)

n ? mol

4. Calculate

NH3

Downwards m 34 g=> divide

M 17 g/mol

n 34/17= 2 molAnswer: 2 mol

New example

You have 0.50 mol of NaCl. How many gram is that?

1. Write formula2. Write table

NaCl

m gM g/moln mol

3. Fill in know and ?4. Calculate

NaClm ? gM 58.5 g/mol (from periodic table)

n 0.50 mol

Upwards: multiply0.5 * 58.5 = 29.25 g Answer: 29 g (significant figures)

Percent Composition of Compounds

Percent composition of a compound is usually the Mass percent of the elements in a compound, e.g. H2O M = 18 g/mol

Mass% H = 2/18 = 0.111 = 11.1%Mass% O = 16/18 = 0.888 = 88.8%(Very seldom is the % composition referred to number of atoms; e.g. H2O have 3 atoms% H= 2/3 = 67% and %O = 1/3 = 33%)

Empirical and Molecular formula• Molecular formula: Shows the actual number of each

atom/element in a compound, e.g.Ethane C2H6

Glucose C6H12O6

• Empirical formula: Shows only the ratio of the elements in a compound, e.g.Ethane CH3

Glucose CH2O

(Formulas of salts is empirical formulas)

Determining the Empirical formula

• You can determine the empirical formula from the percentage composition

1. Use the table2. Assume that you have 100g of the compound3. Calculate number of moles4. Compare Mole-ration. The ratio give the

formula

Example

A compound is found to consist of: 70.58 % C, 5.93 % H, 23.49 % O (=100%)

What’s the formula for this compound?

1. Assume 100 g of the compound.2. Calculate number of moles

C H OMass% 70.58 5.93 23.49

(=100%)

m 70.58 5.93 23.49 g (=100g)

M 12.01 1.01 16.00 g/moln 5.88 5.87 1.47 mol

Check the ratio (divide with the lowest)

C H O 5.88 5.87 1.47 1.47 1.47 1.47

4 : 4 : 1Empirical formula

C4H4O

Molecular formula

If you know the Empirical formula, C4H4O, and you know the Molar mass of the compound, 136 g/mol then you can calculate the Molecular formula.

C4H4O M = 68 g/mol To LowC8H8O2 M = 136 g/mol CorrectC12H12O3 M = 204 g/mol To High

1.3 Chemical Equations

• A chemical reaction describes what is going on in a chemical reaction.

• Shows the Reactant(s)• Shows the Product(s)• Show the Ratio between the compounds (if correct

balanced)

• Can show in which state the compounds are:(s) solid, (l) liquid, (g) gas, (aq) water solution (the most common)

Examples

Reactants ProductsPropane + Oxygen Carbon dioxide + water

Word equation gives only the compounds.Equation must be balanced

C3H8 + 5 O2 3 CO2 + 4 H2ONow we have a balanced equation = same number of all

elements on both sides of the reaction arrow. The green numbers = subscripts; cannot be changed

(a certain compound have a certain formula, if you change the formula you change the compound)

The red numbers = coefficients; changes so that the reaction will be balanced (coefficients is valid only for a specific reaction)

What does an equation tell us?

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)• The state of the compounds• That you need 5 Oxygen molecules/Propane• 1 Propane molecule will produce 3 Carbon

dioxide molecules and 4 Water molecules• Etc• Or multiples thereof

Calculate mass etc in a reactionExamples

If you start with 15 g of propane: Which mass of oxygen do you need and which mass of carbon dioxide and water do you get?

1. Write the balanced reaction equation2. Write the table3. Fill in known and ?

C3H8 + 5 O2 3 CO2 + 4 H2Om 15g ? ? ?

M 44g/mol 32g/mol 44g/mol 18g/mol

n

1. Calculate mol of propane (down =divide)2. Go to next compound on the “mol level”. Check the Ration between the

compounds.

C3H8 + 5 O2 3 CO2 + 4 H2Om 15g ? ? ?

M 44g/mol 32g/mol 44g/mol 18g/mol

n 0.34mol 1:5 1.7mol 5:3 1.02mol 3:4 1.36mol

1. Go upwards = multiply to get the masses

C3H8 + 5 O2 3 CO2 + 4 H2Om 15g 54.5g 44.9g 24.5g

M 44g/mol 32g/mol 44g/mol 18g/mol

n 0.34mol 1:5 1.7mol 5:3 1.02mol 3:4 1.36mol

More examples

Sodium reacts with chlorine and the product is sodium chloride. If you want to make 10 g of sodium chloride how many grams of sodium and chlorine do you need?

1. Write the balanced equation2. Write the table3. Fill in known and ?

2 Na + Cl2 2 NaClm ? ? 10gM 58.5g/moln

1. Mol of NaCl? Down = divide2. Go on “mol level” to other compounds.

Check Ratio

2 Na + Cl2 2 NaClm ? ? 10gM 23g/mol 71g/mol 58.5g/moln 0.17mol 2:1 0.08mol 1:2 0.17mol

1. Up = multiply

2 Na + Cl2 2 NaClm 3.9g 6.1g 10gM 23g/mol 71g/mol 58.5g/moln 0.17mol 2:1 0.08mol 1:2 0.17mol

More examples

Ammonia, NH3, can be made from Nitrogen and Hydrogen.

What mass of ammonia can you get from 168 g of Nitrogen and 24 g of Hydrogen?

1. Write balanced equation2. Write table3. Fill in known and ?4. Calculate mol

N2 + 3 H2 2 NH3 m 168 g 24 g ?

M 28 g/mol 2 g/mol 17 g/moln 6 mol 12 mol

Theoretical ratio between N2 and H2 => 1:3Actual ratio between N2 and H2 => 1:2We need 18mol of H2 if all N2 should react : Nitrogen is in excess. Hydrogen is

limiting reactant, and decides amount of product that can be formed.

N2 + 3 H2 2 NH3 m 168 g 24 g 136g

M 28 g/mol 2 g/mol 17 g/moln 12 mol 3:2 8mol

1.4 Theoretical and Percent yield

The theoretical yield in the example was 136g.In real life nothing is perfect. Some of the

reactants may not react or they react and become something else etc.

Lets say that you only end up with 105g product.Percent yield = 105/136 = 77%

GAS

A gas have under standard condition the mol volume:

22.4 dm3/mol

Reacting gas volumes

For a gas at a constant temperature and pressure, the volume is directly proportional to the number of moles of gas.

The mol ration The volume ratio

If you have 2 dm3 of Hydrogen gas, which volume of Oxygen do you need for complete reaction and which volume of water vapour will you get? (Constant:

Pressure and Volume)

2 H2(g) + O2(g) 2 H2O(g)

2dm3 ? ?Mol Ratio 2 : 1 : 2 Volume Ratio=>

2 H2(g) + O2(g) 2 H2O(g)

2dm3 1 dm3 2 dm3

Under other conditions

The ideal gas equation

pV=nRTp = Pressure; Pa V = Volume; m3 OBS! Usually in chemistry dm3

n = Mole; molT = Temperatur; K OBS! K = 273 + oC R = Gas constant; 8.314 J/mol*K

Example

6.0 g Carbon burns in Oxygen. Give the volume of formed Carbon dioxide.

Temperature = 400KPressure = 1.0 kPa =100 000Pa

1. Balanced Equation2. Table3. Fill in known and ?4. Calculate

C + O2 CO2

m 6.0gM 12g/moln 0.50 mol 1 : 1 0.50 mol

Use the gas law: pV = nRT

V = nRT/p = 0.50*8.314*400/100000 = 0.017m3

V= 17 dm3

1.5 Solutions

Solute + Solvent Solution = Substances = the liquid

You often have to dissolve chemicals to make them react

The composition of Solutions can be given in many ways, e.g.

Mass percent = Mass of substance/Mass of solution

Volume percent = volume of solute/ totale Vol Mol fraction = Xa = na/(na+nb)

Molality = moles of solute/kg of solvent

Gram/dm3 (IB sullabus)

Molarity = moles of solute/ dm3 of solution (IB sullabus)

Concentration in mol/dm3 is often represented by square brackets around the substance under consideration, eg [HCl]

Examples

You dissolve 59.5 g of Potassium Bromide in water to 750 cm3. What’s the concentration/molarity?

1. Write compound formula2. Write the table; add volume and

concentration3. Fill in known and ?

KBrm 59,5 gM 119 g/molnV 0.75 dm3

c ?OBS! Don’t mix up the V in solutions and the V in gas equation.

Gas equations relation to the table is the number of moles; n

KBr Down = dividem 59,5 gM 119 g/moln 0.500 molV 0.75 dm3

c 0.667 mol/dm3

[KBr] = 0.667 M (Capital M as a unit is often used instead of

mol/dm3, don’t mix up with M for Molar mass) Square brackets = concentration of….

More ExamplesHow many grams of NaCl is it in 200 cm3 of a solution with the concentration of 0.50 mol/dm3?

1. Write compound formula2. Write the table; add volume and

concentration3. Fill in known and ?

NaClm ? Up = multiplyM 58.5 g/molnV 0.200 dm3

C 0.50 mol/dm3

___________________________________________________

NaClm 5.9 gM 58.5 g/moln 0.10 molV 0.200 dm3

C 0.50 mol/dm3

More examples

The concentration of chloride ions, if you dissolve 10 g AlCl3 in 0.50 dm3 in water?

1. Write equation2. Write the table3. Fill in known and ?

Salts dissociate in water solutions:

AlCl3 + aq Al3+ (aq) + 3 Cl- (aq)

m 10g

M 133.5 g/mol

n 1 : 3

V

c ?

AlCl3 + aq Al3+ + 3 Cl-

m 10gM 133.5 g/moln 0.075 mol 1 : 3 0.225 mol V 0.5 dm3 0.50 dm3

c 0.15 mol/dm3 0.45 mol/dm3

OBS![Cl-] = 0.45 mol/dm3

[AlCl3] = [Al3+] = 0.15 mol/dm3

More ExamplesYou have 0.25 dm3 of NaOH solution with a

concentration of 0.50 M. Which volume of a 0.10 M Sulphuric acid do you need to neutralise the NaOH solution?

The mass of the formed Sodium sulphate?

1. Write balanced equation2. Write the table3. Fill in known and ?

2 NaOH + H2SO4 Na2SO4 + 2 H2O

M ?M 142 g/MOLn 2:1 1:1V 0.25 dm3 ?C 0.50mol/dm3 0.10 mol/dm3

up =multiply down =divide

2 NaOH + H2SO4 Na2SO4 + 2 H2OM 8.9gM 142 g/moln 0.125mol 2:1 0.0625mol 1:1 0.0625molV 0.25 dm3 0.625 dm3

C 0.50mol/dm3 0.10 mol/dm3

Dilution

You have 0.75 dm3 of 0.3 M HCl. You add 0.25 dm3 of water.

The new concentration?1. Write equation2. Write the table3. Fill in known and ?

HCl(aq) before HCl(aq) after

m Mn 0.225 mol 1:1 0.225 molV 0.75 dm3 1.00 dm3

c 0.3 mol/dm3 0.225 mol/dm3

( New volume = V = 0.75 + 0.25 = 1.00 dm3)