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MASS RELATIONS IN CHEMISTRY;STOICHIOMETRY

[MH5; Ch. 3] • Each element has its own unique mass.• The mass of each element is found on the Periodic Table under the

chemical symbol for the element (it is usually given to severaldecimal places); this may be referred to as the ATOMIC MASS(or ATOMIC WEIGHT).

• Atomic masses are relative measurements; based on how heavy anelement is compared to another element.

• The mass of 1 atom of an element may be expressed in amu, shortfor Atomic Mass Units.

• The modern scale of atomic masses is based on the most commonisotope of carbon; 12 C........

• This isotope is assigned a mass of 12 amu:

Mass of one 12 C atom = 12 amu (exactly) So......... 12

1 amu = 1 x mass of one C atom 12 6

• When determining atomic weight, we take a large number ofatoms with isotopes present in their natural abundances.

• Atomic weights of all elements except 12 C are subject toexperimental error; (12 C has atomic mass of exactly 12 amu).

• For elements that are isotopically pure (consist of one isotopeonly; monoisotopic), we may talk of atomic mass (mass of one atom)

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EXAMPLE: Sodium, 2311 Na (only natural isotope), At. Wt = 22.9898

• So the mass of 1 atom of Na is 22.9898 amu.• Notice that the atomic mass of the element is very close to its

mass number....• For elements that naturally contain more than one isotope, we can

determine an average atomic mass; an average of the isotopicmasses weighted according to their fractional relative naturalabundances.

• This average is numerically the same as atomic weight.......

EXAMPLE 1: Chlorine 35CR isotopic mass: 34.969 amu, 75.77%37CR isotopic mass: 36.966 amu, 24.23%

(Note: these are atom %’s, not mass %’s)What is the “average atomic mass” ?

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EXAMPLE 2: Antimony, Sb, consists of two isotopes; 121Sb (atomic mass 120.90) and 123Sb (atomic mass 122.90)

Calculate the “relative abundances” (in atom %) of these two isotopes.

The Average Atomic Mass for any element is found on the PeriodicTable.

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• How do we calculate the mass of a molecule?• This is called Molecular Weight (or Mass): it is the sum of the

atomic weights of all the atoms in the molecule. EXAMPLE 1: O2

EXAMPLE 2: H2O

EXAMPLE 3: Mg(NO3)2

• We never actually measure the mass of one atom or one moleculeof anything....an amu is an extremely small measurement.

• Chemists like to measure masses in grams............so what is the relationship between amu and grams ?

• Before we can answer this question, we must get a very clear ideaof what a mole is....

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Demystifying The Mole• A mole is a number; a very large number.• It is a word we use to represent a certain number of things.• We use it because it is easier to say the word than to say the

number itself......• We have commonly used names for many groups of items.........

Quantity Number of Items

pair

trio

quartet

dozen

gross

mole

• So, the mole (which we abbreviate as mol) contains Avogadro’sNumber (NA) of items: NA = 6.02 x 10

23

• This is a really strange number, but we use it because of theofficial definition of the mole; which is....................... “the number of atoms in exactly 12 g of the isotope 12 C ”.

• It turns out that this number of atoms is Avogadro’s Number, or6.02 x 1023.

• Therefore, 1 mole of anything contains 6.02 x 1023 of those things.12 g of 12 C is one mole of 12 C and contains NA (Avogadro’s No.)atoms (by definition).

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• Atoms and molecules are really small, and in any sample ofmeasurable size (usually in g or mg) we have a lot of them; so weuse the term “moles” as a sort of shorthand way to express howmany molecules or atoms we are actually talking about.

• Back to the question: Relating the amu to grams......• 12 g of 12 C is one mole of 12 C and contains NA (Avogadro’s

Number) of atoms ......

• This is another very small number......and remember that wenormally do not measure the mass of just one atom or molecule.

• It is much more practical to measure the mass of a larger amountof a substance; so we use the mass of one mole of the substance...

• The MOLAR MASS is the mass in grams of 1 mole of compound.

EXAMPLES: O2:

H2O:

Mg(NO3)2:

• Notice that the number (from the Periodic Table) is the same asthe Molecular Weight; it is the units that are different!!

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• One atom (or molecule) of substance will have a mass given in amu.• One mole of atoms (or molecules) of a substance will have a mass

given in grams. (Molar Mass is usually given units of “ g mol —1 “ )

• Some compounds exist as "hydrated" molecules; for eachmolecule of compound, a certain number of water molecules areassociated with that molecule.

• Think of it as a molecule of compound being surrounded by acertain number of water molecules.

• To determine the Molar Mass of one of these compounds, themass of the water molecules associated with the molecule must beadded to the mass of the molecule.

EXAMPLE: The Molar Mass of CuSO4 C 5 H2O

• We use Molar Mass to convert from moles to g or vice-versa.

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EXAMPLE 1: a) How many moles of CO2 molecules are there in a 55.0 g sample of

CO2 ?

b) How many moles of CO2 molecules are there in a 25.0 g sample of CO2 ?

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EXAMPLE 2: What is the mass of 0.1625 moles of C2H5OH ?

EXAMPLE 3: Suppose we have a sample of a compound with a mass of2.68 g. An experiment has shown that in this sample there are 1.26 x 10–2 moles of the compound in this sample. What is the molarmass of the compound?

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CHEMICAL FORMULAE• Recall that a chemical formula is the ratio (in whole numbers) of

the atoms in a compound...............

EXAMPLE: CH4 (Methane)

1 molecule of CH4 contains:

1 mole of CH4 molecules contains:

How many atoms would this be ??

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PERCENTAGE COMPOSITION• The composition of a compound is often expressed by mass as %’s,

because that is the way elemental analyses are reported.

• % composition by mass is found by using atomic masses and theformula of the compound.

EXAMPLE: Using C8H18, find the % composition; i) by mass ii) by moles i) C8H18 has a Molar Mass of

C8H18 contains 8 C (at 12.01 g mol-1 each) and 18 H (at 1.008 g mol-1)

% C (by mass) =

% H (by mass) =

• We could use this method to determine the MASS FRACTION ofone of the elements in a compound.

• In reporting the composition of C8H18 by mass fractions, we wouldsay that the fraction of C8H18 that is Carbon is:and the fraction that is Hydrogen is:

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ii) Mole percent uses the same principle, but is reporting thecomposition using moles instead of mass.

• In one mole of C8H18 molecules, there are 8 moles of Carbon atomsand 18 moles of Hydrogen atoms.

• The total number of moles of atoms is twenty-six (26).

• The mole % of Carbon =

• The mole % of Hydrogen =

• The term MOLE FRACTION (denoted by X) is usedfrequently.......

• The mole fraction is the mole percent reported as a fraction.• We could say that:

The mole fraction of Carbon in C8H18 is:

The mole fraction of Hydrogen is:

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Can we deduce a formula from % composition ? Yes!!

EXAMPLE: A compound contains 80.1 % C and 19.9 % H by mass. Find the SIMPLEST FORMULA of the compound. (The simplest, orempirical formula shows the simplest ratio of atoms in the compound.)

Given % by mass, we use a 100 g sample of compound; therefore, wehave 80.1 g of C and 19.9 g of H.Find the number of moles of each element first.......

To find the simplest formula, we want a whole number ratio, so divideeach # moles by the smallest # moles.......

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• The MOLECULAR FORMULA of a compound is a multiple of thesimplest formula; for CH3 , we could write the molecular formulaas (CH3)n.

• To find "n" , we need the Molar Mass of the compound's molecularformula........

EXAMPLE: If we know that the Molar Mass of (CH3)n is 30 g mol-1, we can say:

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• Sometimes you will not be given the elemental analysis, only someexperimental data.

• From the experimental data you will need to determine theamounts of each element present in the compound.

EXAMPLE: A compound contains only C , H and O. When burned in excess O2 , a 2.568 g sample of compound yields 5.134 g CO2 and 2.122 g H2O. Ifthe Molar Mass of the compound is 90 ± 5 g mol-1 , find the molecularformula.

• To find molecular formula, we must first find simplest formula.• We need either % composition, or the masses of each element in

the compound.• This will enable us to find the # moles of each element, and

therefore the mole ratio ( the simplest formula).

For C :

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For H :

For O:

– 33 –

• Now find the formula.............

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• Sometimes we do not know the identity of all the elements in acompound, but we do know the percentage of one element in saidcompound......

• We can use this percentage to determine the molar mass, and thenthe identity of the unknown element.

EXAMPLE:An oxide of an unknown metal has the formula X3O4. Analysis showsthat this compound is 27.64% oxygen by mass. Determine the molarmass and identity of X.

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MASS RELATIONS IN REACTIONS: [MH5; 3.4]

• Stoichiometry ; a term used to describe the relationship betweenamounts of reactants and products (in moles and grams) usingbalanced chemical equations.

• In a chemical equation, symbols of elements or compounds representMOLES............

2 S + 3 O2 ! 2 SO3

two moles three moles two moles of of Sulfur of Oxygen Sulfur Trioxide

----REACTANTS---- ----PRODUCTS----

• What is this equation telling us ?

1) 2 moles of Sulfur will react with 3 moles of Oxygen.2) 2 moles of Sulfur will produce 2 moles of Sulfur Trioxide.3) 3 moles of Oxygen will produce 2 moles of Sulfur Trioxide.

Notes:

a) Equations must be balanced in mass of each element, and in charge.

b) By convention, coefficients in an equation are usually integers, thesmallest values possible to balance.

S + O2 ! SO3 S + O2 ! SO3

S + O2 ! SO3

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H2 (g) N2 (g) O2 (g) F2 (g) CR2 (g) Br2 (R) I2 (s)

c) Use a one-way arrow (!) for a reaction going mainly “forwards”:

C + O2 ! CO2

Use a two-way arrow ( º ) for an equilibrium:

2NO + O2 º 2NO2

(In an equilibrium situation, the reaction is proceeding in bothdirections.)

d) Indicate the phase of reactants and products, if required, by thesymbols:

(s) solid (R) liquid (g) gas (aq) aqueous sol’n

2 Na (s) + 2 H2O (R) ! 2 Na+ (aq) + 2 OH- (aq) + H2 (g)

e) Remember the difference between atoms and molecules. Don’t write a symbol for an atom when an element occurs as adiatomic molecule:

Balancing Equations - this is essential !

Some are easy: C(s) + O2(g) ! CO2(g)

Others need work: N2 + H2 ! NH3

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f) Combustion Reactions are important - what happens when octane (a component of gasoline) burns in oxygen?

C8H18 + O2 ! CO2 + H2O

C8H18 + O2 ! CO2 + H2O

Always check your equation AFTER balancing it !!

g) By knowing the ratios by which species react to form products, wecan determine the number of moles of species formed or reacted,given the number of moles of another species present in thereaction.

EXAMPLE 1: If 2.00 g of C2H4 are burnt in excess O2, how many grams of CO2 are formed?

First, balance the equation:

C2H4 + O2 ! CO2 + H2O

• The steps to solving this problem are:

1) Find number of moles of C2H4.

2) Find number of moles of CO2 by relating moles reactant (whichwe know) to moles product (which is what we want).

3) Find mass CO2.

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• The solution is:

1) moles C2H4

2) Relate # moles of reactant to # moles product. (from equation)

3) Mass CO2 =

EXAMPLE 2: White phosphorous, P4, may be burned in oxygen, O2, to produce P4O10.What mass of oxygen is needed to completely burn (or react with) 1.20 g of white phosphorus ?

• First, write the balanced equation:

P4 + O2 ! P4O10

• Then find the number of moles of P4 : (MM = 4 x 30.97 gmol—1)

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• Write what is happening in words:

• Set up a ratio and solve it.

• Finish the calculation....

Note: The Molar Mass you use must match the species in the equation!

• LIMITING REAGENT problems involve reactions where theamount of product formed is limited by the amount of reagentavailable to react.

EXAMPLE 1: You are assembling bicycles. You have 6 frames and 10 wheels. How many bicycles can you build?

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EXAMPLE 2: What mass of AR2(SO4)3 can be prepared by reacting 25.0 g of AR with 100 g of H2SO4 ? The reaction is: AR + H2SO4 ! AR2(SO4)3 + H2

• First things first: Make sure that the equation is balanced!!!• The clue that this is a limiting reagent problem is that the amounts

of both starting materials are given......• These amounts may be given in various ways; they will all enable you

to find the # moles of each reactant.

• Find # moles of each reactant:

TO FIND THE LIMITING REAGENT: • Divide the actual number of moles of each reactant by its

coefficient (the required number of moles) from the balancedequation; then compare the resultant numbers.

• The smallest of these two numbers tells you the limiting reagent !

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• It is important to realize that you only use this step to determinethe limiting reagent; you do not necessarily use these numbers again (although you often see them again!)

• You can only form as much product as you have limiting reagentavailable to react..............so all calculations must relate the amountof limiting reagent to another species.

• Relate # moles of limiting reagent to # moles product.......

• Set up a ratio:

• Finish the calculation:

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YIELD CALCULATIONS

• Yield is usually expressed on a % basis, showing the amount ofproduct obtained as a % of what would be expected from theequation:

moles obtained x 100 % OR mass obtained x 100 %moles expected mass expected

EXAMPLE 1: When 12.0 g iron, Fe, were burned in air, 13.8 g iron oxide, Fe3O4, wereproduced. Calculate the % yield of Fe3O4 in this reaction.

• For % Yield, remember.................

moles obtained x 100 % = mass obtained x 100%moles expected mass expected

• Write the balanced equation:

Fe + O2 ! Fe3O4

• Find number of moles of Fe:

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• From the equation, we would expect to obtain:

• We actually obtained 13.8 g; less than we would expect.........

• The PERCENTAGE YIELD would be:

EXAMPLE 2: A 5.00 g sample of S was chemically treated and several steps laterresulted in the retrieval of 41.6 g of Na2SO4C10 H2O . Calculate the % yield.

• All the intermediate steps may be ignored!• Since the final product contains 1 S atom; one mole S gives one

mole of Na2SO4C10H2O (this statement serves as the balancedequation!!)

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• As usual, start by determining the number of moles......

• The molar mass of Na2SO4C10H2O is:

• We would expect a yield of:• Mass of Na2SO4C10H2O expected:

-45-

EXAMPLE 3:

• We have seen that in most chemical reactions, one of the reactantsis present in a lesser amount; this is called the Limiting Reagent.

• Some reactions take more than one step to occur; we saw that themoles of the various species carry all the way through the entireprocess.

• Here’s an example which incorporates almost all the intricaciesinvolved with stoichiometry.

EXAMPLE:Copper metal is usually obtained from sulfide ores, such as Cu2S. Therefining process takes two steps; the first is “roasting” the Cu2S inoxygen, O2, to obtain copper (I) oxide, Cu2O.In the second step, the Cu2O is reacted with powdered carbon, aprocess which yields carbon monoxide gas and copper metal.

1) 2 Cu2S (s) + 3 O2 (g) ! 2 Cu2O (s) + 2 SO2 (g)

2) Cu2O (s) + C (s) ! 2 Cu (s) + CO (g)

A) What is the overall equation ?

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B) Suppose we react 10.0 grams of Cu2S with 4.50 grams of O2. What yield of Cu2O would be expected in step 1)?

C) If 5.16 g of Cu (s) results from the second step, what is thepercentage yield from the entire process ?

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% PURITY

• Some problems involve the use of reactants that are not 100 % pure.

EXAMPLE 1:A solution of CS2(R) has a purity of 76.5 % (by mass) and has a densityof 1.20 gmL —1. What mass of SO2 (g) could be produced if 15.0 mL ofCS2 were burned in excess oxygen ?

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EXAMPLE 2:An ore contains 35.0 % by mass iron, Fe. What mass of this ore is needed to produce 500 g of Fe2O3 ?

• Write a balanced equation:

Fe + O2 ! Fe2O3

• Find the number of moles of whatever you can......

Moles Fe2O3 =

• From the equation, we know:

– 49 –

• Mass of Fe required:

• Set up a ratio to relate the % purity of the ore to the amount of Fethat is needed: