Chapter 3
Calculations involving Chemical
Formulae and Equations
John A. Schreifels
Chem 211
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Contents
• Mass and Moles of a Substance
– Molecular weight
– Moles
• Determining Chemical Formulae
– Mass % from formula
– Elemental Analysis: gives % C, H, O
– Determining formula from elemental analysis
• Stoichiometry
– Amounts of substances in a reaction
– Limiting reagents
John A. Schreifels
Chem 211
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Atomic and Molecular Weights
• Molecular Weight (Formula weight): Summed masses of all atoms in
the formula unit of the compound.
• How?
•
For the general formula AaBbCcDd...
fm = formula mass
= aMA + bMB + cMC + dMD + ...
• E.g. determine the formula mass (formula weight) of the following:
• NaOH, H2O, CaCO3, C2H6O
• Percentage composition from formula (mass % contribution of each
element to the total mass of the compound). For AaBbCc
a  mA
%A 
 100
FW
• E.g. determine the mass % composition of each element in: NaOH,
H2O, CaCO3, C2H6O
John A. Schreifels
Chem 211
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The Mole
• Since individual atoms are very small, they weigh very
little. It is not convenient deal with masses that small,
since we can typically measure gram amounts in the lab.
• Mole: Number of atoms (or molecules) as there are in
12.00 g Carbon-12; 1 mol = 6.02x1023 and is called
Avogadro's number.
• Mole is used to indicate a certain number of particles
(like a dozen or bushel might be used).
1 mol of
• water occupies approximately 18 mL and has 6.02x1023
molecules.
• gold occupies approximately 10 mL and has 6.02x1023
atoms.
John A. Schreifels
Chem 211
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Molar mass
• 1 mol = FM = 6.02x1023 particles
• The definition of a mol allows us to perform a variety of
conversions.
• Let n = # mol, n’ = # atoms (molecules), m = mass and FM = formula
mass then we can convert to:
• m from n, FM
E.g. Determine the mass of NaCl needed to have 5.0 mol of it.
• n from m, FM
E.g. Determine the number of mol of NaCl in 15.00 g of it.
• n’ from n
E.g. Determine the number of molecules in 3.222 mol of NaCl
• n’ from m, FM
E.g. Determine the number of atoms in 4.32 g of NaCl
• m from n’, FM
E.g. Determine the mass of one formula unit of NaCl.
John A. Schreifels
Chem 211
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EMPIRICAL FORMULA
Empirical Formula: Simplest formula where all coefficients are
integers.
• In earlier example Fe2O3, Fe4O6 Fe6O9 Fe8O12 possible formulas for
iron oxide. Its empirical formula Fe2O3.
• Often obtained from % composition data.
–
–
–
–
Assume 100 g of compoundDetermine mol of each element.
Divide by smallest # of moles.
Multiply by smallest whole number to make all coefficients whole
numbers.
• E.g. Determine the empirical formula of a compound with the
following % compositions:
Mass%O = 34.7%
Mass%C = 52.1%
Mass%H = 13.1%
John A. Schreifels
Chem 211
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COMBUSTION ANALYSIS
• Experimental mass % also determined. For
organic compounds the sample is combusted.
Analyzes the amount of C,H, and O in
compounds.
– C converted to CO2 and mass measured.
– H converted to H2O and mass measured
• E.g.: Combustion analysis of a 1.621 g sample
of ethanol, which contains only C,H,O, was
performed. Calculate the mass % composition of
each element in ethanol.
• Results: Mass CO2 = 3.095 g
Mass of H2O = 1.902 g.
John A. Schreifels
Chem 211
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MOLECULAR FORMULA
• Measure molecular mass
• Multiply empirical mass by integer to
obtain molecular mass.
• Multiply all the co-efficients by this integer.
• E.g. Determine the molecular mass of a
compound with the empirical formula NO2
and a molecular mass of 92.00 g/mol.
John A. Schreifels
Chem 211
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STOICHIOMETRY
The relative amounts of reactants and products in a reaction are given
by the ratio of stoichiometric coefficients. (Conservation of mass).
• E.g. For the reaction :
• 2Na(s) + Cl2(g)  2NaCl(s)
• 2 mol Na = 1 mol of Cl2 = 2 mol of NaCl.
• E.g. 2 Determine # mol of Cl2 needed to react with 4.2 mol of Na.
How many mol of NaCl will form?
1mol Cl2
1 mol Cl2
x
• Mol of Cl2

 x  4.2 mol Na 
4.2 mol Na 2 mol Na
2 mol Na
x  2.1 mol Cl2
x
2 mol NaCl
2 mol NaCl
• Mol NaCl:

 x  4.2 mol Na 
4.2 mol Na
2 mol Na
x  4.2 mol NaCl
• Summary: For aA + bB  cC
John A. Schreifels
Chem 211
2 mol Na
a
c
mol A  mol B   mol A  mol C 
b
a
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STOICHIOMETRY2
• Mass of one reactant can be related to the mass of another reactant
or product. (Law of definite proportions).
E.g. Determine the mass of Na needed to react with 34.45 g of Cl2 and
the maximum mass of NaCl that could be produced.
– Write reaction and express the moles of one compared to the moles of
the other.
2
2
mol Na  mol Cl2   mol NaCl 
1
2
– Substitute for mol in terms of mass and formula mass in each part.
Let n = mol; FM = formula mass then n  m
FM
E.g. 1: Determine the mass of oxygen consumed when it reacts with 10
g CH3CHO.
E.g. 2. Calculate the mass of oxygen needed to react with 100 g Al to
for Al2O3.
John A. Schreifels
Chem 211
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LIMITING REAGENTS
•
Limiting reagent: the reactant which limits the maximum
amount of product that can be produced. It will be completely
consumed in the reaction before any other reactants.
•
Limiting reagent must be known before theoretical yield can be
determined.
E.g. 1 Determine limiting reagent if 3.00 moles of Al react with 2.15
moles of O2 to form Al2O3.
Strategy:
–
–
–
Determine mol of Al2O3 that could be produced from Al
Determine mol of Al2O3 that could be produced from O2
Reactant producing smallest amount of Al2O3 is limiting reagent.
E.g.2 Calculate the theoretical yield when 20 g Al react with 25 g O2.
Strategy:
–
–
Determine expected yield from each reactant.
Compare. If mass of Al2O3 from produced from Al is less than from
O2, then Al is limiting. Otherwise O2 is limiting.
John A. Schreifels
Chem 211
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YIELDS IN CHEMICAL
REACTIONS
• Theoretical yield: the maximum amount of product that can be
obtained from given amounts of reactants.
• Actual Yield : the actual amount of product obtained from a
reaction.
• Theoretical yield calculated from amount of reactants and
stoichiometry
• Actual yield is the amount of product actually obtained.
• % yield is the yield as a percent of the theoretical yield:
Actual Yield
% Yield =
x100%
Theoretical (expected) yield
• E.g. The reaction for the synthesis of acetic acid from methanol is
shown below. What is the % yield if 15.0 g of methanol were
combined with a stoichiometric amount of carbon monoxide to
form19.1 g of product?
CH3OH(l) + CO(g)  HC2H3O2(l)
John A. Schreifels
Chem 211
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Summary
• Moles are used to describe measurable quantities of a
substance (1 mol = 6.02x1023 particles).
• Reactions occur in well defined proportions and are
represented by balanced chemical equations.
• The ratio of stoichiometric coefficients tells how much of
one compound will react if we know the amount of the
other:
? a
n
aA + bB  cC
B b
• Limiting reagent is one that limits the amount of product.
Use it to determine the “theoretical yield”.
• Empirical formula: simplest formula; all integers.
• Molecular formula: actual formula of a compound.
John A. Schreifels
Chem 211
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