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Chapter 4
3 October 2013
Calculations and the Chemical
Equation
Calculations Using the Chemical
Equation (Stoichiometry)
• Calculation quantities of reactants and
products in a chemical reaction has many
applications
• Need a balanced chemical equation for the
reaction of interest
• The coefficients represent the number of
moles of each substance in the equation
1
General Principles
1. Chemical formulas of all reactants and
products must be known
2. Equation must be balanced to obey the law of
conservation of mass
•
Calculations of an unbalanced equation are
meaningless (would lead to wrong answer)
3. Calculations are performed in terms of moles
•
Coefficients in the balanced equation represent
the relative number of moles of products and
reactants
General Principles
3. Calculations are performed in terms of moles
•
Coefficients in the balanced equation represent
the relative number of moles of products and
reactants
2
Using the Chemical Equation
• Examine the reaction:
2 H2 + O2  2 H2O
• Coefficients tell us?
– 2 mol H2 react with 1 mol O2 to produce 2 mol
H2O
• What if 4 moles of H2 react with 2 moles of O2?
– It yields 4 moles of H2O
Using the Chemical Equation
2 H2 + O2  2 H2O
• The coefficients of the balanced equation are
used to convert between moles of substances
• How many moles of O2 are needed to react
with 4.26 moles of H2?
• Use the factor-label method to perform this
calculation
3
Use of Conversion Factors
2 H2 + O2  2 H2O
4.26 mol H 2 
__mol
O2
1
 2.13 mol O2
__
2 mol H 2
• Digits in the conversion factor come
from the balanced equation
Conversion Between Moles
and Grams
• Requires only the formula weight
• Convert 1.00 mol O2 to grams
of
grams of
– Plan the path moles
oxygen
oxygen
– Find the molar mass of oxygen
• 32.0 g O2 = 1 mol O2
– Set up the equation
– Cancel units 1.00 mol O2 x 32.0 g O2
1 mol O2
– Solve equation 1.00 x 32.0 g O2 = 32.0 g O2
4
Conversion of Mole Reactants to
Mole Products
• Use a balanced equation
• C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
• 1 mol C3H8 results in:
– 5 mol O2 consumed
1 mol C3H8 /5 mol O2
– 3 mol CO2 formed
1 mol C3H8 /3 mol CO2
– 4 mol H2O formed
1 mol C3H8 /4 mol H2O
• This can be rewritten as conversion
factors
Calculating Reacting Quantities
• Calculate grams O2 reacting with 1.00 mol C3H8
• Use 2 conversion factors
– Moles C3H8 to moles O2
– Moles of O2 to grams O2
moles
C3H8
moles
oxygen
grams
oxygen
– Set up the equation and cancel units
– 1.00 mol C3H8 x 5 mol O2 x 32.0 g O2 =
1 mol C3H8 1 mol O2
– 1.00 x 5 x 32.0 g O2 = 1.60 x 102 g O2
5
Calculating Grams of Product
from Moles of Reactant
• Calculate grams CO2 from combustion of 1.00
mol C3H8
• Use 2 conversion factors
– Moles C3H8 to moles CO2
– Moles of CO2 to grams CO2
moles
C3H8
moles
CO2
grams
CO2
– Set up the equation and cancel units
– 1.00 mol C3H8 x 3 mol CO2 x 44.0 g CO2 =
1 mol C3H8 1 mol CO2
– 1.00 x 3 x 44.0 g CO2 = 1.32 x 102 g CO2
Relating Masses of Reactants
and Products
• Calculate grams C3H8 required to produce
36.0 grams of H2O
• Use 3 conversion factors
– Grams H2O to moles H2O
– Moles H2O to moles C3H8
– Moles of C3H8 to grams C3H8
grams
H2O
moles
H2O
moles
C3H8
grams
C3H8
– Set up the equation and cancel units
36.0 g H2O x 1 mol H2O x 1 mol C3H8 x 44.0 g C3H8
18.0 g H2O
4 mol H2O 1 mol C3H8
– 36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8
6
Calculating a Quantity of Reactant
• Ca(OH)2 neutralizes HCl
• Calculate grams HCl neutralized by 0.500 mol
Ca(OH)2
– Write chemical equation and balance
• Ca(OH)2(s) + 2HCl(aq)
CaCl2(s) + 2H2O(l)
– Plan the path
moles
Ca(OH)2
moles
HCl
grams
HCl
– Set up the equation and cancel units
0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl
1 mol Ca(OH)2 1 mol HCl
Solve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl
A Visual Example of the Law of
Conservation of Mass
7
General Problem-solving Strategy
Sample Calculation
Na + Cl2  NaCl
1. Balance the equation
2 Na + Cl2  2 NaCl
2. Calculate the moles Cl2 reacting with 5.00 mol
Na
3. Calculate the grams NaCl produced when 5.00
mol Na reacts with an excess of Cl2
4. Calculate the grams Na reacting with 5.00 g Cl2
8
Theoretical and Percent Yield
• Theoretical yield - the maximum amount of
product that can be produced
– Pencil and paper yield
• Actual yield - the amount produced when the
reaction is performed
– Laboratory yield
• Percent yield:

% yield 
actual yield
100%
theoretical yield
125 g CO2 actual
x 100% = 97.4%
132 g CO2 actual
Sample Calculation
If the theoretical yield of iron was 30.0 g
and actual yield was 25.0 g, calculate the
percent yield:
2 Al(s) + Fe2O3(s)  Al2O3(aq) + 2 Fe(aq)
• [25.0 g / 30.0 g] x 100 % = 83.3 %
• Calculate the % yield if 26.8 grams iron
was collected in the same reaction
9