STOICHIOMETRY
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Transcript STOICHIOMETRY
STOICHIOMETRY
Calculations Based on
Chemical Equations
Iron (III) oxide reacts with carbon
monoxide to form iron and carbon dioxide.
Fe2O3
+
3 CO
2 Fe
+
3 CO2
How many CO molecules are required to
react with 25 particles of Fe2O3?
25 form.units Fe2O3 3 moleculesCO
1
1 form.unit Fe2O3
75 molecules CO
Iron (III) Oxide reacts with carbon
monoxide to form iron and carbon dioxide.
Fe2O3
+ 3 CO 2 Fe + 3 CO2
How many iron atoms can be produced by
the reaction of 2.5 x 105 particles of
Fe2O3?
2.510 form.units Fe2O3
2 atomsFe
1
1 form.unit Fe2O3
5
5.0 10 atomsFe
5
STOICHIOMETRIC RELATIONSHIPS
• Formulas can also represent MOLES
of substances involved in chemical
reactions.
• Equations define reaction ratios, i.e.
the molar ratios of reactants and
products
What mass of CO is required to react
with 146 grams of iron (III) oxide?
Fe2O3
+
3CO
2Fe +
3CO2
146 g Fe2O3
3 molCO
28 g CO
1 mol Fe2O3
159.6 g Fe2O3 1 mol Fe2 O3 1 mol CO
76.8g CO
What mass of iron (III) oxide is
required to produce 8.65 grams of
carbon dioxide?
Fe2O3
+
3CO
8.65 g CO 2 1 molCO 2
44 g CO 2
2Fe +
3CO2
1 mol Fe2 O3 159.6g Fe2O3
1 molFe2O3
3 molCO 2
10.5g Fe2O3
Limiting Reactants
(Reagents)
and Percent Yield
Calculations need to be based on the
limiting reactant.
• Example 1: Suppose a box contains 87
bolts, 110 washers and 99 nails. How
many sets of 1 bolt, 2 washers and 1 nail
can you use to create? What is the
limiting factor?
55 sets; washers limit the amount
Calculations need to be based on the
limiting reactant.
• Example 2: What is the maximum mass
of sulfur dioxide that can be produced
by the reaction of 95.6 g carbon
disulfide with 100. g oxygen?
CS2 + O2 CO2 + SO2
Start by balancing the equation…
Calculations need to be based on the
limiting reactant.
• Example 2: What is the maximum mass
of sulfur dioxide that can be produced
by the reaction of 95.6 g carbon
disulfide with 100. g oxygen?
CS2 + 3O2 CO2 + 2SO2
Now solve the problem…
• Example 2: What is the maximum mass of sulfur dioxide that can
be produced by the reaction of 95.6 g carbon disulfide with 100.
g oxygen?
CS2 + 3O2 CO2 + 2SO2
95.6 g CS 2
100.g O2
1m ol CS 2 2m ol SO2 64.1g SO2
161g SO2
76.2 g CS 2 1m ol CS 2 1m ol SO2
1m ol O2 2m ol SO2 64.1g SO2
134g SO2
32.0 g O2 3m ol O2 1m ol SO2
Which reactant is LIMITING?
• Example 2: What is the mass of sulfur dioxide that can be
produced by the reaction of 95.6 g carbon disulfide with 100. g
oxygen?
CS2 + 3O2 CO2 + 2SO2
95.6 g CS 2
100.g O2
1m ol CS 2 2m ol SO2 64.1g SO2
161g SO2
76.2 g CS 2 1m ol CS 2 1m ol SO2
1m ol O2 2m ol SO2 64.1g SO2
134g SO2
32.0 g O2 3m ol O2 1m ol SO2
O2 limits the amount of SO2 that can be
produced. CS2 is in excess.
• Example 2: What is the mass of sulfur dioxide that can be
produced by the reaction of 95.6 g carbon disulfide with 100. g
oxygen?
CS2 + 3O2 CO2 + 2SO2
95.6 g CS 2
1m ol CS 2 2m ol SO2 64.1g SO2
161g SO2
76.2 g CS 2 1m ol CS 2 1m ol SO2
100.g O2
1m ol O2 2m ol SO2 64.1g SO2
134g SO2
32.0 g O2 3m ol O2 1m ol SO2
134 g of SO2 can be produced in this reaction.
Calculations need to be based on
the limiting reactant.
• Example 3: What mass of CO2 could
be formed by the reaction of 8.0 g
CH4 with 48 g O2?
CH4 + O2 CO2 + H2O
Start by balancing the equation…
Calculations need to be based on
the limiting reactant.
• Example 3: What mass of CO2 could
be formed by the reaction of 8.0 g
CH4 with 48 g O2?
CH4 + 2O2 CO2 + 2H2O
Now solve the problem…
• Example 3: What mass of CO2 could be formed by the
reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2 CO2 + 2H2O
8.0 g CH 4
48g O2
1mol CH 4 1mol CO2 44.0 g CO2
22g CO2
16.0 g CH 4 1mol CH 4 1mol CO2
1mol O2 1mol CO2 44.0 g CO2
33g CO2
32.0 g O2 2mol O2
1mol CO2
Which reactant is LIMITING?
• Example 3: What mass of CO2 could be formed by the
reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2 CO2 + 2H2O
8.0 g CH 4
48g O2
1mol CH 4 1mol CO2 44.0 g CO2
22g CO2
16.0 g CH 4 1mol CH 4 1mol CO2
1mol O2 1mol CO2 44.0 g CO2
33g CO2
32.0 g O2 2mol O2
1mol CO2
CH4 limits the amount of CO2 that can be
produced. O2 is in excess.
Many chemical reactions do not go to
completion (reactants are not completely
converted to products).
Percent Yield: indicates what percentage of a
desired product is obtained.
experimental yield
% Yield
100
theoretical yield
• So far, the masses we have calculated from
chemical equations were based on the
assumption that each reaction occurred
100%.
• The THEORETICAL YIELD is the yield
calculated from the balance equation.
• The ACTUAL YIELD is the amount
“actually” obtained in an experiment.
• Look back at Example 2. We found
that 134 g of SO2 could be formed
from the reactants.
• In an experiment, you formed 130 g
of SO2. What is your percent yield?
130g
% Yield
100 97%
134g
Example: A 10.0 g sample of ethanol, C2H5OH, was
boiled with excess acetic acid, CH3COOH, to produce
14.8 g of ethyl acetate, CH3COOC2H5. What percent
yield of ethyl acetate is this?
CH3COOH + C2H5OH CH3COOC2H5 + H2O
10.0g C H OH 1 mol C H OH 1 mol CH COOC H
88 g CH COOC H
2 5
2
5
3
2
5
3
2 5
46g C H OH
1 mol C H OH
1 mol CH COOC H
2 5
2 5
3
2 5
19.1g CH3COOC2H5
Example: A 10.0 g sample of ethanol, C2H5OH, was
boiled with excess acetic acid, CH3COOH, to produce
14.8 g of ethyl acetate, CH3COOC2H5. What percent
yield of ethyl acetate is this?
experimental yield
% Yield
100
theoretical yield
14.8g
% Yield
100
19.1g
% Yield 77.5%