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Lecture 5
Stoichiometry
Tutorial
1) Calculate the mass percent of each element in Na2SO4.
(2 Na atom) (22.99 amu/Na atom)
× 100
(2(22.99) + 32.07 + 4(16.00))amu
(2 Na atom) (22.99 amu/Ca atom)
mass % Na =
× 100
142.05 amu
mass % Na = 32.37% Na
mass % Na =
(1 S atom) (32.07 amu/S atom)
×100
(2(22.99) + 32.07 + 4(16.00))amu
(1 S atom) (32.07 amu/S atom)
mass % S =
×100
142.05 amu
mass % S = 22.58% Na
mass % S =
(4 O atom) (16.00 amu/O atom)
×100
(2(22.99) + 32.07 + 4(16.00))amu
mass % O = 45.05 % O
mass % O =
2) A sample of potassium chlorate (KClO3) is known to contain some impurities. It
is found that K+ makes up 9.50 % of the entire mass of the sample. All of the K+
comes from the KClO3 compound. Find the mass percent of KClO3 in the
sample.
(1 K atom) (39.10 amu/K atom) 9.50% K
=
(39.10 + 35.45 + 3(16.00)) amu % KClO3
39.10 amu of K
9.50% K
=
122.55 amu of KClO3 % KClO3
% KClO3 = 29.78% KClO3
3) How many moles of sodium bicarbonate (NaHCO3) are contained within 18.93g
of sodium bicarbonate?
18.93 g NaHCO3 ×
1 mol NaHCO3
= 0.2253 mol NaHCO3
84.01 g NaHCO3
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4) How many grams of CO2 are contained within 1.45 moles of CO2?
1.45 mol CO 2 ×
44.01 g CO 2
= 63.8 g CO 2
1 mol CO 2
5) How many oxygen atoms are contained in 12.7 g of zinc sulfate, ZnSO4?
12.7 g ZnSO 4 ×
1 mol ZnSO 4
4 mol O
6.022 ×1023 atoms O
×
×
161.46 g ZnSO4 1 mol ZnSO 4
1 mol O
= 1.89 × 1023 atoms O
6) A 5.75 g sample of silicone dioxide reacts with 5.50 g of sodium hydroxide
according to the following chemical equation.
SiO2(s) + 2 NaOH(aq) Æ Na2SiO3(aq) + H2O(l)
a. What is the limiting reactant? Justify your answer.
5.75 g SiO 2 ×
1 mol SiO 2 1 mol Na 2SiO3 122.07 g Na 2SiO3
×
×
= 11.7 g Na 2SiO3
60.09 g SiO 2
1 mol SiO 2
1 mol Na 2SiO3
5.50 g NaOH×
1 mol NaOH 1 mol Na 2SiO3 122.07 g Na 2SiO3
×
×
= 8.39 g Na 2SiO3
40.00 g NaOH 2 mol NaOH
1 mol Na 2SiO3
NaOH is the limiting reactant, as it would produce less Na2SiO3. There is only enough
NaOH to produce 8.38 g Na2SiO3. After 8.38 g Na2SiO3 have been produced, there will
not be any NaOH left to react with the rest of the SiO2.
b. How many grams of Na2SiO3(aq) are produced?
8.39g of Na2SiO3(aq) would be produced in theory.
c. What is the % yield if only 7.24 g Na2SiO3(aq) are produced.
actual yield
× 100
theoretical yield
7.24 g
%Yield =
×100
8.39g
%Yield = 86.3%
%Yield =
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7) A hydrocarbon (a compound that contains only hydrogen and carbon) is found to
be 85.60 % Carbon by mass.
a. Find the empirical formula for the compound.
Step 1) Find moles of C and H Assume you have a 100g sample
(Thus, you have 85.60 g C and 100.00 – 85.60 = 14.40 g H)
85.60 g C×
1 mol C
= 7.127 mol C
12.01g C
14.40 g H ×
1 mol H
= 14.26 mol H
1.01 g H
Step 2) Put both over the least number of moles (7.127 mol)
7.127 mol
=1
7.127 mol
The empirical formula is CH2
Carbon :
Hydrogen :
14.26 mol
=2
7.127 mol
b. If the molar mass of the compound is 28.06 g/mol, find its molecular
formula.
molar mass
empirical formula mass
28.06 g/mol
multiple for empirical formula =
=2
14.03 g/mol
multiple for empirical formula =
As the multiplier for the empirical formula is 2, we must multiply the number of atoms in
the empirical formula by 2.
The molecular formula is C2H4
8) A sample of a hydrocarbon was analyzed and found to contain 12.59 g of carbon
and 1.41 g of hydrogen.
a. Find the empirical formula of the compound. (Hint: as the masses are
known you do not need to assume you have a 100 g sample.)
Step 1) Find the number of moles of C and H
12.59 g C×
1 mol C
= 1.048 mol C
12.01 g C
1.41 g H ×
1 mol H
= 1.40 mol H
1.01 g H
Step 2) Put both over the least number of moles (in this case 1.048 mol)
1.048 mol
=1
1.048 mol
1.40 mol
Hydrogen :
= 1.34
1.048 mol
Carbon :
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Step 3) We must have whole numbers so multiply both by 3
3(1C and 1.34H) = 3C and 4H
The empirical formula is C3H4
b. If the molar mass of the hydrocarbon was found to be 40.07 g/mol, what is
its molecular formula?
molar mass
empirical formula mass
40.07 g / mol
multiple for empirical formula =
=1
40.07 g / mol
multiple for empirical formula =
As the multiplier for the empirical formula is 1, the empirical formula is the same as the
molecular formula.
The molecular formula is also C3H4
3) A 1.357 gram sample of a compound containing only carbon, hydrogen, and
oxygen was burned in excess oxygen gas. The combustion produced 1.989 g of
carbon dioxide and 0.8143g of water.
a. Find the empirical formula of the compound.
Step 1) Find masses of Carbon, Hydrogen, and Oxygen in the compound
Mass Carbon:
1.989 g CO 2 ×
1 mol CO 2
1 mol C 12.01 g C
×
×
= 0.5428 g C
44.01 g CO2 1 mol CO 2 1 mol C
Mass Hydrogen:
0.8143 g H 2O×
1 mol H 2 O
2 mol H 1.008 g H
×
×
=0.09110 g H
18.02 g H 2 O 1 mol H 2 O 1 mol H
Mass Oxygen:
Omass= total mass – (Hmass + Cmass)
Omass= 1.357g – (0.09110g H + 0.5428g C)
Omass= 0.723g O
Step 2) Find moles of Carbon, Hydrogen, and Oxygen
Moles of Carbon:
1 mol C
0.5428g C×
= 0.04520 mol C
12.01g C
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Moles of Hydrogen:
1 mol H
0.09110g H×
=0.09038 mol H
1.008g H
Moles of Oxygen:
1 mol O
0.723g O×
= 0.0452 mol O
16.00g O
Step 3) Put all moles over the smallest number of moles to find empirical formula
0.04520mol
=1
0.00452mol
0.09038mol
Hydrogen :
=2
0.0452mol
0.0452mol
Oxygen :
=1
0.0452mol
Carbon :
The Empirical Formula is CH2O
b. If the molar mass of the compound is found to be 60.06 g/mol, what is its
molecular formula?
molar mass
empirical formula mass
60.06 g/mol
multiple for empirical formula =
=2
30.03 g/mol
The molecular formula is C2H4O2 or CH3COOH
multiple for empirical formula =
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