Stoichiometry
Lecture Examples Answer Key
Ex. 1
Balance the following chemical equations:
→
3 NaBr + 1 H3PO4
2 C3H5N3O9
2 Ca(OH)2
Ex. 2
2 SO2
+
Ex. 5
Ex. 6
+
25 O2(g)
+
3 N2
1 O2
→
+
9 O2
2 H2O + 2 CaSO4
→
18 H2O(l) + 16 CO2(g)
→
2 CaO(s)
Calculate the molecular weight of
–1
(a) N2O5
MW = 108.0104 g mol
–1
(b)
Ca(C2H3O2)2
MW = 158.167 g mol
(c)
[K2(UO2)2(VO4)2•3H2O]
MW = 902.1758 g mol
–1
Calculate the percentage by mass of oxygen in 4a and 4b above.
–1
–1
(a)
[(5 × 15.9994 g mol ) / 108.0104 g mol
(b)
[(4 × 15.9994 g mol ) / 158.167 g mol
(a)
Calculate the number of O atoms in 0.470 mol of C6H12O6.
–1
23
# O atoms = (0.470 mol C6H12O6)(6.022 × 10
= 1.70 × 10
(b)
24
–1
] × 100% = 40.4620%
O atoms
Calculate the total number of ions in 38.1 g of CaF2.
= 8.82 × 10
(a)
] × 100% = 74.0642%
molecules/1 mol)(6 O atoms / 1 molecule)
# ions = (38.1 g CaF2)(1 mol / 78.075 g)( 6.022 × 10
Ex. 7
+
5 H2O
Write the balanced chemical equation for the combination of the metallic element calcium
with the nonmetallic element oxygen, O2.
2 Ca(s) + O2(g)
Ex. 4
6 CO2
1 Na3PO4
Write a balanced chemical reaction for the combustion of octane, C8H18(l), in air.
2 C8H18(l)
Ex. 3
+
→
+
3 HBr
23
23
molecules/1 mol) (3 ions / 1 molecule)
ions
What is the mass in grams of 1 mol of glucose, C6H12O6?
mass = (1 mol)(180.1572 g/1 mol) = 180.1572 g
(b)
What is the mass in grams of 3.52 mol of chromium(III) sulfate decahydrate?
1
[Cr2(SO4)3•10H2O]
mass = (3.52 mol)(572.335 g/1 mol) = 2015 g
Ex. 8
How many moles of chloride ions are in 0.0750 g of magnesium chloride?
–
–
moles Cl = (0.0750 g MgCl2)(1 mol MgCl2/95.211 g)(2 mol Cl /1 mol MgCl2)
= 0.00158 mol Cl
Ex. 9
–
What is the mass, in grams, of 1.75 × 10
mass = (1.75 × 10
= 0.0564 g
Ex. 10
20
20
molecules of caffeine, C8H10N4O2?
molecules)(1 mol/6.022×10
23
What is the molar mass of cholesterol if 0.00105 mol weighs 0.406 g?
Molar mass = 0.406 g / 0.00105 mol = 387 g mol
Ex. 11
molecules)(194.1926 g/1 mol)
–1
Determine the empirical formula of a compound containing 0.104 mol K, 0.052 mol C,
and 0.156 mol O.
K ==> 0.104 mol / 0.052 mol = 3
C ==> 0.052 mol / 0.052 mol = 1
K3CO3
O ==> 0.156 mol / 0.052 mol = 3
Ex. 12
What is the molecular formula of a compound that has an empirical formula, CH2, and
–1
a molar mass of 84 g mol ?
Molar mass of CH2 = 14 g mol
84 g mol
–1
/ 14 g mol
–1
–1
=6
Molecular formula: C6H12
Ex. 13
Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and
1.37 mg of H2O. If the compound contains only carbon and hydrogen, what is its
empirical formula?
moles C = (5.86×10
–3
g CO2)(1 mol CO2/44.0098 g)(1 mol C/1 mol CO2) = 1.33×10
–3
g H2O)(1 mol H2O/18.0152 g)(2 mol H/1 mol H2O) = 1.52×10
moles H = (1.37×10
–4
mol C
–4
mol H
Since toluene contains only C and H (we are told in the problem) we have everything we
need.
C ==> 1.33×10
–4
mol / 1.33×10
–4
mol / 1.33×10
–4
mol = 1
7
×7
–4
H ==> 1.52×10
mol = 1.144
C7H8
2
8
Ex. 14
A sample of 0.600 mol of a metal M reacts completely with excess fluorine to form 46.8 g
of MF2.
-
(a) How many moles of F are in the sample of MF2 that forms?
M + F2 → MF2
-
-
mol F = (0.600 mol M)(1 mol MF2/1 mol M)(2 mol F /1 mol MF2) = 1.20 mol F
(b) How many grams of M are in this sample of MF2?
mass of M in this sample will equal the total mass of the sample minus the mass that
due to F .
is
-
-
-
-
mass of F in MF2 = (1.20 mol F )(18.9984 g /1 mol F ) = 22.8 g F
-
mass of M in MF2 = 46.8 g – 22.8 g F = 24.0 g M
(c) What element is represented by the symbol M?
We can determine what M is if we know it’s molar mass. We can get that by
dividing the mass of M in the sample by the number of moles of M in the sample:
molar mass of M = (24.0 g M)(0.600 mol M) = 40.0 g/mol
From the periodic table we see that M must be Ca.
Ex. 15
Propane, C3H8, is a common fuel used for cooking and home heating. What mass of
O2 is consumed in the combustion of 1.00 g of propane?
4 C3H9 + 21 O2 → 12 CO2 + 18 H2O
mass O2 = (1.00 g C3H9)(1 mol C3H9/45.1041g)(21 mol O2/4 mol C3H9)(31.9988 g O2/ 1 mol)
= 3.72 g O2
Ex. 16
g of
When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction
forms sodium sulfide and water. How many grams of sodium sulfide are formed if 2.50
hydrogen sulfide is bubbled into a solution containing 1.85 g of sodium hydroxide,
assuming that the limiting reactant is completely consumed?
H2S + 2 NaOH → Na2S + 2 H2O
moles H2S = (2.50 g H2S)(1 mol H2S/34.0818 g) = 0.07335 mol H2S
moles NaOH = (1.85 g NaOH)(1 mol NaOH/39.9972 g) = 0.04625 mol NaOH
moles of Na2S formed if H2S is limiting = (0.07335 mol H2S)(1 mol Na2S / 1 mol H2S)
= 0.07335 mol Na2S
moles of Na2S formed if NaOH is limiting = (0.04625 mol NaOH)(1 mol Na2S / 2 mol NaOH)
= 0.023125 mol Na2S
3
The smaller amount of Na2S formed will tell us the limiting reactant.
NaOH is limiting reactant.
mass of Na2S = (0.023125 mol Na2S)(78.0458 g/1 mol) = 1.80 g Na2S
Ex. 17
When ethane, C2H6, reacts with chlorine, Cl2, the main product is C2H5Cl, but other
products are also obtained in small quantities. The formation of these other products
reduces the yield of C2H5Cl. Calculate the percent yield of C2H5Cl if the reaction of
125 g of C2H6 with 255 g of Cl2 produced 206 g of C2H5Cl.
C2H6 + Cl2 → C2H5Cl + HCl
First calculate the theoretical yield:
moles C2H6 = (125 g)(1 mol / 30.0694 g) = 4.157 mol C2H6
moles Cl2 = (255 g)(1 mol / 70.906 g) = 3.596 mol Cl2
moles C2H5Cl if C2H6 is limiting = (4.157 mol C2H6)(1 mol C2H5Cl/ 1 mol C2H6)
= 4.157 mol C2H5Cl
moles C2H5Cl if Cl2 is limiting = (3.596 mol Cl2)(1 mol C2H5Cl/ 1 mol Cl2)
= 3.596 mol C2H5Cl
Cl2 is limiting reactant
theoretical yield = (3.596 mol Cl2H5Cl)(64.5145 g / 1 mol) = 232 g C2H5Cl
% yield = (actual yield / theoretical yield) × 100% = (206 g/232 g) × 100% = 88.8%
Ex. 18
Hydrogen gas has been suggested as a clean fuel because it produces only water
vapor when it burns. If the reaction has a 98.8% yield, what mass of hydrogen
forms 85.0 kg of water?
2 H2 + O2 → 2 H2O
The actual yield is 85.0 kg. Using this and the percentage yield we can find the
theoretical yield:
theoretical yield = 85.0 kg / 0.988 = 86.0 kg
moles H2O = (86.0 kg)(1 kmol / 18.0152 kg) = 4.774 kmol H2O
moles H2 = (4.774 kmol H2O)(2 kmol H2/ 2 kmol H2O) = 4.774 kmol H2
mass H2 = (4.774 kmol H2)(2.0158 kg H2/ 1 kmol H2) = 9.62 kg H2
Ex. 19
During studies of the reaction N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g) a
chemical engineer measured a less-than-expected yield of N2 and discovered that
the following side reaction occurs: 2 N2O4(l) + N2H4(l) → 6 NO(g) + 2 H2O(g).
In one experiment, 10.0 g of NO formed when 100.0 g of each reactant was used.
What is the highest percent yield of N2 that can be expected?
4
Here are the numbers of moles of reactants that we start with:
moles N2O4 in 100.0 g = (100.0 g)(1 mol / 92.011 g) = 1.0868 mol N2O4
moles N2H4 in 100.0 g = (100.0 g)(1 mol / 32.045 g) = 3.1206 mol N2H4
Assuming no side reaction occurs:
moles N2 that can be made from N2O4 =
(1.0868 mol N2O4)(3 mol N2 / 1 mol N2O4) = 3.2604 mol N2 possible
moles N2 that can be made from N2H4 =
(3.1206 mol N2H4)(3 mol N2 / 2 mol N2H4) = 4.6809 mol N2 possible
N2O4 is the limiting reactant
This gives us the theoretical yield of N2 (if side reaction does not occur)
= (3.2604 mol N2)(28.0134 g / 1 mol) = 91.3 g N2
Since we know the side reaction does occur we need to know how much of each reactant
is used up in the reaction (because that means we have that much less reactant available
to make N2):
moles NO in 10.0 g = (10.0 g NO)(1 mol / 30.0061 g) = 0.33327 mol NO
moles N2O4 used to make 10.0 g NO = (0.33327 mol NO)(2 mol N2O4/6 mol NO) =0.1111 mol
moles N2H4 used to make 10.0 g NO =(0.33327 mol NO)(1 mol N2H4/6 mol NO)=0.05554 mol
Now we can calculate the number of moles of each reactant left over to make N2:
moles N2O4 available to make N2 = 1.0868 mol – 0.1111 mol = 0.9757 mol
moles N2H4 available to make N2 = 3.1206 mol – 0.05554 mol = 3.0651 mol
moles N2 that can be made from N2O4 =
(0.9757 mol N2O4)(3 mol N2 / 1 mol N2O4) = 2.9271 mol N2 possible
moles N2 that can be made from N2H4 =
(3.0651 mol N2H4)(3 mol N2 / 2 mol N2H4) = 4.5976 mol N2 possible
N2O4 is the limiting reactant (still)
Maximum yield of N2 possible, since side reaction does occur (we will use this as our
actual yield)
= (2.9271 mol N2)(28.0134 g / 1 mol) = 82.0 g N2
Maximum percentage yield = (82.0 g / 91.3 g) × 100% = 89.9%
5