Unit 6b Question Packet Name ……………………………………………
Stoichiometry Period ………….
Skills
1. calculating Molar Mass (GFM)
2. Mole-Mass calculations
3. Mole-Particle calculations
4. calculating % composition
5. calculating % composition of a Hydrate
6. calculating Mole to Mole Ratios
7. Mass to Mass problems
Skill #1: calculating Molar Mass (GFM) - refer to your notes, the P.T., & RB p. 88
1. Calculate the gram-formula mass for each compound below. Remember GSSC.
a. Fe2(SO4)3 b. C8H18
G: G:
S: add atomic masses on P.T. S: add atomic masses on P.T.
S: Fe 2 x 56 = 112 S: C 8 x 12 = 96
S 3 x 32 = 96 H 18 x 1 = 18
O 12 x 16 = 192 114 g
400 g
e. (NH4)3PO4 f. MgSO4·7H2O
G: G:
S: add atomic masses on P.T. S: add atomic masses on P.T.
S: N 3 x 14 = 42 S: Mg 1 x 24 = 24
H 12 x 1 = 12 S 1 x 32 = 32
P 1 x 31 = 31 O 11 x 16 = 176 O 4 x 16 = 64 H 14 x 1 = 246
149 g 107 g
c. C11H22O11 d. NH4Cl·3H2O
G: G:
S: add atomic masses on P.T. S: add atomic masses on P.T.
S: C 11 x 12 = 132 S: N 1 x 14 = 14
H 22 x 1 = 22 H 10 x 1 = 10
O 11 x 16 = 176 Cl 1 x 35 = 35 330 g O 3 x 16 = 48
107 g
e. (NH4)3PO4 f. MgSO4·7H2O
G: G:
S: add atomic masses on P.T. S: add atomic masses on P.T.
S: N 3 x 14 = 42 S: Mg 1 x 24 = 24
H 12 x 1 = 12 S 1 x 32 = 32
P 1 x 31 = 31 O 11 x 16 = 176 O 4 x 16 = 64 H 14 x 1 = 246
149 g 107 g
Skill #2: Mole-Mass calculations – refer to your notes, Table T, and RB p. 89
2. Determine the mass of each of the following quantities. Remember GSSC.
a. / 2.0 mol of NaCl (GFM = 58 g)G:
S: Table T
S: # moles = given mass
GFM
2.0 = X f
1 58
X = 116 g / c. / 3.25 mol of CuSO4·5H2O
(molar mass = 250. g)
G:
S: Table T
S: # moles = given mass
GFM
3.25 = X f
1 250.
X = 812.5 g
b. / 0.50 mol of H2O (GFM = 18 g)
G:
S: Table T
S: # moles = given mass
GFM
0.5 = X f
1 18
X = 9 g / d. / 0.75 mol of Cu (mass of 1 mol = 64 g)
G:
S: Table T
S: # moles = given mass
GFM
0.75 = X f
1 64
X = 48 g
3. Determine the number of moles in each of the following quantities. Use the GFM’s given in #4 to solve. Remember GSSC.
a. / 35 g of NaClG:
S: Table T
S: # moles = given mass
GFM
X = 35 f
1 58
X = 0.60 mol / c. / 110. g of CuSO4·5H2O
G:
S: Table T
S: # moles = given mass
GFM
X = 110. f
1 250.
X = 0.440 mol
b. / 108 g of H2O
G:
S: Table T
S: # moles = given mass
GFM
X = 108 f
1 18
X = 6.00 mol / d. / 250. g of Cu
G:
S: Table T
S: # moles = given mass
GFM
X = 250. f
1 64
X = 3.91 mol
Skill #3: Mole-Particle calculations – refer to your notes and RB p. 92
4. Determine the number of particles in each of the following quantities. Remember GSSC.
a. / How many molecules in0.50 mol of H2O?
G:
S: Table T
S: # moles = given particles
(6.02 x 10^23)
0.50 = X f
1 (6.02 x 10^23)
X = 3.0 x 1023 molecules / c. / How many molecules in
3.0 mol of NH3?
G:
S: Table T
S: # moles = given particles
(6.02 x 10^23)
3.0 = X f
1 (6.02 x 10^23)
X = 1.8 x 1024 molecules
b. / How many atoms in
1.00 mol of Fe?
G:
S: Table T
S: # moles = given particles
(6.02 x 10^23)
1.00 = X f
1 (6.02 x 10^23)
X = 6.02 x 1023 atoms / d. / How many atoms in
0.75 mol of Cu?
G:
S: Table T
S: # moles = given particles
(6.02 x 10^23)
0.75 = X f
1 (6.02 x 10^23)
X = 4.5 x 1023 atoms
5. Determine the number of moles in each of the following quantities. Remember GSSC.
a. / 1.2 x 1024 atoms of AuG:
S: Table T
S: # moles = given particles
(6.02 x 10^23)
X = (1.20 x 1024) f
1 (6.02 x 10^23)
X = 2.0 mol of Au / c. / 1.8 x 1024 molecules of NH3
G:
S: Table T
S: # moles = given particles
(6.02 x 10^23)
X = (1.8 x 1024) f
1 (6.02 x 10^23)
X = 3.0 mol of NH3
b. / 1.5 x 1023 atoms of Fe
G:
S: Table T
S: # moles = given particles
(6.02 x 10^23)
X = (1.5 x 1023) f
1 (6.02 x 10^23)
X = 0.25 mol of Fe / d. / 4.5 x 1023 molecules of H2O
G:
S: Table T
S: # moles = given particles
(6.02 x 10^23)
X = (4.5 x 1023) f
1 (6.02 x 10^23)
X = 0.75 mol of H2O
Skill #4: calculating % composition – refer to your notes, Table T, and RB p. 90
6. Determine the percent by mass of the given element in the following compounds.
a. / % O in Fe2(SO4)3 (GFM = 400.g)G:
S: Table T
S% O = part/whole x 100 / c. / % O in CuSO4·5H2O (GFM = 250. g)
G:
S: Table T
S: % = part x 100
whole
% O = (9 x 16) x 100
250.
% O = 57.6 %
b. / % H in H2O (GFM = 18 g)
G:
S: Table T
S: % = part x 100
whole
% H = (2 x 1) x 100
18
% H = 11 % / d. / % P in (NH4)3PO4 (GFM = 149 g)
G:
S: Table T
S: % = part x 100
whole
% P = 31 x 100
149
% P = 20.8 %
7. A substance known as heavy water can be obtained from ordinary water and could be a significant source of energy in the future. Heavy water contains deuterium, H-2. Instead of the two hydrogen atoms in a typical water molecule, a heavy water molecule has two deuterium atoms. In 3.78 kilograms of ordinary water, the percent composition by mass of heavy water is approximately 0.0156%.
Calculate the mass of heavy water in a 3.78-kilogram sample of ordinary water. Your response must include both a correct numerical setup and the calculated result. [2]
% heavy = X x 100 0.0156 = X x 100 0.058968 = 100 X
whole 3.78 kg
X = 5.90 x 10-4 kg
8. In a 13.7-g sample of carbon, the percent composition by mass of carbon-14 is approximately 0.211%. Calculate the mass of carbon-14 in this sample.
% C-14 = X x 100 0.211 = X x 100 0.028907 = 100 X
whole 13.7 g
X = 0.0289 g
9. A sample of boron is approximately 3.14% B-6 by mass. The mass of just B-6 in this sample is 0.376 g. Calculate the total mass of the sample.
% B-6 = part x 100 3.14 = 0.376 x 100 3.14 X = 37.6
X X
X = 12.0 g
Skill #5: calculating % composition of a Hydrate – refer to your notes, Table T, and RB p. 90-91
10. Determine the percent by mass of water in the following hydrates.
a. / Na2CO3·10H2O (GFM = 286g)G:
S: Table T
S: % = part x 100
whole
% H2O = (10 x 18) x 100
286
% H2O = 63 % / c. / MgSO4·7H2O (GFM = 246 g)
G:
S: Table T
S: % = part x 100
whole
% H2O = (7 x 18) x 100
246
% H2O = 51 %
b. / Initial mass of hydrate: 9.5 g
Final mass of anhydrous salt: 3.77 g
G:
S: Table T
S: % = part x 100
whole
% H2O = (9.5 – 3.77) x 100
9.5
% H2O = 60. % / d. / Initial mass of hydrate: 5.3 g
Final mass of anhydrous salt: 4.1 g
G:
S: Table T
S: % = part x 100
whole
% H2O = (5.3 – 4.1) x 100
5.3
% H2O = 23 %
Skill #6: Mole to Mole problems – refer to your notes & RB p. 93
11. Calculate the number of moles in each problem.
a. Given the balanced equation: Ca + 2H2O à Ca(OH)2 + H2
What is the total number of moles of 1 = 2
Ca that will react with 6 moles of water? X 6 X = 3 mol Ca
b. Given the balanced equation: 2N2O5 à 4NO2 + O2
What is the total number of moles of 2 = 4
NO2 that will be produced from 5 moles 5 X X = 10 mol NO2
of N2O5?
c. Given the balanced equation: 2Li+ H2SO4 à Li2SO4 + H2
How many moles of lithium must react to 2 = 1
produce 3.3 moles of lithium sulfate? X 3.3 X = 6.6 mol Li
d. Given the balanced equation: 3H2 + N2 à 2NH3
How many moles of ammonia can be 1 = 2
produced by reacting 2.8 moles of nitrogen? 2.8 X X = 5.6 mol NH3
e. Given the balanced equation: 2C2H2 + 5O2 à 4CO2 + 2H2O
How many moles of C2H2 are needed to 2 = 5
completely react with 11.5 moles of O2? X 11.5 X = 4.6 mol C2H2
f. Given the balanced equation: CH4 + 2O2 à CO2 + 2H2O
How many moles of CO2 will be 1 = 2
produced from reacting 7.4 mol of O2? X 7.4 X = 3.7 mol CO2
Skill #7: Mass to Mass problems – refer to your notes & RB p. 94
12. Calculate the number of grams in each problem.
a. Given the balanced equation: 2C2H6 + 7O2 à 4CO2 + 6H2O
What is the total mass of C2H6 that will 2(60) = 6(18)
react to produce 72g of H2O? X 72 X = 80.g C2H6
b. Given the balanced equation: Mg + H2SO4 à MgSO4 + H2
What is the total number of grams of (24) = (120)
MgSO4 produced when 100. g of Mg 100. X X = 500.g MgSO4
is reacted?
c. Given the balanced equation: 2KClO3 à 2KCl + 3O2
What is the total number of grams of 2(122) = 3(32)
KClO3 that will react to produce 32g X 32 X = 81g KClO3
of O2 at STP?
d. Given the balanced equation: 2Li + H2SO4 à Li2SO4 + H2
How many grams of lithium sulfate 2(7) = (110)
are produced from reacting 27.0g of 27.0 X X = 212g Li2SO4
lithium?