Chemistry Unit 7 Practice Test
The Mole Concept
No work shown = no credit. Please write legibly. Numerical answers must be rounded to the
appropriate number of significant figures or decimal points and must contain the appropriate
unit.
1. Calculate the number of representative particles in each of the following quantities. Round all
answers to three significant figures. Each item is worth 4 points (12 points).
(A) 2.35 mol N2
2.35 mol N2 x
6.022x10 23 molecules N2
= 1.42x1024 molecules N2
1mol N2
(B) 0.993 mol CH2Cl2
0.993 mol CH2Cl2 x
6.022x10 23 molecules CH 2 Cl 2
= 5.98x1023 molecules CH2Cl2
1mol CH 2 Cl 2
(C) 38.4 mol Ag
38.4 mol Ag x
6.022x10 23 atoms Ag
= 2.31x1025 atoms Ag
1mol Ag
2. Calculate the mass of each of the following quantities. Round all answers to three significant
figures. Each item is worth 4 points (12 points).
(A) 0.00144 mol Hg
Molar mass of Hg = 200.59 g/mol
0.00144 mol Hg x
200.59 g Hg
= 0.289 g Hg
1mol Hg
(B) 5.00 mol H2O
Molar mass of H2O = 18.0153 g/mol
5.00 mol H2O x
18.0153 g H 2 O
= 90.1 g H2O
1mol H 2 O
(C) 9.15x104 mol NaNO3
Molar mass of NaNO3 = 84.9947 g/mol 9.15x104 mol NaNO3 x
84.9947 g NaNO 3
= 7.78x106 g NaNO3
1mol NaNO 3
3.
Calculate the number of moles in each of the following quantities. Round all answers to three signficant
figures. Each item is worth 4 points (24 points).
(A) 4.11x1026 molecules of glucose
4.11x1026 molecules of glucose x
1 mol glucose
6.022x10 23 glucose molecules
= 682 mol glucose
(B) 0.500 grams of CH3COOH
Molar mass of CH3COOH = 60.0522 g/mol
0.500 g CH3COOH x
1mol CH 3 COOH
= 0.00833 mol CH3COOH
60.0522 g CH 3 COOH
(C) 2.95x1024 atoms of carbon
2.95x1024 C atoms x
1mol C
= 4.90 mol C
6.022x10 23 C atoms
(D) 12.3 grams of water
Molar mass of H2O = 18.0153 g/mol
12.3 g H2O x
1mol H 2 O
= 0.683 mol H2O
18.0153 g H 2 O
(E) 6.39x1022 molecules of hydrazine
6.39x1022 hydrazine molecules x
1 mol hydrazine
6.022x10 23 hydrazine molecules
= 0.106 mol hydrazine
(F) 330. grams of lithium fluoride
1 mol LiF
= 12.7 mol LiF
25.9396 g LiF
4. Calculate the percentage composition of the following compounds. If the compound is a
hydrate, treat water as an element with a molar mass of 18.0153 g/mol. Each item is worth 10
points (30 points).
Molar mass of LiF = 25.9396 g/mol
330. g LiF x
(A) CaCO3
Molar mass of CaCO3:
1 x Ca = 40.078
1 x C = 12.011
3 x O = 47.9982
Total = 100.087 g/mol
%Ca =
%O =
40.078 g
x 100% = 40.043% Ca
100.087 g
47.9982 g
x 100% = 47.9565% O
100.087 g
%C =
12.011 g
x 100% = 12.001% C
100.087 g
(B) BrF3
Molar mass of BrF3:
1 x Br = 79.904
3 x F = 56.995209
Total = 136.899 g/mol
%Br =
79.904 g
x 100% = 58.367% Br
136.899 g
%F =
56.995209 g
x 100% = 41.6330% F
136.899 g
%Cl =
70.906 g
x 100% = 34.877% Cl
203.303 g
(C) MgCl2●6H2O
Molar mass of MgCl2●6H2O:
1 x Mg = 24.305
2 x Cl = 70.906
6 x H2O = 108.092
Total = 203.303 g/mol
24.305 g
x 100% = 11.955% Mg
203.303 g
%Mg =
%H2O =
108.092 g
x 100% = 53.1679% H2O
203.303 g
5. How many grams of iron could be obtained from 445 grams of Fe(NO 3)3●9H2O? (5 points)
Molar mass of Fe(NO3)3●9H2O:
1 x Fe = 55.847
3 x N = 42.0201
9 x O = 143.995
9 x H2O = 162.138
Total = 404.000 g/mol
%Fe =
55.847 g
x 100% = 13.823%
404.000 g
13.823% of 445 g = 0.13823 x 445 g = 61.5 grams of Fe
6. How many grams of water are in 310. grams of CuSO4●5H2O? (5 points)
Molar mass of CuSO4●5H2O:
1 x Cu = 63.55
1 x S = 32.07
4 x O = 64.00
5 x H2O = 90.10
Total = 249.72 g/mol
%H2O =
90.10 g
x 100% = 36.08%
249.72 g
36.08% of 310. g = 0.3608 x 310. g = 112
grams of H2O
7. Chloroform was once used as an anaesthetic until its toxic properties were discovered. The
percentage compositon of chloroform is:
10.061% carbon
0.84436% hydrogen
89.094% chlorine
What is the empirical formula of chloroform? (7 points)
10.061 g C x
1 mol C
= 0.83764 mol C
12.011 g C
0.84436 g H x
1 mol H
= 0.83771 mol H
1.00794 g H
89.094 g Cl x
1mol Cl
= 2.5130 mol Cl
35.453 g Cl
C0.83764H0.83771Cl2.5130  CHCl3
8. Glyceraldehyde is a molecule produced during cellular respiration. Its empirical formula is CH 2O
and it has a molar mass of 90.09 g/mol. What is the molecular formula of glyceraldehyde? (5
points)
Formula mass of CH2O:
1 x C = 12.011
2 x H = 2.01599
1 x O = 15.9994
Total = 30.026 g/mol
90.09 g/mol
Molar mass
=
≈3
Formula mass
30.026 g/mol
Molecular formula: (CH2O)3 = C3H6O3