CHEMISTRY WKST KEY: Ch. 6 Review
p.1
1)
6.022 x 1023
2)
a) 24.31 g Mg
3)
a) ratio of the atoms
4)
a) BaCl2
137.33 g + 2(35.45 g) = 208.23 g
b) P4O10
4(30.97g) + 10(16.00 g) = 283.88 g
c) Pb(C2H3O2)4
207.2 g + 8(12.01 g) + 12(1.01 g) + 8(16.00 g) = 443.40 g
d) Ba(OH)2•8H2O
137.33 g + 2(16.00 g) + 2(1.01 g) + 8(18.02 g) = 315.51 g
5)
6)
b) 6.022 x 1023 Mg atoms
b) mole ratio of the elements
1 mol CO2
) = 2.84 mol CO2
44.01 g CO2
a)
125 g CO2 (
b)
3.221 x 1023 NH3 molecules (
c)
2.0 x 1023 O atoms (
a)
0.13 mol NaCl (
b)
2.89 x 1023 N2 O5 molecules (
c)
1.000 x 1022 Ag atoms (
7)
75 g Fe2 O3 (
8)
a) NaCl
9)
M=
M=
1 mol NH3
6.022 x 1023 NH3 molecules
) = 0.5349 mol NH3
1 molecule C6 H12 O6
1 mol C6 H12 O6
)(
) = 0.055 mol C6 H12 O6
6 O atoms
6.022 x 1023 C6 H12 O6 molecules
58.44 g NaCl
) = 7.6 g NaCl
1 mol NaCl
1 mol N2 O5
23
6.022 x 10
N2 O5 molecules
)(
108.02 g N2 O5
) = 51.8 g N2 O5
1 mol N2 O5
1 mol Ag
23
6.022 x 10
107.87 g Ag
)(
) = 1.791 g Ag
1 mol Ag
Ag atoms
1 mol Fe2 O3
6.022 x 1023 Fe2 O3 molecules
3 O atoms
)(
)(
) = 8.5 x 1023 O atoms
159.70 g Fe2 O3
1 mol Fe2 O3
1 Fe2 O3 molecule
b) H2O
g solute
(molar mass solute)(L soln)
12 g Pb(NO3 )2
(331.22 g Pb(NO3 )2 )(0.8250 L)
= 0.044 M Pb(NO3 )2
CHEMISTRY WKST KEY: Ch. 6 Review
10)
M=
g solute
(molar mass solute)(L soln)

p.2
g solute=(M)(molar mass solute)(L soln)
g solute = (2.25 M)(40.00 g/mol)(1.750 L) = 158 g NaOH
11)
M=
g solute
(molar mass solute)(L soln)
L soln =
12)

L soln =
g solute
(M)(molar mass solute)
205 g NH4 Cl
= 4.7897 L = 5 L
(0.8 M)(53.50 g/mol NH4 Cl)
% element
molar mass element
=
100
molar mass compound
a)
CuSO4  63.55 g + 32.07 g + 4(16.00 g) = 159.62 g
x
4(16.00 g O)
=
100 159.62 g CuSO4
x = 40.10% O
b)
Ba(NO3)2  137.33 g + 2(14.01 g) + 6(16.00 g) = 261.35 g
x
6(16.00 g O)
=
100 261.35 g Ba(NO3 )2
x = 36.73% O
13)
% solute
mass solute
=
100
mass solution
x
60.0 g NaCl
=
100 560.0 g soln
x = 10.7% NaCl
14)
% solute
mass solute
=
100
mass solution

% solute
mass solute
=
100
mass solute + mass solvent
22.25 % Mg(NO3)2
x
=
100
x + 500.0 g H2O
100x = 22.25x + 11125 g
77.75x = 11125 g
x = 143.1 g Mg(NO3)2
CHEMISTRY WKST KEY: Ch. 6 Review
15)
p.3
mass element
molar mass element
=
mass compound molar mass compound
ZnSO4  65.38 g + 32.07 g + 4(16.00 g) = 161.45 g
x
65.38 g Zn
=
50.00 g ZnSO4 161.45 g ZnSO4
161.45x = 3269 g
x = 20.25 g Zn
16)
% element
molar mass element
=
100
molar mass compound
0.342 % Fe 4(55.85 g Fe)
=
100
x
0.342 x = 22340 g
x = 65321.64 g = 65300 g
17)
a) C3H4O3
18)
b) CH
2.62 g S
8.44 g – 2.62 g = 5.82 g Cl
19)
2.62 g S
= 0.0817 mol S
g
32.07 mol
S
S0.0817 Cl 0.164

5.82 g Cl
= 0.164 mol Cl
g
35.45 mol
Cl
S1.00Cl2.01

0.0817
0.0817

SCl2
- assume 100 g sample
60.8 % Na
 60.8 g Na

60.8 g Na
= 2.64 mol Na
g
22.99 mol
Na
28.5 % B
 28.5 g B

28.5 g B
= 2.64 mol B
g
10.81 mol
B
10.5 % H
20)
c) C14H18N2O5
 10.5 g H

10.5 g H
= 10.4 mol H
g
1.01 mol
H
g
C2HCl  60.48 mol
g
181.4 mol
g = 2.999= 3
60.48 mol
C2x3H1x3Cl1x3  C6H3Cl3
Na2.64 B2.64 H10.4
2.64
2.64
2.64
Na1.00B1.00H3.94

NaBH4
CHEMISTRY WKST KEY: Ch. 6 Review
21)
p.4
- assume 100 g sample
42.87 % C
 42.87 g C

42.87 g C
= 3.570 mol C
g
12.01 mol
C
3.598 % H
 3.598 g H

3.598 g H
= 3.56 mol H
g
1.01 mol
H
28.55 % O
 28.55 g O

28.55 g O
= 1.784 mol O
g
16.00 mol
O
25.00 % N
 25.00 g N

25.00 g N
= 1.784 mol N
g
14.01 mol
N
C3.570 H 3.56 O1.784 N1.784
1.784
1.784
1.784
1.784
C2.001H2.00O1.000N1.000
C2H2ON  empirical formula
g
C2H2ON ≈ 56 mol
168 g
= 3
56 g
C2x3H2x3O1x3N1x3

C6H6O3N3  molecular formula
22)
The atomic mass for hydrogen (as well as for all elements) is the average of ALL the hydrogen atoms in the
universe.
23)
The purpose of distilling a solution is to separate the liquid from the solutes dissolved in the liquid.
24)
This was a physical change because the liquid, which was H2O, never changed—it was still H2O as steam and
H2O when converted back into a liquid. The color seen in the solution was from something that was dissolved in
the liquid.
25)
When steam is converted back into liquid water, the water loses energy. This is an exothermic process.
26)
When distilling the colored substance in the solution is left behind in the boiler—only the pure liquid (H2O in our
case) was distilled. When filtering, the filter paper does NOT stop any substance still dissolved in the liquid.
Therefore, the colored substance passes through the filter paper since it was still dissolved in the H2O.
27)
Solubility of the solids in the liquid determines whether or not they can be separated by filtering. One of the solids
in the mixture is soluble in H2O while the other substance was insoluble in H2O. The insoluble solid will be
stopped by the filter paper while the soluble solid will pass through.
28)
You will find the mass of the object by weighing it on a balance. There are two ways to find the volume:
 If the object is small enough, place an known amount of water in a graduated cylinder. Place the object
into the water and see how much the volume of the water in the graduated cylinder increases—this will be
the volume of the object.
 If the object is too large for a graduated cylinder, put water into a container with an overflow spout—fill it
with enough water to have some come out of the spout. After the water quits coming out of the spout,
place a graduated cylinder under the spout, place the object into the water, and catch the water that
comes out of the spout. The volume of water that comes out will be the volume of the object.
CHEMISTRY WKST KEY: Ch. 6 Review
29) 1.00
30)
g
mL
density=
mass
volume
p.5