Name ____Mr. Perfect_____________________________ Date ______Sp 17________
1. Calculate the molarity and osmolarity of a solution containing 9.82 g of magnesium
phosphate dissolved into 1.75 L of water. (10 pts)
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
𝑚𝑜𝑙
9.82 𝑔 × 262 𝑔
1.75 𝐿
= 𝟎. 𝟎𝟐𝟏𝟒 𝑴
Mg3(PO4)2  3Mg2+ + 2PO43- (5 particles)
𝑂𝑠𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = 5 𝑥 0.0214 = 𝟎. 𝟏𝟎𝟕 𝒐𝒔𝒎𝒐𝒍
2. Fill in the missing information in the following table: (12 pts)
Symbol
13 -4
C
Protons
6
Neutrons
7
Electrons
10
Charge
-4
17
20
18
-1
Ne
10
10
10
0
Bk+4
97
150
93
+4
37
Cl-
20
247
3. Write the molecular formula for the following compounds: (8 pts)
a.
Potassium Sulfite
b.
Dinitrogen Tetrahydride
K2SO3
c.
Vanadium (III) Bromite
N2H4
d.
Copper (I) Oxide
V(BrO2)3
Cu2O
4. An element has four naturally occurring isotopes with the following masses and
abundances:
Isotopic Mass (Amu) Fractional Abundance
53.940
0.058
55.935
0.917
56.935
0.022
57.933
0.003
What is the atomic mass of this element? Report answer to 5 significant figures. (2 pts)
Atomic Weight = (53.940 x 0.058) + (55.935 x 0.917) + (56.935 x 0.022) + (57.933 x
0.003) = 55.847 Amu
Chemistry 101 Exam 1
Name ____Mr. Perfect_____________________________ Date ______Sp 17________
5. For each of the following pairs of ions, write the formula of the corresponding
compound. (6 pts)
a. Mg+2 and NO3-
b. Fe+3 and PO4-3
Mg(NO3)2
Soluble
Na+, K+, Li+, NH4+
NO3Cl-, Br-, ISO42-
FePO4
Except
None
None
Ag+, Pb2+, Hg22+
Ca2+, Ag+, Pb2+, Ba2+
c. Al+3 and F-
AlF3
Insoluble
CO32PO43S2OH-
Except
Group 1A, NH4+
Group 1A, NH4+
Group 1A, NH4+
Group 1A, Ca2+, Ba2+
6. Write the balance molecular equation, complete ionic equation, and net ionic equation
for each of the following aqueous reactions and include phase labels. If no reaction
occurs (no precipitate), just write NR after the arrow: (18 pts)
a.
Mg(NO3)2(aq) +
Na2SO4(aq)

MgSO4(aq) + 2NaNO3(aq)
N.R.
b.
Ni(NO3)2(aq)
+
2NaOH(aq)

Ni(OH)2(s) + 2NaNO3(aq)
Ni2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2OH-(aq)  Ni(OH)2(s) + 2Na+(aq) +2NO3(aq)
Ni2+(aq) + 2OH-(aq)  Ni(OH)2(s)
c.
3Mn(NO3)2(aq) +
2Na3PO4(aq) 
6NaNO3(aq) + Mn3(PO4)2(s)
3Mn2+(aq) + 6NO3-(aq) + 6Na+(aq) + 2PO43-(aq)  6Na+(aq) + 6NO3-(aq) + Mn3(PO4)2(s)
3Mn2+(aq) + 2PO43-(aq)  Mn3(PO4)2(s)
Chemistry 101 Exam 1
Name ____Mr. Perfect_____________________________ Date ______Sp 17________
7. Balance the following redox reactions using the half-reaction method and clearly label
the oxidation and reduction steps. (14 pts)
a)
Sb
NO3-
+

Sb4O6
+
NO
[Acidic]
6H2O + 4Sb  Sb4O6 + 12H+ + 12e4 x (3e- + 4H+ + NO3-  NO + 2H2O)
______________________________
6H2O + 4Sb  Sb4O6 + 12H+ + 12e12e- + 16H+ + 4NO3-  4NO + 8H2O
____________________________________
4H+ + 4Sb + 4NO3-  Sb4O6 +4NO + 2H2O
b)
Zn
NO3-
+

Zn(OH)4-2
+
(Oxidation)
(Reduction)
NH3
4 x (4H2O + Zn  Zn(OH)42- + 4H+ + 2e-)
8e- + 9H+ + NO3-  NH3 + 3H2O
__________________________________
[Basic]
(Oxidation)
(Reduction)
16H2O + 4Zn  4Zn(OH)42- + 16H+ + 8e8e- + 9H+ + NO3-  NH3 + 3H2O
________________________________________
7OH- + 6H2O + 4Zn + NO3-  4Zn(OH)42- + NH3
8. An unknown compound yields the following percent composition upon elemental
analysis: 69.6% C, 8.34% H, and 22.1% N. The molecular mass was determined to be
352 g/mol. Write both the empirical formula and molecular formula for this compound.
(10 pts)
Assume a 100 g sample:
69.6 𝑔 ×
𝑚𝑜𝑙
= 5.8 𝑚𝑜𝑙 ÷ 1.38 𝑚𝑜𝑙 = 4.2 × 5 = 21
12 𝑔
8.34 𝑔 ×
𝑚𝑜𝑙
= 8.34 𝑚𝑜𝑙 ÷ 1.38 𝑚𝑜𝑙 = 6 × 5 = 30
1𝑔
22.1 𝑔 ×
𝑚𝑜𝑙
= 1.38 𝑚𝑜𝑙 ÷ 1.38 𝑚𝑜𝑙 = 1 × 5 = 5
14 𝑔
Empirical Formula:
C21H30N5 Empirical Mass = 352 g/mol
Empirical Mass = Molecular Mass thus the Empirical Formula is the same as the
Molecular Formula.
Chemistry 101 Exam 1
Name ____Mr. Perfect_____________________________ Date ______Sp 17________
9. When 20.5 g of methane, CH4, reacts with 45.0 g of chlorine gas, Cl2, the products are
chloromethane, CH3Cl, and hydrogen chloride. Calculate the experimental yield in grams
of chloromethane if the percent yield is 75.0%. (10 pts)
CH4 + Cl2  CH3Cl + HCl
20.5 𝑔 𝐶𝐻4 ×
45.0 𝑔 𝐶𝑙2 ×
𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝐶𝐻3 𝐶𝑙
×
= 1.28 𝑚𝑜𝑙
16 𝑔
1 𝑚𝑜𝑙 𝐶𝐻4
𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝐶𝐻3 𝐶𝑙
×
= 0.63 𝑚𝑜𝑙
71 𝑔
1 𝑚𝑜𝑙 𝐶𝐻4
𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑖𝑠 𝐶𝑙2
50.5 𝑔
= 31.8 𝑔 𝐶𝐻3 𝐶𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑
𝑚𝑜𝑙
𝑥
× 100 = 75.0 %
31.8 𝑔
x = (0.75)(31.8 g) = 23.9 g Experimental Yield
10. A sample of impure magnesium was analyzed by allowing the sample to react with
excess HCl solution:
0.63 𝑚𝑜𝑙 ×
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
After 1.32 g of the impure metal was treated with 0.100 L of 0.750 M HCl, 0.0125 mol of
HCl remained. Assuming the impurities do not react, what is the mass % of Mg in the
sample? Show work to receive full credit. (10 pts)
0.100 L x 0.450 mol/L = 0.075 mol HCl Total
0.075 mol HCl Total – 0.0125 mol HCl remained = 0.0625 mol HCl reacted
0.0625 𝑚𝑜𝑙 𝐻𝐶𝑙 ×
1 𝑚𝑜𝑙 𝑀𝑔 24.3 𝑔
×
= 0.76 𝑔 𝑀𝑔
2 𝑚𝑜𝑙 𝐻𝐶𝑙
𝑚𝑜𝑙
% 𝑀𝑔 =
0.76
× 100 = 𝟓𝟖 %
1.32
11. Extra Credit. Clearly indicate which reagent is the oxidizing agent and which is the
reducing agent in the following redox reaction. (5 pts)
Ag + Cu2+  Ag+
Reducing Oxidizing
Agent
Agent
+ Cu
Chemistry 101 Exam 1