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Problem Set for Topics 2 and 3
Topic 2
1. Freon-12 (𝐶𝐶𝑙2 𝐹2 ) is used as a refrigerant in air conditioners and as a propellant in aerosol cans.
Calculate the number of molecules of Freon-12 in 5.56 mg of Freon-12. What is the mass of
chlorine in 5.56 mg of Freon-12?
Molar mass of CCl2F2 = 12.01 + 2(35.45) + 2(19.00) = 120.91 g/mol
5.56 mg CCl2F2 
1g
1 mol
6.022  10 23 molecules


1000 mg
120.91 g
mol
= 2.77 × 1019 molecules CCl2F2
5.56 × 10−3 g CCl2F2 
1 mol CCl 2 F2
2 mol Cl
35.45 g Cl


120.91 g
1 mol CCl 2 F
mol Cl
= 3.26 × 10−3 g = 3.26 mg Cl
2. Chloral hydrate (𝐶2 𝐻3 𝐶𝑙3 𝑂2 ) is a drug formerly used as a sedative and hypnotic. It is the
compound used to make “Mickey Finns” in detective stories.
a. Calculate the molar mass of chloral hydrate.
2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol
b. What amount (moles) of 𝐶2 𝐻3 𝐶𝑙3 𝑂2 molecules are in 500.0 g chloral hydrate?
500.0 g ×
1 mol
= 3.023 mol C2H3Cl3O2
165.39 g
c. What is the mass in grams of 2 × 10−2 mol chloral hydrate?
2.0 × 10-2 mol ×
165.39 g
= 3.3 g C2H3Cl3O2
mol
d. What number of chlorine atoms are in 5.0 g chloral hydrate?
5.0 g C2H3Cl3O2 ×
1 mol
6.022  10 23 molecules 3 atoms Cl


165.39 g
mol
molecule
= 5.5 × 1022 atoms of chlorine
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e. What mass of chloral hydrate would contain 1.0 g of Cl?
1.0 g Cl ×
1 mol Cl 1 mol C 2 H 3 Cl 3 O2 165.39 g C 2 H 3 Cl 3 O2


= 1.6 g chloral hydrate
35.45 g
3 mol Cl
mol C 2 H 3 Cl 3 O2
f. What is the mass of exactly 500 molecules of chloral hydrate?
500 molecules ×
1 mol
165.39 g

= 1.373 × 10−19 g C2H3Cl3O2
23
mol
6.022  10 molecules
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3. Caffeine, a stimulant found in coffee, tea, and chocolate, contains 49.48% carbon, 5.15%
hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2 g/mol.
Determine the empirical and molecular formulas of caffeine.
Assume 100 g, therefore mass = %
Mass C = 49.48 g
Mass H = 5.15 g
Mass N = 28.87 g
Mass O = 16.49 g
Convert mass to moles
𝑚𝑜𝑙𝐶 = 49.48 𝑔 ∗ (
1 𝑚𝑜𝑙 𝐶
) = 4.120
12.01 𝑔
1 𝑚𝑜𝑙 𝐻
𝑚𝑜𝑙𝐻 = 5.15 𝑔 ∗ (
) = 5.109
1.008 𝑔
𝑚𝑜𝑙𝑁 = 28.87 𝑔 ∗ (
1 𝑚𝑜𝑙 𝑁
) = 2.062
14.00 𝑔
1 𝑚𝑜𝑙 𝑂
𝑚𝑜𝑙𝑂 = 16.49 𝑔 ∗ (
) = 1.031
16.00 𝑔
Normalize the values (divide all by the lowest number of moles)
𝑚𝑜𝑙𝐶 = 3.99747734 ≈ 4
𝑚𝑜𝑙𝐻 = 4.957309384 ≈ 5
𝑚𝑜𝑙𝑁 = 2.000866326 ≈ 2
𝑚𝑜𝑙𝑂 = 1
Therefore, the empirical formula for caffeine is 𝑪𝟒 𝑯𝟓 𝑵𝟐 𝑶
To find the molecular formula (𝐶4 𝐻5 𝑁2 𝑂)𝑛 divide the molar mass by the formula weight to obtain n
𝑛=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
194.2 𝑔/𝑚𝑜𝑙
=
= 2.000412 ≈ 2
𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 97.08 𝑔/𝑚𝑜𝑙
Therefore the molecular formula of caffeine is (𝐶4 𝐻5 𝑁2 𝑂)2 = 𝑪𝟖 𝑯𝟏𝟎 𝑵𝟒 𝑶𝟐
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4. A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by
mass. The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical
and molecular formulas?
Assume 100 g, therefore mass = %
Mass P = 43.64 g
Mass O = 56.36 g
Convert mass to moles
1 𝑚𝑜𝑙 𝑃
𝑚𝑜𝑙𝑃 = 43.64 𝑔 ∗ (
) = 1.409
30.97 𝑔
1 𝑚𝑜𝑙 𝑂
𝑚𝑜𝑙𝑂 = 56.36 𝑔 ∗ (
) = 3.5225
16.00 𝑔
Normalize the values (divide all by the lowest number of moles)
𝑚𝑜𝑙𝑃 = 1
𝑚𝑜𝑙𝑂 = 2.500
Formula coefficients must be whole numbers, therefore multiply all by the lowest common value
𝑚𝑜𝑙𝑃 = 1 ∗ 2 = 2
𝑚𝑜𝑙𝑂 = 2.500 ∗ 2 = 5
Therefore, the empirical formula for the white powder is 𝑷𝟐 𝑶𝟓
To find the molecular formula (𝑃2 𝑂5 )𝑛 divide the molar mass by the formula weight to obtain n
𝑛=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
283.88 𝑔/𝑚𝑜𝑙
=
=2
𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 141.94 𝑔/𝑚𝑜𝑙
Therefore the molecular formula of the powder is (𝑃2 𝑂5 )2 = 𝑷𝟒 𝑶𝟏𝟎
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5. A compound containing only sulfur and nitrogen is 69.6% sulfur by mass; the molar mass is 184
g/mol. What are the empirical and molecular formulas of the compound?
Assume 100 g, therefore mass = %
Mass S = 69.6 g
Mass N = 100 – 69.6 g = 30.4 g
Convert mass to moles
1 𝑚𝑜𝑙 𝑆
𝑚𝑜𝑙𝑆 = 69.6 𝑔 ∗ (
) = 2.17
32.065 𝑔
1 𝑚𝑜𝑙 𝑁
𝑚𝑜𝑙𝑁 = 56.36 𝑔 ∗ (
) = 2.17
14.00 𝑔
Normalize the values (divide all by the lowest number of moles)
𝑚𝑜𝑙𝑠 = 1
𝑚𝑜𝑙𝑁 = 1
Therefore, the empirical formula for the white powder is 𝑵𝑺
To find the molecular formula (𝑁𝑆)𝑛 divide the molar mass by the formula weight to obtain n
𝑛=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
184 𝑔/𝑚𝑜𝑙
=
≈4
𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 46.065 𝑔/𝑚𝑜𝑙
Therefore the molecular formula of the powder is (𝑁𝑆)4 = 𝑵𝟒 𝑺𝟒
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6. An ionic compound 𝑀𝑋3 is prepared according to the following unbalanced chemical equation.
𝑀 + 𝑋2 → 𝑀𝑋3
A 0.105-g sample of 𝑋2 contains 8.92 × 1020 molecules. The compound 𝑀𝑋3 consists of 54.47%
X by mass. What are the identities of M and X, and what is the correct name for 𝑀𝑋3 ? Starting
with 1.00 g each of M and 𝑋2 , what mass of 𝑀𝑋3 can be prepared?
0.105 g
Molar mass X2 =
8.92  10
20
1 mol
molecules 
6.022  1023 molecules
= 70.9 g/mol
The mass of X = 1/2(70.9 g/mol) = 35.5 g/mol. This is the element chlorine.
Assuming 100.00 g of MX3 (= MCl3) compound:
54.47 g Cl ×
1 mol
= 1.537 mol Cl
35.45 g
1.537 mol Cl ×
Molar mass of M =
1 mol M
= 0.5123 mol M
3 mol Cl
45.53 g M
= 88.87 g/mol M
0.5123 mol M
M is the element yttrium (Y), and the name of YCl3 is yttrium(III) chloride.
The balanced equation is 2 Y + 3 Cl2 → 2 YCl3.
Assuming Cl2 is limiting:
1.00 g Cl2 ×
2 mol YCl3
195.26 g YCl3
1 mol Cl 2


= 1.84 g YCl3
70.90 g Cl 2
3 mol Cl 2
1 mol YCl3
Assuming Y is limiting:
1.00 g Y ×
2 mol YCl3
195.26 g YCl3
1 mol Y


= 2.20 g YCl3
88.91 g Y
2 mol Y
1 mol YCl3
Because Cl2, when it all reacts, produces the smaller amount of product, Cl2 is the limiting
reagent, and the theoretical yield is 1.84 g YCl3.
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7. The reaction of ethane gas (𝐶2 𝐻6) with chlorine gas produces 𝐶2 𝐻5 𝐶𝑙 as its main product (along
with HCl). In addition, the reaction invariably produces a variety of other minor products,
including 𝐶2 𝐻4 𝐶𝑙2 , 𝐶2 𝐻3 𝐶𝑙3, and others. Naturally, the production of these minor products
reduces the yield of the main product. Calculate the percent yield of 𝐶2 𝐻5 𝐶𝑙 if the reaction of
300. g of ethane with 650. g of chlorine produced 490. g of 𝐶2 𝐻5 𝐶𝑙.
C2H6(g) + Cl2(g) → C2H5Cl(g) + HCl(g)
If C2H6 is limiting:
300. g C2H6 ×
1 mol C 2 H 6
1 mol C 2 H 5Cl 64.51 g C 2 H 5Cl


= 644 g C2H5Cl
30.07 g C 2 H 6
mol C 2 H 6
mol C 2 H 5Cl
If Cl2 is limiting:
650. g Cl2 ×
1 mol C 2 H 5Cl 64.51 g C 2 H 5Cl
1 mol Cl 2


= 591 g C2H5Cl
70.90 g Cl 2
mol Cl 2
mol C 2 H 5Cl
Cl2 is limiting because it produces the smaller quantity of product. Hence, the theoretical yield for
this reaction is 591 g C2H5Cl. The percent yield is:
percent yield =
490. g
actual
× 100 =
× 100 = 82.9%
591 g
theoretical
Topic 3
8. Calculate the molarity of each of these solutions.
a. A 5.623-g sample of 𝑁𝑎𝐻𝐶𝑂3 is dissolved in enough water to make 250.0 mL of
solution.
5.623 g NaHCO3 ×
M=
1 mol NaHCO 3
= 6.693 × 10 −2 mol NaHCO3
84.01 g NaHCO 3
6.693  10−2 mol
1000 mL
= 0.2677 M NaHCO3

250.0 mL
L
a. A bottle of wine contains 12.5% ethanol (𝐶2 𝐻5 𝑂𝐻) by volume. The density of
𝑔
the ethanol solution is 0.789 𝑚𝐿
.
𝐴𝑠𝑠𝑢𝑚𝑒 𝑎 100 𝑚𝐿 𝑠𝑎𝑚𝑝𝑙𝑒
0.789 𝑔
100 𝑚𝐿 ∗ (
) = 78.9 𝑔
1 𝑚𝐿
%𝑚𝑎𝑠𝑠 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
∗ 100
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
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%𝑚𝑎𝑠𝑠
12.5%
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = (𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) ∗ (
) = (78.9 𝑔) (
)
100
100
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 9.86 𝑔 𝑜𝑓 𝑒𝑡ℎ𝑎𝑛𝑜𝑙
1 𝑚𝑜𝑙
9.86 𝑔 ∗ (
) = 0.214 𝑚𝑜𝑙 𝑒𝑡ℎ𝑎𝑛𝑜𝑙
46.06844 𝑔
0.0214 𝑚𝑜𝑙 𝑒𝑡ℎ𝑎𝑛𝑜𝑙
= 2.14 𝑀
0.1 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
b. A 184.6-mg sample of 𝐾2 𝐶𝑟2 𝑂7 is dissolved in enough water to make 500.0 mL
of solution.
0.1846 g K2Cr2O7 ×
M=
1 mol K 2 Cr2 O 7
= 6.275 × 10 −4 mol K2Cr2O7
294.20 g K 2 Cr2 O 7
6.275  10−4 mol
500.0  10
−3
L
= 1.255 × 10 −3 M K2Cr2O7
c. A 0.1025-g sample of copper metal is dissolved in 35 mL of concentrated 𝐻𝑁𝑂3
to form 𝐶𝑢2+ ions and then water is added to make a total volume of 200.0 mL.
(Calculate the molarity of 𝐶𝑢2+.)
0.1025 g Cu ×
M=
1 mol Cu
= 1.613 × 10 −3 mol Cu = 1.613 × 10 −3 mol Cu2+
63.55 g Cu
1.613  10−3 mol Cu 2+
1000 mL
= 8.065 × 10 −3 M Cu2+

200.0 mL
L
9. Suppose 50.0 mL of 0.250 M 𝐶𝑜𝐶𝑙2 solution is added to 25.0 mL of 0.350 M 𝑁𝑖𝐶𝑙2 solution.
Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume
that the volumes are additive.
Mol CoCl2 = 0.0500 L ×
0.250 mol CoCl 2
= 0.0125 mol
L
Mol NiCl2 = 0.0250 L ×
0.350 mol NiCl 2
= 0.00875 mol
L
Both CoCl2 and NiCl2 are soluble chloride salts by the solubility rules. A 0.0125-mol aqueous sample
of CoCl2 is actually 0.0125 mol Co2+ and 2(0.0125 mol) = 0.0250 mol Cl−. A 0.00875-mol aqueous
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sample of NiCl2 is actually 0.00875 mol Ni2+ and 2(0.00875) = 0.0175 mol Cl−. The total volume of
solution that these ions are in is 0.0500 L + 0.0250 L = 0.0750 L.
M Co 2 + =
M Cl− =
0.0125 mol Co 2+
0.00875 mol Ni 2+
= 0.167 M ; M Ni2 + =
= 0.117 M
0.0750 L
0.0750 L
0.0250 mol Cl − + 0.0175 mol Cl −
= 0.567 M
0.0750 L
10. What mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate
with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in
solution after precipitation is complete.
2 AgNO3(aq) + CaCl2(aq) → 2 AgCl(s) + Ca(NO3)2(aq)
0.1000 L 
0.20 mol AgNO3
2 mol AgCl
143.4 g AgCl


= 2.9 g AgCl
2 mol AgNO3
mol AgCl
L
0.1000 L ×
0.15 mol CaCl 2 2 mol AgCl 143.4 g AgCl


= 4.3 g AgCl
mol CaCl 2
mol AgCl
L
AgNO3 is limiting (it produces the smaller mass of AgCl) and 2.9 g AgCl can form.
The net ionic equation is Ag+(aq) + Cl−(aq) → AgCl(s). The ions remaining in solution are the
unreacted Cl− ions and the spectator ions NO3− and Ca2+ (all Ag+ is used up in forming AgCl). The
moles of each ion present initially (before reaction) can be easily determined from the moles of each
reactant. We have 0.1000 L(0.20 mol AgNO3/L) = 0.020 mol AgNO3, which dissolves to form 0.020
mol Ag+ and 0.020 mol NO3−. We also have
0.1000 L(0.15 mol CaCl2/L) = 0.015 mol CaCl2, which dissolves to form 0.015 mol Ca2+ and 2(0.015) =
0.030 mol Cl−. To form the 2.9 g of AgCl precipitate, 0.020 mol Ag+ will react with 0.020 mol of Cl− to
form 0.020 mol AgCl (which has a mass of 2.9 g).
Mol unreacted Cl− = 0.030 mol Cl− initially − 0.020 mol Cl− reacted
Mol unreacted Cl− = 0.010 mol Cl−
M Cl− =
0.010 mol Cl −
0.010 mol Cl −
= 0.050 M Cl−
=
total volume
0.1000 L + 0.1000 L
The molarities of the spectator ions are:
0.020 mol NO3
0.2000 L
−
= 0.10 M NO3−;
0.015 mol Ca 2+
= 0.075 M Ca2+
0.2000 L
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11. A 2.20-g sample of an unknown acid (empirical formula = 𝐶3 𝐻4 𝑂3 ) is dissolved in 1.0 L of water.
A titration required 25.0 mL of 0.500M NaOH to react completely with all the acid present.
Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula
of the unknown acid?
Let HA = unknown monoprotic acid; HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)
Mol HA present = 0.0250 L ×
0.500 mol NaOH
1 mol HA

L
1 mol NaOH
= 0.0125 mol HA
x g HA
2.20 g HA
=
, x = molar mass of HA = 176 g/mol
mol HA
0.0125 mol HA
Empirical formula weight  3(12) + 4(1) + 3(16) = 88 g/mol.
Because 176/88 = 2.0, the molecular formula is (C3H4O3)2 = C6H8O6.
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12. A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was
then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of
precipitate. Determine the concentration of the NaOH solution.
With the ions present, the only possible precipitate is
Cr(OH)3.
Cr(NO3)3(aq) + 3 NaOH(aq) → Cr(OH)3(s) + 3 NaNO3(aq)
Mol NaOH used = 2.06 g Cr(OH)3 ×
precipitate
1 mol Cr (OH) 3
3 mol NaOH

= 6.00 × 10−2 mol to form
103.02 g
mol Cr (OH) 3
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Mol NaOH used = 0.1000 L ×
MNaOH =
0.400 mol HCl 1 mol NaOH

= 4.00 × 10−2 mol to react with HCl
L
mol HCl
total mol NaOH 6.00  10 −2 mol + 4.00  10 −2 mol
=
= 2.00 M NaOH
0.0500 L
volume
13. For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and
the conjugate acid:
a. 𝐻2 𝑂 + 𝐻2 𝐶𝑂3 ⇌ 𝐻3 𝑂+ + 𝐻𝐶𝑂3 −
b. 𝐶5 𝐻5 𝑁𝐻 + + 𝐻2 𝑂 ⇌ 𝐶5 𝐻5 𝑁 + 𝐻3 𝑂+
c. 𝐻𝐶𝑂3 − + 𝐶5 𝐻5 𝑁𝐻+ ⇌ 𝐻2 𝐶𝑂3 + 𝐶5 𝐻5 𝑁
An acid is a proton (H+) donor, and a base is a proton acceptor. A conjugate acid-base pair differs by
only a proton (H+).
Acid
Base
Conjugate Base
Conjugate Acid
a.
H2CO3
H2O
H3O+
b.
C5H5NH+
H2O
HCO3−
C5H5N
c.
C5H5NH+
HCO3−
C5H5N
H2CO3
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14. Calculate the pH, pOH, and [𝑂𝐻 − ] for each of the following:
d. 0.00040 M 𝐶𝑎(𝑂𝐻)2
Ca(OH)2 → Ca2+ + 2 OH−; Ca(OH)2 is a strong base and dissociates completely.
[OH−] = 2(0.00040) = 8.0 × 10 −4 M; pOH = –log[OH−] = 3.10
pH = 14.00 − pOH = 10.90
e. A solution containing 25 g KOH per liter
25 g KOH
1 mol KOH

= 0.45 mol KOH/L
L
56.11 g KOH
KOH is a strong base, so [OH−] = 0.45 M; pOH = −log (0.45) = 0.35; pH = 13.65
f.
2.0 × 10−2 𝑀 𝐻𝑁𝑂3
[𝐻 + ] = [𝐻𝑁𝑂3 ] = 2.0 × 10−2 𝑀
𝑝𝐻 = − log[𝐻 + ] = − log(2.0 × 10−2 𝑀) = 1.70
𝑝𝑂𝐻 = 14.00 − 𝑝𝐻 = 14.00 − 1.70 = 12.30
[𝑂𝐻 − ] = 10−𝑝𝑂𝐻 = 10−12.30 = 5.0 × 10−13 𝑀
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