Mole Calculations Review Worksheet - answers on next page ... - Free Printable

Let's solve each problem step-by-step using dimensional analysis and proper unit labeling, as instructed.

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1. Calculate the molar mass of each compound.

#### a. LiOH
- Li: 6.94 g/mol
- O: 16.00 g/mol
- H: 1.01 g/mol
Molar mass = 6.94 + 16.00 + 1.01 = 23.95 g/mol

#### b. Barium bromide (BaBr₂)
- Ba: 137.33 g/mol
- Br: 79.90 g/mol × 2 = 159.80 g/mol
Molar mass = 137.33 + 159.80 = 297.13 g/mol

#### c. Mg(C₂H₃O₂)₂
- Mg: 24.31 g/mol
- C₂H₃O₂: (2×12.01) + (3×1.01) + (2×16.00) = 24.02 + 3.03 + 32.00 = 59.05 g/mol
- Two acetate groups: 59.05 × 2 = 118.10 g/mol
Molar mass = 24.31 + 118.10 = 142.41 g/mol

#### d. Ca(NO₃)₂
- Ca: 40.08 g/mol
- NO₃: (14.01 + 3×16.00) = 62.01 g/mol
- Two NO₃ groups: 62.01 × 2 = 124.02 g/mol
Molar mass = 40.08 + 124.02 = 164.10 g/mol

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2. How many molecules are in 45.0 grams CH₄?

Step 1: Molar mass of CH₄
- C: 12.01, H: 1.01 × 4 = 4.04 → Total = 16.05 g/mol

Step 2: Convert grams to moles
$$
\frac{45.0 \text{ g}}{16.05 \text{ g/mol}} = 2.804 \text{ mol}
$$

Step 3: Convert moles to molecules
$$
2.804 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.688 \times 10^{24} \text{ molecules}
$$

✔ Answer: $1.69 \times 10^{24}$ molecules (rounded to 3 sig figs)

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3. How many moles are in 18.8 grams NaOH?

Molar mass NaOH:
- Na: 22.99, O: 16.00, H: 1.01 → 40.00 g/mol

$$
\frac{18.8 \text{ g}}{40.00 \text{ g/mol}} = 0.470 \text{ mol}
$$

✔ Answer: 0.470 mol

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4. A salt shaker containing $9.58 \times 10^{23}$ formula units NaCl contains how many moles?

Use Avogadro’s number:
$$
\frac{9.58 \times 10^{23} \text{ f.u.}}{6.022 \times 10^{23} \text{ f.u./mol}} = 1.590 \text{ mol}
$$

✔ Answer: 1.59 mol (3 sig figs)

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5. Which has the greatest mass?

Compare:
- A: 4.2 mol C
- B: $8.34 \times 10^{24}$ f.u. CaCO₃
- C: 12.6 g Al(NO₃)₃

#### a. 4.2 mol C
- Molar mass C = 12.01 g/mol
- Mass = $4.2 \times 12.01 = 50.44$ g

#### b. $8.34 \times 10^{24}$ f.u. CaCO₃
- Molar mass CaCO₃ = 40.08 + 12.01 + 48.00 = 100.09 g/mol
- Moles = $\frac{8.34 \times 10^{24}}{6.022 \times 10^{23}} = 13.84$ mol
- Mass = $13.84 \times 100.09 = 1385.4$ g

#### c. 12.6 g Al(NO₃)₃ — already given

Clearly, B is largest.

✔ Answer: $8.34 \times 10^{24}$ f.u. CaCO₃ has the greatest mass

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6. Find the number of representative particles in each:

Use: $1 \text{ mol} = 6.022 \times 10^{23}$ particles

#### a. 0.28 mol Ni
$$
0.28 \times 6.022 \times 10^{23} = 1.686 \times 10^{23} \text{ atoms}
$$

#### b. 1.84 mol S
$$
1.84 \times 6.022 \times 10^{23} = 1.108 \times 10^{24} \text{ atoms}
$$

#### c. 1.32 mol F₂
F₂ is diatomic, but "representative particles" = molecules
$$
1.32 \times 6.022 \times 10^{23} = 7.949 \times 10^{23} \text{ molecules}
$$

#### d. 3.15 mol K₃PO₄
Each formula unit is one particle
$$
3.15 \times 6.022 \times 10^{23} = 1.897 \times 10^{24} \text{ formula units}
$$

✔ Answers:
- a. $1.69 \times 10^{23}$ atoms
- b. $1.11 \times 10^{24}$ atoms
- c. $7.95 \times 10^{23}$ molecules
- d. $1.90 \times 10^{24}$ formula units

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7. How many liters would 1.45 moles oxygen gas occupy at STP?

At STP, 1 mol gas = 22.4 L

$$
1.45 \text{ mol} \times 22.4 \text{ L/mol} = 32.48 \text{ L}
$$

✔ Answer: 32.5 L (3 sig figs)

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8. How many grams would 15.8 L of ammonia occupy at STP?

Ammonia = NH₃
Molar mass: 14.01 + 3(1.01) = 17.04 g/mol

First, find moles:
$$
\frac{15.8 \text{ L}}{22.4 \text{ L/mol}} = 0.705 \text{ mol}
$$

Then convert to grams:
$$
0.705 \text{ mol} \times 17.04 \text{ g/mol} = 11.99 \text{ g}
$$

✔ Answer: 12.0 g (3 sig figs)

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9. How many total atoms are in 5.8 mol CaCl₂?

Each CaCl₂ molecule has: 1 Ca + 2 Cl = 3 atoms

So:
- Moles of atoms = $5.8 \text{ mol} \times 3 = 17.4 \text{ mol atoms}$

Now convert to atoms:
$$
17.4 \times 6.022 \times 10^{23} = 1.047 \times 10^{25} \text{ atoms}
$$

✔ Answer: $1.05 \times 10^{25}$ atoms

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10. Calculate percent composition of each element

#### a. Iron(III) oxide → Fe₂O₃
- Fe: 55.85 × 2 = 111.70
- O: 16.00 × 3 = 48.00
- Molar mass = 159.70 g/mol

% Fe = $\frac{111.70}{159.70} \times 100 = 69.94\%$
% O = $\frac{48.00}{159.70} \times 100 = 30.06\%$

#### b. Ca₃(PO₄)₂
- Ca: 40.08 × 3 = 120.24
- P: 30.97 × 2 = 61.94
- O: 16.00 × 8 = 128.00
- Total = 310.18 g/mol

% Ca = $\frac{120.24}{310.18} \times 100 = 38.77\%$
% P = $\frac{61.94}{310.18} \times 100 = 19.97\%$
% O = $\frac{128.00}{310.18} \times 100 = 41.26\%$

#### c. Methane (CH₄)
- C: 12.01, H: 4.04 → 16.05 g/mol
% C = $\frac{12.01}{16.05} \times 100 = 74.83\%$
% H = $\frac{4.04}{16.05} \times 100 = 25.17\%$

#### d. H₂SO₄
- H: 2.02, S: 32.07, O: 64.00 → 98.09 g/mol
% H = $\frac{2.02}{98.09} \times 100 = 2.06\%$
% S = $\frac{32.07}{98.09} \times 100 = 32.69\%$
% O = $\frac{64.00}{98.09} \times 100 = 65.25\%$

✔ All percent compositions calculated.

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11. Adrenaline: molecular mass = 183 g/mol; 59.0% C, 7.1% H, 26.2% O, 7.7% N

Assume 100 g sample:

| Element | Mass (g) | Moles | Divide by smallest |
|--------|----------|-------|--------------------|
| C | 59.0 | 59.0 / 12.01 = 4.91 | 4.91 / 1.11 ≈ 4.42 |
| H | 7.1 | 7.1 / 1.01 = 7.03 | 7.03 / 1.11 ≈ 6.33 |
| O | 26.2 | 26.2 / 16.00 = 1.64 | 1.64 / 1.11 ≈ 1.48 |
| N | 7.7 | 7.7 / 14.01 = 0.55 | 0.55 / 0.55 = 1.00 |

Wait — let's use actual moles:

- C: 59.0 / 12.01 = 4.91
- H: 7.1 / 1.01 = 7.03
- O: 26.2 / 16.00 = 1.64
- N: 7.7 / 14.01 = 0.55

Divide by smallest (0.55):

- C: 4.91 / 0.55 ≈ 8.93 → ~9
- H: 7.03 / 0.55 ≈ 12.78 → ~13
- O: 1.64 / 0.55 ≈ 2.98 → ~3
- N: 0.55 / 0.55 = 1

Empirical formula: C₉H₁₃O₃N

Empirical mass:
- C: 9×12.01 = 108.09
- H: 13×1.01 = 13.13
- O: 3×16.00 = 48.00
- N: 14.01
→ Total = 183.23 g/mol

This matches molecular mass (183 g/mol), so:

✔ Empirical = Molecular = C₉H₁₃O₃N

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12. Compound: 72.2% Mg, 27.8% N

Assume 100 g:

- Mg: 72.2 g → 72.2 / 24.31 = 2.97 mol
- N: 27.8 g → 27.8 / 14.01 = 1.98 mol

Ratio:
Mg : N = 2.97 : 1.98 ≈ 1.5 : 1 → Multiply by 2 → 3:2

✔ Empirical formula: Mg₃N₂

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13. Nicotine: 74.1% C, 8.6% H, 17.3% N; MM = 162 g/mol

100 g sample:

- C: 74.1 g → 74.1 / 12.01 = 6.17 mol
- H: 8.6 g → 8.6 / 1.01 = 8.51 mol
- N: 17.3 g → 17.3 / 14.01 = 1.23 mol

Divide by smallest (1.23):

- C: 6.17 / 1.23 ≈ 5.02 → 5
- H: 8.51 / 1.23 ≈ 6.92 → 7
- N: 1.23 / 1.23 = 1

Empirical formula: C₅H₇N

Empirical mass:
- C: 5×12.01 = 60.05
- H: 7×1.01 = 7.07
- N: 14.01
→ Total = 79.13 g/mol

Now divide molecular mass by empirical:
$$
\frac{162}{79.13} \approx 2.05 \rightarrow \text{round to } 2
$$

Multiply empirical by 2: C₁₀H₁₄N₂

✔ Empirical: C₅H₇N, Molecular: C₁₀H₁₄N₂

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14. How many total ions are in 2.95 grams MgSO₃?

MgSO₃ dissociates:
MgSO₃ → Mg²⁺ + SO₃²⁻ → 2 ions per formula unit

Molar mass MgSO₃:
- Mg: 24.31
- S: 32.07
- O: 48.00
→ Total = 104.38 g/mol

Moles = $\frac{2.95 \text{ g}}{104.38 \text{ g/mol}} = 0.02825 \text{ mol}$

Formula units = $0.02825 \times 6.022 \times 10^{23} = 1.699 \times 10^{22}$

Each gives 2 ions → total ions:
$$
1.699 \times 10^{22} \times 2 = 3.398 \times 10^{22} \text{ ions}
$$

✔ Answer: $3.40 \times 10^{22}$ ions (3 sig figs)

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## ✔ Final Answers Summary:

1.
a. 23.95 g/mol
b. 297.13 g/mol
c. 142.41 g/mol
d. 164.10 g/mol

2. $1.69 \times 10^{24}$ molecules
3. 0.470 mol
4. 1.59 mol
5. $8.34 \times 10^{24}$ f.u. CaCO₃
6.
a. $1.69 \times 10^{23}$ atoms
b. $1.11 \times 10^{24}$ atoms
c. $7.95 \times 10^{23}$ molecules
d. $1.90 \times 10^{24}$ formula units
7. 32.5 L
8. 12.0 g
9. $1.05 \times 10^{25}$ atoms
10. See above calculations
11. Empirical & Molecular: C₉H₁₃O₃N
12. Mg₃N₂
13. Empirical: C₅H₇N, Molecular: C₁₀H₁₄N₂
14. $3.40 \times 10^{22}$ ions

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