Moles calculations worksheet - Free Printable

Let's solve each part of the problem step by step.

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Problem 1: Determine the number of moles present in each of the following

We use the formula:

$$
\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
$$

#### a) 2.3 g of sodium (Na)
- Molar mass of Na = 23.0 g/mol
$$
\text{moles} = \frac{2.3}{23.0} = 0.10 \text{ mol}
$$

#### b) 0.32 g of O₂
- Molar mass of O₂ = 2 × 16.0 = 32.0 g/mol
$$
\text{moles} = \frac{0.32}{32.0} = 0.010 \text{ mol}
$$

#### c) 1.0 g of CH₄
- Molar mass of CH₄ = 12.0 + (4 × 1.0) = 16.0 g/mol
$$
\text{moles} = \frac{1.0}{16.0} = 0.0625 \text{ mol}
$$

#### d) 0.10 g of SO₂
- Molar mass of SO₂ = 32.1 + (2 × 16.0) = 64.1 g/mol
$$
\text{moles} = \frac{0.10}{64.1} ≈ 0.00156 \text{ mol}
$$

#### e) 4.0 g of N₂
- Molar mass of N₂ = 2 × 14.0 = 28.0 g/mol
$$
\text{moles} = \frac{4.0}{28.0} ≈ 0.143 \text{ mol}
$$

#### f) 2.5 g of Na₂CO₃
- Molar mass of Na₂CO₃ = (2×23.0) + 12.0 + (3×16.0) = 46.0 + 12.0 + 48.0 = 106.0 g/mol
$$
\text{moles} = \frac{2.5}{106.0} ≈ 0.0236 \text{ mol}
$$

#### g) 15.6 g of Cu(NO₃)₂
- Molar mass of Cu(NO₃)₂ = 63.5 + 2×(14.0 + 3×16.0) = 63.5 + 2×(62.0) = 63.5 + 124.0 = 187.5 g/mol
$$
\text{moles} = \frac{15.6}{187.5} ≈ 0.0832 \text{ mol}
$$

#### h) 2.7 g of Fe₂O₃
- Molar mass of Fe₂O₃ = (2×55.8) + (3×16.0) = 111.6 + 48.0 = 159.6 g/mol
$$
\text{moles} = \frac{2.7}{159.6} ≈ 0.0169 \text{ mol}
$$

#### i) 3.0 g of (NH₄)₂SO₄
- Molar mass of (NH₄)₂SO₄ = (2×14.0 + 8×1.0) + 32.1 + (4×16.0) = (28.0 + 8.0) + 32.1 + 64.0 = 132.1 g/mol
$$
\text{moles} = \frac{3.0}{132.1} ≈ 0.0227 \text{ mol}
$$

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Problem 2: Work out the mass of each of the following

Use:
$$
\text{mass} = \text{moles} × \text{molar mass}
$$

#### a) 3.0 mol NaOH
- Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol
$$
\text{mass} = 3.0 × 40.0 = 120.0 \text{ g}
$$

#### b) 0.10 mol C₃H₈
- Molar mass of C₃H₈ = (3×12.0) + (8×1.0) = 36.0 + 8.0 = 44.0 g/mol
$$
\text{mass} = 0.10 × 44.0 = 4.4 \text{ g}
$$

#### c) 0.400 mol CuSO₄
- Molar mass of CuSO₄ = 63.5 + 32.1 + (4×16.0) = 63.5 + 32.1 + 64.0 = 159.6 g/mol
$$
\text{mass} = 0.400 × 159.6 = 63.84 \text{ g}
$$

#### d) 100.0 mol SO₃
- Molar mass of SO₃ = 32.1 + (3×16.0) = 32.1 + 48.0 = 80.1 g/mol
$$
\text{mass} = 100.0 × 80.1 = 8010 \text{ g}
$$

#### e) 0.27 mol HNO₃
- Molar mass of HNO₃ = 1.0 + 14.0 + (3×16.0) = 1.0 + 14.0 + 48.0 = 63.0 g/mol
$$
\text{mass} = 0.27 × 63.0 = 17.01 \text{ g}
$$

#### f) 0.85 mol Al₂(SO₄)₃
- Molar mass of Al₂(SO₄)₃ = (2×27.0) + 3×(32.1 + 64.0) = 54.0 + 3×96.1 = 54.0 + 288.3 = 342.3 g/mol
$$
\text{mass} = 0.85 × 342.3 ≈ 290.96 \text{ g}
$$

#### g) 0.600 mol CaCl₂
- Molar mass of CaCl₂ = 40.1 + (2×35.5) = 40.1 + 71.0 = 111.1 g/mol
$$
\text{mass} = 0.600 × 111.1 = 66.66 \text{ g}
$$

#### h) 2.40 mol NH₄NO₃
- Molar mass of NH₄NO₃ = (14.0 + 4×1.0) + (14.0 + 3×16.0) = 18.0 + 62.0 = 80.0 g/mol
$$
\text{mass} = 2.40 × 80.0 = 192.0 \text{ g}
$$

#### i) 2.0 mol CaCO₃
- Molar mass of CaCO₃ = 40.1 + 12.0 + (3×16.0) = 40.1 + 12.0 + 48.0 = 100.1 g/mol
$$
\text{mass} = 2.0 × 100.1 = 200.2 \text{ g}
$$

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Problem 3: Calculate the percentage by mass of carbon in each of the following

Use:
$$
\% \text{C} = \left( \frac{\text{total mass of C in molecule}}{\text{molar mass of compound}} \right) × 100\%
$$

#### a) CO₂
- Molar mass = 12.0 + (2×16.0) = 44.0 g/mol
- Mass of C = 12.0 g
$$
\% \text{C} = \frac{12.0}{44.0} × 100\% ≈ 27.27\%
$$

#### b) C₂H₆
- Molar mass = (2×12.0) + (6×1.0) = 24.0 + 6.0 = 30.0 g/mol
- Mass of C = 24.0 g
$$
\% \text{C} = \frac{24.0}{30.0} × 100\% = 80.0\%
$$

#### c) C₆H₅NO₂ (Nitrobenzene)
- Molar mass = (6×12.0) + (5×1.0) + 14.0 + (2×16.0) = 72.0 + 5.0 + 14.0 + 32.0 = 123.0 g/mol
- Mass of C = 72.0 g
$$
\% \text{C} = \frac{72.0}{123.0} × 100\% ≈ 58.54\%
$$

#### d) C₆H₅COCH₃ (Acetophenone)
Note: This is written as C₆H₅C(O)CH₃, which is C₈H₈O.
- So molecular formula: C₈H₈O
- Molar mass = (8×12.0) + (8×1.0) + 16.0 = 96.0 + 8.0 + 16.0 = 120.0 g/mol
- Mass of C = 96.0 g
$$
\% \text{C} = \frac{96.0}{120.0} × 100\% = 80.0\%
$$

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✔ Final Answers:

#### Problem 1: Moles
a) 0.10 mol
b) 0.010 mol
c) 0.0625 mol
d) 0.00156 mol
e) 0.143 mol
f) 0.0236 mol
g) 0.0832 mol
h) 0.0169 mol
i) 0.0227 mol

#### Problem 2: Mass
a) 120.0 g
b) 4.4 g
c) 63.84 g
d) 8010 g
e) 17.01 g
f) 290.96 g
g) 66.66 g
h) 192.0 g
i) 200.2 g

#### Problem 3: % Carbon by mass
a) 27.27%
b) 80.0%
c) 58.54%
d) 80.0%

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