Moles calculations worksheet - Free Printable

Let’s solve each part step by step. We’ll use the formula:

Number of moles = mass (g) ÷ molar mass (g/mol)
Mass (g) = number of moles × molar mass (g/mol)
Percentage by mass of carbon = (total mass of carbon in compound ÷ molar mass of compound) × 100%

We’ll need atomic masses (rounded to 1 decimal place for simplicity, but we’ll use precise values where needed):

- H = 1.0
- C = 12.0
- N = 14.0
- O = 16.0
- Na = 23.0
- S = 32.1
- Cu = 63.5
- Fe = 55.8
- Ca = 40.1
- Al = 27.0

---

Question 1: Moles from mass

#### a) 2.3 g sodium (Na)
Molar mass Na = 23.0 g/mol
→ moles = 2.3 / 23.0 = 0.10 mol

#### b) 0.32 g O₂
Molar mass O₂ = 2 × 16.0 = 32.0 g/mol
→ moles = 0.32 / 32.0 = 0.010 mol

#### c) 1.0 g CH₄
Molar mass CH₄ = 12.0 + 4×1.0 = 16.0 g/mol
→ moles = 1.0 / 16.0 = 0.0625 mol

#### d) 0.10 g SO₂
Molar mass SO₂ = 32.1 + 2×16.0 = 64.1 g/mol
→ moles = 0.10 / 64.1 ≈ 0.00156 mol (or 1.56 × 10⁻³ mol)

#### e) 4.0 g N₂
Molar mass N₂ = 2×14.0 = 28.0 g/mol
→ moles = 4.0 / 28.0 ≈ 0.143 mol

#### f) 2.5 g Na₂CO₃
Molar mass Na₂CO₃ = 2×23.0 + 12.0 + 3×16.0 = 46.0 + 12.0 + 48.0 = 106.0 g/mol
→ moles = 2.5 / 106.0 ≈ 0.0236 mol

#### g) 15.6 g Cu(NO₃)₂
Molar mass Cu(NO₃)₂ = 63.5 + 2×[14.0 + 3×16.0] = 63.5 + 2×(14.0 + 48.0) = 63.5 + 2×62.0 = 63.5 + 124.0 = 187.5 g/mol
→ moles = 15.6 / 187.5 ≈ 0.0832 mol

#### h) 2.7 g Fe₂O₃
Molar mass Fe₂O₃ = 2×55.8 + 3×16.0 = 111.6 + 48.0 = 159.6 g/mol
→ moles = 2.7 / 159.6 ≈ 0.0169 mol

#### i) 3.0 g (NH₄)₂SO₄
Molar mass (NH₄)₂SO₄ = 2×[14.0 + 4×1.0] + 32.1 + 4×16.0 = 2×18.0 + 32.1 + 64.0 = 36.0 + 32.1 + 64.0 = 132.1 g/mol
→ moles = 3.0 / 132.1 ≈ 0.0227 mol

---

Question 2: Mass from moles

#### a) 3.0 mol NaOH
Molar mass NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol
→ mass = 3.0 × 40.0 = 120 g

#### b) 0.10 mol C₃H₈
Molar mass C₃H₈ = 3×12.0 + 8×1.0 = 36.0 + 8.0 = 44.0 g/mol
→ mass = 0.10 × 44.0 = 4.4 g

#### c) 0.400 mol CuSO₄
Molar mass CuSO₄ = 63.5 + 32.1 + 4×16.0 = 63.5 + 32.1 + 64.0 = 159.6 g/mol
→ mass = 0.400 × 159.6 = 63.84 g → round to 63.8 g

#### d) 100.0 mol SO₃
Molar mass SO₃ = 32.1 + 3×16.0 = 32.1 + 48.0 = 80.1 g/mol
→ mass = 100.0 × 80.1 = 8010 g

#### e) 0.27 mol HNO₃
Molar mass HNO₃ = 1.0 + 14.0 + 3×16.0 = 1.0 + 14.0 + 48.0 = 63.0 g/mol
→ mass = 0.27 × 63.0 = 17.01 g → round to 17.0 g

#### f) 0.85 mol Al₂(SO₄)₃
Molar mass Al₂(SO₄)₃ = 2×27.0 + 3×[32.1 + 4×16.0] = 54.0 + 3×(32.1 + 64.0) = 54.0 + 3×96.1 = 54.0 + 288.3 = 342.3 g/mol
→ mass = 0.85 × 342.3 ≈ 290.955 g → round to 291 g

#### g) 0.600 mol CaCl₂
Molar mass CaCl₂ = 40.1 + 2×35.5 = 40.1 + 71.0 = 111.1 g/mol
→ mass = 0.600 × 111.1 = 66.66 g → round to 66.7 g

#### h) 2.40 mol NH₄NO₃
Molar mass NH₄NO₃ = 14.0 + 4×1.0 + 14.0 + 3×16.0 = 18.0 + 62.0? Wait — better:
N: 2 atoms → 2×14.0 = 28.0
H: 4 atoms → 4×1.0 = 4.0
O: 3 atoms → 3×16.0 = 48.0
Total = 28.0 + 4.0 + 48.0 = 80.0 g/mol
→ mass = 2.40 × 80.0 = 192 g

#### i) 2.0 mol CaCO₃
Molar mass CaCO₃ = 40.1 + 12.0 + 3×16.0 = 40.1 + 12.0 + 48.0 = 100.1 g/mol
→ mass = 2.0 × 100.1 = 200.2 g → round to 200 g

---

Question 3: Percentage by mass of carbon

#### a) CO₂
Molar mass = 12.0 + 2×16.0 = 44.0 g/mol
Carbon mass = 12.0
% C = (12.0 / 44.0) × 100 ≈ 27.3%

#### b) C₂H₆
Molar mass = 2×12.0 + 6×1.0 = 24.0 + 6.0 = 30.0 g/mol
Carbon mass = 24.0
% C = (24.0 / 30.0) × 100 = 80.0%

#### c) C₆H₅NO₂
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
Carbon mass = 72.0
% C = (72.0 / 123.0) × 100 ≈ 58.5%

#### d) C₆H₅C(OCH₃) — wait, this is written as “C₆H₅C OCH₃” — likely means C₆H₅COCH₃ (acetophenone or similar). Let’s interpret as C₈H₈O (since C₆H₅–C–OCH₃ would be C₆H₅ + C + O + CH₃ = C₈H₈O).

Check: C₆H₅ (phenyl) + COCH₃ (acetyl group) → C₈H₈O

Molar mass = 8×12.0 + 8×1.0 + 16.0 = 96.0 + 8.0 + 16.0 = 120.0 g/mol
Carbon mass = 96.0
% C = (96.0 / 120.0) × 100 = 80.0%

*(Note: If it was meant to be something else like benzyl methyl ether C₆H₅CH₂OCH₃, that’s also C₈H₁₀O — but given notation “C₆H₅C OCH₃”, likely missing bond, so standard interpretation is acetophenone C₈H₈O.)*

But let’s double-check the original: “C₆H₅C OCH₃” — probably typo and should be C₆H₅COCH₃, which is C₈H₈O.

Alternatively, if it’s C₆H₅–C≡N? No, says OCH₃. So yes, C₈H₈O.

Final Answer:

---

Final Answer:

1. Number of 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

2. Mass in grams:
a) 120 g
b) 4.4 g
c) 63.8 g
d) 8010 g
e) 17.0 g
f) 291 g
g) 66.7 g
h) 192 g
i) 200 g

3. Percentage by mass of carbon:
a) 27.3%
b) 80.0%
c) 58.5%
d) 80.0%