Free Printable Percent Composition Worksheets - Free Printable

Let’s solve each problem step by step. We’ll find the percent composition of each element in the compound.

Percent Composition Formula:
> % of element = (total mass of element in formula / molar mass of compound) × 100%

We’ll use atomic masses from the periodic table (rounded to 2 decimal places for simplicity):

- K = 39.10 g/mol
- Mn = 54.94 g/mol
- O = 16.00 g/mol
- H = 1.01 g/mol
- Cl = 35.45 g/mol
- Mg = 24.31 g/mol
- N = 14.01 g/mol
- P = 30.97 g/mol
- Fe = 55.85 g/mol
- Ag = 107.87 g/mol

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1. KMnO₄

Molar mass = K + Mn + 4×O = 39.10 + 54.94 + 4×16.00 = 39.10 + 54.94 + 64.00 = 158.04 g/mol

- K: (39.10 / 158.04) × 100 ≈ 24.74%
- Mn: (54.94 / 158.04) × 100 ≈ 34.76%
- O: (64.00 / 158.04) × 100 ≈ 40.50%

Check: 24.74 + 34.76 + 40.50 = 100.00 ✔

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2. HCl

Molar mass = H + Cl = 1.01 + 35.45 = 36.46 g/mol

- H: (1.01 / 36.46) × 100 ≈ 2.77%
- Cl: (35.45 / 36.46) × 100 ≈ 97.23%

Check: 2.77 + 97.23 = 100.00 ✔

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3. Mg(NO₃)₂ → This means 1 Mg, 2 N, and 6 O atoms

Molar mass = Mg + 2×N + 6×O = 24.31 + 2×14.01 + 6×16.00 = 24.31 + 28.02 + 96.00 = 148.33 g/mol

- Mg: (24.31 / 148.33) × 100 ≈ 16.39%
- N: (28.02 / 148.33) × 100 ≈ 18.89%
- O: (96.00 / 148.33) × 100 ≈ 64.72%

Check: 16.39 + 18.89 + 64.72 = 100.00 ✔

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4. (NH₄)₃PO₄ → Break it down:

- N: 3 atoms (from 3 NH₄ groups)
- H: 12 atoms (each NH₄ has 4 H → 3×4=12)
- P: 1 atom
- O: 4 atoms

Molar mass = 3×N + 12×H + P + 4×O
= 3×14.01 + 12×1.01 + 30.97 + 4×16.00
= 42.03 + 12.12 + 30.97 + 64.00 = 149.12 g/mol

- N: (42.03 / 149.12) × 100 ≈ 28.19%
- H: (12.12 / 149.12) × 100 ≈ 8.13%
- P: (30.97 / 149.12) × 100 ≈ 20.77%
- O: (64.00 / 149.12) × 100 ≈ 42.92%

Check: 28.19 + 8.13 + 20.77 + 42.92 = 100.01 → rounding error, acceptable ✔

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5. Fe(OH)₃ → 1 Fe, 3 O, 3 H

Molar mass = Fe + 3×O + 3×H = 55.85 + 3×16.00 + 3×1.01 = 55.85 + 48.00 + 3.03 = 106.88 g/mol

- Fe: (55.85 / 106.88) × 100 ≈ 52.25%
- O: (48.00 / 106.88) × 100 ≈ 44.91%
- H: (3.03 / 106.88) × 100 ≈ 2.84%

Check: 52.25 + 44.91 + 2.84 = 100.00 ✔

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6. AgNO₃ → 1 Ag, 1 N, 3 O

Molar mass = Ag + N + 3×O = 107.87 + 14.01 + 48.00 = 169.88 g/mol

- Ag: (107.87 / 169.88) × 100 ≈ 63.50%
- N: (14.01 / 169.88) × 100 ≈ 8.25%
- O: (48.00 / 169.88) × 100 ≈ 28.25%

Check: 63.50 + 8.25 + 28.25 = 100.00 ✔

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Final Answer:

1. KMnO₄
K = 24.74%
Mn = 34.76%
O = 40.50%

2. HCl
H = 2.77%
Cl = 97.23%

3. Mg(NO₃)₂
Mg = 16.39%
N = 18.89%
O = 64.72%

4. (NH₄)₃PO₄
N = 28.19%
H = 8.13%
P = 20.77%
O = 42.92%

5. Fe(OH)₃
Fe = 52.25%
O = 44.91%
H = 2.84%

6. AgNO₃
Ag = 63.50%
N = 8.25%
O = 28.25%