Worksheet for calculating the percentage composition of elements in various chemical compounds.

Let’s solve each problem step by step. We’ll find the molar mass of each compound first, then calculate the percentage composition for each element using:

> % composition = (mass of element in one mole of compound / molar mass of compound) × 100

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

- N = 14.01 g/mol
- H = 1.01 g/mol
- S = 32.07 g/mol
- O = 16.00 g/mol
- Al = 26.98 g/mol
- C = 12.01 g/mol
- Na = 22.99 g/mol
- Br = 79.90 g/mol
- Cu = 63.55 g/mol

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Problem 1: Ammonium sulfite — (NH₄)₂SO₃

First, count atoms:

- N: 2 (from two NH₄⁺ ions)
- H: 8 (each NH₄ has 4 H → 2×4=8)
- S: 1
- O: 3

Molar mass:

- N: 2 × 14.01 = 28.02
- H: 8 × 1.01 = 8.08
- S: 1 × 32.07 = 32.07
- O: 3 × 16.00 = 48.00
→ Total = 28.02 + 8.08 + 32.07 + 48.00 = 116.17 g/mol

Now percentages:

- % N = (28.02 / 116.17) × 100 ≈ 24.12%
- % H = (8.08 / 116.17) × 100 ≈ 6.96%
- % S = (32.07 / 116.17) × 100 ≈ 27.61%
- % O = (48.00 / 116.17) × 100 ≈ 41.32%

Check: 24.12 + 6.96 + 27.61 + 41.32 = 100.01% → close enough (rounding error)

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Problem 2: Aluminum acetate — Al(C₂H₃O₂)₃

Note: Acetate ion is C₂H₃O₂⁻, and there are 3 of them.

Count atoms:

- Al: 1
- C: 2 × 3 = 6
- H: 3 × 3 = 9
- O: 2 × 3 = 6

Molar mass:

- Al: 1 × 26.98 = 26.98
- C: 6 × 12.01 = 72.06
- H: 9 × 1.01 = 9.09
- O: 6 × 16.00 = 96.00
→ Total = 26.98 + 72.06 + 9.09 + 96.00 = 204.13 g/mol

Percentages:

- % Al = (26.98 / 204.13) × 100 ≈ 13.22%
- % C = (72.06 / 204.13) × 100 ≈ 35.30%
- % H = (9.09 / 204.13) × 100 ≈ 4.45%
- % O = (96.00 / 204.13) × 100 ≈ 47.03%

Check: 13.22 + 35.30 + 4.45 + 47.03 = 100.00% → perfect!

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Problem 3: Sodium bromide — NaBr

Atoms:

- Na: 1
- Br: 1

Molar mass:

- Na: 22.99
- Br: 79.90
→ Total = 22.99 + 79.90 = 102.89 g/mol

Percentages:

- % Na = (22.99 / 102.89) × 100 ≈ 22.34%
- % Br = (79.90 / 102.89) × 100 ≈ 77.66%

Check: 22.34 + 77.66 = 100.00% → good!

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Problem 4: Copper(II) hydroxide — Cu(OH)₂

Atoms:

- Cu: 1
- O: 2 (from two OH groups)
- H: 2 (from two OH groups)

Molar mass:

- Cu: 63.55
- O: 2 × 16.00 = 32.00
- H: 2 × 1.01 = 2.02
→ Total = 63.55 + 32.00 + 2.02 = 97.57 g/mol

Percentages:

- % Cu = (63.55 / 97.57) × 100 ≈ 65.13%
- % O = (32.00 / 97.57) × 100 ≈ 32.80%
- % H = (2.02 / 97.57) × 100 ≈ 2.07%

Check: 65.13 + 32.80 + 2.07 = 100.00% → perfect!

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Final Answer:

1. ammonium sulfite (NH₄)₂SO₃
Molar mass: 116.17 g/mol
% N: 24.12%
% H: 6.96%
% S: 27.61%
% O: 41.32%

2. aluminum acetate Al(C₂H₃O₂)₃
Molar mass: 204.13 g/mol
% Al: 13.22%
% C: 35.30%
% H: 4.45%
% O: 47.03%

3. sodium bromide NaBr
Molar mass: 102.89 g/mol
% Na: 22.34%
% Br: 77.66%

4. copper(II) hydroxide Cu(OH)₂
Molar mass: 97.57 g/mol
% Cu: 65.13%
% O: 32.80%
% H: 2.07%