Question

what is the empirical formula (lowest whole number ratio) of the compounds below?

1. 75% carbon, 25% hydrogen
2. 52.7% potassium, 47.3% chlorine
3. 22.1% aluminum, 25.4% phosphorus, 52.5% oxygen
4. 13% magnesium, 87% bromine
5. 32.4% sodium, 22.5% sulfur, 45.1% oxygen
6. 25.3% copper, 12.9% sulfur, 25.7% oxygen, 36.1% water

Explanation:

Compound 1: 75% C, 25% H

Step1: Assume 100g sample, find moles

Moles of C: $\frac{75}{12.01} \approx 6.245$
Moles of H: $\frac{25}{1.008} \approx 24.80$

Step2: Divide by smallest mole value

Ratio of C: $\frac{6.245}{6.245} = 1$
Ratio of H: $\frac{24.80}{6.245} \approx 4$

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Compound 2: 52.7% K, 47.3% Cl

Step1: Assume 100g sample, find moles

Moles of K: $\frac{52.7}{39.10} \approx 1.348$
Moles of Cl: $\frac{47.3}{35.45} \approx 1.334$

Step2: Divide by smallest mole value

Ratio of K: $\frac{1.348}{1.334} \approx 1$
Ratio of Cl: $\frac{1.334}{1.334} = 1$

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Compound 3: 22.1% Al, 25.4% P, 52.5% O

Step1: Assume 100g sample, find moles

Moles of Al: $\frac{22.1}{26.98} \approx 0.819$
Moles of P: $\frac{25.4}{30.97} \approx 0.820$
Moles of O: $\frac{52.5}{16.00} = 3.281$

Step2: Divide by smallest mole value

Ratio of Al: $\frac{0.819}{0.819} = 1$
Ratio of P: $\frac{0.820}{0.819} \approx 1$
Ratio of O: $\frac{3.281}{0.819} \approx 4$

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Compound 4: 13% Mg, 87% Br

Step1: Assume 100g sample, find moles

Moles of Mg: $\frac{13}{24.31} \approx 0.535$
Moles of Br: $\frac{87}{79.90} \approx 1.089$

Step2: Divide by smallest mole value

Ratio of Mg: $\frac{0.535}{0.535} = 1$
Ratio of Br: $\frac{1.089}{0.535} \approx 2$

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Compound 5: 32.4% Na, 22.5% S, 45.1% O

Step1: Assume 100g sample, find moles

Moles of Na: $\frac{32.4}{22.99} \approx 1.409$
Moles of S: $\frac{22.5}{32.07} \approx 0.702$
Moles of O: $\frac{45.1}{16.00} \approx 2.819$

Step2: Divide by smallest mole value

Ratio of Na: $\frac{1.409}{0.702} \approx 2$
Ratio of S: $\frac{0.702}{0.702} = 1$
Ratio of O: $\frac{2.819}{0.702} \approx 4$

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Compound 6: 25.3% Cu, 12.9% S, 25.7% O, 36.1% H₂O

Step1: Assume 100g sample, find moles

Moles of Cu: $\frac{25.3}{63.55} \approx 0.398$
Moles of S: $\frac{12.9}{32.07} \approx 0.402$
Moles of O: $\frac{25.7}{16.00} \approx 1.606$
Moles of H₂O: $\frac{36.1}{18.02} \approx 2.003$

Step2: Divide by smallest mole value

Ratio of Cu: $\frac{0.398}{0.398} = 1$
Ratio of S: $\frac{0.402}{0.398} \approx 1$
Ratio of O: $\frac{1.606}{0.398} \approx 4$
Ratio of H₂O: $\frac{2.003}{0.398} \approx 5$

Answer:

1. $\text{CH}_4$
2. $\text{KCl}$
3. $\text{AlPO}_4$
4. $\text{MgBr}_2$
5. $\text{Na}_2\text{SO}_4$
6. $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$