Question

In the combustion of 1.99 g of an unknown compound containing only C, H, and o, 2.97 g CO, and 0.911 g H20. are produced. What is the empirical formula for the unknown compound? Assume all carbon goes to produce CO, and all hydrogen goes to produce H20.

Answer 1

The empirical formula of the unknown compound is C₂H₁O₃₀, indicating a ratio of 2 moles of carbon, 1 mole of hydrogen, and 30 moles of oxygen.

1. Moles of CO (carbon dioxide):

Molar mass of CO: 27.9694 g/mol

Moles of CO: 2.97 g / 27.9694 g/mol ≈ 0.106 mol CO

2. Moles of H₂O:

Molar mass of H₂O: 18.01528 g/mol

Moles of H₂O: 0.911 g / 18.01528 g/mol ≈ 0.050 mol H₂O

3. Moles of C and H:

Moles of C ≈ 0.106 mol

Moles of H ≈ 0.050 mol

4. Moles of O:

Moles of O ≈ 1.99 g / (0.106 mol * 12.01 g/mol + 0.050 mol * 1.00794 g/mol)

Moles of O ≈ 1.504 mol

Now, let's find the ratio of moles and round to the nearest whole number:

Moles of C: 0.106 mol / 0.050 mol ≈ 2.12

Moles of H: 0.050 mol / 0.050 mol = 1

Moles of O: 1.504 mol / 0.050 mol ≈ 30.08

Rounding these ratios to the nearest whole number:

Moles of C: 2

Moles of H: 1

Moles of O: 30

The empirical formula is C₂H₁O₃₀. However, express it in the simplest, whole-number ratio by dividing each subscript by the smallest subscript (1 in this case):

`\[ \text{C}_2\text{H}_1\text{O}_(30) / 1 \]`

The simplest, whole-number ratio is C₂H₁O₃₀. Therefore, the empirical formula of the unknown compound is C₂H₁O₃₀.