Question

Upon combustion, a compound containing only carbon and hydrogen produces 2.67 g
`CO_(2)` and 1.10 g
`H_(2)O`. Find the empirical formula of the compound.

Answer 1

`\boxed{\text{CH$_(2)$}}`

Step-by-step explanation:

1. Calculate the mass of each element

`\text{Mass of C} = \text{2.67 g } \text{CO}_(2)* \frac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_(2)}= \text{0.7286 g C}\\\\\text{Mass of H} = \text{1.10 g }\text{H$_(2)$O}* \frac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_(2)$O}}} = \text{0.1231 g H}`

2. Calculate the moles of each element

`\text{Moles of C = 0.7286 g C}*\frac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.060 67 mol C}\\\\\text{Moles of H = 0.1231 g H} * \frac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.1221 mol H}`

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

`\text{C: } (0.06067)/(0.06067)= 1\\\\\text{H: } (0.1221)/(0.06067) = 2.015`

4. Round the ratios to the nearest integer

C:H = 1:2

5. Write the empirical formula

The empirical formula is
`\boxed{\text{CH$_(2)$}}`