Question

Upon combustion, a compound containing only carbon and hydrogen produces 2.67 g"TexFormula1" title="CO_{2}" alt="CO_{2}" align="absmiddle" class="latex-formula"> and 1.10 g$H_{2}O$. Find the empirical formula of the compound.

Answer 1

Answer:

$\boxed{\text{CH _{2} }}$

Explanation:

1. Calculate the mass of each element

$\text{Mass of C} = \text{2.67 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.7286 g C}\\\\\text{Mass of H} = \text{1.10 g }\text{H _{2} O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H _{2} O}}} = \text{0.1231 g H}$

2. Calculate the moles of each element

$\text{Moles of C = 0.7286 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.060 67 mol C}\\\\\text{Moles of H = 0.1231 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.1221 mol H}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.06067}{0.06067}= 1\\\\\text{H: } \dfrac{0.1221}{0.06067} = 2.015$

4. Round the ratios to the nearest integer

C:H = 1:2

5. Write the empirical formula

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