Question

Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of CO2 and 2.58 mg of H2O. What is the empirical formula of the compound

Answer 1

Answer:

$\boxed{\text{C _{3} H _{6} O}}$

Explanation:

Let's call the ethyl butyrate X.

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{6.32 mg } \text{CO}_{2}\times \dfrac{\text{12.01 mg C}}{\text{44.01 mg }\text{CO}_{2}}= \text{1.725 mg C}$

(b) Mass of H

$\text{Mass of H} = \text{2.58 mg }\text{H _{2} O}\times \dfrac{\text{2.016 mg H}}{\text{18.02 mg } \text{{H _{2} O}}} = \text{0.2886 mg H}$

(d) Mass of O

Mass of O = 2.78 - 1.725 - 0.2886 = 0.7667 g

2. Calculate the moles of each element

$\text{Moles of C = 1.725 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{0.1436 mmol C}\\\\\text{Moles of H = 0.2886 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{0.2863 mmol H}\\\\\text{Moles of O = 0.7767 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{0.04792 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.1436}{0.04972}= 2.997\\\\\text{H: } \dfrac{0.2863}{0.04972} = 5.976\\\\\text{Fe: } \dfrac{0.04972}{0.04972} = 1$

4. Round the ratios to the nearest integer

C:H:O = 3:6:1

5. Write the empirical formula $\text{The empirical formula is } \boxed{\textbf{C _{3} H _{6} O}}$