Question

A 2.724 grams sample of diethyl ether was analyzed using combustion analysis. Produced by the combustion were 6.470 g of CO2 and 3.311 g of water. What is the empirical formula of diethyl ether?

Answer 1

$$n(CO_2) = {m \over M} = {6.470g \over 44.01g/mol} = 0.147mol$$

$$n(H_2O) = {3.311g \over 18.01528g/mol} = 0.184mol$$

$$n(C) = n(CO_2) = 0.147mol$$

$$n(H) = 2n(H_2O) = 2*0.184mol = 0.368mol$$

$${n(C) \over n(H)} = {0.147 \over 0.368} = {1 \over 2.5} = {2 \over 5}$$

$$C_2H_5O = C_4H_{10}O$$

ANSWER: C₄H₁₀O