Question

1. a 4.05 g sample of a compound containing c, h, and o was burned completely. the only combustion products were 10.942 g co₂ and 4.476 g h₂o. what is the empirical formula of the compound?

a) c₇h₇o
b) c₇h₁₄o
c) c₆h₁₁o
d) c₆ho
e) ch₂o₂

Explanation:

Step1: Calculate moles of C

The molar mass of $CO_2$ is $M_{CO_2}=44.01\ g/mol$. The moles of $CO_2$ is $n_{CO_2}=\frac{10.942\ g}{44.01\ g/mol}=0.2486\ mol$. Since 1 mole of $CO_2$ contains 1 mole of C, the moles of C, $n_C = 0.2486\ mol$.

Step2: Calculate moles of H

The molar mass of $H_2O$ is $M_{H_2O}=18.02\ g/mol$. The moles of $H_2O$ is $n_{H_2O}=\frac{4.476\ g}{18.02\ g/mol}=0.2484\ mol$. Since 1 mole of $H_2O$ contains 2 moles of H, the moles of H, $n_H=2\times0.2484\ mol = 0.4968\ mol$.

Step3: Calculate mass of C and H

The mass of C is $m_C=n_C\times12.01\ g/mol=0.2486\ mol\times12.01\ g/mol = 2.9867\ g$. The mass of H is $m_H=n_H\times1.008\ g/mol=0.4968\ mol\times1.008\ g/mol = 0.5008\ g$.

Step4: Calculate mass of O

The mass of the sample is 4.05 g. The mass of O, $m_O=4.05\ g-(2.9867\ g + 0.5008\ g)=0.5625\ g$.

Step5: Calculate moles of O

The molar mass of O is 16.00 g/mol. The moles of O, $n_O=\frac{0.5625\ g}{16.00\ g/mol}=0.0352\ mol$.

Step6: Find the mole - ratio

Divide the moles of each element by the smallest number of moles (0.0352 mol). For C: $\frac{0.2486\ mol}{0.0352\ mol}\approx7$. For H: $\frac{0.4968\ mol}{0.0352\ mol}\approx14$. For O: $\frac{0.0352\ mol}{0.0352\ mol}=1$.

Answer:

B. $C_7H_{14}O$