5. a 4.05 g sample of a compound containing c, h, and o was burned completely. the only combustion products…

Explanation:

Step1: Calculate moles of C

First find moles of $CO_2$. Molar mass of $CO_2$ is $M_{CO_2}=12 + 2\times16=44\ g/mol$. Moles of $CO_2$, $n_{CO_2}=\frac{10.942\ g}{44\ g/mol}=0.2487\ mol$. Since 1 mole of $CO_2$ has 1 mole of C, moles of C, $n_C = 0.2487\ mol$.

Step2: Calculate moles of H

Find moles of $H_2O$. Molar mass of $H_2O$ is $M_{H_2O}=2\times1 + 16 = 18\ g/mol$. Moles of $H_2O$, $n_{H_2O}=\frac{4.476\ g}{18\ g/mol}=0.2487\ mol$. Since 1 mole of $H_2O$ has 2 moles of H, moles of H, $n_H=2\times0.2487\ mol = 0.4974\ mol$.

Step3: Calculate mass of C and H

Mass of C, $m_C=n_C\times12\ g/mol=0.2487\ mol\times12\ g/mol = 2.9844\ g$. Mass of H, $m_H=n_H\times1\ g/mol=0.4974\ mol\times1\ g/mol = 0.4974\ g$.

Step4: Calculate mass of O

Mass of O, $m_O=4.05\ g-(2.9844\ g + 0.4974\ g)=0.5682\ g$.

Step5: Calculate moles of O

Molar mass of O is 16 g/mol. Moles of O, $n_O=\frac{0.5682\ g}{16\ g/mol}=0.0355\ mol$.

Step6: Find mole - ratio

Divide moles of C, H and O by the smallest number of moles (0.0355 mol). For C: $\frac{0.2487\ mol}{0.0355\ mol}\approx7$. For H: $\frac{0.4974\ mol}{0.0355\ mol}\approx14$. For O: $\frac{0.0355\ mol}{0.0355\ mol}=1$.

Answer:

$C_7H_{14}O$