Question

1. combustion analysis of 63.8 mg of a c, h, and o - containing compound produced 145.0 mg of co2 and 59.38 mg of h2o. what is the empirical formula for the compound? a) c3h6o b) c5h2o c) c6ho3 d) c3h7o e) cho

Explanation:

Step1: Calculate moles of carbon

The molar mass of $CO_2$ is $M_{CO_2}=44.01\ g/mol$. The mass of $CO_2$ is $m_{CO_2} = 145.0\ mg=0.1450\ g$. Moles of $CO_2$, $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{0.1450\ g}{44.01\ g/mol}$. Since 1 mole of $CO_2$ contains 1 mole of C, moles of C, $n_C=\frac{0.1450\ g}{44.01\ g/mol}\approx0.0033\ mol$.

Step2: Calculate moles of hydrogen

The molar mass of $H_2O$ is $M_{H_2O}=18.02\ g/mol$. The mass of $H_2O$ is $m_{H_2O}=59.38\ mg = 0.05938\ g$. Moles of $H_2O$, $n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{0.05938\ g}{18.02\ g/mol}$. Since 1 mole of $H_2O$ contains 2 moles of H, moles of H, $n_H = 2\times\frac{0.05938\ g}{18.02\ g/mol}\approx0.0066\ mol$.

Step3: Calculate mass of oxygen

Mass of C is $m_C=n_C\times12.01\ g/mol=0.0033\ mol\times12.01\ g/mol\approx0.0396\ g$. Mass of H is $m_H=n_H\times1.008\ g/mol=0.0066\ mol\times1.008\ g/mol\approx0.00665\ g$. The mass of the compound is $m = 63.8\ mg=0.0638\ g$. Mass of O, $m_O=m - m_C - m_H=0.0638\ g-0.0396\ g - 0.00665\ g\approx0.01755\ g$. Moles of O, $n_O=\frac{m_O}{16.00\ g/mol}=\frac{0.01755\ g}{16.00\ g/mol}\approx0.0011\ mol$.

Step4: Find the mole - ratio

Divide the moles of each element by the smallest number of moles ($n_O\approx0.0011\ mol$). For C: $\frac{n_C}{n_O}=\frac{0.0033\ mol}{0.0011\ mol}=3$. For H: $\frac{n_H}{n_O}=\frac{0.0066\ mol}{0.0011\ mol}=6$. For O: $\frac{n_O}{n_O}=1$.

Answer:

A. $C_3H_6O$