Question

part a upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g co₂ and 0.6551 g h₂o. find the empirical formula of the compound. determine the empirical formula of the compound. a chemical reaction does not occur for this question.

Explanation:

Step1: Calculate moles of carbon

The mass of carbon in the $CO_2$ produced is used. The molar - mass of $CO_2$ is $M_{CO_2}=44.01\ g/mol$, and the molar - mass of C is $M_C = 12.01\ g/mol$. In $1.6004\ g$ of $CO_2$, the number of moles of $CO_2$ is $n_{CO_2}=\frac{1.6004\ g}{44.01\ g/mol}=0.03636\ mol$. Since there is 1 mole of C in 1 mole of $CO_2$, the moles of C, $n_C = 0.03636\ mol$.

Step2: Calculate moles of hydrogen

The molar - mass of $H_2O$ is $M_{H_2O}=18.02\ g/mol$, and the molar - mass of H is $M_H = 1.01\ g/mol$. In $0.6551\ g$ of $H_2O$, the number of moles of $H_2O$ is $n_{H_2O}=\frac{0.6551\ g}{18.02\ g/mol}=0.03635\ mol$. Since there are 2 moles of H in 1 mole of $H_2O$, the moles of H, $n_H=2\times0.03635\ mol = 0.0727\ mol$.

Step3: Calculate mass of oxygen

The mass of the sample is $m = 0.8009\ g$. The mass of C is $m_C=n_C\times M_C=0.03636\ mol\times12.01\ g/mol = 0.4367\ g$. The mass of H is $m_H=n_H\times M_H=0.0727\ mol\times1.01\ g/mol = 0.0734\ g$. The mass of O is $m_O=m - m_C - m_H=0.8009\ g-0.4367\ g - 0.0734\ g=0.2908\ g$. The moles of O, $n_O=\frac{m_O}{M_O}=\frac{0.2908\ g}{16.00\ g/mol}=0.01818\ mol$.

Step4: Find the mole - ratio

Divide each number of moles by the smallest number of moles ($n_O = 0.01818\ mol$).
For C: $\frac{n_C}{n_O}=\frac{0.03636\ mol}{0.01818\ mol}=2$
For H: $\frac{n_H}{n_O}=\frac{0.0727\ mol}{0.01818\ mol}=4$
For O: $\frac{n_O}{n_O}=1$

Answer:

$C_2H_4O$