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What is the molecular formula of a compound containing only carbon and hydrogen if combustion of $1.05~\mathrm{g}$ of the compound produces $3.30~\mathrm{g}~\ce{CO2}$ and $1.35~\mathrm{g}~\ce{H2O}$ and its molar mass is about $70~\mathrm{g}$?

Here is my work:

$$ \begin{array}{cccccc} & \ce{C_{a}H_{b}} & \ce{->} & \ce{CO2} & + & \ce{H2O} \\ \text{masses (g)} & 1.05 & & 3.30 & & 1.35 \end{array} $$

\begin{align*} \ce{CO2} &\rightarrow \ce{C} \\ 44~\mathrm{g} &\rightarrow 12~\mathrm{g} \\ 3.30~\mathrm{g} &\rightarrow x \end{align*}

$$ x = 0.9~\mathrm{g},~\text{moles of C} = \frac{0.9}{12} = 0.075 $$

\begin{align*} \ce{H2O} &\rightarrow \ce{2H} \\ 18~\mathrm{g} &\rightarrow 2~\mathrm{g} \\ 1.35~\mathrm{g} &\rightarrow y \end{align*}

$$ y = 0.15~\mathrm{g},~\text{moles of H} = \frac{0.15}{1} = 0.15 $$

$$ \text{empirical formula}~\ce{C_{0.075/0.075}H_{0.15/0.075} -> CH2} $$

$$ \frac{70}{14} = 5 $$

$$ \text{molecular formula is}~\ce{C5H10} $$

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  • $\begingroup$ The molar mass of any compound would not be 70 g, although it might be 70 g/mol. $\endgroup$ – LDC3 Sep 19 '14 at 4:10
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    $\begingroup$ What actually is the question here? $\endgroup$ – Jan Oct 4 '15 at 13:37
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$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -> $a$CO2 + $\frac{b}{2}$ H2O}$$

Suppose you had $n$ moles of hydrocarbon, then we have $a\cdot n$ moles of $\ce{CO2}$ and $\frac{b}{2}\cdot n$ moles of $\ce{H2O}$ dividing their moles we'll get $2\frac{a}{b}$: $$2\frac ab=\frac{330/44}{135/18}=\frac{7.5}{7.5}=1\implies \frac ab=\frac12$$ So the empirical formula is $\ce{CH2}$ Now for actual formula $\frac{70}{14}=5$ Yes the formula is $\ce{C5H10}$.

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