The problem is:

A quantity of $\pu{35.2 g}$ if a certain hydrocarbon gas, occupies $\pu{13.2 L}$, measured at $\pu{1 atm}$ and at $\pu{323 K}$. Knowing that 85.5% of is C, find the molecular formula for the hydrocarbon.

Attempt. So I tried with the first step doing $0.85\cdot \pu{35.2 g} = \pu{29.92 g}$ of C and I'm pretty sure that's the wrong step but I'm not fully sure, so anyways, I continue. After finding the mass of C on the hydrocarbure, I subtracted $35.2 - 29.92 = \pu{5.28 g}$ of $\ce{H}$ to get the mass of $\ce{C}$ on the hydrocarbon. Now basically:

$$\pu{29.92 g} \text{ of C}\cdot \frac{\pu{1 mol}~\ce{C}}{\pu{12 g}~\ce{C}} = 2.5$$

$$\pu{5.28 g} \text{ of H}\cdot \frac{\pu{1 mol}~\ce{H}}{\pu{1 g}~\ce{H}} = 5.28$$

and this is why I think I'm wrong, I can't get a relationship of natural numbers for the empirical formula. May anyone give some input on this?

Continuation. So bravely approximating to the empirical formula by diving by the smaller one, i get $\ce{(C1H2)_n}$ so the empirical molar mass call it, $M_\text{empirical} = 12 + 2 = \pu{14g/mol}$. Getting the $n$: $$n = \frac{M_\text{real}}{M_\text{empirical}} = \frac{M_\text{real}}{\pu{14g/mol}}$$ and to get the $M_\text{real}$, we go to $$PV = nRT \implies PV= \frac{m}{M_\text{real}}RT \implies M_\text{real} = \frac{mRT}{PV}$$ hence $$n=\frac{\frac{mRT}{PV}}{\pu{14g/mol}} = \frac{\frac{\pu{35.2 g}\cdot \pu{0.082 L atm//K mol}\cdot \pu{323 K}}{\pu{1 atm}\cdot \pu{13.2 L}}}{\pu{14g/mol}} = 5$$ then I get $\ce{(C1H2)_5}$ hence $F_\text{molecular}=\ce{C5H10}$. Yes, right?

  • $\begingroup$ You don't need to approximate it to $1:2$ so early. You could always leave it as $\ce{C_{2.5n}H_{5.28n}}$, which means a molar mass of $(2.5n \cdot \pu{12 g/mol}) + (5.28n \cdot \pu{1 g/mol})$. $\endgroup$
    – orthocresol
    Dec 13, 2020 at 0:02


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