# An unknown alkane (not a cycloalkane) yielded 12.1g of CO2 and 5.4g of H2O when combusted in excess oxygen. What is the molecular formula for the alkane?

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Feb 25, 2018

If you have ${\text{12.1 g CO}}_{2}$, then:

$\text{12.1 g CO"_2 xx ("1 mol")/("44.009 g CO"_2) xx "1 mol C"/("1 mol CO"_2) = "0.2749 mols C}$

$\text{5.4 g H"_2"O" xx ("1 mol")/("18.015 g H"_2"O") xx "2 mol H"/("1 mol H"_2"O") = "0.5995 mols H}$

All the carbon came from the alkane, and all the hydrogen also came from the alkane. Therefore, this gives the carbon-to-hydrogen ratio.

$\text{0.5995 mols H"/"0.2749 mols C} = 2.18045$

This ratio is given as $\text{mols H"/"mols C} = \frac{2 n + 2}{n}$, so...

$2.18045 = \frac{2 n + 2}{n} = 2 + \frac{2}{n}$

$\frac{2}{n} = 0.18045$

$n = \frac{2}{0.18045} = \textcolor{g r e e n}{11.08}$

So maybe it is $\textcolor{b l u e}{{\text{C"_11"H}}_{24}}$...? hard to say, you can't really round this without noticeable error.

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Feb 25, 2018

It isn't possible

#### Explanation:

5,4 g of water MM= 18 g/mol) are 0,3 mol.
12,1 g of $C {O}_{2}$ (MM= 44 g/mol)= 0,3 mol
so y= z and the number of atoms of H will be the double of the atoms of C.
Hence it is impossible for an alkane but it possible for cycloalkane or alkene whose formula is ${C}_{n} {H}_{2} n$

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