# 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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#### Explanation

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If you have

#"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"#

#"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.

#"0.5995 mols H"/"0.2749 mols C" = 2.18045#

This ratio is given as

#2.18045 = (2n+2)/(n) = 2 + 2/n#

#2/n = 0.18045#

#n = 2/0.18045 = color(green)(11.08)#

So maybe it is

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#### Answer:

It isn't possible

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5,4 g of water MM= 18 g/mol) are 0,3 mol.

12,1 g of

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

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