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 #"12.1 g CO"_2#, then:

#"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 #"mols H"/"mols C" = (2n+2)/n#, so...

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

#2/n = 0.18045#

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

So maybe it is #color(blue)("C"_11"H"_24)#...? hard to say, you can't really round this without noticeable error.

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Feb 25, 2018

Answer:

It isn't possible

Explanation:

5,4 g of water MM= 18 g/mol) are 0,3 mol.
12,1 g of #CO_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_nH_2n#

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