Question #b2435

Without more data it'd be nearly impossible to carry out a quantitative combustion analysis. The method we'll use is simple trial and error since this is relatively trivial,

We know that one carbon atom is approximately #(12g)/"mol"# and that one hydrogen atom is approximately #(1g)/"mol"# in the context of a mole of a molecular compound.

Hence,

#6C*(12g)/C + 14H*(1g)/H approx 86g#

Which is characteristic of an alkane with the molecular formula #C_6H_14#,

Moreover, its empirical formula (the lowest ratio of atoms) is #C_3H_7#.

Clearly this is guesswork. If I had more data, or the actual hydrocarbon compound I could play around (safely!) in the lab and find out.

Assuming we are talking about a simple alkane here..

An alkane must have two #CH_3# groups

The relative molecular mass for a #CH_3# group is (12 x 1) + (3 x 1) = 15

Two #CH_3# groups means 30.

Therefore we can subtract 30 from 86 to see what we are left with. 86 - 30 = 56

a #CH_2# group has a relative molecular mass of 14 (12 + 2).

#56/14 = 4#. Therefore we have four #CH_2# groups in our hydrocarbon.

So we now know that we have;

#CH_3CH_2CH_2CH_2CH_2CH_3#

In total, 6 carbons (6 x 12 = 72) and 14 hydrogens (14 x 1). (72 + 14 = 86 #gmol^-1#).

The molecular formula is; #C_6H_14#

An empirical formula is the SIMPLEST ratio of atoms within in a molecule. In this case;

#C_3H_7#