Question

Question #b2435

Answer 1

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.

Answer 2

The molecular formula is; `C_6H_14` and the empirical formula is `C_3H_7`.

Explanation:

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`