A hydrocarbon diffuses twice as fast as sulphur dioxide. Calculate the molar mass of the hydrocarbon. (Given: Molar mass of S = 32.06 g/mol, O = 16.00 g/mol)

Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$

Where $r_1$ and $r_2$ are the rates of diffusion of gas 1 and gas 2, and $M_1$ and $M_2$ are their respective molar masses.

The molar mass of sulphur dioxide ($\text{SO}_2$):

$$\text{Molar mass of S} = 32.06 \text{ g/mol}$$ $$\text{Molar mass of O} = 15.999 \text{ g/mol}$$ $$\text{Molar mass of SO}_2 = 32.06 + (2 \times 15.999) = 64.058 \text{ g/mol}$$

Let the hydrocarbon be gas 1 and sulphur dioxide be gas 2.
Given that the hydrocarbon diffuses twice as quickly as sulphur dioxide:

$$r_{\text{hydrocarbon}} = 2 \times r_{\text{SO}_2}$$

So,

$$\frac{r_{\text{hydrocarbon}}}{r_{\text{SO}_2}} = 2$$

Using Graham's Law:

$$2 = \sqrt{\frac{M_{\text{SO}_2}}{M_{\text{hydrocarbon}}}}$$ $$2 = \sqrt{\frac{64.058 \text{ g/mol}}{M_{\text{hydrocarbon}}}}$$

Square both sides of the equation:

$$2^2 = \frac{64.058 \text{ g/mol}}{M_{\text{hydrocarbon}}}$$ $$4 = \frac{64.058 \text{ g/mol}}{M_{\text{hydrocarbon}}}$$

Rearrange to solve for $M_{\text{hydrocarbon}}$:

$$M_{\text{hydrocarbon}} = \frac{64.058 \text{ g/mol}}{4}$$ $$M_{\text{hydrocarbon}} = 16.0145 \text{ g/mol}$$

The general formula for the hydrocarbon is $\text{C}_n\text{H}_{2n+2}$.
The molar mass of carbon ($\text{C}$) is $12.011 \text{ g/mol}$.
The molar mass of hydrogen ($\text{H}$) is $1.008 \text{ g/mol}$.
So, the molar mass of $\text{C}_n\text{H}_{2n+2}$ is:

$$(n \times 12.011) + ((2n+2) \times 1.008) = 16.0145$$ $$12.011n + 2.016n + 2.016 = 16.0145$$ $$14.027n + 2.016 = 16.0145$$ $$14.027n = 16.0145 - 2.016$$ $$14.027n = 13.9985$$ $$n = \frac{13.9985}{14.027} \approx 0.9979 \approx 1$$

Since 'n' must be an integer, $n=1$.

Substitute $n=1$ into the formula $\text{C}_n\text{H}_{2n+2}$:

$$\text{C}_1\text{H}_{(2 \times 1)+2} = \text{CH}_4$$

The molecular formula of the hydrocarbon is $\text{CH}_4$.