A gaseous hydrocarbon on combustion gives 0.72 g of water and 3.08 g $$CO_2$$. What is the empirical formula of the hydrocarbon?

We begin by assuming that the unknown gaseous hydrocarbon has a general formula $$C_xH_y$$. On complete combustion it produces only carbon dioxide and water according to the schematic reaction

$$C_xH_y + O_2 \;\rightarrow\; x\,CO_2 + \dfrac{y}{2}\,H_2O$$

The experiment tells us that $$0.72\;{\rm g}$$ of water and $$3.08\;{\rm g}$$ of carbon dioxide are obtained. To find the numbers of C and H atoms present in the original hydrocarbon, we must first convert these masses into moles because stoichiometric relations operate on moles.

The molar mass of water is $$18\;{\rm g\;mol^{-1}}$$ (since $$H_2O$$ has $$2\times1+16=18$$), while that of carbon dioxide is $$44\;{\rm g\;mol^{-1}}$$ (since $$CO_2$$ has $$12+2\times16=44$$).

Using the formula $$n=\dfrac{m}{M}$$ where $$n$$ is moles, $$m$$ is mass and $$M$$ is molar mass, we have:

$$n(H_2O)=\dfrac{0.72\;{\rm g}}{18\;{\rm g\;mol^{-1}}}=0.04\;{\rm mol}$$

$$n(CO_2)=\dfrac{3.08\;{\rm g}}{44\;{\rm g\;mol^{-1}}}=0.07\;{\rm mol}$$

Each mole of $$CO_2$$ contains exactly one mole of carbon atoms, so the moles of carbon originally present in the hydrocarbon are

$$n(C)=n(CO_2)=0.07\;{\rm mol}$$

Each mole of $$H_2O$$ contains two moles of hydrogen atoms, so the moles of hydrogen atoms produced are

$$n(H)=2 \times n(H_2O)=2 \times 0.04=0.08\;{\rm mol}$$

Therefore the mole ratio of carbon to hydrogen in the burned hydrocarbon is

$$C:H = 0.07:0.08$$

To convert this into the simplest whole-number ratio, we divide each value by the smaller one, $$0.01$$ is convenient:

$$\dfrac{0.07}{0.01}=7, \qquad \dfrac{0.08}{0.01}=8$$

Thus the simplest integer ratio is $$C_7H_8$$.

Hence, the correct answer is Option B.