25 g of an unknown hydrocarbon upon burning produces 88 g of CO$$_2$$ and 9 g of H$$_2$$O. This unknown hydrocarbon contains:

We are told that complete combustion of 25 g of an unknown hydrocarbon gives 88 g of CO$$\_2$$ and 9 g of H$$\_2$$O. From these products we will back-calculate the masses of carbon and hydrogen that must have been present in the original sample.

First, we look at the carbon that appears in CO$$\_2$$. The molar mass of CO$$\_2$$ is $$44\;\text{g mol}^{-1}$$, made up of $$12\;\text{g}$$ of carbon and $$32\;\text{g}$$ of oxygen. The fraction of the mass of CO$$\_2$$ that is due to carbon is therefore

$$\frac{12}{44}.$$

The given mass of CO$$\_2$$ is 88 g. Hence the mass of carbon in those 88 g is

$$\text{mass of C} \;=\; \frac{12}{44}\times 88 =\;12\times\frac{88}{44} =\;12\times 2 =\;24\;\text{g}.$$

Now we examine the hydrogen that appears in H$$\_2$$O. The molar mass of H$$\_2$$O is $$18\;\text{g mol}^{-1}$$, which contains $$2\;\text{g}$$ of hydrogen and $$16\;\text{g}$$ of oxygen. Thus the fraction of the mass of H$$\_2$$O that corresponds to hydrogen is

$$\frac{2}{18}.$$

The mass of water produced is 9 g, so the mass of hydrogen present in it is

$$\text{mass of H} \;=\; \frac{2}{18}\times 9 =\;2\times\frac{9}{18} =\;2\times\frac{1}{2} =\;1\;\text{g}.$$

Therefore, in the 25 g of the original unknown hydrocarbon, we had 24 g of carbon and 1 g of hydrogen.

Among the given options, this matches

A. 24 g of carbon and 1 g of hydrogen.

Hence, the correct answer is Option A.