Animated Solution for Chemistry - Basic Concepts in Chemistry: 25 g of an unknown hydrocarbon upon burning produces 88 g of ${\text{CO}}_2$ and 9 g of ${\text{H}}_2\text{O}$. This unknown hydrocarbon contains

Imagine you are a chemical detective handed a mysterious vial containing an unknown hydrocarbon. You are told it weighs exactly $25\text{ g}$. Your mission? To find out exactly how much carbon and hydrogen are locked inside its molecular structure.

How do we unlock this secret? We burn it! This process is known as Combustion Analysis (historically related to Liebig's method). When a pure hydrocarbon (a compound made entirely of carbon and hydrogen, represented as $C_x{H}_y$) is burned in the presence of excess oxygen ($O_2$), it undergoes a complete transformation.

The carbon atoms have nowhere to go but to bond with oxygen, forming carbon dioxide ($C{O}_2$). Similarly, the hydrogen atoms pair up with oxygen to form water ($H_{2}O$). The beauty of this reaction is its absolute predictability.

The Principle of Atomic Conservation

The most powerful tool in our stoichiometric arsenal is the Principle of Atomic Conservation (POAC). It states that atoms are neither created nor destroyed in a chemical reaction.

This means that every single atom of carbon that was originally in our $25\text{ g}$ hydrocarbon sample is now trapped inside the $88\text{ g}$ of $C{O}_2$ gas produced. If we can figure out the mass of carbon inside that $C{O}_2$, we have found the mass of carbon in our original sample!

To extract the mass of carbon from the $C{O}_2$, we use the concept of molar mass fractions.

We know that one mole of $C{O}_2$ has a molar mass of $44\text{ g/mol}$ (since Carbon is $12$ and two Oxygens are $16\times 2=32$). Out of this $44\text{ g}$, exactly $12\text{ g}$ is pure carbon. Therefore, the fraction of carbon in any sample of $C{O}_2$ is always $\frac{12}{44}$.

Let's apply this to our specific yield of $88\text{ g}$:

Since $44$ goes into $88$ exactly $2$ times, the math becomes beautifully simple:

Our unknown hydrocarbon contains exactly $24\text{ g}$ of carbon.

Now, we apply the exact same logic to the hydrogen. All the hydrogen from the hydrocarbon is now swimming in the $9\text{ g}$ of water ($H_{2}O$) produced.

Water has a molar mass of $18\text{ g/mol}$ (two Hydrogens at $1$ each, plus one Oxygen at $16$). Notice that there are two hydrogen atoms in every water molecule. So, out of the $18\text{ g}$ total mass, $2\text{ g}$ is hydrogen. The mass fraction of hydrogen in water is $\frac{2}{18}$.

Let's calculate the mass of hydrogen in our $9\text{ g}$ yield:

Simplifying this, $9$ goes into $18$ exactly $2$ times, leaving us with:

Our unknown hydrocarbon contains exactly $1\text{ g}$ of hydrogen.

We have found $24\text{ g}$ of carbon and $1\text{ g}$ of hydrogen. Let's do a quick sanity check.

This perfectly matches the initial mass of our hydrocarbon sample! This confirms that our compound was indeed a pure hydrocarbon with no other hidden elements like oxygen.

Thus, the correct answer is $24\text{ g}$ of carbon and $1\text{ g}$ of hydrogen.