Question

Complete combustion of $1.80 g$ of an oxygen containing compound $\left( C _{ x } H _{ y } O _{2}\right)$ gave $2.64 g$ of $CO _{2}$ and $1.08 g$ of $H _{2} O$. The percentage of oxygen in the organic compound is:

• A. 51.63
• B. 63.53
• C. 53.33
• D. 50.33

Answer 1

The Correct Option is C

To determine the percentage of oxygen in the given organic compound \(\left( C _{ x } H _{ y } O _{2}\right)\), we can use the data from the complete combustion process. Let's break down the problem into clear steps:

1. Calculate the amount of carbon in the compound:

Given that the complete combustion of the compound produces 2.64 g of \(CO_2\).

The molar mass of \(CO_2\) is \(44 \, \text{g/mol}\), and for one mole of \(CO_2\), there is one mole of carbon, which has a molar mass of \(12 \, \text{g/mol}\).

The mass of carbon in \(2.64 \, \text{g}\) of \(CO_2\) can be calculated as follows:

\text{Mass of C in CO}_2 = \dfrac{12}{44} \times 2.64 \, \text{g} = 0.72 \, \text{g}

2. Calculate the amount of hydrogen in the compound:

Given that the compound produces 1.08 g of \(H_2O\).

The molar mass of \(H_2O\) is \(18 \, \text{g/mol}\), and for one mole of \(H_2O\), there are two moles of H, which have a molar mass of \(2 \, \text{g/mol}\).

The mass of hydrogen in \(1.08 \, \text{g}\) of \(H_2O\) can be calculated as follows:

\text{Mass of H in H}_2\text{O} = \dfrac{2}{18} \times 1.08 \, \text{g} = 0.12 \, \text{g}

3. Calculate the amount of oxygen in the compound:

The initial mass of the compound is 1.80 g, which includes carbon, hydrogen, and oxygen. Using the law of conservation of mass:

\text{Mass of Oxygen} = 1.80 \, \text{g} - (0.72 \, \text{g} + 0.12 \, \text{g}) = 0.96 \, \text{g}

4. Calculate the percentage of oxygen in the compound:

\text{Percentage of Oxygen} = \left( \dfrac{0.96}{1.80} \right) \times 100\% = 53.33\%

Hence, the percentage of oxygen in the compound is 53.33%. This corresponds to option 53.33.