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 find the percentage of oxygen in the compound \(C_{x}H_{y}O_{2}\) that upon combustion produces \(2.64 \, \text{g}\) of \(CO_{2}\) and \(1.08 \, \text{g}\) of \(H_{2}O\), follow these steps:

1. Calculate the moles of \(CO_{2}\) produced:

\(\text{Molar mass of } CO_{2} = 44 \, \text{g/mol}\)

\(\text{Moles of } CO_{2} = \frac{2.64 \, \text{g}}{44 \, \text{g/mol}} = 0.06 \, \text{mol}\)

1. Calculate the moles of carbon present:

Each mole of \(CO_{2}\) contains 1 mole of carbon, so:

\(\text{Moles of carbon} = 0.06 \, \text{mol}\)

1. Calculate the mass of carbon:

\(\text{Mass of carbon} = 0.06 \, \text{mol} \times 12 \, \text{g/mol} = 0.72 \, \text{g}\)

1. Calculate the moles of \(H_{2}O\) produced:

\(\text{Molar mass of } H_{2}O = 18 \, \text{g/mol}\)

\(\text{Moles of } H_{2}O = \frac{1.08 \, \text{g}}{18 \, \text{g/mol}} = 0.06 \, \text{mol}\)

1. Calculate the moles of hydrogen present:

Each mole of \(H_{2}O\) contains 2 moles of hydrogen, so:

\(\text{Moles of hydrogen} = 0.06 \, \text{mol} \times 2 = 0.12 \, \text{mol}\)

1. Calculate the mass of hydrogen:

\(\text{Mass of hydrogen} = 0.12 \, \text{mol} \times 1 \, \text{g/mol} = 0.12 \, \text{g}\)

1. Determine the mass of oxygen in the compound:

\(\text{Total mass of compound} = 1.80 \, \text{g}\)

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

1. Calculate the percentage of oxygen:

\(\text{Percentage of oxygen} = \left( \frac{0.96 \, \text{g}}{1.80 \, \text{g}} \right) \times 100 = 53.33\%\)

Therefore, the percentage of oxygen in the organic compound is 53.33%.