0.92 g of an organic compound was analysed by combustion method. The mass of the U-tube increased by 1.08 g. what is the percentage of hydrogen in the compound?

Options

A. 0.1304
B. 0.5217
C. 0.6521
D. 0.113

Explanation

To find the percentage of hydrogen in the organic compound, we can follow these steps:

### Step 1: Understand the combustion analysis
When an organic compound is combusted, it produces water (H2O) and carbon dioxide (CO2). The increase in mass of the U-tube corresponds to the mass of water formed during the combustion.

### Step 2: Identify the given data
- Mass of the organic compound = 0.92 g
- Increase in mass of the U-tube (mass of water formed) = 1.08 g

### Step 3: Calculate the moles of water produced
The molar mass of water (H2O) is 18 g/mol. We can calculate the number of moles of water produced using the formula:

\[ \text{Moles of H2O} = \frac{\text{mass of H2O}}{\text{molar mass of H2O}} = \frac{1.08 \text{ g}}{18 \text{ g/mol}} = 0.06 \text{ mol} \]

### Step 4: Determine the moles of hydrogen in water
Each molecule of water contains 2 hydrogen atoms. Therefore, the moles of hydrogen can be calculated as:

\[ \text{Moles of H} = 2 \times \text{Moles of H2O} = 2 \times 0.06 \text{ mol} = 0.12 \text{ mol} \]

### Step 5: Calculate the mass of hydrogen
To find the mass of hydrogen, we use the molar mass of hydrogen (H), which is approximately 1 g/mol:

\[ \text{Mass of H} = \text{Moles of H} \times \text{molar mass of H} = 0.12 \text{ mol} \times 1 \text{ g/mol} = 0.12 \text{ g} \]

### Step 6: Calculate the percentage of hydrogen in the compound
Now, we can calculate the percentage of hydrogen in the organic compound using the formula:

\[ \text{Percentage of H} = \left( \frac{\text{mass of H}}{\text{mass of compound}} \right) \times 100 = \left( \frac{0.12 \text{ g}}{0.92 \text{ g}} \right) \times 100 \]

Calculating this gives:

\[ \text{Percentage of H} = \left( \frac{0.12}{0.92} \right) \times 100 \approx 13.04\% \]

### Final Answer
The percentage of hydrogen in the compound is approximately **13.04%**.

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