Complete combustion of 0.858 g of compound X gives 2.63 g of `CO_(2)` and 1.28 g of` H_(2)`O. The lowest molecular mass X can have

Options

A. 43 g
B. 86 g
C. 129 g
D. 172 g

Answer

To find the lowest molecular mass of compound X based on the combustion data provided, we can follow these steps:

### Step 1: Calculate the number of moles of CO₂ produced.
The molar mass of CO₂ (carbon dioxide) is 44 g/mol.

Using the formula:
\[ \text{Number of moles of CO₂} = \frac{\text{mass of CO₂}}{\text{molar mass of CO₂}} = \frac{2.63 \, \text{g}}{44 \, \text{g/mol}} \]

Calculating this gives:
\[ \text{Number of moles of CO₂} = 0.0598 \, \text{mol} \approx 0.06 \, \text{mol} \]

### Step 2: Calculate the number of moles of H₂O produced.
The molar mass of H₂O (water) is 18 g/mol.

Using the formula:
\[ \text{Number of moles of H₂O} = \frac{\text{mass of H₂O}}{\text{molar mass of H₂O}} = \frac{1.28 \, \text{g}}{18 \, \text{g/mol}} \]

Calculating this gives:
\[ \text{Number of moles of H₂O} = 0.0711 \, \text{mol} \approx 0.07 \, \text{mol} \]

### Step 3: Determine the number of moles of carbon and hydrogen.
- The number of moles of carbon in CO₂ is equal to the number of moles of CO₂ produced:
\[ \text{Moles of C} = 0.06 \, \text{mol} \]

- The number of moles of hydrogen in H₂O is double the number of moles of H₂O produced (since each molecule of water contains 2 hydrogen atoms):
\[ \text{Moles of H} = 2 \times 0.07 \, \text{mol} = 0.14 \, \text{mol} \]

### Step 4: Find the simplest whole number ratio of carbon to hydrogen.
The ratio of moles of carbon to hydrogen is:
\[ \text{C:H} = 0.06 : 0.14 \]

To simplify this ratio, divide both values by 0.06:
\[ \text{C:H} = 1 : \frac{0.14}{0.06} \approx 1 : 2.33 \]

To convert this to whole numbers, multiply by 3:
\[ \text{C:H} = 3 : 7 \]

### Step 5: Write the empirical formula.
From the ratio, the empirical formula of compound X is \(C_3H_7\).

### Step 6: Calculate the molecular mass of the empirical formula.
The molecular mass of \(C_3H_7\) can be calculated as follows:
\[ \text{Molecular mass} = (3 \times 12 \, \text{g/mol}) + (7 \times 1 \, \text{g/mol}) = 36 \, \text{g/mol} + 7 \, \text{g/mol} = 43 \, \text{g/mol} \]

### Conclusion
The lowest molecular mass of compound X is **43 g/mol**.

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