A certain compound has the molecular formula `X_(4)O_(6)`. If `10 g` of `X_(4) O_(6)` has `5.72 g X`, the atomic mass of `X` is

Options

A. 32 amu
B. 37 amu
C. 42 amu
D. 98 amu

Explanation

To find the atomic mass of the element \( X \) in the compound \( X_4O_6 \), we can follow these steps:

### Step 1: Determine the mass of oxygen in the compound
The total mass of the compound \( X_4O_6 \) is given as 10 g, and the mass of \( X \) is given as 5.72 g. We can find the mass of oxygen by subtracting the mass of \( X \) from the total mass of the compound.

\[ \text{Mass of } O = \text{Total mass} - \text{Mass of } X = 10 \, \text{g} - 5.72 \, \text{g} = 4.28 \, \text{g} \]

### Step 2: Calculate the number of moles of \( X \) and \( O \)
Next, we need to calculate the number of moles of \( X \) and \( O \) using their respective masses and molar masses. The molar mass of oxygen \( O \) is \( 16 \, \text{g/mol} \).

\[ \text{Number of moles of } O (N_2) = \frac{\text{Mass of } O}{\text{Molar mass of } O} = \frac{4.28 \, \text{g}}{16 \, \text{g/mol}} = 0.2675 \, \text{mol} \]

### Step 3: Use the molecular formula to find the number of moles of \( X \)
From the molecular formula \( X_4O_6 \), we know that the ratio of moles of \( X \) to moles of \( O \) is \( \frac{4}{6} \) or \( \frac{2}{3} \). Let \( N_1 \) be the number of moles of \( X \).

Using the ratio:

\[ \frac{N_1}{N_2} = \frac{4}{6} \implies N_1 = \frac{4}{6} \times N_2 = \frac{4}{6} \times 0.2675 \, \text{mol} = 0.1783 \, \text{mol} \]

### Step 4: Calculate the molar mass of \( X \)
Now we can find the molar mass of \( X \) using the mass of \( X \) and the number of moles we just calculated.

\[ \text{Molar mass of } X = \frac{\text{Mass of } X}{N_1} = \frac{5.72 \, \text{g}}{0.1783 \, \text{mol}} \approx 32.0 \, \text{g/mol} \]

### Conclusion
The atomic mass of \( X \) is approximately \( 32 \, \text{g/mol} \).

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