A 2g sample of xenon reacts with fluorine. The mass of the compound produced is 3.158g . The empirical formula of the compound is : (Given : Atomic mass of `"Xe" = 131` g/mole)

Options

A. `XeF_2`
B. `XeF_4`
C. `XeF_5`
D. `XeF_6`

Answer

To determine the empirical formula of the compound produced from the reaction of xenon (Xe) with fluorine (F), we can follow these steps:

### Step 1: Calculate the mass of fluorine that reacted
We know the mass of the compound produced is 3.158 g, and the mass of xenon used is 2 g. To find the mass of fluorine, we subtract the mass of xenon from the mass of the compound:

\[ \text{Mass of Fluorine} = \text{Mass of Compound} - \text{Mass of Xenon} \]
\[ \text{Mass of Fluorine} = 3.158 \, \text{g} - 2 \, \text{g} = 1.158 \, \text{g} \]

### Step 2: Convert the masses to moles
Next, we need to convert the mass of xenon and fluorine to moles using their respective atomic masses. The atomic mass of xenon (Xe) is given as 131 g/mol, and the atomic mass of fluorine (F) is 19 g/mol.

**For xenon:**
\[ \text{Moles of Xe} = \frac{\text{Mass of Xe}}{\text{Molar Mass of Xe}} = \frac{2 \, \text{g}}{131 \, \text{g/mol}} \approx 0.01527 \, \text{mol} \]

**For fluorine:**
\[ \text{Moles of F} = \frac{\text{Mass of F}}{\text{Molar Mass of F}} = \frac{1.158 \, \text{g}}{19 \, \text{g/mol}} \approx 0.06094 \, \text{mol} \]

### Step 3: Determine the mole ratio
Now we find the simplest whole number ratio of moles of xenon to moles of fluorine. We do this by dividing the number of moles of each element by the smallest number of moles calculated.

\[ \text{Ratio of Xe} = \frac{0.01527}{0.01527} = 1 \]
\[ \text{Ratio of F} = \frac{0.06094}{0.01527} \approx 4 \]

### Step 4: Write the empirical formula
From the mole ratio, we find that for every 1 mole of xenon, there are approximately 4 moles of fluorine. Therefore, the empirical formula of the compound is:

\[ \text{Empirical Formula} = \text{XeF}_4 \]

### Final Answer
The empirical formula of the compound is **XeF₄**.

---