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 find the empirical formula of the compound formed by the reaction of xenon (Xe) with fluorine (F), we can follow these steps:

### Step 1: Calculate the moles of xenon (Xe)
Given:
- Mass of xenon = 2 g
- Atomic mass of xenon = 131 g/mol

Using the formula for moles:
\[ \text{Moles of Xe} = \frac{\text{mass of Xe}}{\text{molar mass of Xe}} = \frac{2 \text{ g}}{131 \text{ g/mol}} \approx 0.0153 \text{ moles} \]

### Step 2: Calculate the mass of fluorine (F) used
Given:
- Total mass of the compound = 3.158 g
- Mass of xenon = 2 g

To find the mass of fluorine:
\[ \text{Mass of F} = \text{Total mass of compound} - \text{Mass of Xe} = 3.158 \text{ g} - 2 \text{ g} = 1.158 \text{ g} \]

### Step 3: Calculate the moles of fluorine (F)
Given:
- Atomic mass of fluorine = 19 g/mol

Using the formula for moles:
\[ \text{Moles of F} = \frac{\text{mass of F}}{\text{molar mass of F}} = \frac{1.158 \text{ g}}{19 \text{ g/mol}} \approx 0.0609 \text{ moles} \]

### Step 4: Determine the simplest mole ratio
Now we have:
- Moles of xenon (Xe) = 0.0153
- Moles of fluorine (F) = 0.0609

To find the simplest ratio, divide both by the smaller number of moles (moles of Xe):
\[ \text{Ratio of Xe} = \frac{0.0153}{0.0153} = 1 \]
\[ \text{Ratio of F} = \frac{0.0609}{0.0153} \approx 4 \]

### Step 5: Write the empirical formula
From the mole ratio, we find that for every 1 mole of xenon, there are 4 moles of fluorine. Thus, the empirical formula of the compound is:
\[ \text{Empirical formula} = \text{XeF}_4 \]

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

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