A given sample of pure compound contains `9.81 g` of Zn, `1.8xx10^(23)` atoms of chromium, and `0.60` mol of oxygen atoms. What is the simplest formula?

Options

A. `ZnCr_(2)O_(7)`
B. `ZnCr_(2)O_(4)`
C. `ZnCrO_(4)`
D. `ZnCrO_(6)`

Answer

To find the simplest formula of the compound containing zinc (Zn), chromium (Cr), and oxygen (O), we will follow these steps:

### Step 1: Calculate the number of moles of Zinc (Zn)
Given:
- Mass of Zn = 9.81 g
- Molar mass of Zn = 63.5 g/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \]
\[ \text{Moles of Zn} = \frac{9.81 \text{ g}}{63.5 \text{ g/mol}} \approx 0.154 \text{ mol} \]

### Step 2: Calculate the number of moles of Chromium (Cr)
Given:
- Number of atoms of Cr = \(1.8 \times 10^{23}\)
- Avogadro's number = \(6.022 \times 10^{23}\) atoms/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{Number of atoms}}{\text{Avogadro's number}} \]
\[ \text{Moles of Cr} = \frac{1.8 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.299 \text{ mol} \approx 0.3 \text{ mol} \]

### Step 3: Calculate the number of moles of Oxygen (O)
Given:
- Moles of O = 0.60 mol

### Step 4: Find the simplest ratio of moles
Now we have:
- Moles of Zn = 0.154 mol
- Moles of Cr = 0.3 mol
- Moles of O = 0.60 mol

To find the simplest ratio, we will divide each by the smallest number of moles (0.154 mol):
\[ \text{Ratio of Zn} = \frac{0.154}{0.154} = 1 \]
\[ \text{Ratio of Cr} = \frac{0.3}{0.154} \approx 1.94 \approx 2 \]
\[ \text{Ratio of O} = \frac{0.60}{0.154} \approx 3.90 \approx 4 \]

### Step 5: Write the simplest formula
Based on the ratios:
- Zn: 1
- Cr: 2
- O: 4

The simplest formula is:
\[ \text{ZnCr}_2\text{O}_4 \]

### Final Answer:
The simplest formula of the compound is **ZnCr₂O₄**.

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