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. (a)`ZnCr_(2)O_(7)`
B. (b)`ZnCr_(2)O_(4)`
C. (c )`ZnCrO_(4)`
D. (d)`ZnCrO_(6)`

Explanation

To determine the simplest (empirical) formula of the compound containing zinc (Zn), chromium (Cr), and oxygen (O), we need to follow these steps:

### Step 1: Calculate the number of moles of zinc (Zn)
We are given the mass of zinc:
- Mass of Zn = 9.81 g
- Atomic mass of Zn = 65.39 g/mol

Using the formula for moles:
\[ \text{Number of moles of Zn} = \frac{\text{mass}}{\text{molar mass}} = \frac{9.81 \, \text{g}}{65.39 \, \text{g/mol}} \approx 0.15 \, \text{mol} \]

### Step 2: Calculate the number of moles of chromium (Cr)
We are given the number of atoms of chromium:
- Number of atoms of Cr = \(1.8 \times 10^{23}\)
- Avogadro's number = \(6.022 \times 10^{23} \, \text{atoms/mol}\)

Using the formula for moles:
\[ \text{Number of moles of Cr} = \frac{\text{number of atoms}}{\text{Avogadro's number}} = \frac{1.8 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.30 \, \text{mol} \]

### Step 3: Determine the number of moles of oxygen (O)
We are given the number of moles of oxygen:
- Number of moles of O = 0.60 mol

### Step 4: Find the simplest ratio of moles
Now we have:
- Moles of Zn = 0.15
- Moles of Cr = 0.30
- Moles of O = 0.60

To find the simplest ratio, we divide each by the smallest number of moles (0.15):
\[ \text{Ratio of Zn:Cr:O} = \frac{0.15}{0.15} : \frac{0.30}{0.15} : \frac{0.60}{0.15} = 1 : 2 : 4 \]

### Step 5: Write the empirical formula
From the ratio, we can write the empirical formula:
- Zn: 1
- Cr: 2
- O: 4

Thus, the empirical formula is:
\[ \text{Empirical Formula} = \text{ZnCr}_2\text{O}_4 \]

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

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