A compound is analyzed and found to consist of 50.4% Ce,15.1% N
and 34.5% O by mass. What is the correct formula for
the compound ? (Atomic wt. of Ce = 140)

Options

A. `Ce_(2)(NO_(3))_(2)`
B. `Ce_(2)(NO_(2))_(3)`
C. `Ce(NO_(3))_(2)`
D. `Ce(NO_(2))_(3)`

Explanation

To determine the correct formula for the compound consisting of cerium (Ce), nitrogen (N), and oxygen (O) based on their mass percentages, we can follow these steps:

### Step 1: Write down the mass percentages and atomic weights
- Cerium (Ce): 50.4% (atomic weight = 140 g/mol)
- Nitrogen (N): 15.1% (atomic weight = 14 g/mol)
- Oxygen (O): 34.5% (atomic weight = 16 g/mol)

### Step 2: Convert mass percentages to moles
To find the number of moles of each element, we divide the mass percentage by the atomic weight.

- Moles of Ce = \( \frac{50.4 \, \text{g}}{140 \, \text{g/mol}} = 0.36 \, \text{mol} \)
- Moles of N = \( \frac{15.1 \, \text{g}}{14 \, \text{g/mol}} = 1.07 \, \text{mol} \)
- Moles of O = \( \frac{34.5 \, \text{g}}{16 \, \text{g/mol}} = 2.15 \, \text{mol} \)

### Step 3: Find the simplest mole ratio
Next, we divide the number of moles of each element by the smallest number of moles calculated.

- For Ce: \( \frac{0.36}{0.36} = 1 \)
- For N: \( \frac{1.07}{0.36} \approx 2.97 \) (approximately 3)
- For O: \( \frac{2.15}{0.36} \approx 5.94 \) (approximately 6)

### Step 4: Write the empirical formula
From the simplest mole ratios, we can deduce the empirical formula:
- Ce: 1
- N: 3
- O: 6

Thus, the empirical formula is \( \text{CeN}_3\text{O}_6 \).

### Step 5: Finalize the formula
We can also express this formula in a more compact form as \( \text{CeN}\text{O}_2 \) thrice, which confirms that the formula is \( \text{CeN}_3\text{O}_6 \).

### Conclusion
The correct formula for the compound is \( \text{CeN}_3\text{O}_6 \).

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