What is the empirical formula of a compound containing 60.0% of nitrogen and 40.0% of Oxygen by mass ?

Options

A. `NO_2`
B. `NO_3`
C. `N_3O`
D. `N_2O`

Explanation

To find the empirical formula of a compound containing 60.0% nitrogen and 40.0% oxygen by mass, follow these steps:

### Step 1: Convert Percentages to Masses
Since we are given percentages, we can assume a 100 g sample of the compound. This makes the calculations straightforward:
- Mass of Nitrogen (N) = 60.0 g
- Mass of Oxygen (O) = 40.0 g

### Step 2: Calculate the Number of Moles of Each Element
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- For Nitrogen:
- Molar mass of N = 14 g/mol
\[ \text{Moles of N} = \frac{60.0 \, \text{g}}{14 \, \text{g/mol}} \approx 4.29 \, \text{moles} \]

- For Oxygen:
- Molar mass of O = 16 g/mol
\[ \text{Moles of O} = \frac{40.0 \, \text{g}}{16 \, \text{g/mol}} = 2.5 \, \text{moles} \]

### Step 3: Determine the Simplest Ratio of Moles
Next, we need to find the simplest whole number ratio of the moles of each element. To do this, we divide the number of moles of each element by the smallest number of moles calculated.

- The smallest number of moles is 2.5 (for O).
- For Nitrogen:
\[ \text{Ratio of N} = \frac{4.29}{2.5} \approx 1.72 \]
- For Oxygen:
\[ \text{Ratio of O} = \frac{2.5}{2.5} = 1 \]

### Step 4: Convert Ratios to Whole Numbers
To convert the ratios to whole numbers, we can multiply both ratios by a common factor. In this case, we can multiply by 2 to get whole numbers:
- Nitrogen: \(1.72 \times 2 \approx 3.44\) (rounding gives us 3)
- Oxygen: \(1 \times 2 = 2\)

Thus, the simplest whole number ratio of N to O is approximately 3:2.

### Step 5: Write the Empirical Formula
From the ratios, we can write the empirical formula:
\[ \text{Empirical formula} = N_3O_2 \]

### Final Answer
The empirical formula of the compound is \(N_3O_2\).

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