A substance on analysis, gave the following percentage composition : Na = 43.4 %, C = 11.3%, O = 45.3 %. Calculate the empirical formula. (Na = 23, C = 12, O = 16).

Answer

To calculate the empirical formula from the given percentage composition of a substance, we can follow these steps:

### Step 1: Convert the percentage composition to grams
Assume we have 100 grams of the substance. This means:
- Sodium (Na) = 43.4 g
- Carbon (C) = 11.3 g
- Oxygen (O) = 45.3 g

### Step 2: Convert grams to moles
Using the molar masses provided:
- Molar mass of Na = 23 g/mol
- Molar mass of C = 12 g/mol
- Molar mass of O = 16 g/mol

Now, we can calculate the number of moles for each element:
- Moles of Na = \( \frac{43.4 \text{ g}}{23 \text{ g/mol}} = 1.89 \text{ moles} \)
- Moles of C = \( \frac{11.3 \text{ g}}{12 \text{ g/mol}} = 0.94 \text{ moles} \)
- Moles of O = \( \frac{45.3 \text{ g}}{16 \text{ g/mol}} = 2.83 \text{ moles} \)

### Step 3: Divide by the smallest number of moles
The smallest number of moles calculated is for Carbon, which is 0.94 moles. Now, we divide the number of moles of each element by 0.94:
- For Na: \( \frac{1.89}{0.94} \approx 2.01 \) (approximately 2)
- For C: \( \frac{0.94}{0.94} = 1 \)
- For O: \( \frac{2.83}{0.94} \approx 3.01 \) (approximately 3)

### Step 4: Write the empirical formula
Using the ratios obtained:
- Sodium (Na) = 2
- Carbon (C) = 1
- Oxygen (O) = 3

Thus, the empirical formula is \( \text{Na}_2\text{C}\text{O}_3 \).

### Final Answer:
The empirical formula of the substance is \( \text{Na}_2\text{C}\text{O}_3 \).

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