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 solve the problem of finding the empirical formula from the given percentage composition of sodium (Na), carbon (C), and oxygen (O), we will follow these steps:

### Step 1: Convert Percentage to Moles
We will convert the percentage composition of each element to moles using the formula:

\[ \text{Moles of element} = \frac{\text{Percentage of element}}{\text{Atomic mass of element}} \]

#### For Sodium (Na):
\[ \text{Moles of Na} = \frac{43.4}{23} \approx 1.89 \text{ moles} \]

#### For Carbon (C):
\[ \text{Moles of C} = \frac{11.3}{12} \approx 0.94 \text{ moles} \]

#### For Oxygen (O):
\[ \text{Moles of O} = \frac{45.3}{16} \approx 2.83 \text{ moles} \]

### Step 2: Determine the Simplest Mole Ratio
Next, we will find the simplest ratio of the moles by dividing each mole value by the smallest number of moles calculated.

The smallest number of moles is for Carbon (0.94 moles).

#### For Sodium:
\[ \text{Ratio of Na} = \frac{1.89}{0.94} \approx 2 \]

#### For Carbon:
\[ \text{Ratio of C} = \frac{0.94}{0.94} = 1 \]

#### For Oxygen:
\[ \text{Ratio of O} = \frac{2.83}{0.94} \approx 3 \]

### Step 3: Write the Empirical Formula
Now that we have the simplest whole number ratios, we can write the empirical formula.

The ratios we found are:
- Sodium (Na): 2
- Carbon (C): 1
- Oxygen (O): 3

Thus, the empirical formula is:
\[ \text{Empirical Formula} = \text{Na}_2\text{C}\text{O}_3 \]

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

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