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------------------------------------------------ Calculate the empirical formula for:

a. [tex]39.3 \% \, \text{Na}, \, 60.7 \% \, \text{Cl}[/tex]

To determine the empirical formula based on the given percentages of elements, we can follow these steps:

1. Determine the Moles of Each Element:

- For sodium (Na):
- The percentage composition is 39.3%.
- Atomic mass of sodium (Na) is approximately 22.99 g/mol.
- Calculate moles of Na:
[tex]\[
\text{Moles of Na} = \frac{39.3}{22.99} \approx 1.709
\][/tex]

- For chlorine (Cl):
- The percentage composition is 60.7%.
- Atomic mass of chlorine (Cl) is approximately 35.45 g/mol.
- Calculate moles of Cl:
[tex]\[
\text{Moles of Cl} = \frac{60.7}{35.45} \approx 1.712
\][/tex]

2. Determine the Simplest Whole Number Ratio:

- To find the ratio, divide each element's moles by the smallest number of moles calculated:
- The smallest number of moles is 1.709 (for Na).
- Ratio of Na:
[tex]\[
\text{Ratio of Na} = \frac{1.709}{1.709} = 1.0
\][/tex]
- Ratio of Cl:
[tex]\[
\text{Ratio of Cl} = \frac{1.712}{1.709} \approx 1.002
\][/tex]

3. Convert the Ratios to Whole Numbers:

- Since the ratios are close to whole numbers (1 for Na and approximately 1 for Cl), we round them to the nearest whole numbers:
- Na: 1
- Cl: 1

4. Write the Empirical Formula:

- The simplest whole number ratio of moles of Na to Cl is 1:1.
- Therefore, the empirical formula is [tex]\( \text{NaCl} \)[/tex].

This empirical formula suggests that sodium and chlorine are present in equal amounts, consistent with the structure of common table salt.