High School

Calculate the empirical formula for:

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

b. [tex]56.5 \% \, \text{K}, \, 8.7 \% \, \text{C}, \, 34.8 \% \, \text{O}[/tex]

Answer :

Let's find the empirical formulas for the given compounds using their percentages.

### Part (a)

For the compound containing sodium (Na) and chlorine (Cl), we have:

- 39.3% Na
- 60.7% Cl

Step 1: Determine the moles of each element.

To find moles, divide the percentage by the atomic mass of the element. The atomic masses are approximately:

- Na: 22.99 g/mol
- Cl: 35.45 g/mol

Calculate moles:

- Moles of Na = 39.3 / 22.99 ≈ 1.71
- Moles of Cl = 60.7 / 35.45 ≈ 1.71

Step 2: Determine the simplest whole number ratio.

Find the smallest number of moles and divide all mole values by this number:

- Both Na and Cl have approximately 1.71 moles.
- Divide both by 1.71:
- Na: 1.71 / 1.71 = 1
- Cl: 1.71 / 1.71 = 1

The empirical formula is NaCl.

### Part (b)

For the compound containing potassium (K), carbon (C), and oxygen (O), we have:

- 56.5% K
- 8.7% C
- 34.8% O

Step 1: Determine the moles of each element.

The atomic masses are:

- K: 39.10 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol

Calculate moles:

- Moles of K = 56.5 / 39.10 ≈ 1.45
- Moles of C = 8.7 / 12.01 ≈ 0.72
- Moles of O = 34.8 / 16.00 ≈ 2.18

Step 2: Determine the simplest whole number ratio.

Find the smallest number of moles and divide all mole values by this number (smallest is approximately 0.72):

- K: 1.45 / 0.72 ≈ 2
- C: 0.72 / 0.72 = 1
- O: 2.18 / 0.72 ≈ 3

The empirical formula is K2CO3.

These calculations give us the empirical formulas NaCl and K2CO3 for parts (a) and (b), respectively.

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