Calculate the empirical formula of compound containing 25percent iron 24percent in sulphur and the rest being oxygen ?

To calculate the empirical formula of the compound, we need to determine the ratio of each element present in the compound.

Given:
- Iron (Fe) = 25%
- Sulphur (S) = 24%
- Oxygen (O) = 100% - 25% - 24% = 51%

Step 1: Convert the percentages to moles by assuming a 100g sample of the compound.
- Moles of Fe = 25g / 55.85 g/mol = 0.447 mol
- Moles of S = 24g / 32.06 g/mol = 0.748 mol
- Moles of O = 51g / 16.00 g/mol = 3.188 mol

Step 2: Determine the simplest whole number ratio of the elements.
- Divide each mole value by the smallest mole value (0.447 mol in this case):
Fe: 0.447 mol / 0.447 mol = 1
S: 0.748 mol / 0.447 mol ≈ 1.67
O: 3.188 mol / 0.447 mol ≈ 7.13

Step 3: Round the ratios to the nearest whole number:
Fe: 1
S: 2
O: 7

Therefore, the empirical formula of the compound is FeSO7, which can be simplified to FeSO4.

Step 1: Convert the percentages to grams.

Assume we have 100 g of the compound:

- Iron: 25 g
- Sulfur: 24 g
- Oxygen: 100 g - 25 g - 24 g = 51 g

Step 2: Convert the grams to moles.

- Iron: 25 g / 55.85 g/mol = 0.448 mol
- Sulfur: 24 g / 32.07 g/mol = 0.749 mol
- Oxygen: 51 g / 16.00 g/mol = 3.188 mol

Step 3: Find the simplest whole-number ratio of the moles.

Divide each mole value by the smallest mole value (0.448):

- Iron: 0.448 mol / 0.448 = 1
- Sulfur: 0.749 mol / 0.448 = 1.67
- Oxygen: 3.188 mol / 0.448 = 7.12

Round these values to the nearest whole numbers to get the simplest ratio:

- Iron: 1
- Sulfur: 2
- Oxygen: 7

Therefore, the empirical formula of the compound is FeS2O7.