Question

What is the empirical formula of a compound that is made from 0.76 g of phosphorus atoms and 0.324 g of flooring Adam's

Answer 1

[AI-generated answer — Bot GPT 3.5]

To find the empirical formula of the compound, we first need to determine the number of moles of each element.

The molar mass of phosphorus (P) is 30.97 g/mol, and the molar mass of fluorine (F) is 19.00 g/mol.

Number of moles of phosphorus:
0.76 g / 30.97 g/mol = 0.025 mol

Number of moles of fluorine:
0.324 g / 19.00 g/mol = 0.017 mol

Next, we need to determine the ratio of the elements in the compound by dividing the number of moles of each element by the smallest number of moles:

Phosphorus: 0.025 mol / 0.017 mol = 1.47
Fluorine: 0.017 mol / 0.017 mol = 1

The ratio of phosphorus to fluorine in the compound is approximately 1.47:1. To get whole numbers, we can round these values:

Phosphorus: 1.47 ≈ 1
Fluorine: 1

So, the empirical formula of the compound is PF.