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
Answers
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.
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.