A 45.8 mg sample of phosphorus reacts with selenium to form 133 mg of the compound. Part A What is the empirical formula of the phosphorus selenide? Express your answer as a chemical formula. nothing

P4Se3

Explanation:

Applying the Law of Conservation of Mass:

(133 mg total) - (45.8 mg P) = 87.2 mg Se

Dividing by the molar mass,

Phosphorus:-(45.8 mg P) / (30.97376 g P/mol) = 1.4786 mmol P

Selenium:-(87.2 mg Se) / (78.96 g Se/mol) = 1.1044 mmol Se

Divide by the smaller number of millimoles:

(1.4786 mmol P) / 1.1044 mmol = 1.339

(1.1044 mmol Se) / 1.1044 mmol = 1.000

multiply by 3, to get a whole number interger. then round to the nearest whole numbers to find the empirical formula.

Therefore the empirical formula is P4Se3

it takes on an electronical energy