A45.2 mg sample of phosphorus reacts with selenium to form 131.6 mg of the phosphorus selenide. determine the empirical formula of phosphorus selenide. what will be the simplest molecular formula of this compound

Phosphorus:

mass = 45.2 mg

molar mass = 30.97 mg / mmol

moles = 45.2 mg / (30.97 mg / mmol) = 1.46 mmol ≈ 1.5

Selenium:

mass = 131.6 mg – 45.2 mg = 86.4 mg

molar mass = 78.96 mg / mmol

moles = 86.4 mg / (78.96 mg / mmol) = 1.10 mmol ≈ 1

P1.5Se1

To get the empirical formula, we multiply with 2 to get whole numbers:

P3Se2

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.