What is the empirical formula of the phosphorus selenide?

A 45.2 mg sample of phosphorous reacts with selenium to form 131.6 mg of the selenide.

atomic mass of phosphorous = 31g

atomic mass of selenium = 79g

45.2/31 = 1.45 which is nearly 1.5 mol of P

131.6 – 45.2 = 86.4g

86.4/79 =1.09 which is nearly 1 mol of Se

so the empirical formula of the phosphorus selenide is P3Se2
Phosphorus Selenide
Find the moles of each and compare to get the ratio of the atoms. P = 40.9 / 31 = 1.32 moles. Se = 119 / 79 = 1.50 These are nearly the same and close enough for this type of question. The formula is PSe P=Se
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What is the empirical formula of the phosphorus selenide?

A 45.2 mg sample of phosphorous reacts with selenium to form 131.6 mg of the selenide.
P4Se3