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Multiple Choice
A 45.2 mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. Determine the empirical formula of phosphorus selenide.
A
P2Se3
B
PSe
C
P4Se3
D
PSe2
Verified step by step guidance
1
Convert the mass of phosphorus from milligrams to grams by dividing by 1000. This gives you the mass in grams, which is necessary for calculating moles.
Calculate the moles of phosphorus using its molar mass. The molar mass of phosphorus (P) is approximately 30.97 g/mol. Use the formula: \( \text{moles of P} = \frac{\text{mass of P in grams}}{\text{molar mass of P}} \).
Determine the mass of selenium in the compound by subtracting the mass of phosphorus from the total mass of the selenide. This gives you the mass of selenium in milligrams, which you then convert to grams.
Calculate the moles of selenium using its molar mass. The molar mass of selenium (Se) is approximately 78.96 g/mol. Use the formula: \( \text{moles of Se} = \frac{\text{mass of Se in grams}}{\text{molar mass of Se}} \).
Determine the simplest whole number ratio of moles of phosphorus to moles of selenium by dividing each by the smallest number of moles calculated. This ratio gives you the subscripts for the empirical formula of phosphorus selenide.