A sample containing phosphorus weighs 1.342 g. The phosphorus containing compound in this sample is P4010....

Question

Question

A sample containing phosphorus weighs 1.342 g. The phosphorus containing compound in this sample is P4010. A gravimetric expe

Answer 1

Answer #1

Mass of Precipitate Ca3(PO4)2 = 0.752 g

Molecular Mass of Precipitate Ca3(PO4)2 = 2*Ca + 2*P + 8*O = 310.174 g/mol

Moles of Precipitate Ca3(PO4)2 = Mass / Molecular Mass = 0.752 / 310.174 = 0.00242 Moles

1 Mole of Ca3(PO4)2 contains 2 moles of P (Phosphorus) hence

0.00242 Moles of Ca3(PO4)2 will contains = 0.00242 *2 = 0.00484 moles of P (Phosphorus).

Since Mass of 1 mole of Phosphorus is 30.97 grams

Hence Mass of 0.00484 will be = 0.00484*30.97 = 0.149 grams

Given the total mass of the compound = 1.342 grams

Mass of Phosphorus = 0.149 grams

Mass percentage of Phosphorus = Mass of Phosphorus*100/Mass of the compound

Mass percentage of Phosphorus = 0.149 *100 / 1.342 = 11.16 Percent

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Answer 2

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