Respuesta :

1)number mole in
50.8 g Zn *(1 mol Zn/65.4g)=0.777 mol Zn
2)number mole in
16.0 g P*(1 mol P/31.0 g)= 0.516 mol P
3)number mole in
33.2 g O*(1 mol O/16.0 g) = 2.075 mol O

0.777 mol Zn :  0.516 mol P : 2.075 mol O =
= (0.777 mol Zn/0.516) :  (0.516 mol P/0.516) : (2.075 mol O/0.516)=
=1.5 mol Zn :  1 mol P : 4 mol O=3 Zn : 2 P : 8O
Zn3P2O8
or
the same is Zn3(PO4)2

Eduard22sly

The empirical formula of the compound is Zn₃P₂O₈ or Zn₃(PO₄)₂

Data obtained from the question

  • Zinc (Zn) = 50.8
  • Phosphorus (P) = 16
  • Oxygen (O) = 33.2
  • Empirical formula =?

How to determine the empirical formula

The empirical formula of the compound can be obtained as illustrated below:

Divide by their molar mass

Zn = 50.8 / 65 = 0.782

P = 16 / 31 = 0.516

O = 33.2 / 16 = 2.075

Divide by the smallest

Zn = 0.782 / 0.516 = 1.5

P = 0.516 / 0.516 = 1

O = 2.075 / 0.516 = 4

Multiply by 2

Zn = 1.5 × 2 = 3

P = 1 × 2 = 2

O = 4 × 2 = 8

Thus, the empirical formula of the compound is Zn₃P₂O₈ or Zn₃(PO₄)₂

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