Determine the empirical formulas for compounds with the following percent compositions: (a) 43.6% phosphorus and 56.4% oxygen (b) 28.7% K, 1.5% H, 22.8% P, and 47.0% O

a. P₂O₅

b .KH₂PO₄

Explanation:

In order to determine empirical formula of a compound, we divide the percent (as an integer number of mass) by the molar mass of the element and afterwards we divide, by the lowest value.

Centesimal composition by percent means the mass of the element in 100 g of compound

a. 43.6 g of P / Molar mass P → 43.6 g/ 30.97 g/mol = 1.41 moles

56.4 g of O / Molar mass O →  56.4 g / 16 g/mol = 3.52 moles

So, 1.41 moles of P / 1.41 = 1 mol of P

3.52 moles of O / 1.41 = 2.5 moles of O

We finally have 1 mol of P and 2.5 moles of O, bu we can't have rational numbers, so to have an integer we multiply by 2 (x2)

1 mol of P . 2 = 2 P

2.5 moles of O . 2 = 5 O

Empirical formula is P₂O₅

b. 28.74 g / 39.1 g/mol = 0.735 moles K

1.5 g / 1g/mol = 1.5 moles H

22.8 g / 30.97 g/mol = 0.736 moles P

47 g / 16 g/mol = 2.93 moles O

0.735 moles K / 0.735 = 1 mol K

1.5 moles H / 0.735 = 2 mol H

0.736 moles P / 0.735 = 1 mol P

2.93 moles O / 0.735 = 4 mol O

Empirical formula is KH₂PO₄

the answer is c.the valence of the atoms combine .