Question

what is the empirical formula the compound that is 22.6% phosphorus in 77.4% oxygen?

Answer 1

Given

Phsophorous : 22.6 %                                                         Oxygen: 77.4%

Step 1:

Divide the percentage with their mass numbers

22.6 / 30.9 = 0.7313

77.4 / 15.9994 = 4.8376

Step 2:

Divide the number with small result

0.7313 / 0.7313 = 1

4.8375 / 0.7313 = 6.614 which is approximately 7

therefore emperical formula of the compound is PO₇.