Apure sample of a new chemical compound was analyzed and was found to have the following mass percentages: al 31.5 %; o 56.1 %; s 12.4 %.
which of these could be the empirical formula of the compound?

al5o28s7

al3o9s

alo2s2

al4o14s7

alo6s1.5

The answer to your question is empirical formula   Al₃O₉S

Explanation:

Data

Al = 31.5 %

O = 56.1 %

S = 12.4 %

Process

1.- Look for the atomic masses of the elements

Al = 27 g

O = 16

S = 32

2.- Represent the percentages as grams

Al = 31.5 g

O = 56.1 g

S = 12.4 g

3.- Convert these masses to moles

                               27 g of Al 1 mol

                               31.5 g  x

                                x = 1.17 moles

                               16 g of O  1 mol

                               56.1 g of O  x

                                  x = 3.5 mol

                               32 g of S  1 mol

                               12.4 g of S   x

                                x = 0.39 moles

4.- Divide by the lowest number of moles

Al =   1.17 / 0.39  = 3

O =    3.5 / 0.39 = 8.9 ≈ 9

S  =   0.39 / 0.39 = 1

5.- Write the empirical equation

                                Al₃O₉S

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