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A pure 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

Respuesta :

joseaaronlara

Answer:

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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