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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

User Marksyzm
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1 Answer

2 votes

Answer:

The answer to your question is empirical formula Al₃O₉S

Step-by-step 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

User CharlesC
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