Question

Select the correct answer. what is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur? a. als b. al4s6 c. als2 d. al2s3 e. als3

Answer 1

The empirical formula of the compound would be  $Al_2S_3$

Empirical formula

The compound is composed of aluminum and sulfur:

Al = 35.94                                S = 64.06

Find the mole of each component

Al = 35.94/26.98                       S = 64.06/32

     = 1.33                                        = 2.00

Divide by the smallest:

Al = 1.33/1.33                                   S = 2/1.33

    = 1                                                  = 1.5

Thus, the empirical formula of the compound is $Al_2S_3$

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Answer 2

The empirical formula of the compound from the calculations done is AlS2.

What is empirical formula?

The term empirical formula refers to the simplest formula of a compound. It shows the ratio of atoms of elements in the molecule.

Now, we divide each percentage with the relative atomic mass of the element;

Al - 35.94/27     S - 64.06/32

1.33                           2.00

Divide through by the lowest ratio;

1.33/1.33                 2.00/1.33

A - 1                       S - 2

The empirical formula of the compound is AlS2.

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