Answer:
Al2S3
Explanation:
Firstly, we need to know their atomic masses. The atomic mass of sulphur is 32 while that of aluminum is 27.
So firstly, we divide the percentage compositions with their atomic masses.
S = 64.02/32 = 2.000625
Al = 35.98/27 = 1.3326
We then proceed to divide each by the smallest of the two.
S = 2.000625/1.3326 = 1.5
Al = 1.3326/1.3326 = 1
Let’s multiply each by 2 to yield 3 and 2 respectively.
The empirical formula is thus Al2S3
The empirical formula for the compound that contains 35.98% aluminum and 64.02% sulphur is Al₂S₃
The empirical formula of the compound can be obtained as illustrated below:
Divide by their molar mass
Al = 35.98 / 27 = 1.333
S = 64.02 / 32 = 2
Divide by the smallest
Al = 1.333 / 1.333 = 1
S = 2 / 1.333 = 1.5
Multiply by 2 to express in whole number
Al = 1 × 2 = 2
S = 1.5 × 2 = 3
Thus, the empirical formula of the compound is Al₂S₃
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