select the correct answer.\nwhat is the empirical formula of a compound with 35.94% aluminum and 64.06%…

select the correct answer.\nwhat is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur?\na. als\nb. al₄s₆\nc. als₂\nd. al₂s₃\ne. als₃

select the correct answer.\nwhat is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur?\na. als\nb. al₄s₆\nc. als₂\nd. al₂s₃\ne. als₃

Answer

Explanation:

Step1: Assume 100 g of the compound

So, there are 35.94 g of Al and 64.06 g of S.

Step2: Calculate the moles of each element

The molar - mass of Al is (M_{Al}=26.98\ g/mol), and the molar - mass of S is (M_{S}=32.07\ g/mol). The number of moles of Al, (n_{Al}=\frac{35.94\ g}{26.98\ g/mol}\approx1.33\ mol). The number of moles of S, (n_{S}=\frac{64.06\ g}{32.07\ g/mol}\approx2.0\ mol).

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles ((n_{Al} = 1.33\ mol)). For Al: (\frac{1.33\ mol}{1.33\ mol}=1). For S: (\frac{2.0\ mol}{1.33\ mol}\approx1.5). Multiply each ratio by 2 to get whole - numbers. The ratio of Al to S becomes (Al:S = 2:3).

Answer:

D. (Al_{2}S_{3})