what is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur?\na. ais\nb. al₄s₆\nc…

Answer:

D. $Al_2S_3$

Explanation:

Step1: Assume 100g of the compound

So, we have 35.94g of Al and 64.06g of S.

Step2: Calculate moles of each element

Moles of Al = $\frac{35.94g}{26.98g/mol}\approx1.332mol$ (molar - mass of Al = 26.98g/mol). Moles of S = $\frac{64.06g}{32.07g/mol}\approx2.0mol$ (molar - mass of S = 32.07g/mol).

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles. $\frac{1.332mol}{1.332mol}=1$ for Al and $\frac{2.0mol}{1.332mol}\approx1.5$ for S. Multiply both by 2 to get whole - number ratios. We get 2 for Al and 3 for S. So the empirical formula is $Al_2S_3$.