Question

27 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

Explanation:

Step1: Assume 100g of the compound

Assume the mass of the compound is 100g. So, there are 35.94g of Al and 64.06g of S.

Step2: Calculate the moles of each element

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

Step3: Find the mole - ratio

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

Answer:

D. \(Al_2S_3\)