Question

a compound is found to contain 31.14 % sulfur and 68.86 % chlorine by weight. what is the empirical formula for this compound? to answer the question, enter the elements in the order presented above. 3 item attempts remaining. use the references to access important values if needed for this question.

Explanation:

Step1: Assume 100 g of the compound

Since the percentages are by - weight, in 100 g of the compound, there are 31.14 g of sulfur (S) and 68.86 g of chlorine (Cl).

Step2: Calculate the moles of each element

The molar mass of S is $M_{S}=32.07\ g/mol$, so the moles of S, $n_{S}=\frac{31.14\ g}{32.07\ g/mol}\approx0.971\ mol$. The molar mass of Cl is $M_{Cl}=35.45\ g/mol$, so the moles of Cl, $n_{Cl}=\frac{68.86\ g}{35.45\ g/mol}\approx1.942\ mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles. $\frac{n_{S}}{n_{S}} = 1$ and $\frac{n_{Cl}}{n_{S}}=\frac{1.942\ mol}{0.971\ mol}=2$.

Answer:

SCl₂