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 given by mass, assuming 100 g of the compound means 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 approximately $M_{S}=32.07\ g/mol$, and the molar - mass of Cl is approximately $M_{Cl}=35.45\ g/mol$.
The number of moles of S, $n_{S}=\frac{31.14\ g}{32.07\ g/mol}\approx0.971\ mol$.
The number of 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 (0.971 mol in this case).
For S: $\frac{0.971\ mol}{0.971\ mol}=1$.
For Cl: $\frac{1.942\ mol}{0.971\ mol}=2$.

Answer:

SCl₂