Question

A 0.500-g sample of chromium metal reacted with sulfur powder to give 0.963 g of product. Calculate the empirical formula of the chromium sulfide.

Answer 1

Answer: The empirical formula for the given compound is $Cr_2S_3$

Explanation : Given,

Mass of product = 0.963 g

Mass of Cr = 0.500 g

Mass of S = 0.963 g  - 0.500 g = 0.463 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Cr =$\frac{\text{Given mass of Cr}}{\text{Molar mass of Cr}}=\frac{0.500g}{52g/mole}=0.00962moles$

Moles of S = $\frac{\text{Given mass of S}}{\text{Molar mass of S}}=\frac{0.463g}{32g/mole}=0.0145moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00962 moles.

For Cr = $\frac{0.00962}{0.00962}=1$

For S = $\frac{0.0145}{0.00962}=1.5$

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of Cr : S = 1 : 1.5

The ratio of Cr : S = 2 : 3

Step 3: Taking the mole ratio as their subscripts.

The ratio of Cr : S = 2 : 3

Hence, the empirical formula for the given compound is $Cr_2S_3$