Question

What is the empirical formula for a compound which contains 53.73% Fe and 46.27% S

Answer 1

Answer: The empirical formula for the given compound is
`Fe_2S_3`

Explanation : Given,

Percentage of Fe = 53.73 %

Percentage of S = 46.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Fe = 53.73 g

Mass of S = 46.27 g

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

Step 1: Converting the given masses into moles.

Moles of Fe =
`\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}=(53.73g)/(56g/mole)=0.959moles`

Moles of S =
`\frac{\text{Given mass of S}}{\text{Molar mass of S}}=(46.27g)/(32g/mole)=1.44moles`

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 0.959 moles.

For Fe =
`(0.959)/(0.959)=1`

For S =
`(1.44)/(0.959)=1.50`

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

Fe : S = 2 : 3

Step 3: Taking the mole ratio as their subscripts.

Fe : S = 2 : 3

Hence, the empirical formula for the given compound is
`Fe_2S_3`