Question

A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula of the compound?
A.SF2

B.SF3

C.SF4

D.SF5

E.SF6

Answer 1

Answer: The empirical formula will be $SF_6$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of S= 3.21 g

Mass of F = 11.4 g

Step 1 : convert given masses into moles.

Moles of S =$\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{3.21g}{32g/mole}=0.1moles$

Moles of F =$\frac{\text{ given mass of F}{\text{ molar mass of F}}= \frac{11.04g}{19g/mole}=0.6moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For S= $\frac{0.1}{0.1}=1$

For F = $\frac{0.6}{0.1}=6$

The ratio of S: F= 1: 6

Hence the empirical formula is $SF_6$