Question

A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. What is the empirical formula of the compound?

Answer 1

`SF_(6)`

Step-by-step explanation:

Let the compound be x.

Given the following data;

Sulfur, S = 3.21g

Fluorine, F = 11.4g

Atomic mass of sulfur = 32.07g

Atomic mass of fluorine = 19g

Amount of moles for sulfur;

3.21*(1/32.07) = 0.10mol

Amount of moles for fluorine;

11.4*(1/19) = 0.60mol

We then divide by the smallest to find the ratio;

(0.10/0.10) = 1 Mol of sulfur.

(0.60/0.10) = 6 Mol of fluorine.

Therefore, the ratio of sulfur to fluorine is 1:6.

Compound x =
`SF_(6)`

Hence, the empirical (simplest) formula of the compound is
`SF_(6)`