Question

A chemist found that 4.69 g of sulfur combined with fluorine to produce 15.81 g of a gas. what is the empirical formula of the gas? 1. s2f6 2. s3f2 3. sf4 4. sf7 5. sf5 6. s4f

Answer 1

Mass of sulfur combined - 4.69 g
Mass of gas produced is 15.81 g, therefore mass of fluorine is (15.81-4.69) = 11.12 g
Number of sulfur moles - 4.69 g/32 g/mol = 0.15 mol
Number of fluorine moles - 11.12 g/ 19 g/mol = 0.585 mol
divide both by least number of moles 
S - 0.15/0.15 = 1
F - 0.585/0.15 = 3.9 rounded off is 4
ratio of S to F = 1:4Therefore formula of the gas is SF₄

Answer 2

Answer: The empirical formula for the given compound is $SF_4$

Explanation:

We are given:

Mass of sulfur = 4.69 g

Mass of gas containing fluorine and sulfur = 15.81 g

Mass of fluorine = (15.81 - 4.69) = 11.12 g

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

• Step 1: Converting the given masses into moles.

Moles of Sulfur =$\frac{\text{Given mass of Sulfur}}{\text{Molar mass of Sulfur}}=\frac{4.69g}{32g/mole}=0.146moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{11.12g}{19g/mole}=0.585moles$

• 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 0.146 moles.

For Sulfur = $\frac{0.146}{0.146}=1$

For Fluorine = $\frac{0.585}{0.146}=4$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of S : F = 1 : 4

Hence, the empirical formula for the given compound is $SF_4$