Question

A compound contains 30. percent sulfur and 70. percent fluorine by mass. the empirical formula of the compound is

Answer 1

Answer : The empirical formula of a compound is, $SF_4$

Solution : Given,

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 = 30 g

Mass of F = 70 g

Molar mass of S = 32 g/mole

Molar mass of F = 19 g/mole

Step 1 : convert given masses into moles.

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{30g}{32g/mole}=0.9375moles$

Moles of F = $\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{70g}{19g/mole}=3.684moles$

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

For S = $\frac{0.9375}{0.9375}=1$

For F = $\frac{3.684}{0.9375}=3.92\approx 4$

The ratio of S : F = 1 : 4

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $S_1F_4$  = $SF_4$

Therefore, the empirical formula of a compound is, $SF_4$

Answer 2

Assume 100 g of compound. This turns percent to mass. Calculate moles: 

S ---> 30/32 = 0.9375 
F ---> 70 / 19 = 3.6842 

get whole number ratio: 

0.9375 / 0.9375 = 1 
3.6842 / 0.9375 = 3.93 = 4 

Answer : SF4