Question

What is the empirical formula of a compound containing 60.0% sulfur and 40.0% oxygen by mass?

Answer 1

The empirical formula will be $S_{3}O_{4}$

Answer 2

Answer: The empirical formula is $S_3O_4$

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 = 60 g

Mass of O = 40 g

Step 1 : convert given masses into moles.

Moles of S =$\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{60g}{32g/mole}=1.875moles$

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{40g}{16g/mole}=2.5mole[/tex]

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

For S = $\frac{1.875}{1.875}=1$

For O =$\frac{2.5}{1.875}=1.33$

Converting them into simple whole number ratios by multiplying by 3.

The ratio of S:O= 1 : 1.33= 3 : 4

Hence the empirical formula is $S_3O_4$