Derive the molecular formula of a substance with a composition of sodium 32.29% sulfur 22.54% oxygen 45.07%.

Given:

w% (Na) = 32.3%

w% (S) = 22.54%

w% (O) = 45.07%

Find:

Formula -?

Solution:

1) Find the molecular weight of a substance by the formula:

w% (element) = Ar (element): Mr (substance) * 100%

Mr (substance) = Ar (element): w% (element) * 100%

Substitute sodium data:

Mr (substances) = 23: 32.3% * 100% = 71.

2) Knowing the mass fraction of sodium, we find its molecular weight in the composition and the number of atoms:

71 – 100%

x – 32.3%,

x = 32.3% * 71: 100% = 23.

N (Na) = 23: Mr (Na) = 23: 23 = 1 atom.

3) Knowing the mass fraction of sulfur, we find its molecular weight in the composition and the number of atoms:

71 – 100%

x – 22.54%,

x = 22.54% * 71: 100% = 16.

N (S) = 16: Mr (S) = 16: 32 = 0.5 atoms.

4) Then the molecular weight of oxygen is 71 – 16 – 23 = 32.

N (O) = 32: Mr (O) = 32: 16 = 2 atoms.

We got a substance with the formula NaS0.5O2, the substance cannot have such a formula, so we multiply the indices by 2. We get Na2SO4 – sodium sulfate.

Answer: Na2SO4 – sodium sulfate.