Question

A substance, on analysis, gave the following percentage composition.

Na=43.4 %, C= 11.3%, O = 45.3%

Calculate its empirical formula.​

Answer 1

Moles Of Na =

% of Na / Atomic Mass of Na = 43.4 / 23 = 1.89

Moles of C =

% of C / Atomic Mass of C = 11.3 / 12 = 0.94

Moles of O =

% of O / Atomic Mass of O = 45.3 / 16 = 2.83