Compound 1: 76.0% ru and 24.0% o (by mass), compound 2: 61.2% ru and 38.8% o (by mass). part a what is the empirical formula for compound 1?

To solve for the empirical formula, we write first all the data.

Given:
Compound 1: 76 wt% Ru and 24wt% O
Compound 2: 61.2 wt% Ru and 38.8  wt% O

Required: Empirical Formula of Compound 1

Solution:
Assume total mass of the compound is 100 g

Solving for Compound 1,

76 g Ru x 1 mol Ru      = 0.75195 mol Ru
               101.07 g Ru

24 g O x 1 mol O    = 1.5 mol O
                16 g O

Then, divide each mole with the smallest number of moles calculated

Ru = 0.75195 mol/0.75195 mol = 1
O = 1.5 mol/0.75195 mol = 2

Therefore, the empirical formula for Compound 1 is RuO2.

ANSWER: RuO2

Hi + naoh -> h2o + nal