Answer 1

Given: Atomic mass of Na = 23 u, C = 12 u, and O = 16 u
Percentage of Na, C and O = 43.4%, 11.3% and 45.3% respectively.

To find: The empirical formula of the compound

Calculation:

Moles of Na = `"% of Na"/"Atomic mass of Na"=43.4/ 23` = 1.89 mol

Moles of C = `"% of C"/"Atomic mass of C"=11.3/12` = 0.94 mol

Moles of O = `"% of O"/"Atomic mass of O"=45.3/16` = 2.83 mol

Hence, the ratio of number of moles of Na : C : O is

`1.89/0.94` = 2, `0.94/0.94` = 1 and `2.83/0.94` = 3

Hence, empirical formula is Na₂CO₃.

Empirical formula of the compound = Na₂CO₃