1. Determine the empirical formula of a compound
containing 59.0% Na and 41.0% O by mass.
(a) NaO
(b) Na2O
(c) NaO2
(d) Na2O3
(e) none of these

NaO

Explanation:

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number .

Given data:

Percentage of sodium = 59.0%

Percentage of oxygen = 41%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 59.0 / 23 = 2.6

Number of gram atoms of O = 41 / 16 = 2.6

Atomic ratio:

            Na                      :            O            

           2.6/2.6               :          2.6/2.6

              1                       :              1      

Na  : O = 1:1

Empirical formula is NaO.

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