A compound contains 46% Sodium, 12% Carbon and 42% Oxygen. The molecular mass of the compound is 106 u. Draft a table with these data and find the Empirical formula and Molecular formula of this compound. Given that atomic masses are Na=23 u, C=12u and O=16u.

Empirical formula is Na₂CO₃

Molecular formula is Na₂CO₃

Explanation:

Percentage compositions of the elements;

Sodium = 46%

Carbon = 12 %

Oxygen = 42%

To determine empirical formula, we calculate the mole ratio first.

mole ratio =  % mass/atomic mass; where the atomic masses of the elements are :  Na = 23 u, C = 12 u and O = 16 u.

mole ratio: Na = 46/23; C = 12/12; O = 42/16

mole ratio (Na : C : O) = 2 : 1 : 2.6

approximate mole ratio = 2 : 1 :3

Therefore, empirical formula of compound is Na₂CO₃

To determine the molecular formula;

molecular formula or mass = n(empirical formula or mass)

n = molecular mass/empirical mass

molecular mass = 106 u,

empirical mass = 23*2 + 12 + 16*3 = 106 u

then, n = 106/106 = 1

therefore, molecular formula = 1 * (Na₂CO₃) = Na₂CO₃

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