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

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)