What is the percentage composition of NaHCO3

Na is 27.365 %.

H is 1.20 %.

C is 14.30 %.

O is 57.13 %.

Explanation:

Percent composition = (mass/molar mass) x 100

Na = 22.99g/mol

H = 1g/mol

C = 12.01g/mol

O = 16g/mol

Na = (1.0 x 22.99) / (84.01) x 100 = 27.365 %.

H = (1.0 x 1.01) / (84.01) x 100 = 1.20 %.

C = (1.0 x 12.01) / (84.01) x 100 = 14.30 %.

O = (3.0 x 16.00) / (84.01) x 100 = 57.13 %

The % of Na is 27.365 %.

The % of H is 1.20 %.

The % of C is 14.30 %.

The % of O is 57.13 %.

Explanation:

The composition of NaHCO₃ shows that it contains 1.0 atom of Na, 1.0 atom of H and 3.0 atoms of O. The molar mass of NaHCO₃ = ∑ (no. of atoms x Atomic mass of the element). Atomic mass of Na = 22.99 g/mole, atomic mass of H = 1.01 g/mole, atomic mass of C = 12.01 g/mole, and atomic mass of O = 16.00 g/mole. Then, the molar mass of NaHCO₃ = (1.0 x 22.99) + (1.0 x 1.01) + (1.0 x 12.01) + (3.0 x 16.00) = 84.01 g/mole. To get the percentage of each element using the relation: % composition = [ (no. of atoms x its atomic mass) / (molar mass of the compound) ] x 100. The % of Na = (1.0 x 22.99) / (84.01) x 100 = 27.365 %. The % of H = (1.0 x 1.01) / (84.01) x 100 = 1.20 %. The % of C = (1.0 x 12.01) / (84.01) x 100 = 14.30 %. The % of O = (3.0 x 16.00) / (84.01) x 100 = 57.13 %.