Respuesta :
Answer:- Na = 27.37%, H = 1.20%, C = 14.30% and O = 57.14%
Solution:- For the percentage composition of a compound, the atomic mass of each atoms times its subscript is divided by the molar mass of the compound and multiplied by 100.
The given compound is [tex]NaHCO_3[/tex].
mass of Na = 22.99 g
mass of H = 1.008 g
mass of C = 12.01 g
mass of O = 3(16) = 48 g
Molar mass of compound = 22.99 g + 1.008 g + 12.01 g + 48 g = 84.008 g
percentage of Na = [tex](\frac{22.99}{84.008})100[/tex]
= 27.37%
percentage of H = [tex](\frac{1.008}{84.008})100[/tex]
= 1.20%
percentage of C = [tex](\frac{12.01}{84.008})100[/tex]
= 14.30%
percentage of O = [tex](\frac{48}{84.008})100[/tex]
= 57.14%
Answer:
Na = 27.37%, H = 1.20%, C = 14.30% and O = 57.14%
Explanation:
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