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
.
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 =
![((22.99)/(84.008))100](https://img.qammunity.org/2020/formulas/chemistry/high-school/nkqy0qui8vkxoo7190xrtaos431ej06uhy.png)
= 27.37%
percentage of H =
![((1.008)/(84.008))100](https://img.qammunity.org/2020/formulas/chemistry/high-school/j9cljyaqrjky9ir1pb66mrjl9a01j7ysjb.png)
= 1.20%
percentage of C =
![((12.01)/(84.008))100](https://img.qammunity.org/2020/formulas/chemistry/high-school/4gtdahb8i7x61w9j45m1age9vwsqc9wyh1.png)
= 14.30%
percentage of O =
![((48)/(84.008))100](https://img.qammunity.org/2020/formulas/chemistry/high-school/3nz8bb3l2raehyjiqm1x67ip2ajnsyuqjh.png)
= 57.14%