Question

James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent of each element in the compound. He easily calculated the first three elements, but was stumped when he got to oxygen. What is the percent oxygen in the compound? Round to the nearest whole number.

Answer 1

Answer:

57.14 % ≈ 57 %

Explanation:

First step is to find the molar mass of the compound. This is done by adding the atomic masses of the elements present.

The atomic masses of the elements are given below;

Na = 22.99 g/mol

H = 1.01 g/mol

C = 12.01 g/mol

O = 16.00 g/mol

NaHCO3 = (1 mol * Na) + (1 mol * H) + ( 1 mol * C) + (3 mol * O)

NaHCO3 = (1 * 22.99) + (1 * 1.01) + (1 * 12.01) + (3 * 16)

NaHCO3 = 22.99 + 1.01 + 12.01 + 48

NaHCO3 = 84.01g

Formular for percentage composition;

Percentage composition = (Mass of element / Mass of Compound) * 100

Percentage composition = (48 / 84.01) * 100

Percentage composition = 57.14 %

Rounding up to the nearest whole number = 57 %

Answer 2

This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = 57.14%