James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o
f 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.
The percent of oxygen is approximately 57.14% or rounded to 57%. The first step is determining the molar mass of the compound by adding the atomic weights of the constituent elements. The given atomic weights are as follows: Na = 22.99 g/mol, H = 1.01 g/mol, C = 12.01 g/mol, O = 16.00 g/mol. The molar composition of NaHCO3 is calculated as follows: NaHCO3 = (1 mol * Na) + (1 mol * H) + (1 mol * C) + (3 mol * O) = (1 * 22.99) + (1 * 1.01) + (1 * 12.01) + (3 * 16) = 84.01 g. The formula for percentage composition is expressed as (Mass of element / Mass of Compound) * 100: thus, the percentage composition for oxygen is (48 / 84.01) * 100, resulting in 57.14%. Finally, rounding to the nearest whole number yields 57%.
The question can be easily resolved if we know the abundance of other elements. You would only need to deduct their total from 100. Since this information is missing, the calculation proceeds as follows: 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. The total mass then is 84 g. Thus, the percentage of O is calculated as % O = 48/84 * 100 = 57.14%.
