During the combustion of phosphorus in air, the air volume decreases from 200.0 cm3 to 158.2 cm3. What is the percentage of oxygen in the air?

During the combustion of phosphorus in air, the air volume decreases from 200.0 cm3 to 158.2 cm3. What is the percentage of oxygen in the air?

• A. 20.9%
• B. 15.8%
• C. 41.8%
• D. 18.0%

Answer: (A)

When a substance burns in air, the oxygen from the air is consumed, so the volume of air decreases while nitrogen and other gases largely stay the same. If temperature and pressure are constant, gas volumes reflect the amounts of gas present. The decrease in air volume is 200.0 cm3 − 158.2 cm3 = 41.8 cm3, which is the volume of oxygen that was used up. To find the oxygen fraction in the original air, divide this by the original air volume: 41.8 / 200.0 = 0.209. Multiply by 100 to get a percentage: 20.9%. So the oxygen content of the air is 20.9%.

When a substance burns in air, the oxygen from the air is consumed, so the volume of air decreases while nitrogen and other gases largely stay the same. If temperature and pressure are constant, gas volumes reflect the amounts of gas present.

The decrease in air volume is 200.0 cm3 − 158.2 cm3 = 41.8 cm3, which is the volume of oxygen that was used up. To find the oxygen fraction in the original air, divide this by the original air volume: 41.8 / 200.0 = 0.209. Multiply by 100 to get a percentage: 20.9%.

So the oxygen content of the air is 20.9%.