Chemistry - Experiment to show that air contains about 20% oxygen
Solution 1:
One problem with using combustion reactions to determine the percent of oxygen in air is the additional products, namely $\ce{CO2}$ and water vapor. If performing a water displacement reaction, these gases add to the uncertainty and imprecision of this method.
Water displacement by a combustion reaction can, however, be used if an appropriate combustion reaction is chosen. A more reasonable experiment to determine the percent oxygen in air is to use the oxidation of iron $$\ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}$$ The benefits of this reaction are (a) only one product is formed and (b) the product is a solid so it will not influence the water displacement in a significant way.
This experiment has been reported in the Journal of Chemical Education and a brief search of the web found an adaptation of this experiment if you don't have access to the journal. Briefly, one measures the volume of a test tube, adds a known amount of steel wool into the tube and inverts the tube in a beaker of water. The oxidation of the iron is catalyzed by first rinsing it in a dilute acid (vinegar) solution. Knowing the density of iron one can calculate the volume of the test tube occupied by the solid and then determine the volume of air before and after the reaction is completed.
There are a number of possible errors that can be introduced into this experiment, but unlike the combustion of a hydrocarbon, these errors are much more manageable for an undergraduate or high school chemistry experiment. Most significantly, the reaction occurs near room temperature so the temperature change of the gas can be ignored.
Solution 2:
The paraffin wax is an alkane ($\ce{C_nH_{2n+2}}$); its combustion reaction is (complete combustion): $$\ce{2C_nH_{2n+2} + $(3n+1)$ O2 -> $2(n+1)$ H2O + $2n$ CO2}$$ or (incomplete combustion): $$\ce{2C_nH_{2n+2} + $(2n+1)$ O2 -> $2(n+1)$ H2O + $2n$ CO}$$ In the two cases the quantity (in moles) of $\ce{O2}$ is greater than the quantity of $\ce{CO2}$ or $\ce{CO}$.
Moreover the gaseous $\ce{H2O}$ becomes liquid.
Solution 3:
I have done this demonstration in class to talk about gas laws, but not to talk about the composition of the gas.
What is happening is the candle inside the glass is obviously hot and the flame with heat up the surrounding air. As temperature of a gas increases, the volume of the gas increases. This is why you might see some bubbles coming out of the bottom of the glass if you don't have a good seal. The candle will continue to burn until the oxygen is consumed from the air. Then when the candle goes out the temperature of the air will decrease. As the temperature decreases the gas will contract, which causes a vacuum inside of the glass (or an area of low pressure). Water will move from high pressure to low pressure because the pressure of the air is higher on the water outside of the glass than the pressure on the water inside of the glass.