# Experiment to show that air contains about 20% oxygen

I remember that in my primary classed I had an experiment in science to show air contains 20% oxygen. The experiment involved taking a trough of water, lighting a candle in the middle of the trough and inverting a glass jar over the candle so that air supply for the candle is cut of. When candle extinguishes, water level in the glass jar rises. I have seen this happen.

My question is why should the water raise in the jar, since burning of candle should produce an equivalent amount of carbon dioxide, so the molar amount of gas in jar is constant? Also the heat from the candle should expand the gasses in the jar, water level should go down I think.

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• This is not a good experiment because it makes a bunch of assumptions about combustion products. The water should condense, the carbon dioxide will be less than the oxygen into the combustion, but some of that will dissolve in the water. A better test might be to freeze out the nitrogen and see what is left. – Olin Lathrop Jul 7 '13 at 12:48
• @OlinLathrop The oxygen condenses before the nitrogen (at about 90 K). – dmckee Jul 7 '13 at 23:10

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.

• You mean that the rise in water level is due to the water produced during combustion? – Gautam Jul 7 '13 at 11:09
• It's due to 2 reasons: first, the quantity of gases present at the end of the reaction ($CO_2$ and $CO$) are less than the initial quantity of $O_2$ and so the pressure of the gas decreases and the water rises. Second, a small amount of $H_2O$ is added to the liquid water already present but I think this effect is very small compared to the first one. – Paracosmiste Jul 7 '13 at 11:48
• @Gautam Moreover, I read now the following "It can not be that the oxygen depletion is responsible for the water raising, because the water does not rise immediately. The water rises only after the candle dims. If gas would be going away, this would lead to a steady rise of the water level, not the rapid rise at the end, when the candle goes out". – Paracosmiste Jul 7 '13 at 12:05
• @Gautam It is not "only" the oxygen depletion which matters. There are two effects which matter: the chemical process of the burning as well as a physical process from the temperature change. These effects cancel each other initially. Since these effect hide each other partially, they are more difficult to detect. (source: math.harvard.edu/~knill/pedagogy/waterexperiment/index.html) – Paracosmiste Jul 7 '13 at 12:06
• I would also expect that the heat of the flame of the candle will heat up the gas and expand, so some of the gas will escape through the bottom. So when the gas cools down again, the pressure will drop and water will be sucked up. – fibonatic Jul 7 '13 at 13:43

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.

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.

I doubt that the CO2 dissolves quickly enough. My gut feeling is that this experiment is not working as suggested.

I think it is a physical, not chemical, effect. When you put the glass above the candle, there is already a column of hot air. When that air cools, the water rises. It would be a worthwhile experiment to ignite the candle electrically.

A better experiment would use steel wool, like bobthechemist suggested.

Btw. it is very easy to electrically ignite steel wool.

• This is better off as a comment under the main post IMHO. – paracetamol Sep 19 '17 at 8:39