This page shows how to do simple calculations involving gases. I am assuming you have already read and understood the pages about moles and calculations from equations involving masses. It is pointless reading this page unless you are happy with those.
Avogadro's Hypothesis is sometimes known as Avogadro's Law. What Avogadro says is:
In other words, if you had a litre of any gas at the same temperature and pressure it will always contain the same number of molecules whether you were talking about hydrogen, or ammonia, or carbon dioxide, or chlorine, or methane . . . or any other gas you care to name. You can make use of this in simple reactions involving gases - for example the combustion of hydrogen and oxygen to make water.
It doesn't matter whether the water is formed as steam or liquid water. If you want to maximise the bang, you would need twice as many molecules of hydrogen as of oxygen. What Avogadro tells us is that that means making a mixture with twice the volume of hydrogen as of oxygen. I love this piece of video - this is the third time I have used it in this 14 to 16 section!
This is the most useful thing to come out of Avogadro's Hypothesis. The thinking goes like this: 1 mole of any gas contains the same number of molecules. (That comes from the definition of a mole.) If you have the same number of molecules of any gas they must occupy the same volume at the same temperature and pressure. (That comes from Avogadro's Hypothesis.)
The volume occupied by 1 mole of any gas is called the The molar volume varies with temperature and pressure, but at this level you will almost always be given the value at room temperature and pressure (rtp) which is taken to be about 20°C and 1 atmosphere pressure. The number will usually be quoted as 24 dm The cubic decimetre (dm There are 1000 cm | |

Note: A decimetre is a tenth of a meter - 10 cm. A cubic decimetre is the volume of a cube 10 cm x 10 cm x 10 cm - 1000 cm^{3}. | |

The usual example used to illustrate is the effect of excess dilute hydrochloric acid on calcium carbonate because the numbers are so easy.
I want to repeat the example I used when we were talking about calculations from equations using masses, but this time calculate the volume of carbon dioxide produced. I am going to start by using the same technique as I used in that example. Calcium carbonate and dilute hydrochloric acid react together according to this equation:
What volume of carbon dioxide (measured at rtp) would you get if you added excess hydrochloric acid to 10 g of calcium carbonate? (RAMs: C = 12; O = 16; Ca = 40. Molar volume = 24 dm Start by writing down what the equation says in terms of moles.
Now convert the amounts in moles into grams or dm
So the equation is saying that:
Therefore 10 g of CaCO | |

Note: If you need to, put in another step by calculating what 1 g would give and then multiply that by 10. | |

Now let's look at the same example but using the other method of doing these calculations. Calcium carbonate and dilute hydrochloric acid react together according to this equation:
What volume of carbon dioxide (measured at rtp) would you get if you added excess hydrochloric acid to 10 g of calcium carbonate? (RAMs: C = 12; O = 16; Ca = 40. Molar volume = 24 dm This time, instead of interpreting the equation in terms of moles, you start with what you know most about and work out how many moles of it you have. You know the mass of calcium carbonate, and can easily work out the mass of 1 mole.
The equation says:
Now use the molar volume:
It is a toss-up which of these methods you use, but my recommendation would be that you use the second one. The reason is that that is the approach you have to take to do titration calculations coming up on a later page.
I'm going to reverse this now, using a slightly more complicated relationship and starting with the gas. I'm only going to do this using the second method - not because you can't do it the first way, but because I want to shift the focus on to the second method. Hydrogen gas is produced when you drop lithium metal into water.
Assuming you have an excess of water, what is the maximum mass of lithium you could use to avoid over-filling a 100 cm ^{3} mol^{-1} at rtp.)
Start with what you know most about and work out how many moles of it you have. You know the volume of the syringe you can fill with hydrogen (100 cm
(Again, put in an extra step working out how many moles there are in 1 cm
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Some comments: You may have noticed that I have switched between "mole" and its abbreviation "mol" in these calculations. It really doesn't matter. Normally you use the full word in a piece of text and mol in a calculation. It is a bit like "grams" and "g". But I repeat, it doesn't matter.
In this, and the wherever it is relevant in future calculations, I am expressing small numbers in scientific notation (for example, 4.17 x 10 Your calculator will show this as 4.17 Notice that I have only quoted the answer to 2 significant figures. The 24 dm | |

The calculations we have just looked at are the most likely sort of questions that you could be asked involving molar volume, but there are a few other examples you might get.
You can calculate the molar volume if you know the density of a gas at a particular temperature and pressure. At 0°C and 1 atmosphere pressure, the density of oxygen is 1.429 g dm | |

Note: You read g dm^{-3} as "grams per cubic decimetre" | |

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Note: Again, if necessary, work out what volume 1 g would occupy by dividing 1 by 1.429, and then multiply that by 32 to find the volume occupied by 32 g. | |

You may well find this value for the molar volume quoted at "standard temperature and pressure (stp)" - 0°C and 1 atmosphere pressure.
Calculate the volume of 1.42 g of chlorine, Cl
As always, if it helps you can go via the volume occupied by 1 g and then multiply that by 1.42.
Calculate the density of ammonia gas, NH
You need to practise this by doing as many similar calculations as possible. Find out what sort of questions you might get in an exam by looking at past papers and mark schemes. You will find links to the main UK Exam Boards on this page. **Where would you like to go now?**-
**To the calculations menu . . .**
© Jim Clark 2021 |