Chemical quantities (AS)
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Calculating volumes of gases
This tutorial is designed for those who find gas-volume calculations difficult. We suggest a four step approach but note that there are alternative approaches to steps [3] and [4]. We are not trying to confuse you if you already have a method which ‘works’.
Steps needed
[1] Construct a balanced equation for the reaction if necessary (none is needed in the second worked example below). Sometimes you only need to show the reactants and products that affect the volume of gas formed, but if so, you must show these in the right amounts as in the full balanced equation. You can call this a ‘skeleton’ equation.
[2] Work out the molar masses of any substance for which the mass is given. For the gas mentioned in the question, simply use the information that one mole of any gas has a volume of about 24 dm3 (24000 cm3) at room temperature and pressure (see page 18 of the Nuffield Chemistry Students’ Book).
By the way, you do not need to remember the value for the molar volume for gases - the information will be quoted in any examination question where it is needed.
[3] Use the equation to convert the relevant numbers of moles to masses in grams or to volumes in dm3 (or cm3).
[4] Use arithmetic proportion to scale these quantities into those required by the question.
Calculations involving volumes usually involve simpler arithmetic than those involving masses, so they are often easier, despite what you might think!
Worked example 1
Question
Calculate the volume of nitrogen dioxide, measured at room temperature and pressure, formed when 10.0 g of hydrated calcium nitrate,
Ca(NO3)2.4H2O, are completely decomposed by heating.
Answer
Step 1
A 'skeleton equation' is all that’s needed here, because although calcium oxide, oxygen and water vapour are also formed (see Nuffield Chemistry Students’ Book Topic 4, page 91) they are not relevant in this question.
Ca(NO3)2.4H2O - - - > 2NO2(g)
Notice that we use a dotted arrow, because it’s only a partial (or skeleton) equation.
Notice too that we need a '2' in front of the ‘NO2’ in order to balance the number of nitrogen atoms on both sides of the equation.
Step 2
Because the mass of calcium nitrate is mentioned, its molar mass is of relevance:
Mr[Ca(NO3)2.4H2O] = 40 + (2 x 14) + (6 x 16) + (4 x 18) = 236 g mol-1
Since it is the volume of nitrogen dioxide asked for we need not waste our time working out its molar mass, we simply need to realize that one mole of this gas occupies a volume of 24 dm3 (as does one mole of any gas) at room temperature and pressure (r.t.p.)
Step 3
The equation tells us that one mole of hydrated calcium nitrate forms two moles of nitrogen dioxide gas.
That is 236 g of hydrated calcium nitrate form (2 x 24 =) 48 dm3 of nitrogen dioxide at room temperature and pressure.
Step 4
Look at the question again.
We are concerned with 10.0 g of hydrated calcium nitrate (not 236 g),
so the volume of nitrogen dioxide formed (measured at room temperature and pressure) = (10.0/236) x 24 = 1.016 dm3 (1.02 dm3 to 3 significant figures)
Worked example 2
Question
What is the volume of 1.0 g of methane, measured at room temperature and pressure?
Answer
Step 1
You can see straightaway that no equation is relevant here, because there’s no reaction involved! However, we do need to know the formula for methane, CH4
Step 2
Since the mass of methane is mentioned, we must work out its molar mass:
Mr[CH4(g)] = 12 + (4 x 1) = 16 g mol-1
Since the volume of methane is being asked for we need to realize that one mole of this gas occupies a volume of 24 dm3 at room temperature and pressure.
Step 3
Combining the two ideas in step 2 we can say that 16 g of methane (1 mole) has a volume of 24 dm3 at room temperature and pressure.
Step 4
Look at the question again.
It refers to 1.0 g of methane (not 16 g),
so 1.0 g of methane must have a volume of (1.0/16) x 24 = 1.5 dm3 at room temperature and pressure.
To test your understanding of these types of questions, you will find some examples in recent unit test 1 papers.
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updated: 12 January 2007
