Chemical quantities (AS)

Read our general notes on Risk Assessment

Calculating reacting masses from balanced equations

There have been several questions to this web site asking for help on calculating reacting masses from balanced equations, and we thought that it would be useful to produce a tutorial taking you through the various steps and providing a couple of worked examples.
 
Note that this topic is part of the specification for AS Unit Test 1 (UT1) if you are taking the Nuffield exams.

Before we start, it should emphasised that steps [3] and [4] below represent only one way of arriving at the correct solution. There are alternative routes to the final answer after stage [2] has been reached, and we do not wish to confuse you if you are happy with the way you do things already. If, however, you find this type of calculation hard, try out our method below and see if it works for you!
 
Four steps to the answer
[1] Construct a balanced equation for the reaction. Sometimes a “skeleton equation” will do, but, if so, it must be balanced (see worked example 2).
[2] Work out (or look up) the molar masses of the relevant substances.
[3] Use the equation to convert the relevant numbers of moles to masses in grams.
[4] Use arithmetic proportion to scale these quantities into those required by the question.
 
Worked example 1
 
Question
Calculate the mass of magnesium carbonate which needs to be heated in order to produce 2.0 g of magnesium oxide.
 
Answer
Step 1
In this example the equation is an easy one:
MgCO₃ -> MgO + CO₂.
 
Try another tutorial if you find balancing equations hard.
 
Step 2
Notice from the question that it is the magnesium carbonate and the magnesium oxide which are of relevance, and not the carbon dioxide, so we shan’t be wasting our time working out the molar mass of CO₂.
M_r[MgCO₃(s)] = 24+12+(3x16) = 84 g mol^-1 and
M_r[MgO(s)] = 24+16 = 40 g mol^-1
 
Step 3
The equation tells us that one mole of magnesium carbonate forms one mole of magnesium oxide. That is 84 g of magnesium carbonate decompose to give 40 g of magnesium oxide
 
Step 4
Now the question asks about the formation of 2.0 g of magnesium oxide (not 40 g),
so the mass of magnesium carbonate needed = (2/40) x 84 = 4.2 g
 
Worked Example 2
 
Question
Calculate the mass of hydrated sodium sulphate, Na₂SO₄.10H₂O, which could theoretically be made starting from 4.0 g of sodium hydroxide, by neutralizing it with dilute sulphuric acid.
 
Answer
Step 1
You can see in this case that the equation is going to more difficult to balance, because it involves the formation of a hydrated salt, Na₂SO₄.10H₂O.
 
But this really isn’t a problem! We needn’t worry in this question about the sulphuric acid and water required, since they aren’t a requirement - i.e. they’re not relevant here. All we need to concentrate on are the sodium hydroxide (formula: NaOH) and the hydrated sodium sulphate (formula given in question: Na₂SO₄.10H₂O), and so a balanced “skeleton equation” will do.
2NaOH - - - > Na₂SO₄.10H₂O
 
Notice that we use a dotted arrow, because it’s only a partial (or skeleton) equation.
 
Notice also that there must be 2NaOH, so that the number of sodiums balance on both sides. It is true that sulphur and extra hydrogen and oxygen atoms have appeared on the right-hand side of the skeleton equation, but these will be provided by the H₂SO₄ and the H₂O in the dilute sulphuric acid, and remember that we’re not going to worry ourselves about them since they’re not relevant in this particular question!
 
Step 2
M_r [NaOH] = 23+16+1 = 40 g mol^-1and M_r [Na₂SO₄.10H₂O] = (2x23)+32+(4x16)+(10x18) = 322 g mol^-1
 
Step 3
The equation tells us that two moles of sodium hydroxide form one mole of hydrated sodium sulphate.
i.e. (2 x 40 =) 80 g of sodium hydroxide form 322 g of hydrated sodium sulphate
 
Step 4
Now look at the question again. It refers to 4.0 g of sodium hydroxide (not 80 g),
so the mass of hydrated sodium sulphate formed = (4.0/80) x 322 = 16.1 g (16 g to 2 significant figures)
 
Incidentally, although 16 g of this hydrated salt could be made (and is the correct answer to this question) you should realize that in practice the yield obtained will be far lower. You should refer to pages 12-14 of the Nuffield Chemistry Students’ Book where practical details of forming pure, dry samples of salts are given, and work out for yourself why theoretical yields of salts are never obtained in practice. It is important that you should do this. Apart from anything else it is a popular question in a unit test 1 paper!
 
You should now try to tackle some of the Review and Examination questions on pages 21-25 (Nuffield Chemistry Students’ Book) to give you further practice. You will also find other questions to tackle in recent unit test 1 papers.
 

 

---

Rate this page or react
Share your views on this page, 112 ratings so far 4 , rated at

---

updated: 03 November 2003