GCSE Chemistry Calculations Part 14 Water of crystallisation, % yield and % purity at Doc Brown's Chemistry Clinic. GCE-AS-A2 KS4 IGCSE IB K12 for notes, examples and self-assessment quizzes suitable for all teaching, learning and revising styles docb4

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14. Other GCSE chemical calculations

* % purity of a product * % reaction yield * atom economy * dilution of solutions *

* water of crystallisation * how much of a reactant is needed? *

14.1 Percentage purity

Purity is very important e.g. for analytical standards in laboratories or pharmaceutical products where impurities could have dangerous side effects in a drug or medicine. However in any chemical process it is almost impossible to get 100.00% purity and so samples are always analysed in industry to monitor the quality of the product.
% purity is the percentage of the material which is the actually desired chemical in a sample of it.
Example 14.1.1
- A 12g sample of a crystallised pharmaceutical product was found to contain 11.57g of the active drug.
- Calculate the % purity of the sample of the drug.
- % purity = actual amount of desired material x 100 / total amount of material
- % purity = 11.57 x 100 / 12 = 96.4% (to 1dp)
Example 14.1.2
- Sodium chloride was prepared by neutralising sodium hydroxide solution with dilute hydrochloric acid. The solution was evaporated to crystallise the salt.
- The salt is required to be completely anhydrous, that is, not containing any water.
- The prepared salt was analysed for water by heating a sample in an oven at 110^oC to measure the evaporation of any residual water.
- The following results were obtained and from them calculate the % purity of the salt.
- Mass of evaporating dish empty = 51.32g.
- Mass of impure salt + dish = 56.47g
- Mass of dish + salt after heating = 56.15g
- Therefore the mass of original salt = 56.47 - 51.32 = 5.15g
- and the mass of pure salt remaining = 56.15 - 51.32 = 4.83g
- % salt purity = 4.83 x 100 / 5.15 = 93.8% (to 1dp)
AS example of purity-titration calculation based on an aspirin titration. This sort of analysis is called an 'assay'.

14.2a Percentage yield

The % yield of a reaction is the percentage of the product obtained compared to the theoretical maximum as calculated from the balanced equation.
In carrying out a chemical preparation, the aim is to work carefully and recover as much of the desired reaction product as you can, and as pure as is possible and practicable.
However in any chemical process it is almost impossible to get 100% of the product because of several reasons:
1. the reaction might not be 100% completed because it is reversible reaction and an equilibrium is established.
2. You always get losses of product as it is separated from the reaction mixture by filtration, distillation, crystallisation or whatever method is required.
3. Some of the reactants may react in another way to give a different product to the one you want.
4. The aim is to work carefully and recover as much of the desired reaction product, and as pure as is possible and practicable
% yield = actual amount of desired chemical obtained x 100 / maximum theoretical amount formed
Example 14.2a.1
- Magnesium metal dissolves in hydrochloric acid to form magnesium chloride.
- Mg_(s) + 2HCl_(aq) ==> MgCl_2(aq) + H_2(g)
- Atomic masses : Mg = 24 and Cl = 35.5, and formula mass MgCl₂ = 24 + (2 x 35.5) = 95
- (a) What is the maximum theoretical mass of magnesium chloride which can be made from 12g of magnesium?
  - Reacting mass ratio calculation from the balanced equation:
  - 1 Mg ==> 1 MgCl₂, so 24g ==> 95g or 12g ==> 47.5g MgCl₂
- (b) If only 47.0g of purified magnesium chloride was obtained after crystallising the salt from the solution, what is the % yield of the salt preparation?
  - % yield = actual amount obtained x 100 / maximum theoretical amount possible
  - % yield = 47.0 x 100 / 47.5 = 98.9% (to 1dp)
Example 14.2a.2
- 2.8g of iron was heated with excess sulphur to form iron sulphide.
- Fe + S ==> FeS
- The excess sulphur was dissolved in a solvent and the iron sulphide filtered off, washed with clean solvent and dried.
- If 4.1g of purified iron sulphide was finally obtained, what was the % yield of the reaction?
- 1st a reacting mass calculation of the maximum amount of FeS that can be formed:
  - Relative atomic/formula masses: Fe =56, FeS = 56 + 32 = 88
  - This means 56g Fe ==> 88g FeS, or by ratio, 2.8g Fe ==> 4.4g FeS
  - because 2.8 is 1/20th of 56, so theoretically you can get 1/20th of 88g of FeS.
- 2nd the % yield calculation itself.
  - % yield = actual amount obtained x 100 / maximum theoretical amount possible
  - % yield = 4.1 x 100 / 4.4 = 93.2% (to 1dp)
Example 14.2a.3
- (a) Theoretically how much iron can be obtained from 1000 tonne of pure haematite ore, formula Fe₂O₃ in a blast furnace?
- If we assume the iron(III) oxide ore (haematite) is reduced by carbon monoxide, the equation is:
- Fe₂O_3(s) + 3CO_(g) ==> 2Fe_(l) + 3CO_2(g)
- (atomic masses: Fe = 56, O = 16)
- For every Fe₂O₃ ==> 2Fe can be extracted, formula mass of ore = (2 x 56) + (3 x 56) = 160
- Therefore reacting mass ratio is: 160 ==> 112 (from 2 x 56)
- so, solving the ratio, 1000 ==> 112 / 160 = 700 tonne copper = max. can be extracted

(b) If in reality, only 670 tonne of iron is produced what is the % yield of the overall blast furnace process?
% yield = actual yield x 100 / theoretical yield
% yield = 670 x 100 / 700 = 95.7%
In other words, 4.3% of the iron is lost in waste e.g. in the slag.

14.2b Atom economy

The atom economy of a reaction is a theoretical measure of the amount of starting materials that ends up as 'desired' reaction product. The greater the atom economy of a reaction, the more 'efficient' or 'economic' it is likely to be, though this is a gross simplification when complex and costly chemical synthesis are looked at. It can be defined numerically in words in several ways, all of which amount to the same theoretical % number!

Atom economy =	mass of atoms desired product	mass of useful product	total M_r of useful product
	------------------------------------- x 100 =	---------------------------- x 100 =	-------------------------------- x 100
	total mass of reactant atoms	mass of all products	total M_r of all products

Example 14.2b.1

This is illustrated by using the blast furnace reaction from example 14.2a.3 above.

Fe₂O_3(s) + 3CO_(g) ===> 2Fe_(l) + 3CO_2(g)

Using the atomic masses of Fe = 56, C = 12, O = 16, we can calculate the atom economy for extracting iron.

the reaction equation can be expressed in terms of theoretical reacting mass units

[(2 x 56) + (3 x 16)] + [3 x (12 + 16)] ===> [2 x 56] + [3 x (12 + 16 + 16)]

[160 of Fe₂O₃] + [84 of CO] ===> [112 of Fe] + [132 of CO₂]

so there are a total of 112 mass units of the useful/desired product iron, Fe

out of a total mass of reactants or products of 112 + 132 = 244.

Therefore the atom economy = 112 x 100 / 244 = 45.9%

Note: It doesn't matter whether you use the total mass of reactants or the total mass products in the calculations, they are the same from the law of conservation of mass.

14.3 Dilution calculations

volumetric apparatus

It is important to know how to accurately dilute a more concentrated solution to a specified solution of lower concentration. It involves a bit of logic using ratios of volumes. The diagram above illustrates some of the apparatus that might be used when dealing with solutions.
Example 14.3.1
A purchased standard solution of sodium hydroxide had a concentration of 1.0 mol/dm³. How would you prepare 100 cm³ of a 0.1 mol/dm³ solution to do a titration of an acid?
- The required concentration is 1/10th of the original solution.
- To make 1dm³ (1000 cm³) of the diluted solution you would take 100 cm³ of the original solution and mix with 900 cm³ of water.
- The total volume is 1dm³ but only 1/10th as much sodium hydroxide in this diluted solution, so the concentration is 1/10th, 0.1 mol/dm³.
- To make only 100 cm³ of the diluted solution you would dilute 10cm³ by mixing it with 90 cm³ of water.
- How to do this in practice is described at the end of Example 14.3.2 below and a variety of accurate/inaccurate apparatus is illustrated above.
Example 14.3.2
- Given a stock solution of sodium chloride of 2.0 mol/dm³, how would you prepare 250cm³ of a 0.5 mol/dm³ solution?
- The required 0.5 mol/dm³ concentration is 1/4 of the original concentration of 2.0 mol/dm³.
- To make 1dm³ (1000 cm³) of a 0.5 mol/dm3 solution you would take 250 cm³ of the stock solution and add 750 cm³ of water.
- Therefore to make only 250 cm³ of solution you would mix 1/4 of the above quantities i.e. mix 62.5 cm³ of the stock solution plus 187.5 cm³ of pure water.
- This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker.
- It can be done much more accurately by using a burette or a pipette to measure out the stock solution directly into a 250 cm³ graduated-volumetric flask.
- Topping up the flask to the calibration mark (meniscus should rest on it). Then putting on the stopper and thoroughly mixing it by carefully shaking the flask holding the stopper on at the same time!
14.4 Water of crystallisation
Example 14.4.1: Calculate the % of water in hydrated magnesium sulphate MgSO₄.7H₂O
- Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1
- relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246
- 7 x 18 = 126 is the mass of water
- so % water = 126 x 100 / 246 = 51.2%
Example 14.4.2 The % water of crystallisation and the formula of the salt are calculated as follows:
- Suppose 6.25g of blue hydrated copper(II) sulphate, CuSO₄.xH₂O, (x unknown) was gently heated in a crucible until the mass remaining was 4.00g. This is the white anhydrous copper(II) sulphate.
- The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.25-4.00 = 2.25g
- The % water of crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%
- [ Ar's Cu=64, S=32, O=16, H=1 ]
- The mass ratio of CuSO₄ : H₂O is 4.00 : 2.25
- To convert from mass ratio to mole ratio, you divide by the molecular mass of each 'species'
- CuSO₄ = 64 + 32 + (4x18) = 160 and H₂O = 1+1+16 = 18
- The mole ratio of CuSO₄ : H₂O is 4.00/160 : 2.25/18
- which is 0.025 : 0.125 or 1 : 5, so the formula of the hydrated salt is CuSO₄.5H₂O
14.5 Calculation of quantities required for a reaction, % yield and atom economy
Suppose you want to make 50g of the blue hydrated copper(II) sulphate crystals
- All of section 14.5 is based on this quantity of 50g of the familiar blue crystals. The preparation is briefly described in the GCSE Acids, Bases and Salts Notes.
- Your reactants are dilute sulphuric acid and the black solid copper(II) oxide.
- You can use copper(II) carbonate, but this is not a pure simple compound and the predictive nature of the calculations will not be as good.
- copper(II) oxide + sulphuric acid ==> copper(II) sulphate + water
- (i) CuO_(s) + H₂SO_4(aq) ==> CuSO_4(aq)+ H₂O_(l)

How much copper(II) oxide is needed?
- A 'non-moles' calculation first of all, involving a reacting mass calculation.
- The crucial change overall is CuO ==> CuSO₄.5H₂O
- (Note: In reacting mass calculations you can often ignore other reactant/product masses)
- Atomic masses: Cu = 64, S = 32, H = 1, O = 16
- Formula masses are for: CuO = 64 + 16 = 80, CuSO4 = 64 + 32 + (4 x 16) = 160
- and CuSO₄.5H₂O = 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250
- The crucial reacting mass ratio is: 80 ==> 250 since formula ratio is 1:1 in the equation.
- Therefore, theoretically, to make 50g of the crystals (¹/₅th of 250),
- you need ¹/₅th of 80g of copper(II) oxide,
- and 80/5 = 16g of copper(II) oxide is required.
- However, in reality, things are not so simple because the method involves adding excess copper(II) oxide to the dilute sulphuric acid. (see salt preparation method (b))
- So in practice you would need to take more of the CuO to get anything like 50g of the salt crystals.
- There is another way to calculate the quantities required based on the acid.
How much dilute sulphuric acid (of concentration 1 mol dm^-3) is required?
- Mol = mass in g / formula mass,
- so moles of CuSO₄.5H₂O required = 50/250 = 0.2 mol (see basics of moles)
- Therefore 0.2 mol of H₂SO₄ is required (¹/₅th mol), since the mole ratio CuO : H₂SO₄ is 1 : 1 in the equation.
- 1dm³ (1000 cm³) of a 1 mol dm^-3 solution of contains 1 mole (by definition - see molarity page)
- Therefore 1/5th of 1dm³ is required, so 200 cm³ of 1 mol dm^-3 dilute sulphuric acid is required,
- or 100 cm³ of 2M dilute sulphuric acid. (2M is old fashioned notation for 2 mol dm^-3 as seen on many laboratory bottles!)
- You would then add copper(II) oxide in small amounts until no more dissolves in the warm-hot acid and the excess black powder is filtered off. There is no need to weigh out an exact amount of copper oxide.
Suppose after carrying out the preparation you finally crystallise 34g of pure the blue crystals of CuSO₄.5H₂O and weigh the product when dry.
What is the 'atom economy' of the preparation? (you need to refer to equations (i)/(ii) at the start of section 14.5
- Atom economy = useful theoretical products x 100/mass of all reactants
- based on equation (i) Atom economy = mass CuSO₄ x 100 / (mass CuO + H₂SO₄)
- = 160 x 100 / (80 + 98) = 16000/178 = 89.9%
- based on equation (ii) the atom economy is 100% if you include water as a 'reactant', can you see why?
- What is the % yield? i.e. comparing what you actually get with the maximum possible, i.e. a 'reality check'!
  - % yield = mass of product obtained x 100 / theoretical mass from the equation
  - % yield = 34 x 100 / 50 = 68%

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docb4_73calcs14other updated Jan 17th 2008