14. Other GCSE chemical calculations

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

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

14.1 Percentage purity of a chemical reaction product

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

o A 12.00g sample of a crystallised pharmaceutical product was found to contain 11.57g of the active drug.

o Calculate the % purity of the sample of the drug.

o % purity = actual amount of desired material x 100 / total amount of material

o % purity = 11.57 x 100 / 12 = 96.4% (to 1dp)

Example 14.1.2

o Sodium chloride was prepared by neutralising sodium hydroxide solution with dilute hydrochloric acid. The solution was

evaporated to crystallise the salt.

o The salt is required to be completely anhydrous, that is, not containing any water.

o The prepared salt was analysed for water by heating a sample in an oven at 110oC to measure the evaporation of any residual

water.

o The following results were obtained and from them calculate the % purity of the salt.

o Mass of evaporating dish empty = 51.32g.

o Mass of impure salt + dish = 56.47g

o Mass of dish + salt after heating = 56.15g

o Therefore the mass of original salt = 56.47 - 51.32 = 5.15g

o and the mass of pure salt remaining = 56.15 - 51.32 = 4.83g

o % salt purity = 4.83 x 100 / 5.15 = 93.8% (to 1dp)

  Advanced AS Q17 & Q20 examples of purity-titration calculation based on   an organic acid titrations

o which are like the assay calculation sketched out below ...

o

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14.2a Percentage yield of the product of a reaction

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

o Magnesium metal dissolves in hydrochloric acid to form magnesium chloride.

o Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g)

o Atomic masses : Mg = 24 and Cl = 35.5, and formula mass MgCl2 = 24 + (2 x 35.5) = 95

o (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

o (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

o 2.8g of iron was heated with excess sulphur to form iron sulphide.

o Fe + S ==> FeS

o The excess sulphur was dissolved in a solvent and the iron sulphide filtered off, washed with clean solvent and dried.

o If 4.1g of purified iron sulphide was finally obtained, what was the % yield of the reaction?

o 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.

o 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

o (a) Theoretically how much iron can be obtained from 1000 tonne of pure haematite ore, formula Fe2O3 in a blast furnace?

o If we assume the iron(III) oxide ore (haematite) is reduced by carbon monoxide, the equation is:

o Fe2O3(s) + 3CO(g) ==> 2Fe(l) + 3CO2(g)

o (atomic masses: Fe = 56, O = 16)

o For every Fe2O3 ==> 2Fe can be extracted, formula mass of ore = (2 x 56) + (3 x 56) = 160

o Therefore reacting mass ratio is: 160 ==> 112 (from 2 x 56)

o so, solving the ratio, 1000 ==> 112 / 160 = 700 tonne copper = max. can be extracted

o (b) If in reality, only 670 tonne of iron is produced what is the % yield of the overall blast furnace process?

o % yield = actual yield x 100 / theoretical yield

o % yield = 670 x 100 / 700 = 95.7%

o In other words, 4.3% of the iron is lost in waste e.g. in the slag.

Example 14.2a.4

o Given the atomic masses: Mg = 24 and O = 16,

o and the reaction between magnesium to form magnesium oxide is given by the symbol equation

o 2Mg(s) + O2(g) ==> 2MgO(s)

o (a) What mass of magnesium oxide can be made from 1g of magnesium?

2Mg ==> 2MgO

in terms of reacting masses (2 x 24) ==> {2 x (24 +16)}

so 48g Mg ==> 80g MgO (or 24g ==> 40g, its all the same)

therefore solving the ratio

1g Mg ==> w g MgO, using the ratio 48 : 80

w = 1 x 80 / 48 = 1.67g MgO

o (b) Suppose the % yield in the reaction is 80%. That is only 80% of the magnesium oxide formed is actually recovered as useful

product. How much magnesium needs to be burned to make 30g of magnesium oxide?

This is a bit tricky and needs to done in two stages and can be set out in several ways.

Now 48g Mg ==> 80g MgO (or any ratio mentioned above)

so y g Mg ==> 30g MgO

y = 30 x 48 / 80 = 18g Mg

BUT you only get back 80% of the MgO formed,

so therefore you need to take more of the magnesium than theoretically calculated above.

Therefore for practical purposes you need to take, NOT 18g Mg, BUT ...

... since you only get 80/100 ths of the product ...

... you need to use 100/80 ths of the reactants in the first place

therefore Mg needed = 18g x 100 / 80 = 22.5g Mg

CHECK: reacting mass calculation + % yield calculation CHECK:

22.5 Mg ==> z MgO, z = 22.5 x 80 / 48 = 37.5g MgO,

but you only get 80% of this,

so you actually get 37.5 x 80 / 100 = 30g

This means in principle that if you only get x% yield ...

... you need to take 100/x quantities of reactants to compensate for the losses.

Below is an example of a more advanced level yield calculation

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14.2b The Atom economy of a chemical reaction

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 of desired product mass of useful product total Mr of useful product

------------------------------------------------ x 100 = ----------------------------------- x 100 = ------------------------------------- x 100

total mass of reactant atoms mass of all products total Mr of all products

Example 14.2b.1

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

Fe2O3(s) + 3CO(g) ===> 2Fe(l) + 3CO2(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.

See also    Example 6.6 in chemical calculations section 6

 

14.3 Dilution of solutions calculations

calculating dilutions - volumes involved etc.

It is important to study Calculations Part 11 Introducing Molarity, volumes and the concentration of solutions before tackling

this section and note the triangle of relationships you need to know!

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?

o The required concentration is 1/10th of the original solution.

o 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.

o 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.

o To make only 100 cm3 of the diluted solution you would dilute 10cm3 by mixing it with 90 cm3 of water.

o How to do this in practice is described at the end of Example 14.3.2 below and a variety of accurate/'less inaccurate' apparatus is

illustrated above.

Example 14.3.2

o Given a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?

o The required 0.5 mol/dm3 concentration is 1/4 of the original concentration of 2.0 mol/dm3.

o 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.

o 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.

o This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker.

o 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.

o 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!

Example 14.3.3

 In the analytical laboratory of a pharmaceutical company a laboratory assistant

was asked to make 250 cm3 of a 2.0 x 10-2 mol dm-3 (0.02M) solution of paracetamol (C8H9NO2).

o (a) How much paracetamol should the laboratory assistant weigh out to make up the solution? Atomic masses: C = 12, H

= 1, N = 14, O = 16

method (i): Mr(paracetamol) = (8 x 12) + (9 x 1) + (1 x 14) + (2 x 16) = 151

1000 cm3 of 1.0 molar solution needs 151g

1000 cm3 of 2.0 x 10-2 molar solution needs 151 x 2.0 x 10-2/1 = 3.02g

(this is just scaling down the ratio from 151g : 1.0 molar)

Therefore to make 250 cm3 of the solution you need 3.02 x 250/1000 = 0.755 g

o method (ii): Mr(paracetamol) = 151

moles = molarity x volume in dm3

mol paracetamol required = 2.0 x 10-2 x 250/1000 = 5.0 x 10-3 (0.005)

mass = mol x Mr = 5.0 x 10-3 x 151 = 0.755 g

o (b) Using the 2.0 x 10-2 molar stock solution, what volume of it should be added to a 100cm3 volumetric

flask to make 100 cm3 of a 5.0 x 10-3 mol dm-3 (0.005M) solution?

The ratio of the two molarities is stock/diluted = 2.0 x 10-2/5.0 x 10-3 = 4.0 or a dilution factor of 1/4 (0.02/0.005).

Therefore 25 cm3 (1/4 of 100) of the 2.0 x 10-2 molar solution is added to the 100 cm3 volumetric flask prior to making it up

to 100 cm3 with pure water to give the 5.0 x 10-3 mol dm-3 (0.005M) solution.

There are more questions involving molarity in    section 7. introducing molarity  and    section 12. on dilution

Example 14.3.4

o You are given a stock solution of concentrated ammonia with a concentration of 17.9 mol dm-3 (conc. ammonia! ~18M!)

o (a) What volume of the conc. ammonia is needed to make up 1dm3 of 1.0 molar ammonia solution?

Method (i) using simple ratio argument.

The conc. ammonia must be diluted by a factor of 1.0/17.9 to give a 1.0 molar solution.

Therefore you need 1.0/17.9 x 1000 cm3 = 55.9 cm3 of the conc. ammonia.

If the 55.9 cm3 of conc. ammonia is diluted to 1000 cm3 (1 dm3) you will have a 1.0 mol dm-3 (1M) solution.

Method (ii) using molar concentration equation - a much better method that suits any kind of dilution calculation

involving molarity.

molarity = mol / volume (dm3), therefore mol = molarity x volume in dm3

Therefore you need 1.0 x 1.0 = 1.0 moles of ammonia to make 1 dm3 of 1.5M dilute ammonia.

Volume = mol / molarity

Volume of conc. ammonia needed = 1.0 / 17.9 = 0.0559 dm3 (55.9 cm3) of the conc. ammonia is required,

and, if this is diluted to 1 dm3, it will give you a 1.0 mol dm-3 dilute ammonia solution.

o (b) What volume of conc. ammonia is needed to make 5 dm3 of a 1.5 molar solution?

molarity = mol / volume (dm3), therefore mol = molarity x volume in dm3

Therefore you need 1.5 x 5 = 7.5 moles of ammonia to make 5 dm3 of 1.5M dilute ammonia.

Volume (of conc. ammonia needed) = mol / molarity

Volume of conc. ammonia needed = 7.5 / 17.9 = 0.419 dm3 (419 cm3) of the conc. ammonia is required,

and, if this is diluted to 5 dm3, it will give you a 1.5 mol dm-3 dilute ammonia solution.

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14.4 Water of crystallisation in a crystallised salt

Example 14.4.1: Calculate the % of water in hydrated magnesium sulphate MgSO4.7H2O salt crystals

o Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1

o relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246

o 7 x 18 = 126 is the mass of water

o 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:

o 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.

o The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.25-4.00 = 2.25g

o The % water of crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%

o [ Ar's Cu=64, S=32, O=16, H=1 ]

o The mass ratio of CuSO4 : H2O is 4.00 : 2.25

o To convert from mass ratio to mole ratio, you divide by the molecular mass of each 'species'

o CuSO4 = 64 + 32 + (4x18) = 160 and H2O = 1+1+16 = 18

o The mole ratio of CuSO4 : H2O is 4.00/160 : 2.25/18

o 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 chemical reaction, % yield and atom economy

Suppose you want to make 50g of the blue hydrated copper(II) sulphate crystals

o 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.

o Your reactants are dilute sulphuric acid and the black solid copper(II) oxide.

o 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.

o copper(II) oxide + sulphuric acid ==> copper(II) sulphate + water

o (i) CuO(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l)

o on crystallisation you get the blue hydrated crystals of formula CuSO4.5H2O

o so strictly speaking, after evaporation-crystallisation the equation is

o (ii) CuO(s) + H2SO4(aq) + 4H2O(l) ==> CuSO4.5H2O(s)

o How much copper(II) oxide is needed?

o A 'non-moles' calculation first of all, involving a reacting mass calculation.

o The crucial change overall is CuO ==> CuSO4.5H2O

o (Note: In reacting mass calculations you can often ignore other reactant/product masses)

o Atomic masses: Cu = 64, S = 32, H = 1, O = 16

o Formula masses are for: CuO = 64 + 16 = 80, CuSO4 = 64 + 32 + (4 x 16) = 160

o and CuSO4.5H2O = 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250

o The crucial reacting mass ratio is: 80 ==> 250 since formula ratio is 1:1 in the equation.

o Therefore, theoretically, to make 50g of the crystals (1/5th of 250),

o you need 1/5th of 80g of copper(II) oxide,

o and 80/5 = 16g of copper(II) oxide is required.

o 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))

o So in practice you would need to take more of the CuO to get anything like 50g of the salt crystals.

o There is another way to calculate the quantities required based on the acid.

o 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.

o Suppose after carrying out the preparation you finally crystallise 34g of pure the blue crystals of CuSO4.5H2O and weigh

the product when dry.

o 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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