Section 4: Chemical CalculationsTopic Page Numbers

Relative Atomic Mass 2

Relative Molecular Mass 4

Calculating the Formula of Simple Compounds 7

Calculating the Mass of Products/Reactants from an Equation 11 - 12

Percentage Yield 18

Atom Economy 23-24

Comparing Atom Economy for Methods of Titanium Extraction 27-28

1

BY THE END OF THIS TOPIC YOU SHOULD BE ABLE TO:

Describe relative atomic mass (Ar) and know that different elements have different massesCalculate the relative molecular (formula) mass (Mr) of a compound by using its formulaCalculate the formula of a simple compound from given dataBe able to calculate reacting masses of reactants or products using a balanced symbol equationDescribe and calculate the percentage yield of a reactionCalculate the atom economy of a reaction from a given equationAssess and compare the level of waste in chemical reactions using atom economy

2

Relative Atomic MassDifferent elements have different numbers of protons, neutrons and electrons.

The mass of an element comes from the neutrons and protons in its nucleus (an electron’s mass is negligible).

The relative atomic mass is the largest number in the element’s box on the periodic table.

It is usually at the top of the box.

It is often written as Ar. It tells us how heavy an element is compared to

other elements.

Because it is a way of comparing elements it doesn’t have any units.

3

Relative Atomic Mass Questions

Question 1:

Write down the Ar for the following elements:

Osmium LithiumSodium NitrogenPlatinum HeliumTitanium CarbonChlorine HydrogenArgon Sulphur

Question 2:

Circle the element that is heavier:

a) Carbon / Oxygenb) Sodium / Lithiumc) Gold / Copperd) Hydrogen / Heliume) Lead / Tin

Question 3:

Write down the elements with the following Ar values:

Ar Element

1121623

222

Relative Molecular Mass The relative molecular mass can be worked out

by adding up the relative atomic masses of the elements in the molecule.

4

It can also be written as Mr or relative formula mass

Example 1: Calculating the relative molecular mass of CH4

CH4 contains 1 x C and 4 x H atoms (remember the small number applies to the H only)

Ar for C = 12Ar for H = 1

So Mr = (1 x 12) + (4 x1) = 16

Calculate the Mr of the following substances:(the first 2 have been done for you)

(a) H2O = 1 + 1 + 16

= 18

(b) CO2 = 12 + 16 + 16 = 44

(c) NaCl =

(d) O2 = (e) H2 = (f) HCl =

(g) NaOH = (h) KOH = (i) MgO =

(j) CO = (k) C2H4 = (l) C2H6 =

(m)CH4 = (n) C4H10 = (o) C5H12 =

(p) C6H6 = (q) S8 = (r) Al2O3 =

Example 2: Calculating the relative molecular mass of Al(OH)3

Al(OH)3 contains 1 x Al, 3 x O and 3 x H (remember the small number applies to everything inside the bracket)

Ar for Al = 27Ar for O = 16Ar for H = 1

So Mr = (1 x 27) + (3 x 16) + (3 x 1)

5

= 78.

Now try these examples – the first one has been done for you.

(a) Ca(OH)2 = 40 + (16 + 1)2 = 74

(b) (NH4)2SO4 =

(b) (NH4)2Cr2O7 = (d) Fe(OH)2 =

(e) Fe(OH)3 = (f) (NH4)2S203 =

(g) (NH4)3PO4 = (h) C6H2CH3(NO2)3 =

6

Fill in this worksheet…

7

Calculating the Formula of Simple Compounds

You can use relative atomic masses (Ar) to calculate the formula of compounds when you know the mass of reactants you used.

Example: 3.2g of sulphur reacts with oxygen to produce 6.4g of sulphur oxide. What is the formula of the oxide?

Step 1 – Write down the masses of reactants:

S = 3.2g O = 6.4 - 3.2 = 3.2g

Step 2 – Divide Masses by Ar

3.2 = 0.1 3.2 = 0.2 32 16

Step 3 – Divide everything by the smallest number to get the ratio

0.1 = 1 0.2 = 2 0.1 0.1 Ratio S : O = 1:2

Step 4 - Write out the formula

* If you are given % just use the % as the mass in grams.

Answer these questions to calculate the formula of simple compounds:

Step Process 1 Write down masses of reactants*2 Divide by Ar3 Divide by smallest number to get

ratio4 Write down formula

8

SO2

Use this table as a prompt until you are confident with the method.

1. 10.2g of an oxide of aluminium contains 5.4g of aluminium. Calculate the formula of the oxide (i.e. empirical formula)

2. A compound contains 29.11% sodium, 40.51% sulphur and 30.38% oxygen by mass. Calculate its simplest formula (empirical formula)

3. 14.2g of a compound that contains 3 elements (sodium, sulphur and oxygen). The mass of sodium is 4.6g and sulphur is 3.2g of sulphur. The rest is oxygen. Calculate the mass of oxygen, and then calculate the simplest formula of the compound.

4. A compound contains 40.21% of potassium, 26.80% chromium and 32.99% oxygen by mass. Calculate its empirical formula

9

5. A sample of a compound containing 3 elements (potassium, chromium, oxygen) weighs 22.62g. This amount contains 6g of potassium and 8g of chromium and the rest is oxygen. Calculate the mass of oxygen and then calculate the simplest formula (empirical formula)

10

Calculating the Mass of Products from a Reaction Equation

You can use the reaction equation and information about relative atomic mass to calculate the mass of products you would expect to get.

Example N2 + 3H2 → 2NH3

If we have 7g of N2, how much NH3 can we make?

Step 1Write down the Mr or Ar underneath each chemical and multiply by any balancing number

N2 + 3H2 → 2NH3

28 3x2 = 6 2x17 = 34

This means that for every 28g of N2 we would expect to get 34g of NH3.

Step 2 Decide what fraction or multiple of the Mr the reacting amount is:By examination 7 is 28/4

Step 3 Work out how much product you would get

34 / 4 =

Excess means you have more than enough of that reactant so it won’t affect your calculationCalculating the Mass of Reactants from a

Reaction EquationThe previous method can also be used to work out the mass of reactants used in a chemical reaction.

Example

11

8.5g

The reaction between iron (Fe) and sulphur (S) to make iron (II) sulphide (FeS).

Fe + S FeS

How much Fe is needed to make 176g of FeS?

Step 1Write down the Mr or Ar underneath each chemical and multiply by any balancing number

Fe + S FeS56 32 88

This means that for every 88g of FeS made, we need 56g of Fe and 32g of S

Step 2 Decide what fraction or multiple of the Mr the reacting amount is:By examination: 176 is 2 x 88

Step 3 Work out how much reactant you would need

56 x 2 =

Q. How much S would be needed to make 176g of FeS?Now try these questions yourself. Remember to start by writing the Mr or Ar under each formula in the equation, and multiplying by the balancing number.5.6g of iron react with excess sulphur to make WHAT MASS of iron(II) sulphide?Fe + S FeS

12

112g

3.65g of HCl react with excess sodium hydroxide. What mass of water is formed?HCl + NaOH NaCl + H2O

8g of hydrogen are exploded in excess oxygen. What mass of water is formed?2H2 + O2 2H2O

If 3.2 g of oxygen is reacted with excess hydrogen, what mass of water is formed?2H2 + O2 2H2O

What mass of hydrogen and oxygen are needed to make exactly 3.6g of water (2 answers needed)2H2 + O2 2H2O

When sulphur (found in coal) burns it forms acid rain. What mass of sulphur dioxide is formed from 32 tonnes of sulphur and excess oxygen?S + O2 SO2

13

When a hydrocarbon burns in excess oxygen, 2 products are formed (water and carbon dioxide). What mass of the greenhouse gas (CO2) forms when 10g of heptane (C7H16), found in petrol, are burned in excess O2?C7H16 + 11O2 7CO2 + 8H2O

14

Extension QuestionsTry these questions…

123g of sulphur are burned in excess oxygen. What mass of sulphur dioxide forms?S + O2 SO2

29 tonnes of iron (III) oxide are reacted in the blast furnace with excess carbon monoxide. What mass of iron is produced?Fe2O3 + 3CO 2Fe + 3CO2

15

2.567g of pure sodium hydroxide are reacted with excess dilute sulphuric acid. What mass of water is formed?2NaOH + H2SO4 Na2SO4 + 2H2O

What mass of lead nitrate has to be thermally decomposed to give 50g of lead oxide?2Pb(NO3)2 2PbO + 4NO2 + O2

16

What mass of water is formed when 1kg (1000g) of octane (equivalent to just over a litre of petrol) is burned in excess oxygen?2 C8H18 + 25 O2 16 CO2 + 18 H2O

What mass of carbon dioxide is formed on a typical 25 mile car journey? (Assume the petrol consists of 2kg {2000g} of pure heptane and that half a gallon of it [2.25 litres] reacts with excess oxygen in the air).C7H16 + 11 O2 7 CO2 + 8 H2O

17

Percentage YieldIn most chemical reactions you do not get the maximum expected amount of product. This can be to do with the chemistry of the reaction or the way the experiment took place.

Example

Fe + S → FeS

56 g of iron reacts to give 80 g of iron sulphide

Step 1: Calculate the theoretical yield (the maximum amount of product we could expect) in the same way you did before.

Fe + S → FeS 56 32 88

So 56g of iron should give us 88 g of iron sulphide.

Step 2:Calculate the percentage yield:

= 80 x 100 = 90% 88

18

Look at your answers to the theoretical yield questions a few pages back and answer these related questions…3.65g of HCl react with excess sodium hydroxide. What is the % yield if 1.5g of water is formed?HCl + NaOH NaCl + H2O

8g of hydrogen are exploded in excess oxygen. What is the % yield if only 65g of water is formed?2H2 + O2 2H2O

If 3.2 g of oxygen is reacted with excess hydrogen to give 3.2g of water. What is the theoretical yield?2H2 + O2 2H2O

19

When sulphur (found in coal) burns it forms acid rain. What is the theoretical yield when 32 tonnes of sulphur are burned to give 63 tonnes of sulphur dioxide?S + O2 SO2

When a hydrocarbon burns in excess oxygen, 2 products are formed (water and carbon dioxide). What is the percentage yield when 10g of heptane, found in petrol, is burned in excess O2 to give 25g of the greenhouse gas (CO2)? C7H16 + 11O2 7CO2 + 8H2O

20

1 tonne of nitrogen is reacted with excess hydrogen to give 0.25 tonnes of ammonia (NH3). What is the percentage conversion?N2 + 3H2 2NH3

Try this typical GCSE question…

21

22

Atom Economy This is a measure of how successful a reaction has

been in converting raw materials into a useful product.

It looks similar to percentage yield, but it’s not the same thing.

If most of what you produce in a reaction is the desired product then there will, by definition, be very little waste.

That is, the “Atom Economy (Percentage)” will be high (just below 100).

In an ideal world, all industrial chemical processes should have an atom economy of 100% since this would mean there would be no waste (unwanted material).

In practice, much “unwanted” material that used to be dumped/buried/burned is now used.

For example, the waste slag from a Blast Furnace is nowadays turned into insulation blocks for new homes.

23

Atom Economy = Mass of useful product x100 Total mass of reactants used

Calculating Atom EconomyMost chemical reactions produce useful products and waste products. Atom economy allows us to calculate the percentage of reactants that are converted into useful products.

Example 1:What is the atom economy for making hydrogen by reacting coal with steam?

C(s) + 2H2O(g)    →    CO2(g) + 2H2(g)

Step 1: Work out the Mr and Ar values for the reactants and the useful products.

C(s) + 2H2O(g)    →    CO2(g) + 2H2(g)Mr / Ar 2(2+16) 2x2 12 36 4

Step 2: Work out the percentage of useful product using the following equation:

= 4 x 100 = 8.3% 36 + 12

So this reaction is not very economical and probably would not be used to make hydrogen industrially.

24

Atom Economy = Mass of useful product x100 Total mass of reactants used

Example 2:

Lithium hydroxide will react with nitric acid to produce lithium nitrate plus water. Suppose that we are trying to make lithium nitrate, and that the water we also make is the waste material. Firstly we write an equation…LiOH + HNO3 LiNO3 + H2O

Then we write the Mr (Relative Formula Masses) underneathLiOH + HNO3 LiNO3 + H2O 24 63 69 18

Atom economy = (69/87) x 100 = 79.3%

(If you are wondering where 87 comes from, it’s 24 + 63)Calculate the atom economy percentage for the Blast Furnace reaction. The desired product is iron and the waste product is carbon dioxide

Fe2O3 + 3CO 2Fe + 3CO2

25

Calculate the atom economy percentage for the Haber Process. The desired product is ammonia (NH3).

N2 + 3H2 2NH3

Calculate the atom economy percentage for the Oswald Process (the desired product is nitric acid).

NH3 + 2O2 HNO3 + H2O

26

Try this typical GCSE question…

27

Comparing the Atom Economy for 2 methods of Titanium (Ti) extraction

Titanium is a very useful metal. It is light (low density), strong, hard and resists

corrosion.

For these reasons it has many uses including in the aerospace industry & for artificial hip joints.

Although there is plenty of titanium in the Earth’s crust it is an expensive metal due to extraction costs and the low atom economy.

There are 2 methods for extracting Titanium – the traditional method and the modern electrolytic method.

Method 1 – The Traditional Method

The following reaction is used:

TiO2 + 2Cl2 + C + 4Na CO2 + Ti + 4NaCl

Calculate the atom economy for this reaction: (The desired product is Ti (CO2 & NaCl are waste products)

28

TiO2 + 2Cl2 + C + 4Na CO2 + Ti + 4NaCl

Do you agree that this is rather wasteful? Your answer should have been 14.7%. This means that 100 – 14.7 = 85.3% is potentially

waste material (or, at least, not very useful).

Method 2 – The Modern Electrolytic Method

More recently a better process has been developed using direct electrolysis.

This will hopefully enable titanium to be produced at half its current price, or even less.

You do not need to know details of the process, but it can be summarised as follows:

TiO2 Ti + O2

Notice that the only by-product is oxygen, which is, of course, very useful.

Can YOU now work out the atom economy percentage for this process?

It ought to be much more efficient than 14.7%.

29

TiO2 Ti + O2

Relevant Internet and School Network Links

In school, try these web links and click on the white arrow in the green square to view:

http://www.yenka.com/freecontent/item.action?quick=13i#http://www.yenka.com/freecontent/item.action?quick=13j#http://www.yenka.com/freecontent/item.action?quick=u0

School network Links:

SUBJECTS > SCIENCE > CHEMISTRY > ABSORB CHEMISTRY FOR GCSE > INTRODUCING MOLES

SUBJECTS > SCIENCE > CHEMISTRY > ABSORB CHEMISTRY FOR GCSE > MOLES IN CHEMICAL EQUATIONS

SUBJECTS > SCIENCE > FOOTPRINTS > MATERIALS & THEIR PROPERTIES > ELEMENTS, COMPOUNDS & MIXTURES

SUBJECTS > SCIENCE > CHEMISTRY > ABSORB CHEMISTRY FOR GCSE > EXTRACTION OF TITANIUM

SUBJECTS > SCIENCE > CHEMISTRY > ABSORB CHEMISTRY FOR GCSE > MOLES & CHEMICAL FORMULAE

Useful websites:

http://web.visionlearning.com/MW_calculator.shtmlhttp://www.tufts.edu/as/wright_center/fellows/george/georgepage3.htm

Please note – some of these websites include extension materials and useful concepts (such as “the mole”) which can be extremely useful but you do not necessarily need to know for your exam!

30