Calculations in ChemistryIGCSE Chemistry

Mole concept

Relative atomic mass and Relative molecular massWhy atomic mass or molecular

mass has no unit?Atoms and molecules are too

small to weigh.Relative atomic mass and

Relative molecular mass are comparative values.

This shows how many times an atom or a molecule heavier than th of a Carbon atom.

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The mole….Atomic mass or molecular mass with g

after the number is called a mole.Example: Atomic mass of Oxygen is 16.16g of oxygen(O) is 1 mole of oxygen atoms.Molecular mass of Oxygen O2 is 32.32g of Oxygen (O2) is 1 mole of oxygen

molecules.Molecular mass of water is 18.18g of water is called 1 mole of water.

Number of moles =

Mass (g) =Number of moles X Molar mass

Molarmass

gmass )(

Questions1. How many moles are there in 4g

of Sodium hydroxide?Number of moles =

= 0.10moles2. Find the number of moles present

in 1.28g of oxygen molecules.Number of moles =

= 0.04 moles

g

g

40

4

g

g

32

28.1

3. What is the mass of 0.01 moles of sodium hydroxide?

Mass = 0.01 X 40g = 0.40g

4. How many grams 1.5 moles of sulphuric acid weighs?

Mass = 1.5 X 98g = 147g

Try the following questions:1. What is the mass of 0.02 moles of carbon dioxide gas?0.88g2. How many moles are present in 1g of calcium

carbonate,CaCO3?0.01moles3. Find the number of moles present in 3g of potassium

hydroxide, KOH.0.054moles4. Calculate the mass of 100 moles of copper oxide, CuO8000g5. 5. 2g of sodium hydroxide,NaOH or 3g of potassium

hydroxide, KOH contains more number of moles? Show your working clearly.

NaOH = 0.13 moles KOH = 0.054 moles. So NaOH

Avogadro constantOne mole of any substance

contains 6.02 X 10 atoms or molecules.6.02 X 10 is a constant and is

called Avogadro constant or Avogadro Number.

In other words, Avogadro Number is the number of atoms or number of molecules present in ONE mole of a substance.

Avogadro constant is used to find number of particles or number of moles.

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Number of moles = Number of molecules Avogadro constant

Number of molecules = Number of moles X Avogadro constant

CalculationsRemember always change mass

or number of molecules into moles.

Examples:How many molecules of Carbon

dioxide are there in 0.44g of the gas?

Answer:Number of moles = 0.44g/44g = 0.01molesNumber of molecules = 0.01 X 6.02 X

10 = 6.02 X 10

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Calculate how many oxygen atoms are present in 1.96g of sulphuric acid?Answer:

Moles of sulphuric acid = 1.96g/98g = 0.02 molesNumber of molecules of sulphuric

acid = 0.02 X 6.02 X 10 =1.20 X 10Each sulphuric acid contains 4

oxygen atoms.So number of oxygen atoms 0.02

moles of H2SO4 is= 4 X 1.20 X 10 = 4.80 X

10

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Mole ratioWhen chemical reactions take place, the

reactants and products are in a simple ratio (mole ratio)

For example, Zinc and hydrochloric acid react according to the following reaction:

Zinc + Hydrochloric acid Zinc chloride + Hydrogen gas

Zn + 2HCl ZnCl2 + H2

It means, zinc, Hydrochloric acid, Zinc chloride and Hydrogen moles are in the ratio 1:2:1:1

One mole of Zinc reacts with 2 moles of Hydrochloric acid to produce, One mole of Zinc chloride and 1 mole of Hydrogen gas.

The reactants and products are always in a ratio by moles and not by mass.

Find how many grams of copper oxide can be made from 3.2g of copper.

Answer:2Cu + O2 2CuOThe balanced equation shows that two moles of

copper reacts with 1 mole of oxygen to form 2 moles of copper(II) oxide.

So change 3.2g to molesMoles of copper = 3.2g/64g =0.05molesCopper :Copper oxide is 1:1 ratio. So moles of copper(II) oxide = 0.05molesMass (g) = 0.05 X 80g = 4g

When Posphorus burns in oxygen, Phosporus(V)oxide is produced. In an experiment, 8g of phsphorus(V)oxide is produced. Find the mass of Phosphorus reacted with oxygen.Answer:Equation: 4P + 5O2 2P2O5

Mole ratio is 4:5:2Moles of phosphorus(V)oxide= 8g/142g = 0.056molesMole ratio Phosphorus oxide:phosphorus 2:4Moles of phosphorus burnt = 0.056 X 2 = 0.112Mass of phosphorus = 0.112 X 31g = 3.472g

Limiting reagent and excess

When there is a chemical reaction, the reactant which is used up completely is the limiting reagent.

The chemical which is left some unreacted at the end is called excess.

When the limiting reagent is finished, the reaction stops.

All mole ratios should be done using the limiting reagent.

Example 13Mg + Al2O3 3MgO + 2AlIf 2 moles of Magnesium is used with 1 moles of

aluminium oxide, which is the limiting agent? Which chemical is in excess? How many moles in excess?

Mole ratio of Mg and Al2O3 is 3:1It means for 2 moles of Mg, 0.67 moles of Al2O3

is enough. But we have 1 mole of Al2O3. So Al2O3 is excess and Mg is the limiting reagent.

Moles of Al2O3 (excess) left after reaction is

1.00-0.67 = 0.33moles.

If mass is given, change all masses into moles first.Example 22Cu + O2 2CuOIn a reaction, 1.92g of copper is oxidised with 0.64g of oxygen.(a) Find which chemical is the limiting agent(b) How many g of the other chemical (excess) remaining at

the end?(c) What mass of Copper (II)oxide produced?AnswerChange mass in to molesNumber of moles of copper = 1.92g/64g = 0.03 molesNumber of moles of oxygen = 0.64g/32g = 0.02 molesCu : O2 mole ratio is 2:1It means, for 0.03 moles of copper only 0.015 moles of oxygen is enough. But we have 0.02 moles. Copper is the limiting reagent and oxygen is excess.

Excess oxygen = 0.020 – 0.015 = 0.005 moles Mass of excess oxygen = 0.005 X 32g = 0.16gMoles of copper oxide (make mole ratio with

limiting reagent) 2:2 That is 0.03molesMass of copper oxide = 0.03 X 80g = 2.40gNow try out the following:2Na + S Na2SIf 0.46g of sodium and 0.30g of sulphur reacted,(a) find the limiting reagent and excess reagent.Sulphur(b) find the mass of excess reagent left.0.0012 moles Na which is 0.028g(c) find what mass of sodium sulfide formed.Moles of Na2S is 0.0094 means 0.73g

Molar volumeGases have volume. Volume of a

gas depends on two factors :Temperature and PressureVolume decreases when pressure

is increased.Volume increases when the

temperature is increased.

Molar VolumeOne mole of any gas at Room Temperature

and Pressure has a volume of 24dm^3 (24000 cm^3)

Room temperature is considered as 20 degree C and atmospheric pressure is 1 atmosphere (760 mm of mercury)It means 1 mole of oxygen gas is 24000

cm33 moles of oxygen gas is 3X24000 cm3 at

RTP.24dm3(1mole) is called molar volume of

gases

Moles X molar volume = volume of a gasMoles = volume of a gas/molar volume

Remember to convert volume or mass to moles first.Example 1What is the volume of 1.1g of carbon dioxide gas at RTP?First convert mass in to molesMoles of CO2 = 1.1g/44g = 0.025molesVolume = moles X molar volume = 0.025 X 24000 cm3 = 600cm3

Example 2Find out the mass of 8000 cm3 of

sulphur dioxide gas, SO2 at RTP.First change the volume into

moles.Moles = Volume/molar volume 8000 cm3/24000 cm3 = 0.33 molesNow change the moles in to massMass = moles X molar mass 0.33 X 64g = 21.12g

% YieldWe can calculate and predict how much of a

product is formed during a reaction using the mole ratio.

How're in a real experiment, the predicted (calculated)amount may be different that what we get in actual experiment.

This can be due to: 1. Some reactants may be left unreacted.2. Change of substances in to other forms

such as gases3.Some reactants may be lost to surrounding

by overflowing, splashing etc4. Experimental errors and low quality

measuring devices.

%yield = Actual yield X 100

Theoretical yieldExample 1: In an experiment, 20g of Calcium carbonate

when heated gave 10g of calcium oxide is formed. Find the percentage yield of calcium oxide.

CaCO3 CaO + CO2%yield = Actual yield X 100 Theoretical yieldTo work out the above problem, we need the

actual yield and theoretical yield(Calculated yield).

Actual yield is given as 10g of Calcium oxide.Theoretical yield should be calculated by using

mole ratio method.

CaCO3 CaO + CO2We can find the mass of calcium oxide formed

from the mass of calcium carbonate given.But first convert mass in to moles.Moles of CaCO3 = 20g/100g = 0.20moles Mole ratio CaCO3:CaO is 1:1So, moles of CaO is also 0.20Mass of CaO = 0.20 X 56g = 11.20gNow insert the values% yield = Actual yield X 100% Theoretical yieldThat is 10g X 100% = 89.28% 11.20g

Example 22Na + 2H2O 2NaOH + H2115g of sodium was reacted with

water which produced 4.2g of hydrogen gas. What is the % yield of hydrogen gas?

AnswerMoles of sodium = 5.00 molesMoles of H2 = 2.50 moles

(2:1)Mass of H2 = 5.00g% yield = 84%

End of part 1