2.1.3 (e) Mole Calculations

Syllabus

(e) calculations, using amount of substance in mol, involving:

(i) mass

(ii) gas volume

(iii) solution volume and concentration

What does this mean?

In the first instance you need to be able to convert masses, gas volumes and volumes & concentrations of solutions into moles.

And vice versa.

i) Mass to moles and back

You probably know how to do this but, if not,

a) Given a mass.

Make sure it is in grams (g)
Calculate the Molar Mass of the substance
Divide Mass by Molar Mass

eg. How many moles in 2.2 kg of CO₂?

2.2 kg = 2200 g

M_r(CO₂) = 12 + 16 + 16 = 44 g/mol

Moles = Mass ÷ M_r=2200 ÷ 44 = 500 moles

b) Given a number of moles

Calculate the Molar Mass of the substance
Multiply Moles by Molar Mass

eg. What mass is 1.5 x10^-3 moles of Magnesium Oxide (MgO)?

M_r(MgO) = 24 + 16 = 40 g/mol

Mass = Moles x M_r=1.5 x10^-3 x 40=0.06 moles

ii) Gas Volume to moles and back

a) Given a gas volume.

Make sure Volume is in cubic decimetres (dm³)
Look up the Molar Gas Volume (24 dm³ unless stated otherwise)
Divide Volume by Molar Gas Volume

eg. How many moles in 960 cm³ of CO₂?

960 cm³ = 0.96 dm³

Molar Gas Volume = 24 dm³

Moles = Volume ÷ Molar Gas Volume = 0.96 ÷ 24 = 0.04 moles

b) Given a number of moles

Look up the Molar Gas Volume (24 dm³ unless stated otherwise)
Multiply Moles by Molar Gas Volume
Convert to appropriate unit (A level papers like to make you convert from dm³ to cm³ and back)

eg. What Volume, in cm³, is occupied by 2.5 moles of Oxygen (or any other gas)?

Molar Gas Volume = 24 dm³

Volume= Moles x Molar Gas Volume =2.5 x 24= 60 dm³

1 dm³ = 1000 cm³

60 dm³ = 60,000 cm³

^Video

iii) Volume & concentration to moles and back

a) Given a volume and a concentration of a solution.

Make sure Volume is in cubic decimetres (dm³)
1. Make sure Concentration is in mol/dm³ (occasionally you may need to convert from g/dm³)
Multiply Volume by Concentration

eg. How many moles in 200 cm³ of 0.4 g/dm³ NaOH?

200 cm³ = 0.2 dm³

M_r(NaOH) = 23+16+1= 40

g/dm³ ÷ M_r= mol/dm³

0.4÷ 40 = 0.01 mol/dm³

Moles = Volume x Concentration (mol/dm³) = 0.2 X 0.01 = 0.002 moles

b) Given a number of moles and a concentration

Make sure Concentration is in mol/dm³
Divide Moles by Concentration
Convert Volume to appropriate unit (usually dm³)

eg. What Volume, in cm³, is occupied by 2.4 moles of a 0.1 mol/dm³ solution?

Volume= Moles ÷ Concentration = 2.4 ÷ 0.1= 24 dm³

1 dm³ = 1000 cm³

24 dm³ = 24,000 cm³

c) Given a number of moles and a volume

Make sure volume is in dm³
Divide Moles by Volume

eg. What concentrationsolution is formed when 1.8 moles of a substance dissolves in 500 cm³?

500 cm³ = 0.5 dm³

Concentration= Moles ÷ Volume =1.8 ÷ 0.5= 3.6 mol/dm³

^{"What is Concentration?" PowerPoint}

Google Presentation

Stoichiometry

You know this - you just probably didn't call it stoichiometry at GCSE.

In an equation there are generally coefficients - big numbers.

If we have the means to calculate the number of moles of Nitrogen that reacted and it came to 0.5 moles we could instantly say that it would take 3 x 0.5 = 1.5 moles of Hydrogen to react with.

And 2 x 0.5 = 1 mole of Ammonia would be made.

That's Stoichiometry.

Using moles PowerPoint

Google Presentation

Mixed Questions

More usually an A level paper will mix two of the above.

A useful acronym is MR MA- Moles, Ratio, Moles, Answer

eg. What mass of Magnesium Oxide is needed to exactly neutralise 300 cm³ of a 0.5 mol/dm³ solution of Hydrochloric acid?

a) Moles: There has to enough information to find the number of moles of something.

Moles = Concentration x Volume

Moles (HCl) = 0.5 x 0.3 = 0.15 moles

a) Ratio: The exam will probably include this. Very rarely you have to write your own.

MgO_(s) + 2 HCl_(aq) → MgCl_{2 (aq)} + H₂O_(l)

c) Moles: Look at the coefficients (big numbers) in the equation to see the ratio to use.

Don't assume 1:1

This is called stoichiometry

HCl :MgO

2 : 1

1 : 0.5

0.15 : 0.075

d) Answer: The question always then asks you to convert the number of moles to something else

Mass = Moles x M_r

Mr (MgO) = 24 + 16 = 40

Mass (MgO) = 0.075 x 40 = 3g

Exam style Questions

1. (a) Define the term relative atomic mass.

.....................………………………....................................................................................................................……

.....................………………………....................................................................................................................……(2)

(b) How would you calculate the mass of one mole of atoms from the mass of a single atom?

.....................………………………....................................................................................................................……(1)

(c) Sodium hydride reacts with water according to the following equation.

NaH_(s) + H₂O _(l) → NaOH_(aq) + H_{2 (g)}

A 1.00 g sample of sodium hydride was added to water & the resulting solution was diluted to a volume of exactly 250 cm³

(i) Calculate the concentration in moldm^-3, of sodium hydroxide solution formed.

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(ii) Calculate the volume of hydrogen gas evolved, measured at Standard Temperature and Pressure.

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(iii) Calculate the volume of 0.112 M hydrochloric acid which would react exactly with a 25.0 cm³ sample of sodium hydroxide solution.

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(8)

2. (a) Sodium carbonate forms a number of hydrates of general formula Na2CO3.xH2O

A 3.01 g sample of one of these hydrates was dissolved in water and the solution made up to 250 cm3.

In a titration, a 25.0 cm3 portion of this solution required 24.3 cm3 of 0.200 mol–1 dm–3 hydrochloric acid for complete reaction.

The equation for this reaction is shown below.

Na2CO3 + 2HCl--> 2NaCl + H2O + CO2

(i) Calculate the number of moles of HCl in 24.3 cm3 of 0.200 mol dm–3 hydrochloric acid.

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(ii) Deduce the number of moles of Na2CO3 in 25.0 cm3 of the Na2CO3 solution.

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(iii) Hence deduce the number of moles of Na2CO3 in the original 250 cm3 of solution.

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(iv) Calculate the Mr of the hydrated sodium carbonate.

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.....................………………………....................................................................................................................……(5)

(b) In an experiment, the Mr of a different hydrated sodium carbonate was found to be 250.

Use this value to calculate the number of molecules of water of crystallisation, x, in this hydrated sodium carbonate, Na2CO3.xH2O

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.....................………………………....................................................................................................................……(3)

Answers

1.

(a) Mean (1) Mass of atom/Mass of ¹²C × 12 (1)

(b) Multiply by Avogadro’s number (1)

(c) (i) Moles NaH = 1/24 (1)

moles NaOH = Moles NaH = 0.0417 (1)

= 0.167 moldm^–3 (1)

(allow 0.166 – 0.168)

(ii) pV = nRT (1)

= 0.0417 x 24 (1) allow consequential

= 1.0008 m³ (1)(iii) vol HCl = (moles/0.112) × 1000 = (0.00417/0.112) × 1000 (1)

= 37.3 cm³ (1)

(allow 37.1 to 38.0 and conseq.)

2. (a)

(i) 4.86 × 10^–3 (1)

(ii) 2.43 × 10^–3 (1)

(mark conseq on (a)(i))

(iii) 2.43 × 10^–2 (1)

(mark conseq on (a)(ii))

(iv) 3.01/2.43 × 10^–2 (1)

(mark conseq on (a)(iii))

124 (1)

(Do not allow 124 without evidence of appropriate calculation in (a)(iii))

(b) Mr(Na₂CO₃) = 106 (1)

Mr (xH₂O) = 250 –106 = 144 (mark conseq on M1) (1)

x = 8 (mark conseq on M2) (1)

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