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
Mr(CO2) = 12 + 16 + 16 = 44 g/mol
Moles = Mass ÷ Mr=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)?
Mr(MgO) = 24 + 16 = 40 g/mol
Mass = Moles x Mr =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)
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
Mr(NaOH) = 23+16+1= 40
g/dm3 ÷ Mr = 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
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
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) → MgCl2 (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 Mr
Mr (MgO) = 24 + 16 = 40
Mass (MgO) = 0.075 x 40 = 3g
Exam style Questions
1. (a) Define the term relative atomic mass.
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(b) How would you calculate the mass of one mole of atoms from the mass of a single atom?
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(c) Sodium hydride reacts with water according to the following equation.
NaH (s) + H2O (l) → NaOH (aq) + H2 (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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(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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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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