Doc Brown's Chemistry KS4 science GCSE/IGCSE/GCE-AS O Level  Revision Notes

7. Introducing 'moles' and their the connection with mass and formula mass

This page describes and explains the concept of the mole with lots of fully worked out examples of mole calculations. You should learn the formula connecting moles, mass and formula/molecular mass and methods for solving mole based problems.

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The mole is most simply expressed as the relative 'formula mass in g' or the 'molecular mass in g' of the defined chemical 'species', and that is how it is used in most chemical calculations.

The molar mass is M_r g mol^-1 i.e. the molar mass in grams

M_r is 'shorthand' for relative formula mass or molecular mass in amu (atomic mass units).

The term relative molecular mass (sum of the atomic masses of the atoms in a single molecule of the substance) is usually applied to definite molecular species

e.g. molecular mass 18 for the water molecule H₂O, 17 for the ammonia molecule NH₃

16 for the methane molecule CH₄ and 180 for the glucose sugar molecule C₆H₁₂O₆

(atomic masses for these examples H = 1, O = 16, N = 14, C = 12, S = 32, Na = 23, Cl = 35.5)

The term relative formula mass (sum of the atomic masses of the atoms in a specified formula) can be used for ANY specified formula of ANY chemical substance, though it is most often applied to ionic substances.

e.g. molar mass of ionic sodium chloride NaCl or Na⁺Cl^- is 58.5g

the molar mass of ammonium sulfate (ionic salt) (NH₄)₂SO₄ or (NH₄⁺)₂(SO₄^2-) =  130g

or 18g for the molar mass of water H₂O, and

17g is the molar mass of the ammonia molecule NH₃ etc. as for molecular mass

BUT in most cases either term for M_r is ok to use, and if in doubt, just call it M_r

Every mole of any substance contains the same number of the defined species.

The actual particle number is known and is called the Avogadro Constant, denoted N_A).

It is equal to 6.023 x 10²³ 'defined species' per mole i.e. 6.023 x 10²³ mol^-1

This means there are that many atoms in 12g of carbon (C = 12)

or that many molecules in 18g water (H₂O = 1+1+16 = 18, H = 1; O = 16) *.

* This is about 18cm³, so picture this number of molecules in a nearly full 20cm³ measuring cylinder or a 100ml beaker less than ¹/₅th full!

The Avogadro number is 6.023 x 10²³ = 602 300 000 000 000 000 000 000 atoms or molecules per mole!

= six hundred and two thousand and three hundred million million million 'particles' per mole !

A thimble full of water is about 1cm³, 1 mole of water = 18g and ~ 18cm³ because the density of water is ~1.0 g/cm³

Therefore in a thimble full of water there are ~6.023 x 10²³/18 = ~3.3 x 10²² = 33 000 000 000 000 000 000 000 molecules!

= thirty three thousand million million million molecules of water!

So, just think how many molecules of water are in your body!

However, the real importance of the mole is that it allows you to compare ratios of the relative amounts of reactants and products, or the element composition of a compound, at the atomic and molecular level.  If you have a mole ratio for A:B of 1:3, it means 1 particle of A to 3 particles of B irrespective of the atomic or formula masses of A and B. (see also section 6. for reacting masses not using moles)

Important Note. Relative atomic mass is just a number based on the carbon-12 relative atomic mass scale. Molar mass is a term used to describe the mass of one mole i.e. the relative atomic/formula/molecular mass in grams (g).

Examples:

• Example 7.1.1

  • Consider the formation of 1 mole of ammonia, NH₃,

  • consists of 1 mole of nitrogen atoms combined with 3 moles of hydrogen atoms.

  • Or you could say 2 moles of ammonia is formed from 1 mole of nitrogen molecules (N₂) combining with 3 moles of hydrogen molecules (H₂).

  • N₂(g) + 3H₂(g) ==> 2NH₃(g)

  • You can then think in any ratio you want e.g. 0.05 mol nitrogen combines with 0.15 mol hydrogen to form 0.10 mol of ammonia.

  • So, you can calculate using any ratio on the basis of the 1:3:2 ratio (or 1:3 ==> 2) of the reactants and products in the balanced symbol equation.

• Example 7.1.2

  • Consider the formation of 1 or 2 moles of aluminium oxide,

  • Al₂O₃, consists of 2 moles of aluminium atoms combined with 3 moles of oxygen atoms (or 1.5 moles of O₂ molecules) to form 1 mole of aluminium oxide.

    • 2Al(s) + ³/₂O₂(g) ==> Al₂O₃(s)

  • To avoid fractions in equations you can say 4 moles of aluminium atoms combine with exactly 3 moles of oxygen molecules to form 2 moles of aluminium.

    • 4Al(s) + 3O₂(g) ==> 2Al₂O₃(s)

    • So the simplest whole number reacting molar ratio is 4:3:2 (or 4:3 ==> 2)

• Example 7.1.3

  • How do you go from a reacting mole ratio to  reacting mass ratio?

  • You read the equation in relative numbers of moles and convert the moles into mass.

    • mass moles x formula mass - see triangle below

    • e.g. the formation of copper(II) chloride from copper(II) oxide and hydrochloric acid.

    • CuO(s) + 2HCl(aq) ==> CuCl₂(aq) + H₂O(l)

    • 1 mole + 2 moles ==> 1 mole + 1 mole

    • Atomic masses: Cu = 64, O = 16, H = 1, Cl =35.5

    • Therefore ...

    • (64 + 16)g CuO + 2x(1 + 35.5)g HCl ==> (64 + 2 x 35.5) CuCl₂ + (2 x 1 + 16)g H₂O

    • 80g CuO + 73g HCl ==> 135g CuCl2 + 18g H₂O

    • So, from a mole ratio of 1:2 ==> 1:1 you get a mass ratio of 80:73 ==> 135:18

• Example 7.1.4

  • This can be useful for calculating the quantities of chemicals you need for e.g. the chemical preparation of a compound.

  • Using the concept of mole ratio and the exemplar reactions above ...

  • (a) Calculate how many grams of copper(II) oxide you need to dissolve in hydrochloric acid to make 0.25 moles of copper(II) chloride?

  • From the equation, 1 mole of copper oxide makes 1 mole of copper chloride,

  • therefore you need 0.25 moles of CuO

  • since mass = molar mass x formula mass

  • you need 0.25 x 80 = 20g of CuO

  • (b) What mass of aluminium metal do you need to make 0.1 moles of aluminium oxide?

  • 4Al(s) + 3O₂(g) ==> 2Al₂O₃(s) and the atomic mass of aluminium is 27

  • 4 moles of aluminium makes 2 moles of aluminium oxide, (ratio 4:2 or 2:1)

  • therefore 0.2 moles of aluminium metal makes 0.1 moles of aluminium oxide (keeping the ratio of 2:1)

  • mass of aluminium metal needed = 0.2 x 27 = 5.4g of Al

  • Note that you can pick out the ratio you need to solve a problem - you DON'T need all the numbers of the full molar ratio, all you do is pick out the relevant ratio!

For calculation purposes learn the following formula for 'Z' and use a triangle if necessary.

For a substance 'Z' (1) mole of Z = g of Z / atomic or formula mass of Z, (2) or g of Z = mole of Z x atomic or formula mass of Z (3) or atomic or formula mass of Z = g of Z / mole of Z where Z represents atoms, molecules or formula of the particular element or compound defined in the question and all masses quoted in grams (g).

• Example 7.2.1: How many moles of potassium ions and bromide ions in 0.25 moles of potassium bromide?

  • 1 mole of KBr contains 1 mole of potassium ions (K⁺) and 1 mole of bromide ions (Br^-).

  • So there will be 0.25 moles of each ion.

• Example 7.2.2: How many moles of calcium ions and chloride ions in 2.5 moles of calcium chloride?

  • 1 mole of CaCl₂ consists of 1 mole of calcium ions (Ca²⁺) and 2 moles of chloride ion (Cl^-).

  • So there will be 2.5 x 1 = 2.5 moles of calcium ions and 2.5 x 2 = 5 moles chloride ions.

• Example 7.2.3: How many moles of lead and oxygen atoms are needed to make 5 moles of lead dioxide?

  • 1 mole of PbO₂ contains 1 mole of lead combined with 2 moles of oxygen atoms (or 1 mole of oxygen molecules O₂).

  • So 1 x 5 = 5 mol of lead atoms and 2 x 5 = 10 mol of oxygen atoms (or 5 mol oxygen molecules) are needed.

• Example 7.2.4: How many moles of aluminium ions and sulphate ions in 2 moles of aluminium sulphate?

  • 1 mole of Al₂(SO₄)₃ contains 2 moles of aluminium ions (Al³⁺) and 3 moles of sulphate ion (SO₄^2-).

  • So there will be 2 x 2 = 4 mol aluminium ions and 2 x 3 = 6 mol of sulphate ion.

• Example 7.2.5: How many moles of chlorine gas in 6.5g? A_r(Cl) = 35.5)

  • chlorine consists of Cl₂ molecules, so M_r = 2 x 35.5 = 71

  • moles chlorine = mass / M_r = 6.5 / 71 =  0.0944 mol

• Example 7.2.6: How many moles of iron in 20g? (Fe = 56)

  • iron consists of Fe atoms, so moles iron = mass/A_r = 20/56 = 0.357 mol Fe

• Example 7.2.7: How many grams of propane C₃H₈ are there in 0.21 moles of it? (C = 12, H = 1)

  • M_r of propane = (3 x 12) + (1 x 8) = 44, so g propane = moles x M_r = 0.21 x 44 = 9.24g

• Example 7.2.8: 0.25 moles of molecule X was found to have a mass of 28g. Calculate its molecular mass.

  • M_r = mass X / moles of X = 28 / 0.25 = 112 

• Example 7.2.9: What mass and moles of magnesium chloride is formed when 5g of magnesium oxide is dissolved in excess hydrochloric acid?

  • reaction equation: MgO + 2HCl ==> MgCl₂ + H₂O

  • means 1 mole magnesium oxide forms 1 mole of magnesium chloride (1 : 1 molar ratio)

  • formula mass MgCl₂ = 24+(2x35.5) = 95, 

  • MgO = 24+16 = 40, 1 mole MgO = 40g, so 5g MgO = 5/40 = 0.125 mol

  • which means 0.125 mol MgO forms 0.125 mol MgCl₂,

  • Mass = moles x formula mass = 0.125 x 95 = 11.9g MgCl₂ 

• Example 7.2.10: What mass and moles of sodium chloride is formed when 21.2g of sodium carbonate is reacted with excess dilute hydrochloric acid?

  • reaction equation: Na₂CO₃ + 2HCl ==> 2NaCl + H₂O + CO₂ 

  • means 1 mole sodium carbonate gives 2 moles of sodium chloride (1:2 ratio in equation)

    • Formula mass of Na₂CO₃ = (2x23) + 12 + (3x16) = 106

    • Formula mass of NaCl = 23 + 35.5 = 58.5

  • moles Na₂CO₃ = 21.2/106 = 0.2 mole

  • therefore 2 x 0.2 = 0.4 mol of NaCl formed.

  • mass of NaCl formed = moles x formula mass = 0.4 x 58.5 = 23.4g NaCl

Using the Avogadro Constant, you can actually calculate the number of particles in known quantity of material.

• Example 7.3.1: How many water molecules are there in 1g of water, H₂O ?

  • formula mass of water = (2 x 1) + 16 = 18

  • every mole of a substance contains 6 x 10²³ particles of 'it' (the Avogadro Constant).

  • moles water = 1 / 18 = 0.0556

  • molecules of water = 0.0556 x 6 x 10²³ = 3.34 x 10²²

  • Since water has a density of 1g/cm³, it means in every cm³ or ml there are

  • 33 400 000 000 000 000 000 000 individual H₂O molecules or particles!

• Example 7.3.2: How many atoms of iron (Fe = 56) are there in an iron filing of mass 0.001g ?

  • 0.001g of iron = 0.001 / 56 = 0.00001786 mol

  • atoms of iron in the nail = 0.00001786 x 6 x 1023 = 1.07 x 10¹⁹ actual Fe atoms

  • (10.7 million million million atoms!)

• Example 7.3.2: (a) How many particles of 'Al₂O₃' in 51g of aluminium oxide?

  • Atomic masses: Al =27, O = 16, f. mass Al₂O₃ = (2x27) + (3x16) = 102

  • moles 'Al₂O₃' = 51/102 = 0.5 mol

  • Number of 'Al₂O₃' particles = 0.5 x 6 x 10²³ = 3 x 10²³

  • (b) Aluminium oxide is an ionic compound. Calculate the number of individual aluminium ions (Al³⁺) and oxide ions (O^2-) in the same 51g of the substance.

  • For every Al₂O₃ there are two Al³⁺ and three O^2- ions.

  • So in 51g of Al₂O₃ there are ...

  • 0.5 x 2 x 6 x 10²³ =  6 x 10²³ Al³⁺ ions, and

  • 0.5 x 3 x 6 x 10²³ =  9 x 10²³ O^2- ions.

More advanced use of the mole and Avogadro Number concepts (for advanced level students only)

• You can have a mole of whatever you want in terms of chemical species e.g.

• In terms of electric charge, 1 Faraday = 96500 C (coulombs) = 6 x 10²³ electrons

• If you have 2.5 moles of the ionic aluminium oxide (Al₂O₃) you have ...

  • 2 x 2.5 = 5 moles of aluminium ions (Al³⁺) and 3 x 2.5 = 7.5 mol of oxide ions (O^2-)

• When you write ANY balanced chemical equation, the balancing numbers, including the un-written 1, are the reacting molar ratio of reactants and products.

 QA7.1 This question involves using the mole concept and the Avogadro Constant in a variety of situations.