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INTRODUCTION to the MOLE and mole CALCULATIONS and EQUATIONS

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

study examples carefully7. Introducing 'moles' and their the connection with mass and formula massstudy the mole examples carefullyAND how to use reacting masses and the mole concept to deduce a symbol equation

Quantitative chemistry calculations online Help for problem solving in doing mole calculations, using experiment data, making predictions. Practice revision questions on mole calculations, moles mass and molecular/formula mass formula. 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. The ideas are extended to show how to deduce a balanced symbol equations by converting masses of reactants and products to moles and from the whole number ratio deduce the balanced equation. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level GCE AS-A2-IB courses. These revision notes and practice questions on mole calculations in chemistry and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (91) chemistry science courses. These revision notes and practice questions as an introduction to the mole and using the mole in chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (91) chemistry science courses.

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Self-assessment Quizzes on basic mole calculations

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(a) WHAT IS THE MOLE CONCEPT? and WHAT IS ONE MOLE OF A SUBSTANCE?

Summary of main points about defining and using the mole concept

For the purposes of calculations chemical amounts are primarily measured in moles (but sometimes just masses are ok).

The symbol for the unit mole is mol and the mass of one mole of a substance in grams is numerically equal to its relative formula mass (can be atomic mass or molecular mass - see below).

This might be an atomic mass of an element e.g. Fe = 56, so 1 mol = 56g

or the molecular mass of an element e.g. O = 16, so 1 mole of oxygen molecules O2 = 2 x 16 = 32g

or the molecular mass of a covalent compound e.g. C = 12, H = 1, so 1 mol of propane C3H8 = (3 x 12) + 8 = 44g

or formula mass of an ionic compound e.g. K = 39, O = 16, so 1 mol of potassium oxide K2O = 39 + 2 x 16) = 71g

 In the latter you can consider 1 mole of K2O made up of 2 mol of potassium ions and 1 mol oxide ion.

One mole of a substance contains the same number of the stated particles, atoms, molecules or ions as one mole of any other substance. So all the three described above have the same number of defined particles.

The number of atoms, molecules or ions in a mole of a given substance is the Avogadro constant. The value of the Avogadro constant is 6.02 x 1023 per mole.

So the four examples described above equate to 6.02 x 1023 Fe atoms,  6.02 x 1023 O2 molecules,

C3H8 molecules and  6.02 x 1023 'K2O's.

The latter consists of 2 x 6.02 x 1023 = 12.04 x 1023 = 1.204 x 1024 potassium ions

and 6.02 x 1023 oxide ions. You must be versatile in your mol thinking!

study the mole examples carefullyYou should understand and how to use measurement of the amounts in moles and apply this knowledge to calculations involving atoms, molecules, ions, electrons, formulae and equations.

This is a lot to take in, so I've described in details lots of examples explained below on how to use the 'mol'

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The 2nd part of the heading is 'easy', the first part is a bit more 'abstract' to get your head round!

study the mole examples carefullyThe mole concept is an invaluable way of solving many quantitative problems in chemistry!

Its a very important way of doing chemical calculations!

The theoretical basis is explained in section (b).

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 mass of one mole of a substance is sometimes referred to as the molar mass.

The atomic/formula mass in grams = one mole of the defined substance.

If your are dealing with individual atoms, one mole of equals the relative atomic mass in grams.

This can be expressed as a simple formula ...

moles of species = (actual mass of species in g) / (atomic/formula mass of species)

therefore (using triangle on right if necessary)

mass of species in g = moles species x atomic/formula mass of species

atomic/formula mass of species = mass of species in g / moles of species

Note these equations are for either an element or a compound,

but, whatever, you must clearly define the chemical species you mean for any mole calculation e.g.

Al metal element atom, H2O covalent molecule, an element O2 molecule, Na+Cl- ionic compound or just any compound formula like CuSO4 etc. etc.

This specificity cannot be overemphasised and you should understand that the measurement of amounts in moles can apply to atoms, molecules, ions, electrons, formulae and equations !!!

 

Mr is 'shorthand' for relative formula mass or molecular mass in amu (atomic mass units, u or dalton Da) and you must be able to work these out correctly from a given formula (Calculating relative formula/molecular mass of a compound or element molecule).

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.

Using the following atomic masses: H = 1, O = 16, N = 14, C = 12, Na = 23, Al = 27, Cl = 35.5, S = 32, Ca = 40

and the three formulae above relating moles, mass and formula mass ...

 

1 mole of carbon atoms = 12 g, 1 mole of carbon dioxide (12 + 2 x 16) = 44, 1 mole CO2 = 44 g

the number of C atoms = the number of CO2 molecules = 6.0 x 1023 specified particles

 

molecular mass Mr 18 for the water molecule H2O, 17 for the ammonia molecule NH3

so 1 mole of water is 18g, 0.333 mole = 0.333 x 18 = 6g

for ammonia 1mol = 17g, 34g = 34/17 = 2 mols ammonia

 

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

so 0.5 mol methane = 0.5 x 16 8g, 72g = 72/16 = 4.5 mols methane

for glucose 18g = 18/180 = 0.10 moles, 0.05 mole = 0.05 x 180 = 9.0g glucose

 

The element nitrogen consists of N2 molecules (Mr = 28),

molar mass = 28g, 0.25 moles = 0.25 x 28 = 7.0g

 

Relative atomic mass of iron Ar = 56, 7g = 7/56 = 0.125 mol Fe (Relative atomic mass explained)

 

The mass of 1 mole of sodium chloride NaCl? Mr(NaCl) = 23 + 35.5 = 58.5 g

 

The mass of 1 mole of calcium carbonate? Mr(CaCO3) = 40 + 12 + (3 x 16) = 100 g

 

So, these calculations are quite simple, but they are often just one part of solving a more complex problem involving moles!

 

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. mass of 1 mole of ionic sodium chloride NaCl or Na+Cl- is 58.5g (from 23 + 35.5)

BUT, each mole of NaCl consists of 1 mole of sodium ions and 1 mole of chloride ions

 

mass of 1 mole of ammonium sulfate (ionic salt) (NH4)2SO4 or (NH4+)2(SO42-) =  130g

BUT 1 mole of the salt consists of 2 moles of ammonium ions and 1 mole of sulfate ion

 

mass of 1 mole of aluminium oxide Al2O3 = (2 x 27) + (3 x 16) = 102 g

BUT being ionic the formula can be written as (Al3+)2(O2-)3, 1 mole of Al2O3 consists of 2 moles of aluminium ions and 3 moles of oxide ions

The mole concept can be extended to reading equations in terms of moles e.g.

MgCO3(s) + 2HCl(aq) ===> MgCl2(aq) + H2O(l) + CO2(g)

1 mol of magnesium carbonate reacts with 2 mol of hydrochloric acid to give 1 mol of magnesium chloride plus 1 mol of water and 1 mol of carbon dioxide

 

CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(l)

1 mol of methane requires 2 mol of oxygen for complete combustion to 1 mol of carbon dioxide and 2 mol of water

 

2Al(s) + 6HCl(aq) ===> 2AlCl3(aq) + 3H2(g)

2 mol of aluminium reacts with 6 mol of hydrochloric acid to form 2 mol of aluminium chloride plus 3 mol of hydrogen gas

 

2C2H6(g) + 7O2(g) ==> 4CO2(g) + 6H2O(l)

2 mol of ethane requires 7 mol of oxygen for complete combustion giving 4 mol of carbon dioxide and 6 mol of water

 

This reading of equations in terms of moles of reactants and products is important in calculations AND the process can be reversed, in other words, given the masses of reactants and products, you can convert them to moles and work out the balanced symbol equation.


There is a periodic table of atomic masses near the end of the page for doing mole calculations

(b) EXAMPLES OF MOLE CALCULATIONS A variety to illustrate the introduction section (a)

study the mole examples carefullyREMINDER For a substance 'Z' i.e. a specifically defined chemical species

(cover over the one you want and what's left is what you need to do)

(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).

Set 1 of mole calculations - using the simple mole formula

(see triangle relating mass, moles and relative formula mass)

  • Simple mole calculations Example 7.1.1

    • (a) How many moles of hydrogen (H2) in 10 g of the gas?

      • molecular mass = 2 x 1 = 2

      • moles = mass / molecular mass = 10 / 2 = 5 mol H2

    • (b) What is the mass of 5 moles of iron?

      • Atomic mass of iron (Fe) = 56

      • mass = moles x atomic mass = 5 x 56 = 280 g Fe

    • (c) How many moles of copper sulfate (CuSO4) in 2.6 g of the salt?

      • formula mass = 63.5 + 32 + (4 x 16) =  159.5

      • moles = mass / formula mass = 2.6 / 159.5 = 0.0163 mol CuSO4 (3sf, 4dp)

    • (d) What is the mass of 3.5 moles of magnesium oxide (MgO)?

      • formula mass MgO = 24 + 16 = 40

      • mass = moles x formula mass = 3.5 x 40 = 140 g MgO

    • (e) 0.15 moles of a compound Y had a mass of 21.0 g, calculate the formula mass of Y.

      • formula mass = mass / moles = 21.0 / 0.15 = 140 (Mr of Y)

    • (f) 0.0025 moles of a compound Z had a mass of 0.5125 g, calculate the formula mass of Z.

      • formula mass = mass / moles = 0.5125 / 0.0025 = 205 (Mr of Z)

  • study the mole examples carefullyMole calculation Example 7.1.2

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

    • which 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).

      • The latter is a better way to look at ammonia formation because nitrogen and hydrogen exist as diatomic molecules and NOT individual atoms.

    • 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 mole ratio on the basis of the 1 : 3 : 2 ratio (or 1 : 3 ==> 2) of the reactants and products in the balanced symbol equation.

  • study the mole examples carefullyMole calculation Example 7.1.3

    • 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 awkward 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 oxide.

      • 4Al(s) + 3O2(g) ==> 2Al2O3(s)

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

  • Mole calculation Example 7.1.4

    • 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 on right

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

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

      • This equation is read as 1 mole of copper oxide reacts with 2 moles of hydrochloric acid to produce 1 mole of copper chloride and 1 mole of water

      • basic mole ratio: 1 mole + 2 moles ==> 1 mole + 1 mole

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

      • Therefore ...

      • (64 + 16)g CuO + [2 x (1 + 35.5)]g HCl ==> [64 + (2 x 35.5)]g 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

  • study the mole examples carefullyMole calculation Example 7.1.5

    • 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 = mass of 1 mole 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 atoms combines with 3 moles of oxygen molecules to form 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!


study the mole examples carefully

Set 2 of mole calculations - relating a formula and equations to the mole concept

  • Mole calculation Example 7.2.1

    • How many moles of potassium ions and bromide ions are there 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.

  • study the mole examples carefullyMole calculation Example 7.2.2

    • How many moles of calcium ions and chloride ions are there 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.

  • study the mole examples carefullyMole calculation 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.

  • study the mole examples carefullyMole calculation Example 7.2.4

    • How many moles of aluminium ions and sulphate ions are 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 (SO42-).

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

  • study the mole examples carefullyMole calculation Example 7.2.5

    • How many moles of chlorine gas are there in 6.5g of the gas? Ar(Cl) = 35.5)

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

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

  • study the mole examples carefullyMole calculation Example 7.2.6

    • How many moles of iron in 20g of the metal? (Fe = 56)

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

  • study the mole examples carefullyMole calculation Example 7.2.7

    • How many grams of propane C3H8 are there in 0.21 moles of the gas? (C = 12, H = 1)

      • Mr of propane = (3 x 12) + (1 x 8) = 44

      • so g propane = moles x Mr = 0.21 x 44 = 9.24g

  • study the mole examples carefullyMole calculation Example 7.2.8

    • 0.25 moles of molecule X was found to have a mass of 28g.

      • Calculate its molecular mass.

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

  • Mole calculation Example 7.2.9

    • What mass and moles of magnesium chloride are 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 

  • Mole calculation 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


Set 3 of mole calculations - deducing a balanced symbol equation from the masses of reactants and products

Since you can interpret equations in terms of moles and masses of reactants and products, you can also work in the other direction.

e.g. if you know reactant and product masses, you can convert them to moles and from the simplest whole number ratio work out a balanced chemical equation.

Abbreviations used: Ar = atomic mass, Mr = formula mass


Equation - mole calculation Example 7.3.1 to work out an equation from reacting masses using moles

It was found that 11.15 g of lead(II) oxide PbO reacted with 0.30 g of carbon to produce 10.35 g of lead

Given the atomic masses: Pb = 207,  O = 16, C = 12, formula mass PbO = 207 + 16 = 223

(a) Convert the reacting masses given into moles

mol = mass / Ar or Mr

mol PbO = 11.15 / 223 = 0.05, mol C = 0.30 / 12 = 0.025, mol Pb = 10.35 / 207 = 0.05

(b) Convert the mole ratio from (a) into the simplest whole number (integer) ratio

PbO : C ==> Pb is 0.05 : 0.025 ==> 0.05 moles

scaling up to the simplest whole number ratio is 2 : 1 ==> 2 moles (divide the 1st ratios by 0.025, always try dividing by the lowest number first)

(c) Deduce the balanced equation for the reduction of lead(II) oxide to lead using carbon

From the whole number ratio above we can then write

2PbO  +  C  ===> 2Pb

BUT, the oxygen must have combined with the carbon to give carbon dioxide because there are two atoms of oxygen in the equation to one atom of carbon, so we can write the full equation

2PbO  +  C  ===> 2Pb  +  CO2

 


Equation - mole calculation Example 7.3.2 to work out an equation from reacting masses using moles

(a) 8.4 g of iron was heated in air to form an oxide, until there was no longer any gain in weight. The final mass of the iron oxide was 11.6 g. There was only one product of the reaction. Atomic masses: Fe = 56,  O = 16

(i) Calculate the mass of oxygen that combined with the iron.

mass of oxygen used = mass of iron oxide - mass of iron that reacted

mass of oxygen (O2) = 11.6 - 8.4 = 3.2 g

(ii) Calculate the moles of iron and oxygen that combined to give the iron oxide.

mol Fe = 8.4 / 56 = 0.15,  mol O2 = 3.2 / 32 = 0.10

Note: O is 16, BUT it is oxygen molecules that are in air, formula mass O2 = 32

(iii) Convert the Fe : O2 ratio to the simplest whole number ratio

Fe : O2 is 0.15 : 0.10, diving by 0.05 gives a reacting mole ratio of 3 : 2 for Fe : O2

(iv) We can therefore write the left-hand side of the equation as

3Fe  +  2O2  ===> ?

Since there was only one product of the reaction, the empirical formula of this particular iron oxide must be composed of three iron atoms and four atoms of oxygen, so the full equation is

3Fe  +  2O2  ===> Fe3O4

(b) It was found that 14.0 g of lithium (Li) reacted with 16.0 g of oxygen (O2) to form 30 g of lithium oxide (Li2O).

Convert the masses to moles and deduce the balanced equation from the mole ratios calculated.

Relative masses:  Li = 7,   O = 16,   O2 = 32,    Li2O = (2 x 7) + 16 = 30

mol Li = 14/7 = 2.0,     mol O2 = 16/32 = 0.5,    mol Li2O = 30/30 = 1.0

To get a set of whole numbers for the ratios you just double everything giving the following ratios of 4Li : 1 O2 : 2 Li2O, so the balanced equation is:

4Li(s)  +  O2(g)  ===>  2Li2O(s)

 

 


Example 7.3.3 to work out an equation from reacting masses using moles (typical GCSE question)

It was found that 7.95 g copper oxide (CuO) was reduced by 0.6 g of carbon to form 6.35 g of copper and releasing 2.2 g of carbon dioxide.

Convert the reactant and product masses to moles and deduced the balanced equation.

Atomic masses: Cu = 63.5,  C = 12,  O = 16   (I've set out the solution to the problem as a table of 'logic')

reactant => product ? CuO ?C ==> ?Cu ?CO2
atomic/formula mass 63.5 + 16 = 79.5 12 ==> 63.5 44
mass in g 7.95 0.6 ==> 6.35 2.2
moles = mass/formula mass 7.95/79.5 = 0.10 0.6/12 = 0.05 ==> 6.35/63.5 = 0.1 2.2/44 = 0.05
simplest whole number mol ratio 0.10/0.05 = 2 mol  0.05/0.05 = 1 mol   0.1/0.05 = 2 mol 0.05/0.05 = 1 mol
Therefore the equation must be

2CuO(s)  +  C(s)  ===> 2Cu(s)  +  CO2(g)

Notes:

(i) Always try dividing by the lowest ratio number first to try to deduce the whole number ratio to deduce the balanced equation

(ii) If you have done the calculation correctly the equation should be balanced, so always do an atom count!

(ii) For GCSE students the numbers are often 'perfect' but A level students may to make a reasonable judgement from ratios like 2.02 : 0.99 ==> 2.01 : 0.97 etc.

(iv) Don't forget that state symbols might be required, especially at A Level.

 

 


Example 7.3.4 to work out an equation from reacting masses using moles (more an A level question?)

7.75 g of solid phosphorus consumed 10 g of oxygen gas to form 17.75 g of a solid phosphorus oxide.

Convert the reactant and product masses to moles and deduced the balanced equation.

Atomic masses: P = 31,  O = 16

(a) convert the masses of phosphorus and oxygen to moles of atoms and deduce the empirical formula of the phosphorus oxide. Remember empirical formula is based on the simplest whole number ratio expressed in the formula.

7.75/31 = 0.25 mol of P atoms,  10/16 = 0.625 mol O atoms

(this is a slightly awkward ratio to deal with, so it needs a bit of trial and error patience to solve it!)

Dividing by the smallest number gives 0.25/0.25 : 0.625/0.25

gives 1 P : 2.5 O, then just multiplying by 2 gives 2 P : 5 O

So the empirical formula is P2O5

(b) From experiments it was further deduced that the molecular mass of the phosphorus oxide was 284.

(i) Deduce the correct molecular formula of the phosphorus oxide

The empirical formula mass of P2O5 = (31 x 2) + (16 x 5) = 142

Since 284/142 = 2, means the true molecular formula = 2 x P2O5 = P4O10

(ii) From the mole ratios and your answer to (i) deduce the correctly balanced symbol equation.

From your answer in (a) you need a ratio of 2 : 5 for P : O, but you must allow for the formation of P4O10 molecules and oxygen exists as diatomic molecules (O2).

Therefore to make a P4O10 you need 4 P atoms and 10 O atoms BUT as 5 O2 molecules

So the equation is: 4P(s)  +  5O2(g)  ===>  P4O10(s)

(don't forget that state symbols might be required)

 


Example 7.3.5 to work out an equation from reacting masses using moles (very much an A level question)

Iodine monofluoride gas (IF) is very unstable and above 0oC it decomposes into solid iodine (I2) and liquid iodine pentafluoride (IF5).

It was found that 14.6 g of iodine monofluoride decomposed to form 10.16 g of iodine and 4.44 g of iodine pentafluoride.

Atomic masses: I = 127,  F = 19

(a) Calculate the formula masses of the reactants and products.

ICl = 127 + 19 = 146,    I2 = 2 x 127 = 254,     IF5 = 127 + (5 x 19) = 222

(b) Convert the reactant and product masses to mol

mol IF = 14.6/146 = 0.1,     mol I2 = 20.32/254 = 0.04,    mol IF5 = 4.44/222 = 0.02

(c) From the mole ratios of reactants and products deduce the balanced symbol equation for the decomposition of iodine monofluoride.

Initial mol ratio: IF : I2 : IF5   is   0.1 : 0.04 : 0.02,

dividing by 0.02 gives a whole number ratio of 0.1/0.02 = 5 (IF),   0.04/0.02 = 2 (I2),   0.02/0.02 = 1 (IF5)

Therefore the equation is: 5IF(g)  ===>  2I2(s)  +  IF5(l)

 

 


(c) THE AVOGADRO CONSTANT

If you don't need to know about the Avogadro Constant, you can skip this section.

One mole of a substance contains exactly the same number of the stated particles, atoms, molecules or ions as one mole of ANY other substance (must be specifically defined e.g.

iron atoms Fe, water molecules H2O, glucose molecules C6H12O6, hydroxide ion OH-, iron(III) ion Fe3+, electron e-

The number of atoms, molecules or ions in a mole of a given substance is the Avogadro constant. The value of the Avogadro constant is 6.02 x 1023 per mole. You should understand that the measurement of amounts in moles can apply to atoms, molecules, ions, electrons, formulae and interpreting equations

So, 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 NA).

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

This means in a mole of any defined species e.g. atom, molecule, ion etc. there are the Avogadro constant number of them.

e.g. this means there are that many atoms in 12g of carbon (C = 12) = 6.023 x 1023 atoms of carbon

or that many molecules in 18g water (H2O = 1+1+16 = 18, H = 1; O = 16) * = 6.023 x 1023 molecules of water

* 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!

AND just think how useful the 'mole' is, to make life 'simple' in calculations! (well sort of!)

 

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.

1 mole of any defined chemical species has an identical number of that species, and that number is the Avogadro Number.

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.

It also means that you can read equations in terms of a mole ratio (see section (b) questions)

e.g. 2NaOH + H2SO4 ==> Na2SO4 + 2H2O

can be read as 2 moles of sodium hydroxide neutralises 1 mole of sulfuric acid to form 1 mole of the salt sodium sulfate and 2 moles of water,

BUT the equation can be read in terms of any molar quantities, as long as you keep the ratios the same!

e.g. by only taking 1/20th of a mole of sodium hydroxide you can deduce (yes predict!)

0.05 moles of NaOH reacts with 0.025 moles of H2SO4 to form 0.025 moles of Na2SO4 and 0.05 moles of H2O

What more, since you can convert moles to mass, you can do deduce the mass of product formed or the mass of reactants need.

Also, since you can go from mass to moles, you can deduce equations from measuring reacting masses.

(see also section 6. for reacting masses not using moles)


 

 

(d) More advanced calculations using the Avogadro Constant

This is for more advanced students and illustrates the concepts in the introduction section (b)

where you can actually calculate the number of particles in known quantity of material !

  • study the mole examples carefullyAvogadro constant Mole calculation Example 7.4.1

    • 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!

  • study the mole examples carefullyAvogadro number calculation Example 7.4.2

    • 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!)

  • study the mole examples carefullyAvogadro constant calculation Example 7.4.3

    • (a) How many particles of 'Al2O3' are there 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 (O2-) in the same 51g of the substance.

      • For every Al2O3 there are two Al3+ and three O2- ions.

      • So in 51g of Al2O3 there are ...

      • 0.5 x 2 x 6 x 10236 x 1023 Al3+ ions, and

      • 0.5 x 3 x 6 x 10239 x 1023 O2- ions.


 

(e) 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.

  • 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 (O2-)

    • Using the Avogadro constant you can convert the number of mol to actual particle numbers.

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

  • This also applies to half equations e.g. electrolysis examples ...

    • Al3+  +  3e-  ===> Al, is read as 1 mol of aluminium ions is reduced by 3 mol of electrons to 1 mol of aluminium atoms.

    • 2Cl-  ==>  Cl2  +  2e-, is read as 2 mol chloride are oxidised to release 1 mol of chlorine gas and 2 mol of electrons.

  • See separate page for Electrolysis products calculations (negative cathode and positive anode products)


 

Extra Advanced Level Chemistry Questions - more suitable for Advanced AS-A2 students which can be completely tackled after ALSO studying section 9 on the molar volume of gases and ANSWERS to QA7.1

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

The Avogadro Constant = 6.02 x 1023 mol-1

The molar volume for gases is 24.0dm3 at 298K/101.3kPa.

Atomic masses: Al = 27, O = 16, H = 1, Cl = 35.5, Ne = 20, Na = 23, Mg = 24.3, C = 12

Where appropriate assume the temperature is 298K and the pressure 101.3kPa.

Calculate ....

(a) how many oxide ions in 2g of aluminium oxide?

(b) how many molecules in 3g of hydrogen?

(c) how many molecules in 1.2 cm3 of oxygen?

(d) how many molecules of chlorine in 3g?

(e) how many individual particles in 10g of neon?

(f) the volume of hydrogen formed when 0.2g of sodium reacts with water.

(g) the volume of hydrogen formed when 2g of magnesium reacts with excess acid.

(h) the volume of carbon dioxide formed when the following react with excess acid

(1) 0.76g of sodium carbonate

(2) 0.76g sodium hydrogencarbonate

(i) the volume of hydrogen formed when excess zinc is added to 50 cm3 of hydrochloric acid, concentration 0.2 mol dm-3.

(j) the volume of carbon dioxide formed when excess calcium carbonate is added to 75 cm3 of 0.05 mol dm-3 hydrochloric acid.

 ANSWERS to QA7.1

Self-assessment Quizzes for GCSE or A Level (the basics) on mole calculations

Type in answer click me for QUIZ! QUIZ   or   multiple choice click me for QUIZ! QUIZ


Above is typical periodic table used in GCSE science-chemistry specifications in doing mole chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses


 

OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

  2. Calculating relative formula/molecular mass of a compound or element molecule

  3. Law of Conservation of Mass and simple reacting mass calculations

  4. Composition by percentage mass of elements in a compound

  5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

  6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

  7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass) (this page)

  8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

  9. Moles and the molar volume of a gas, Avogadro's Law

  10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

  13. Electrolysis products calculations (negative cathode and positive anode products)

  14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials


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