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CHEMISTRY NOTES FOR GCSE
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CHEMICAL CALCULATIONS
12.1. Explaining relative atomic mass (Ar)*
Every atom has its own unique atomic mass based on a standard comparison or relative scale e.g. it has
been based on hydrogen H = 1, oxygen O = 16. The relative atomic mass scale is now based on an isotope
of carbon, carbon-12,
, which is given the value of 12.0000 amu. (* The letter A on its own
usually means the mass number of a particular isotope)
However there are complications due to isotopes and so very accurate atomic masses are not whole
numbers. Isotopes are atoms of the same element with different masses due to different numbers of
neutrons. The very accurate atomic mass scale is based on a specific isotope of carbon, carbon-12, 12C
= 12.0000 units exactly, for most purposes C = 12 is used for simplicity.
For example
,
and
are the three isotopes of hydrogen, though the vast majority of
hydrogen atoms have a mass of 1. When their accurate isotopic masses, and their % abundance are taken
into account the average accurate relative mass for hydrogen = 1.008, but for most purposes H = 1 is
good enough!
The strict definition of relative atomic mass (Ar) is that it equals average mass of all the isotopic
atoms present in the element compared to 1/12th the mass of a carbon-12 atom.
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Example 12.1.1: bromine consists of 50% 79Br and 50% 81Br, calculate the Ar of bromine.
o Ar = [ (50 x 79) + (50 x 81) ] /100 = 80
o So the relative atomic mass of bromine is 80 or Ar(Br) = 80
o Note the full working shown. Yes, ok, you can do it in your head BUT many students
ignore the %'s and just average all the isotopic masses (mass numbers) given.
o
Example 12.1.2: chlorine consists of 75% chlorine-35 and 25% chlorine-37.
o Think of the data based on 100 atoms, so 75 have a mass of 35 and 25 atoms have a mass
of 37.
o The average mass = [ (75 x 35) + (25 x 37) ] / 100 = 35.5
o So the relative atomic mass of chlorine is 35.5 or Ar(Cl) = 35.5
The mass number for any isotope is the sum of the protons and neutrons in the nucleus, and is always a
whole number.
(Relative isotopic mass = the accurate mass of a single isotope of an element compared to 1/12th the
mass of a carbon-12 atom e.g. the accurate mass of
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is 58.9332!
M. ALI. BLL, 2010
CHEMISTRY NOTES FOR GCSE
[email protected],
[email protected]
12.2. Relative formula mass or molecular mass (Mr)
If all the individual atomic masses of all the atoms in a formula are added together you have calculated
the relative formula mass* (for ionic compounds) or molecular mass (for covalent elements or
compounds), Mr. * can be used for any element or compound. Whereas relative atomic mass (1. above)
only applies to a single atom but anything with at least two atoms requires the term relative
formula/molecular mass. The most common error is to use atomic/proton numbers instead of atomic
masses, unfortunately, except for hydrogen, they are different!
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Example 12.2.1: the diatomic molecules of the elements hydrogen H2 and chlorine Cl2
o relative atomic masses, Ar: H = 1, Cl = 35.5
o Formula masses, Mr, are H2 = 2 x 1 = 2, Cl2 = 2 x 35.5 = 71 respectively.
o
Example 12.2.2: the element phosphorus consists of P4 molecules.
o Mr of phosphorus = 4 x its atomic mass = 4 x 31 = 124
o
Example 12.2.3: The compound water H2O
o relative atomic masses are H=1 and O=16
o Mr = (1x2) + 16 = 18 (molecular mass of water)
o
Example 12.2.4: The compound sulphuric acid H2SO4
o relative atomic masses are H=1, S=32 and O=16
o Mr = (1x2) + 32 + (4x16) = 98 (molecular mass of sulphuric acid)
o
Example 12.2.5: The compound calcium hydroxide Ca(OH)2 (ionic)
o relative atomic masses are Ca=40, H=1 and O=16
o Mr = 40 + 2 x (16+1) = 74
12.3. Conservation of mass calculations
When elements and compounds react to form new products, mass cannot be lost or gained - "The Law of
Conservation of Mass" - states that mass cannot be created or destroyed, but changed into
different forms, so in a chemical change the total mass of reactants must equal the total mass of
products. By using this law, together with atomic and formula masses, you can calculate the quantities of
reactants and products involved in a reaction and the simplest formula of a compound
NOTE: (1) the symbol equation must be correctly balanced to get the right answer! (2) There are good
reasons why, when doing a real chemical preparation-reaction to make a substance you will not get 100%
of what you theoretically calculate.
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Example 12.3.1: Magnesium + Oxygen ==> Magnesium oxide
o 2Mg + O2 ==> 2MgO (atomic masses required: Mg=24 and O=16)
o think of the ==> as an = sign, so the mass changes in the reaction are:
o (2 x 24) + (2 x 16) = 2 x (24 + 16)
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CHEMISTRY NOTES
o
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48 + 32 = 2 x 40 and so 80 mass units of reactants = or produces 80 mass units of
products (you can work with any mass units such as g, kg or tonne (1 tonne = 1000 kg)
o
Example 12.3.2: iron + sulphur ==> iron sulphide
o Fe + S ==> FeS (atomic masses: Fe = 56, S = 32)
o If 59g of iron is heated with 32g of sulphur to form iron sulphide, how much iron is left
unreacted? (assuming all the sulphur reacted)
o From the atomic masses, 56g of Fe combines with 32g of S to give 88g FeS.
o This means 59 - 56 = 3g Fe unreacted.
o
Example 12.3.3: When limestone (calcium carbonate) is strongly heated, it undergoes thermal
decomposition to form lime (calcium oxide) and carbon dioxide gas.
o CaCO3  CaO + CO2 (relative atomic masses: Ca = 40, C = 12 and O = 16)
o Calculate the mass of calcium oxide and the mass of carbon dioxide formed by
decomposing 50 tonnes of calcium carbonate.
o (40 + 12 + 3x16)  (40 + 16) + (12 + 2x16)
o 100  56 + 44, scaling down by a factor of two, 50  28 + 22
o so decomposing 50 tonnes of limestone produces 28 tonnes of lime and 22 tonnes of
carbon dioxide gas.
12.4 . Percentage Composition by mass of compoundsThe % by mass composition of a compound in terms of its constituent elements is calculated as follows:
a. calculate the formula or molecular mass of the compound
b. calculate the mass of the element in the compound, taking into account the number of
c.
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atoms of the element in the compound formula
calculate (b) as a percentage of (a)
Example 12.4.1: Calculate the % of copper in copper sulphate, CuSO4
o Relative atomic masses: Cu = 64, S = 32 and O = 16
o relative formula mass = 64 + 32 + 4x16 = 160
o only one copper atom of relative atomic mass 64
o % Cu = 64 x 100 / 160 = 40% copper by mass in the compound
o
Example 12.4.2: Calculate the % of oxygen in aluminium sulphate, Al2(SO4)3
o Relative atomic masses: Al = 27, S = 32 and O = 16
o relative formula mass = 2x27 + 3x(32 + 4x16) = 342
o there are 4 x 3 = 12 oxygen atoms, each of relative atomic mass 16, giving a total mass of
oxygen in the formula of 12 x 16 = 192
o % O = 192 x 100 / 342 = 56.1% oxygen by mass in aluminium sulphate
o
Example 12.4.3: Calculate the % of water in hydrated magnesium sulphate MgSO 4.7H2O
o Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1
o relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246
o 7 x 18 = 126 is the mass of water
o so % water = 126 x 100 / 246 = 51.2 %
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12.5. Simple empirical formula from reacting masses (easy start, no moles!)
The EMPIRICAL FORMULA of a compound can be worked out by knowing the exact masses of the
elements that combine to form a given mass of a compound. The following examples illustrate the ideas
using numbers more easily appreciated than in real experiments. Note: Empirical formula means the
simplest whole number ratio formula found by experiment.
In real laboratory experiments only a fraction of a gram or a few grams of elements would be used, and
a more 'tricky' mole calculation method is required than shown here. However the examples below show
in principal how formulae are worked out from experiments.
Any calculation method must take into account the different relative atomic masses of the elements in
order to get to the actual ratio of the atoms in the formula. For example, just because 10g of X
combines with 20g of Y, does not mean that the formula of the compound is XY 2 !
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Example 12.5.1: It is found that 207g of lead combined with 32g of sulphur to form 239g of
lead sulphide. From the data work out the formula of lead sulphide. (Relative atomic masses: Pb =
207 and S = 32)
o In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of 1
atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass), so the formula is simply
PbS
o
Example 12.5.2: It is found that 207g of lead combined with oxygen to form 239g of a lead
oxide. From the data work out the formula of the lead oxide. (Relative atomic masses: Pb = 207
and O = 16)
In this case, you first have to work out the amount of oxygen combined with the lead.
o
This is 239 - 207 = 32. In atomic ratio terms, the 207 is equivalent to 1 atom of lead
and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16), so the formula is
simply PbO2
o
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Example 12.5.3: It is found that 54g of aluminium forms 150g of aluminium sulphide. Work out
the formula of aluminium sulphide. (Relative atomic masses: Al = 27 and S = 32).
o Amount of sulphur combined with the aluminium = 150 - 54 = 96g
o By atomic ratio, the 54 of aluminium is equivalent to 2 atoms of aluminium and the 96 of
sulphur is equivalent to 3 atoms of sulphur. Therefore the atomic ratio is 2 to 3, so the
formula of aluminium sulphide is Al2S3
12.6. Reacting mass chemical calculations (not using moles)
You can use the ideas of relative atomic, molecular or formula mass AND the law of conservation of
mass to do quantitative calculations in chemistry. Underneath an equation you can add the appropriate
atomic or formula masses. This enables you to see what mass of what, reacts with what mass of other
reactants. It also allows you to predict what mass of products are formed (or to predict what is needed
to make so much of a particular product). You must take into account the balancing numbers in the
equation (e.g. 2Mg), as well of course, the numbers in the formula (e.g. O2).
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NOTE
(1) the symbol equation must be correctly balanced to get the right answer!
(2) There are good reasons why, when doing a real chemical preparation-reaction to make a substance
you will not get 100% of what you theoretically calculate.
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Example 12.6.1: 2Mg + O2 ==> 2MgO
o (atomic masses Mg =24, O = 16)
o converting the equation into reacting masses gives ... (2 x 24) + (2 x 16) ==> 2 x (24 +
16)
o and this gives a basic reacting mass ratio of 48g Mg + 32g O2 ==> 80g MgO
o The ratio can be used, no matter what the units, to calculate and predict quite a lot! and
you don't necessarily have to work out and use all the numbers in the ratio.
o What you must be able to do is solve a ratio!
o e.g. 24g Mg will make 40g MgO, why?, 24 is half of 48, so half of 80 is 40.
o
Example 12.6.2: 2NaOH + H2SO4 ==> Na2SO4 + 2H2O
o (atomic masses Na = 23, O = 16, H = 1, S = 32)
o mass ratio is: (2 x 40) + (98) ==> (142) + (2 x 18) = (80) + (98) ==> (142) + (36),
 but pick the ratio needed to solve the particular problem.
o (a) calculate how much sodium hydroxide is needed to make 5g of sodium sulphate.
o 142g Na2SO4 is formed from 80g of NaOH
o 5g Na2SO4 is formed from 5g x 80 / 142 = 2.82 g of NaOH by scaling down from 142 =>
5
o (b) calculate how much water is formed when 10g of sulphuric acid reacts.
o 98g of H2SO4 forms 36g of H2O
o 10g of H2SO4 forms 10g x 36 / 98 = 3.67g of H2O by scaling down from 98 => 10
o
Example 12.6.3: 2CuO(s) + C(s) ==> 2Cu(s) + CO2(g)
o (atomic masses Cu=64, O=16, C=12)
o Formula Mass ratio is 2 x (64+16) + (12) ==> 2 x (64) + (12 + 2x16)
o = Reacting mass ratio 160 + 12 ==> 128 + 44 (in the calculation, impurities are
ignored)
o (a) In a copper smelter, how many tonne of carbon (charcoal, coke) is needed to
make 16 tonne of copper?
o 12 of C makes 128 of Cu
o scaling down numerically: mass of carbon needed = 12 x 16 / 128 = 1.5 tonne of C
o (b) How many tonne of copper can be made from 640 tonne of copper oxide ore?
o 160 of CuO makes 128 of Cu (or direct from formula 80 CuO ==> 64 Cu)
o scaling up numerically: mass copper formed = 128 x 640 / 160 = 512 tonne Cu
o
Example 12.6.4: What mass of carbon is required to reduce 20 tonne of iron(II) oxide ore
if carbon monoxide is formed in the process as well as iron?
o (atomic masses: Fe = 56, O = 16)
o reaction equation: Fe2O3 + 3C ==> 2Fe + 3CO
o formula mass Fe2O3 = (2x56) + (3x16) = 160
o 160 mass units of iron oxide reacts with 3 x 12 = 36 mass units of carbon
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o
o
o
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So the reacting mass ratio is 160 : 36
So the ratio to solve is 20 : x, scaling down, x = 36 x 20/160 = 4.5 tonne carbon
needed.
Note: Fe2O3 + 3CO ==> 2Fe + 3CO2 is the other most likely reaction that reduces the iron
ore to iron.
o
Example 12.6.5: (a) Theoretically how much copper can be obtained from 2000 tonne of
pure chalcopyrite ore, formula CuFeS2 ?
o (atomic masses: Cu = 64, Fe = 56, S = 32)
o For every CuFeS2 ==> Cu can be extracted, f. mass of ore = 64 + 56 + (2x32) = 184
o Therefore reacting mass ratio is: 192 ==> 64
o so, solving the ratio, 2100 ==> 64 / 184 = 695.7 tonne copper = max. can be extracted
o (b) If only 670.2 tonne of pure copper is finally obtained after further purification
by electrolysis, what is the % yield of the overall process?
o % yield = actual yield x 100/theoretical yield
o % yield = 670.2 x 100 / 695.7 = 96.3%
o
Example 12.6.6: A sample of magnetite iron ore contains 76% of the iron oxide compound
Fe3O4 and 24% of waste silicate minerals. (a) What is the maximum theoretical mass of
iron that can be extracted from each tonne (1000 kg) of magnetite ore by carbon
reduction? [ Atomic masses: Fe = 56, C = 12 and O = 16 ]
o The reduction equation is: Fe3O4 + 2C ==> 3Fe + 2CO2
o Before doing the reacting mass calculation, you need to do simple calculation to take into
account the lack of purity of the ore.
o 76% of 1 tonne is 0.76 tonne (760 kg).
o For the reacting mass ratio: 1 Fe3O4 ==> 3 Fe (you can ignore rest of equation)
o Therefore in reacting mass units: (3 x 56) + (4 x 16) ==> 3 x 56
o so 232 Fe3O4 ==> 168 Fe
o 0.76 Fe3O4 tonne ==> x tonne Fe
o solving the ratio, x = 0.76 x 168/232 = 0.55
o = 0.55 tonne Fe (550 kg)/tonne (1000 kg) of magnetite ore
o (b) What is the atom economy of the carbon reduction reaction?
o You can use some of the data from part (a).
o % atom economy = total mass of useful product x 100 / total mass of reactants
o = 168 x 100 / (232 + 2x12) = 168 x 100 / 256 = 65.6%
o (c) Will the atom economy be smaller, the same, or greater, if the reduction
involves carbon monoxide (CO) rather than carbon (C)? explain?
o The atom economy will be smaller because CO is a bigger molecular/reactant mass than C
and 4 molecules would be needed per 'molecule' of Fe3O4, so the mass of reactants is
greater for the same product mass of iron (i.e. bottom line numerically bigger, so %
smaller). This is bound to be so because the carbon in CO is already chemically bound to
some oxygen and can't remove as much oxygen as carbon itself.
o Fe3O4 + 4CO ==> 3Fe + 4CO2
o so the atom economy = 168 x 100 / (232 + 4x28) = 48.8 %
o
Example 12.6.7: On analysis, a sample of hard water was found to contain 0.056 mg of
calcium hydrogen carbonate per cm3 (0.056 mg/ml). If the water is boiled, calcium
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hydrogencarbonate Ca(HCO3)2, decomposes to give a precipitate of calcium carbonate
CaCO3, water and carbon dioxide.
o (a) Give the symbol equation of the decomposition complete with state symbols.
 Ca(HCO3)2(aq) ==> CaCO3(s) + H2O(l) + CO2(g)
o (b) Calculate the mass of calcium carbonate in grammes deposited if 2 litres (2 dm 3,
2000 cm3 or ml) is boiled in a kettle. [ atomic masses: Ca = 40, H = 1, C = 12, O = 16 ]
 the relevant ratio is based on: Ca(HCO3)2 ==> CaCO3
 The formula masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40 + 12 + 16x3)
respectively
 there the reacting mass ratio is 162 units of Ca(HCO3)2 ==> 100 units of CaCO3
 the mass of Ca(HCO3)2 in 2000 cm3 (ml) = 2000 x 0.056 = 112 mg
 therefore solving the ratio 162 : 100 and 112 : z mg CaCO3
 where z = unknown mass of calcium carbonate
 z = 112 x 100/162 = 69.1 mg CaCO3
 since 1g = 1000 mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate
o (c) Comment on the result, its consequences and why is it often referred to as
'limescale'?
 This precipitate of calcium carbonate will cause a white/grey deposit to be
formed on the side of the kettle, especially on the heating element. Although
0.0691 g doesn't seem much, it will build up appreciably after many cups of tea!
The precipitate is calcium carbonate, which occurs naturally as the rock
limestone, which dissolved in rain water containing carbon dioxide, to give the
calcium hydrogen carbonate in the first place. Since the deposit of 'limestone'
builds up in layers it is called 'limescale'.
12.7. Moles, mass, formula mass and the ratio of reacting masses
The mole is most simply expressed as the 'formula mass in g' of the defined chemical 'species', and
that is how it is used in most chemical calculations.
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 and is equal to 6.023 x 1023
'defined species' per mole. This means there are that many atoms in 12g of carbon (C = 12) or that
many molecules of water in 18g* (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!
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.
Important Note. Relative is just a number based on the carbon-12 relative atomic mass scale. Molar
mass is a term used to descsribe the mass of one mole i.e. the relative atomic/formula/molecular mass in
grams (g).
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Examples:
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Example 12.7.1.1: 1 mole of ammonia, NH3,
o consists of 1 mole of nitrogen atoms combined with 3 moles of hydrogen atoms.
o Or you could say 2 moles of ammonia is formed from 1 mole of nitrogen molecules (N 2)
and 3 moles of hydrogen molecules (H2).
o
Example 12.7.1.2: 1 mole of aluminium oxide,
o Al2O3, consists of 2 moles of aluminium atoms combined with 3 moles of oxygen
atoms
o (or 1.5 moles of O2 molecules).
For calculation purposes learn the following formula for 'Z' and use a triangle if necessary.
(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.
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Example 12.7.2.1: How many moles of potassium ions and bromide ions in 0.25 moles of
potassium bromide?
o 1 mole of KBr contains 1 mole of potassium ions (K +) and 1 mole of bromide ions (Br-).
o So there will be 0.25 moles of each ion.
o
Example 12.7.2.2: How many moles of calcium ions and chloride ions in 2.5 moles of calcium
chloride?
o 1 mole of CaCl2 consists of 1 mole of calcium ions (Ca2+) and 2 moles of chloride ion (Cl-).
o So there will be 2.5 x 1 = 2.5 moles of calcium ions and 2.5 x 2 = 5 moles chloride ions.
o
Example 12.7.2.3: How many moles of lead and oxygen atoms are needed to make 5 moles
of lead dioxide?
o 1 mole of PbO2 contains 1 mole of lead combined with 2 moles of oxygen atoms (or 1 mole
of oxygen molecules O2).
o 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.
o
Example 12.7.2.4: How many moles of aluminium ions and sulphate ions in 2 moles of
aluminium sulphate?
o 1 mole of Al2(SO4)3 contains 2 moles of aluminium ions (Al3+) and 3 moles of sulphate ion
(SO42-).
o So there will be 2 x 2 = 4 mol aluminium ions and 2 x 3 = 6 mol of sulphate ion.
o
Example 12.7.2.5: How many moles of chlorine gas in 6.5g? Ar(Cl) = 35.5)
o chlorine consists of Cl2 molecules, so Mr = 2 x 35.5 = 71
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o moles chlorine = mass / Mr = 6.5 / 71 = 0.0944 mol
o
Example 12.7.2.6: How many moles of iron in 20g? (Fe = 56)
o iron consists of Fe atoms, so moles iron = mass/Ar = 20/56 = 0.357 mol Fe
o
Example 12.7.2.7: How many grams of propane C3H8 are there in 0.21 moles of it? (C = 12,
H = 1)
o Mr of propane = (3 x 12) + (1 x 8) = 44, so g propane = moles x M r = 0.21 x 44 = 9.24g
o
Example 12.7.2.8: 0.25 moles of molecule X was found to have a mass of 28g. Calculate its
molecular mass.
o Mr = mass X / moles of X = 28 / 0.25 = 112
o
Example 12.7.2.9: What mass and moles of magnesium chloride is formed when 5g of
magnesium oxide is dissolved in excess hydrochloric acid?
o reaction equation: MgO + 2HCl ==> MgCl2 + H2O
o means 1 mole magnesium oxide forms 1 mole of magnesium chloride (1 : 1 molar ratio)
o formula mass MgCl2 = 24+(2x35.5) = 95,
o MgO = 24+16 = 40, 1 mole MgO = 40g, so 5g MgO = 5/40 = 0.125 mol
o which means 0.125 mol MgO forms 0.125 mol MgCl2,
o Mass = moles x formula mass = 0.125 x 95 = 11.9g MgCl2
o
Example 12.7.2.10: What mass and moles of sodium chloride is formed when 21.2g of
sodium carbonate is reacted with excess dilute hydrochloric acid?
o reaction equation: Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
o 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
o moles Na2CO3 = 21.2/106 = 0.2 mole
o therefore 2 x 0.2 = 0.4 mol of NaCl formed.
o 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.
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Example 12.7.3.1: How many water molecules are there in 1g of water, H 2O ?
o formula mass of water = (2 x 1) + 16 = 18
o every mole of a substance contains 6 x 1023 particles of 'it' (the Avogadro Constant).
o moles water = 1 / 18 = 0.0556
o molecules of water = 0.0556 x 6 x 1023 = 3.34 x 1022
o Since water has a density of 1g/cm3, it means in every cm3 or ml there are
o 33 400 000 000 000 000 000 000 individual H 2O molecules or particles.
o
Example 7.3.2: How many atoms of iron (Fe = 56) are there in an iron filing of mass
0.001g ?
o 0.001g of iron = 0.001 / 56 = 0.00001786 mol
o atoms of iron in the nail = 0.00001786 x 6 x 1023 = 1.07 x 10 19 actual Fe atoms
o (10.7 million million million atoms!)
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o
Example 7.3.2: (a) How many particles of 'Al2O3' in 51g of aluminium oxide?
o Atomic masses: Al =27, O = 16, f. mass Al2O3 = (2x27) + (3x16) = 102
o moles 'Al2O3' = 51/102 = 0.5 mol
o Number of 'Al2O3' particles = 0.5 x 6 x 1023 = 3 x 1023
o (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.
o For every Al2O3 there are two Al3+ and three O2- ions.
o So in 51g of Al2O3 there are ...
o 0.5 x 2 x 6 x 1023 = 6 x 1023 Al3+ ions, and
o 0.5 x 3 x 6 x 1023 = 9 x 1023 O2- ions.
Extra Advanced Questions
QA7.1 This question involves using the mole concept in a variety of situations.
The Avogadro Constant = 6.02 x 1023 mol-1. The molar volume for gases is 24dm3 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 and (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 cm 3 of 0.05
mol dm-3 hydrochloric acid.
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12.8. Using moles to calculate empirical formula and molecular formula
The basis of this method, is that a mole of defined species has the same number of defined 'particles'
in it. So calculating the mole ratio of atoms in the combination, gives you the actual atomic ratio in the
compound. This ratio is then expressed in the simplest whole number atomic ratio, which is the
empirical formula. (it means the formula 'from experiment', see examples 1 and 2).
However, there is one problem to resolve for covalent molecular compounds. The molecular formula is
the summary of all the atoms in one individual molecule - so the molecular formula might not be the
same as the empirical formula! In order to deduce the molecular formula from the empirical formula, you
ALSO need to know the molecular mass of the molecule from another data source. See example 3, which
also illustrates the calculation using % element composition AND the method is no different for more
than two elements!
Examples of where the empirical formula is the same as the molecular formula ...
water H2O, methane CH4, propane C3H8 (these molecular formula cannot be 'simplified')
Examples of where the molecular formula is different from the empirical formula in () ...
e.g. ethane C2H6 (CH3), phosphorus(V) oxide P4O10 (P2O5), benzene C6H6 (CH)
Three examples are set out below to illustrate all the situations. The relative atomic masses of the
elements (Ar) are given in the tabular format method of solving the problem.
Example 12.8.1: 1.15g of sodium reacted with 0.8g of sulphur. Calculate the empirical formula of sodium
sulphide.
RATIOS ...
Sodium Na (Ar = 23)
Sulphur S (Ar = 32)
Comments and tips
Reacting mass
1.15g
0.80g
not the real atom
ratio
moles (mass in g / Ar)
1.15 / 23 = 0.05 mol
0.8 / 32 = 0.025 mol
simplest whole number
ratio by trial and
error
0.05 / 0.025 = 2
0.025 / 0.025 = 1
or 0.05 x 40 = 2
or 0.025 x 40 = 1
can now divide by
smallest ratio number
or scale up by x factor
to get simplest whole
number ratio
therefore the simplest ratio = empirical formula for sodium sulphide = Na2S
Example 12.8.2: 1.35g of aluminium was heated in oxygen until there was no further gain in weight. The
white oxide ash formed weighed 2.55g. Deduce the empirical formula of aluminium oxide. Note: to get
the mass of oxygen reacting, all you have to do is to subtract the mass of metal from the mass of the
oxide formed.
RATIOS ...
Aluminium Al (Ar = 27)
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Oxygen O (Ar = 16)
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Reacting mass
1.35g
2.55 - 1.35 = 1.2g
moles (mass in g / Ar)
1.35 / 27 = 0.05 mol
1.2 / 16 = 0.075 mol
0.05 / 0.05 = 1
0.075 / 0.05 = 1.5
(then x 2 = 2)
(then x 2 = 3)
or 0.05 x 40 = 2
or 0.075 x 40 = 3
simplest whole number
ratio by trial and
error
not the real atom
ratio
can now divide by
smallest ratio number
or scale up by x factor
to get simplest whole
number ratio
therefore the simplest ratio = empirical therefore the formula for aluminium oxide = Al2O3
Example 12.8.3: A chlorinated hydrocarbon compound when analysed, consisted of 24.24% carbon,
4.04% hydrogen, 71.72% chlorine. The molecular mass was found to be 99 from another experiment.
Deduce the empirical and molecular formula. (you can 'treat' the %'s as if they were grams, and it all
works out like examples 1 and 2)
RATIOS ...
Carbon (Ar = 12)
Hydrogen (Ar = 1)
Chlorine (Ar =
35.5)
Comments and
tips
Reacting mass or
% mass
24.24
4.04
71.72
just think of it as
based on 100g
4.04 / 1 = 4.04
mol
71.72 / 35.5 =
2.02 mol
4.04 / 2.02 = 2
2.02 / 2.02 = 1
molar ratio (mass 24.24 / 12 = 2.02
in g / Ar)
mol
simplest whole
number ratio
2.02 / 2.02 = 1
can now divide by
smallest ratio
number
therefore the simplest ratio = empirical formula for the chlorinated hydrocarbon = CH2Cl
BUT the molecular mass is 99, and the empirical formula mass is 49.5 (12+2+35.5)
AND 99 / 49.5 = 2, and so the molecular formula must be 2 x CH2Cl = C2H4Cl2
12.9. The molar gas volume in calculations
Avogadro's Law states that equal volumes of gases under the same conditions of temperature and
pressure contain the same number of molecules. So the volumes have equal moles of separate particles
in them. One mole of any gas (or the formula mass in g), at the same temperature and pressure occupies
the same volume . This is 24dm3 (24 litres) or 24000 cm3, at room temperature and pressure.
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Some handy relationships for substance Z below:
moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)
o mass of Z in g = moles of Z x atomic or formula mass of Z
o atomic or formula mass of Z = mass of Z / moles of Z
o 1 mole = formula mass of Z in g.
gas volume of Z = moles of Z x molar volume
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o moles of Z = gas volume of Z / molar volume
Note (i): In the following examples, assume you are dealing with room temperature and pressure
i.e. 25oC and 1 atmosphere pressure.
Note (ii):
o Apart from solving the problems using the mole concept (method (a) below, and reading
any equations involved in a 'molar way' ...
o It is also possible to solve them without using the mole concept (method (b) below). You
still use the molar volume itself, but you think of it as the volume occupied by the
formula mass of the gas in g and never think about moles!
Example 12.9.1: What is the volume of 3.5g of hydrogen? [Ar(H) = 1]
o common thinking: hydrogen exists as H2 molecules, so Mr(H2) = 2, so 1 mole or formula
mass in g = 2g
o method (a)
 so moles of hydrogen = 3.5/2 = 1.75 mol H2
 so volume H2 = mol H2 x molar volume = 1.75 x 24 = 42 dm3 (or 42000 cm3)
o method (b):
 2g occupies 24 dm3, so scaling up for the volume of hydrogen ...
 3.5 g will have a volume of 3.5/2 x 24 = 42 dm3 (or 42000 cm3)

Example 12.9.2: Given the equation MgCO3(s) + H2SO4(aq) ==> MgSO4(aq) + H2O(l) +CO2(g)
o What mass of magnesium carbonate is needed to make 6 dm 3 of carbon dioxide?
[Ar's: Mg = 24, C = 12, O = 16, H =1 and S = 32]
o method (a):
 since 1 mole = 24 dm3, 6 dm3 is equal to 6/24 = 0.25 mol of gas
 From the equation, 1 mole of MgCO3 produces 1 mole of CO2, which occupies a
volume of 24 dm3.
 so 0.25 moles of MgCO3 is need to make 0.25 mol of CO2
 formula mass of MgCO3 = 24 + 12 + 3x16 = 84,
 so required mass of MgCO3 = mol x formula mass = 0.25 x 84 = 21g
o method (b):
 converting the equation into the required reacting masses ..
 formula masses: MgCO3 = 84 (from above), CO2 = 12 + 2x16 = 44
 MgCO3 : CO2 equation ratio is 1 : 1
 so 84g of MgCO3 will form 44g of CO2
 44g of CO2 will occupy 24dm3
 so scaling down, 6 dm3 of CO2 will have a mass of 44 x 8/24 = 11g
 if 84g MgCO3 ==> 44g of CO2, then ...
 21g MgCO3 ==> 11g of CO2 by solving the ratio, scaling down by factor of 4

Example 12.9.3: 6g of a hydrocarbon gas had a volume of 4.8 dm3. Calculate its molecular
mass.
o method (a):
 1 mole = 24 dm3, so moles of gas = 4.8/24 = 0.2 mol
 molecular mass = mass in g / moles of gas
 Mr = 6 / 0.2 = 30
 i.e. if 6g = 0.2 mol, 1 mol must be equal to 30g by scaling up
o method (b):
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 6g occupies a volume of 4.8 dm3
 the formula mass in g occupies 24 dm3
 so scaling up the 6g in 4.8 dm3
 there will be 6 x 24/4.8 = 30g in 24 dm3
 so the molecular or formula mass = 30

Example 12.9.4: Given the equation ... (and Ar's Ca = 40, H = 1, Cl = 35.5)
o Ca(s) + 2HCl(aq) ==> CaCl2(aq) + H2(g)
o What volume of hydrogen is formed when (i) 2g of calcium is dissolved in excess
hydrochloric acid?, (ii) 0.25 moles of hydrochloric acid reacts with calcium?
o (i) method (a):
 3g Ca = 3/40 = 0.075 mol Ca
 from 1 : 1 ratio in equation, 1 mol Ca produces 1 mol H 2
 so 0.075 mol Ca produces 0.075 mol H2
 so volume H2 = 0.075 x 24 = 1.8 dm3 (or 1800 cm3)
o (i) method (b):
 from equation 1 Ca ==> 1 H2 means 40g ==> 2g
 so scaling down: 3g Ca will produce 2 x 3/40 = 0.15g H 2
 2g H2 has a volume 24 dm3, so scaling down ...
 0.15g H2 has a volume of (0.15/2) x 24 = 1.8 dm3 (or 1800 cm3)
o (ii) method (a) only:
 from equation: 2 moles HCl ==> 1 mole H 2 (mole ratio 2:1)
 so 0.25 mol HCl ==> 0.125 mol H2, volume 1 mole gas = 24 dm3
 so volume H2 = 0.125 x 24 = 3 dm3
Example 12.9.5: Given the equation ... (and Ar's Mg = 24, H = 1, Cl = 35.5)
o Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g)
o How much magnesium is needed to make 300 cm3 of hydrogen gas?
o method (a)
 300 cm3 = 300/24000 = 0.0125 mol H2 (since 1 mol of any gas = 24000 cm3)
 from the equation 1 mole Mg ==> 1 mole H2
 so 0.0125 mole Mg needed to make 0.0125 mol H 2
 so mass of Mg = mole Mg x Ar(Mg)
 so mass Mg needed = 0.0125 x 24 = 0.3g
o method (b)
 reaction ratio in equation is 1 Mg ==> 1 H2,
 so reacting mass ratio is 24g Mg ==> 2g H2,
 2g H2 has a volume of 24000 cm3 (volume of formula mass in g)
 so scaling down: mass Mg needed = 24 x (300/24000) = 0.3g

Example 12.9.6: A small teaspoon of sodium hydrogencarbonate (baking soda) weighs 4.2g.
Calculate the moles, mass and volume of carbon dioxide formed when it is thermally
decomposed in the oven. Assume room temperature for the purpose of the calculation.
o 2NaHCO3(s) ==> Na2CO3(s) + H2O(g) + CO2(g)
o Formula mass of NaHCO3 is 23+1+12+(3x16) = 84 = 84g/mole
o Formula mass of CO2 = 12+(2x16) = 44 = 44g/mole (not needed by this method)
 or a molar gas volume of 24000 cm3 at RTP (definitely needed for this method)
o In the equation 2 moles of NaHCO3 give 1 mole of CO2 (2:1 mole ratio in equation)
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o
o
o
Moles NaHCO3 = 4.2/84 = 0.05 moles ==> 0.05/2 = 0.025 mol CO2 on decomposition.
Mass = moles x formula mass, so mass CO2 = 0.025 x 44 = 1.1g CO2
Volume = moles x molar volume = 0.025 x 24000 = 600 cm3 of CO2
12.10. Reacting gas volume ratios of reactants or products
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Avogadro's Law states that 'equal volumes of gases at the same temperature and pressure
contain the same number of molecules' or moles of gas.
This means the molecule ratio of the equation automatically gives us the gas volumes ratio of
reactants and products, if all the gas volumes are measured at the same temperature and
pressure.
This calculation only applies to gaseous reactants or products AND if they are all at the
same temperature and pressure.
Example 12.10.1: HCl(g) + NH3(g) ==> NH4Cl(s)
o 1 volume of hydrogen chloride will react with 1 volume of ammonia to form solid
ammonium chloride
o e.g. 25cm3 + 25cm3 ==> products or 400dm3 + 400 dm3 ==> products (no gas formed)
o
Example 12.10.2: N2(g) + 3H2(g) ==> 2NH3(g)
o 1 volume of nitrogen reacts with 3 volumes of hydrogen to produce 2 volumes of ammonia
o e.g. 50 cm3 nitrogen reacts with 150 cm3 hydrogen (3 x 50) ==> 100 cm3 of ammonia (2 x
50)
o
Example 12.10.3: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
o (a) What volume of oxygen is required to burn 25cm3 of propane, C3H8.
 Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5 for burning the fuel
propane.
 so actual ratio is 25 : 5x25, so 125cm3 oxygen is needed.
o (b) What volume of carbon dioxide is formed if 5dm3 of propane is burned?
 Theoretical reactant-product volume ratio is C3H8 : CO2 is 1 : 3
 so actual ratio is 5 : 3x5, so 15dm3 carbon dioxide is formed.
o © What volume of air (1/5th oxygen) is required to burn propane at the rate of 2dm3 per
minute in a gas fire?
 Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5
 so actual ratio is 2 : 5x2, so 10dm3 oxygen per minute is needed,
 therefore, since air is only 1/5th O2, 5 x 10 = 50dm3 of air per minute is required.
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12.11. Molarity and the concentration of solutions
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Revise section 12.7. moles and mass before proceeding in this section 11 and eventually you
may need to be familiar with the use of the apparatus illustrated above, some of which
gives great accuracy when dealing with solutions and some do not.
It is very useful to be know exactly how much of a dissolved substance is present in a solution of
particular concentration or volume of a solution. So we need a standard way of comparing the
concentrations of solutions.
The concentration of an aqueous solution is usually expressed in terms of moles of dissolved
substance per cubic decimetre, mol dm-3, this is called molarity, sometimes denoted in
shorthand as M.
o Note: 1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing cm3/1000 gives dm3, which is
handy to know since most volumetric laboratory apparatus is calibrated in cm 3 (or ml),
but solution concentrations are usually quoted in molarity, that is mol/dm 3 (mol/litre).
Equal volumes of solution of the same molar concentration contain the same number of
moles of solute i.e. the same number of particles as given by the chemical formula.
You need to be able to calculate
o the number of moles or mass of substance in an aqueous solution of given volume and
concentration
o the concentration of an aqueous solution given the amount of substance and volume of
water.
o use the equation (1) molarity of Z = moles of Z / volume in dm3
o You may also need to know that ...
 (2) molarity x formula mass of solute = solute concentration in g/dm3,
 and dividing this by 1000 gives the concentration in g/cm3, and
 (3) (concentration in g/dm3) / formula mass = molarity in mol/dm3,
 both equations (2) and (3) result from equations (1) and (4), work it out
for yourself.
o and don't forget by now you should know:
 (4) moles Z = mass Z / formula mass of Z
 and (5) 1 mole = formula mass in grams
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Example:
How many grams of carbon are required to
react completely with 100 g Fe2O3?
 Fe2O3(s) + 3C(s) 2Fe(l) + 3CO(g)
 Step 1: Write a balanced chemical
equation (or check to see that a given
equation is balanced). In this case, a
balanced chemical equation was given.
Organize the information in the problem.
It's often helpful to write the amounts
given underneath the balanced chemical
equation.
 Fe2O3(s) + 3C(s) ------------2Fe(l) +
3CO(g)
 100 g
?g
 Step 2: Convert grams of a given
substance to moles. Remember that
substances react in terms of their mole
ratios, not their mass ratios. To convert
grams of Fe2O3 to moles, we need to know
the molar mass of this compound.
 2 x 56. g for each mol Fe + 3 x 16.0 g for
each mol O
 = 160 g/mol
Step 3: Use coefficients in the balanced chemical equation to find the mole ratio. Relate moles
of what you were given to moles of what you are determining using the mole ratio.
Step 4: Convert moles to grams using molar mass as a conversion factor. It's always a good idea
to check to make sure you have answered the question you were asked. Here you were asked to
calculate grams of carbon. Another step or two would be necessary if you had been asked to
report your answer in some other unit, such as kg.
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Example 12.11.1
o What mass of sodium hydroxide (NaOH) is needed to make up 500 cm 3 (0.5 dm3) of
a 0.5M solution? [Ar's: Na = 23, O = 16, H = 1]
o 1 mole of NaOH = 23 + 16 + 1 = 40g
o for 1000 cm3 (1 dm3) of 0.5M you would need 0.5 moles NaOH
o which is 0.5 x 40 = 20g
o however only 500 cm3 of solution is needed compared to 1000 cm3
o so scaling down: mass NaOH required = 20 x 500/1000 = 10g
o
Example 12.11.2
o How many moles of H2SO4 are there in 250cm3 of a 0.8M sulphuric acid solution?
What mass of acid is in this solution? [Ar's: H = 1, S = 32, O = 16]
o formula mass of sulphuric acid = 2 + 32 + (4x16) = 98, so 1 mole = 98g
o if there was 1000 cm3 of the solution, there would be 0.8 moles H2SO4
o but there is only 250cm3 of solution, so scaling down ...
o moles H2SO4 = 0.8 x (250/1000) = 0.2 mol
o mass = moles x formula mass, which is 0.2 x 98 = 19.6g of H2SO4
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Example 12.11.3
o 5.95g of potassium bromide was dissolved in 400cm3 of water. Calculate its
molarity. [Ar's: K = 39, Br = 80]
o moles = mass / formula mass, (KBr = 39 + 80 = 119)
o mol KBr = 5.95/119 = 0.05 mol
o 400 cm3 = 400/1000 = 0.4 dm3
o molarity = moles of solute / volume of solution
o molarity of KBr solution = 0.05/0.4 = 0.125M
o
Example 12.11.4
o What is the concentration of sodium chloride (NaCl) in g/dm3 and g/cm3 in a 1.50
molar solution?
o At. masses: Na = 23, Cl = 35.5, formula mass NaCl = 23 + 35.5 = 58.5
o Therefore concentration = 1.5 x 58.5 = 87.8 g/dm3, and
o concentration = 87.75 / 1000 = 0.0878 g/cm3
o
Example 12.11.5
o A solution of calcium sulphate (CaSO4) contained 0.5g dissolved in 2dm 3 of water.
Calculate the concentration in (a) g/dm3, (b) g/cm3 and (c) mol/dm3.
o (a) concentration = 0.5/2 = 0.25 g/dm3, then since 1dm3 = 1000 cm3
o (b) concentration = 0.25/1000 = 0.00025 g/cm3
o (c) At. masses: Ca = 40, S = 32, O = 64, f. mass CaSO4 = 40 + 32 + (4 x 16) = 136
 moles CaSO4 = 0.5 / 136 = 0.00368 mol
 concentration CaSO4 = 0.00368 / 2 = 0.00184 mol/dm3
12.12(a). Volumetric calculations e.g. acid-alkali titrations
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Titrations can be used to find the concentration of an acid or alkali from the relative volumes
used and the concentration of one of the two reactants. The method and apparatus used are
briefly described at the end of this page.
You should be able to carry out calculations involving neutralisation reactions in aqueous solution
given the balanced equation or from your own practical results.
The examples in section 12.7. moles and mass. and section 12.11. concentration will help
you follow the calculations below.
o Note again: 1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing cm3/1000 gives dm3.
o and other useful formulae or relationships are:
 moles = molarity (mol/dm3) x volume (dm3=cm3/1000),
 molarity (mol/dm3) = mol / volume (dm3=cm3/1000),
 1 mole = formula mass in grams.
o In most volumetric calculations of this type, you first calculate the known moles of one
reactant from a volume and molarity. Then, from the equation, you relate this to the
number of moles of the other reactant, and then with the volume of the unknown
concentration, you work out its molarity.
o
Example 12.12.1: Given the equation NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l)
25 cm3 of a sodium hydroxide solution was pipetted into a conical flask and titrated with
0.2M hydrochloric acid. Using a suitable indicator it was found that 15 cm3 of acid was
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required to neutralise the alkali. Calculate the molarity of the sodium hydroxide and
concentration in g/dm3.
o moles HCl = (15/1000) x 0.2 = 0.003 mol
o moles HCl = moles NaOH (1 : 1 in equation)
o so there is 0.003 mol NaOH in 25 cm3
o scaling up to 1000 cm3 (1 dm3), there are ...
o 0.003 x (1000/25) = 0.12 mol NaOH in 1 dm3
o molarity of NaOH is 0.12M or mol dm-3
o since mass = moles x formula mass, and Mr(NaOH) = 23 + 16 + 1 = 40
o concentration in g/dm3 is 0.12 x 40 = 4.41g/dm3
o
Example 12.12.2: Given the equation 2KOH(aq) + H2SO4(aq) ==> K2SO4 + 2H2O(l)
20 cm3 of a sulphuric acid solution was titrated with 0.05M potassium hydroxide. If the
acid required 36 cm3 of the alkali KOH for neutralisation what was the concentration of
the acid?
o mol KOH = 0.05 x (36/1000) = 0.0018 mol
o mol H2SO4 = mol KOH / 2 (because of 1 : 2 ratio in equation above)
o mol H2SO4 = 0.0018/2 = 0.0009 (in 20 cm3)
o scaling up to 1000 cm3 of solution = 0.0009 x (1000/20) = 0.045 mol
o mol H2SO4 in 1 dm3 = 0.045, so molarity of H2SO4 = 0.045M or mol dm-3
o since mass = moles x formula mass, and Mr(H2SO4) = 2 + 32 + (4x16) = 98
o concentration in g/dm3 is 0.045 x 98 = 4.41g/dm3
How to carry out a titration?
Fig. 12.
the typical apparatus (1)-(6) used in manipulating
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Fig. 12. a brief three stage description of titrating an acid with an alkali:
1. An accurate volume of acid is pipetted into the conical flasks using a suction bulb for
2.
3.
health and safety reasons. Universal indicator is then added, which turns red in the acid.
The alkali, of known accurate concentration, is put in the burette and you can
conveniently level off the reading to zero (the meniscus on the liquid surface should rest
on the zero -- graduation mark). The alkali is then carefully added by running it out of
the burette in small quantities, controlling the flow with the tap, until the indicator
seems to be going yellow-pale green. The conical flask should be carefully swirled after
each addition of alkali to ensure all the alkali reacts.
Near the end of the titration, the alkali should added drop-wise until the universal
indicator goes green. This is called the end-point of the titration and the green means
that all the acid has been neutralised. The volume of alkali needed to titrate-neutralise
the acid is read off from burette scale, again reading the volume value on the underside
of the meniscus. The calculation can then be done to work out the concentration of the
alkali.
12.12(b) Reactions with Limiting Amount of Reactants
In actual chemical reactions, one or more reactants may be in excess. The limiting reactant will be
consumed completely and limit the amount of product formed. The following exercise provides a
simplified view of how limiting reactants affect a chemical reaction.
Example:
30 g NO2 and 10 g H2O react as shown below.
3 NO2(g) + H2O(l) ---------------- 2 HNO3(l) + NO(g)
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What is the limiting reactant?
When two or more reactants are present, one reactant must be limiting. To determine which reactant is
limiting, we need to look at the mole ratio of the reactants involved.
The mole ratio of the two reactants is
0.652 mol NO2/0.555 mol H2O = 1.17
According to the stoichiometry of the balanced equation, the mole ratio should be
3 mol NO2/1 mol H2O = 3.
We see that there is not enough NO2, and thus NO2 is the limiting reactant. H2O is the excess
reactant.
b. What amount of HNO3 forms under these conditions?
Once the limiting reactant is consumed, no additional product can be formed. We therefore use the
limiting reactant to calculate the amount of product.
c. What amount of NO2 and H2O remain?
All of the limiting reactant is consumed, so no NO2 remains. Stoichiometry will allow us to calculate the
amount of H2O remaining by first determining how much H2O reacts.
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CHEMISTRY NOTES
12.13. Electrolysis product calculations (negative cathode and positive anode
products)
Relative atomic masses needed: Na = 23, Cl = 35.5, H = 1, Cu = 63.5, Al = 27, O = 16 and the
molar volume of any gas is 24 dm3 at room temperature and pressure. The common electrode
equations you may come across are listed below.
electrode involved: (-) negative
cathode or (+) positive anode for the
Electrode Equation below
sodium (-) Na+(l) + e- ==> Na(l)
chlorine (+) 2Cl
-
(l/aq)
-
- 2e ==> Cl2(g)
hydrogen (-) 2H+(l) + 2e- ==> H2(g)
2+
copper (-) Cu
(aq)
-
+ 2e ==> Cu(s)
moles of electrons
involved (mass of
product formed)
example of industrial process where
this electrode reaction happens
1 (23g) = 1.0 mol Na electrolysis of molten chloride salts to
metal per mol e-s
make chlorine and the metal
2 (71g) = 0.5 mol Cl2
electrolysis of molten chloride salts or
gas (12 dm3)
their aqueous solution to make chlorine
released per mol e-s
2 (2g) = 0.5 mol H2
gas (12 dm3)
released per mol e-s
electrolysis of many salt solutions to
make hydrogen
2 (63.5g) = 0.5 mol
deposition of copper in its electrolytic
Cu deposited per mol
purification or electroplating
e-s
copper (+) Cu(s) - 2e- ==> Cu2+(aq)
2 (63.5g) = 0.5 mol
Cu dissolves per mol
e-s
dissolving of copper in its electrolytic
purification or electroplating
aluminium (-) Al3+(l) + 3e- ==> Al(l)
3 (27g) = 0.33 mol
Al metal per mol e-s
extraction of aluminium in the
electrolysis of its molten oxide ore
4 (32g) = 0.25 mol
O2 (12 dm3) gas
released per mol e-s
electrolysis of molten oxides
2-
oxygen (+) 2O
(l)
-
- 4e ==> O2(g)
4 (32g) = 0.25 mol
oxygen (+) 4OH-(aq) - 4e- ==> 2H2O(l) +
O2 gas (6 dm3)
O2(g)
released per mol e-s
electrolysis of many salt solutions such
as sulphates, sulphuric acid etc. gives
oxygen (but chloride salts ==> chlorine)
Part one: The (+) anode and (-) cathode electrode product ratio

The amount of material in moles formed at the electrode in electrolysis depends on three
factors.
o The charge on the ion. (compare the effect of one mole of electrons in the table above
and see examples 13.1.1 to 13.1.5 below in Part one)
o The current flow. (the current flowing in amperes, A, see examples in Part two)
o The time duration of the electrolysis. (time in seconds, minutes or hours, see examples
in Part two)
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If you know how much of a substance is made at one electrode, you can theoretically calculate
the amount of substance formed at the other electrode.
The basis of these calculations is the ratio of the electrons involved in both electrode reactions
(hence the introductory table of electrode equations above).
These electrode equations in the table above are referred to in the examples below.
In studying the examples below you must refer to the electrode equations in the table above,
o and remember 1 mole = formula mass in grams and 1 mol of gas = 24 dm3 at room
temperature/pressure.
o
Example 12.13.1.1: The electrolysis of brine, aqueous sodium chloride solution, NaCl (aq)
produces hydrogen gas, H2(g) at the -ve electrode and chlorine gas, Cl2(g) at the positive
electrode. Atomic masses: H = 1, Cl = 35.5
o 2H+(l) + 2e- ==> H2(g) and 2Cl-(l/aq) - 2e- ==> Cl2(g)
o 2 electrons are involved in both the formation of a hydrogen molecule [M r(H2) = 2] or a
chlorine molecule [Mr(Cl2) = 71].
o The ratio of the products for H2(g) : Cl2(g) is 1 mol : 1 mol or 24dm3 : 24 dm3 or 2g :
71g
o
Example 12.13.1.2: The electrolysis of molten aluminium oxide Al2O3 is a more complicated
affair.
o Its best to think of the ratio effect of a current of 12 moles of electrons passing
through the electrolyte.
 4Al3+(l) + 12e- ==> 4Al(l) and 6O2-(l) - 12e- ==> 3O2(g)
 It takes 3 moles of electrons to form 1 mole of Al from 1 mole of Al 3+ ions.
 and 4 moles of electrons to form 1 mole of O2 molecules from 2 moles of O2- ions.
 Atomic masses: Al = 27, O = 16
o The ratio of the products from 12 moles of electrons is therefore

Al(l) : O2(g) is 4 mol : 3 mol or 108g : 96g or 72dm3.

Example 12.13.1.3: In the electrolysis of dilute sulphuric acid, 36 cm3 of hydrogen, H2 was
formed at the negative electrode (cathode). What volume of oxygen, O 2 would be formed at the
positive electrode (anode)?
o 2H+(l) + 2e- ==> H2(g) and 4OH-(aq) - 4e- ==> 2H2O(l) + O2(g)
o It takes an electron transfer of 2 electrons to form each hydrogen molecule from 2
hydrogen, H+ ions and the transfer of 4 electrons to make 1 molecule of oxygen from 4
hydroxide, OH- ions.
o Therefore, from the same amount of electrons (current), the ratio of hydrogen : oxygen
formed is 2 : 1
o so the volume of oxygen formed is 18 cm3. (36 : 18 have the ratio 2 : 1)
o
Example 12.13.1.4: In the electrolysis of copper sulphate solution using carbon electrodes,
what mass and volume of oxygen would be formed at the positive electrode if 254g of copper
was deposited on the negative electrode? Atomic masses: Cu = 63.5, O = 16.
o Cu2+(aq) + 2e- ==> Cu(s) and 4OH-(aq) - 4e- ==> 2H2O(l) + O2(g)
o It takes a transfer of 2 moles of electrons to form 1 mole of solid copper (63.5g) from 1
mole of copper(II) ions, Cu2+
o and a transfer of 4 moles of electrons to form 1 mole of oxygen from 4 moles of
hydroxide, OH- ions.
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o Therefore the expected mole ratio of Cu(s) : O2(g) from the electrolysis is 2 : 1
o The moles of Cu deposited = 254/63.5 = 4 moles
o so moles oxygen formed = 2 moles, since Mr(O2) = 2 x 16 = 32
o mass of oxygen formed = 2 x 32 = 64g, volume of oxygen = 2 x 24 = 48 dm 3
o
Example 12.13.1.5: In the industrial manufacture of aluminium by electrolysis of the molten
oxide (plus cryolite) 250kg of aluminium are formed. What volume of oxygen would be
theoretically formed at room temperature and pressure?
o [ Ar(Al) = 27 and 1 mole of gas at RTP = 24 dm3 (litres) ]
o Aluminium oxide is Al2O3, so on splitting in electrolysis the atomic ratio for Al : O is 2 :
3,
o and a mole ratio of Al : O2 or 4 : 3
o 4Al3+(l) + 12e- ==> 4Al(l) and 6O2-(l) - 12e- ==> 3O2(g)
o Note: It takes 12 electrons added to four Al3+ ions to make four atoms of Al, and 12
electrons removed from six oxide ions, O2-, to form six oxygen atoms, which combine to
form three O2 molecules (see next line).
o BUT, oxygen exists as O2 molecules, so the mole ratio of Al atoms : O2 molecules is 4 :
3
o 250kg Al = 250000g, Al = 250000/27 moles = 9259.26 moles Al metal.
o Therefore scaling for moles O2 = 9259.6 x 3/ 24 = 6944.44 moles O 2 molecules.
o Since volume of 1 mole of gas at RTP = 24 dm3 (litres)
o Volume of oxygen formed = 6944.44 x 24 = 166667 dm3
12.14.1 Percentage purity
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Purity is very important e.g. for analytical standards in laboratories or pharmaceutical products
where impurities could have dangerous side effects in a drug or medicine. However in any
chemical process it is almost impossible to get 100.00% purity and so samples are always
analysed in industry to monitor the quality of the product.
% purity is the percentage of the material which is the actually desired chemical in a
sample of it.
Example 12.14.1.1
o A 12g sample of a crystallised pharmaceutical product was found to contain 11.57g of the
active drug.
o Calculate the % purity of the sample of the drug.
o % purity = actual amount of desired material x 100 / total amount of material
o % purity = 11.57 x 100 / 12 = 96.4% (to 1dp)
o
Example 12.14.1.2
o Sodium chloride was prepared by neutralising sodium hydroxide solution with dilute
hydrochloric acid. The solution was evaporated to crystallise the salt.
o The salt is required to be completely anhydrous, that is, not containing any water.
o The prepared salt was analysed for water by heating a sample in an oven at 110 oC to
measure the evaporation of any residual water.
o The following results were obtained and from them calculate the % purity of the salt.
o Mass of evaporating dish empty = 51.32g.
o Mass of impure salt + dish = 56.47g
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o
o
o
o
Mass of dish + salt after heating = 56.15g
Therefore the mass of original salt = 56.47 - 51.32 = 5.15g
and the mass of pure salt remaining = 56.15 - 51.32 = 4.83g
% salt purity = 4.83 x 100 / 5.15 = 93.8% (to 1dp)
12.14.2a Percentage yield
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The % yield of a reaction is the percentage of the product obtained compared to the
theoretical maximum as calculated from the balanced equation.
In carrying out a chemical preparation, the aim is to work carefully and recover as much of the
desired reaction product as you can, and as pure as is possible and practicable.
However in any chemical process it is almost impossible to get 100% of the product because of
several reasons:
1. the reaction might not be 100% completed because it is reversible reaction and an
equilibrium is established.
2. You always get losses of product as it is separated from the reaction mixture by
filtration, distillation, crystallisation or whatever method is required.
3. Some of the reactants may react in another way to give a different product to the one
you want.
4. The aim is to work carefully and recover as much of the desired reaction product, and as
pure as is possible and practicable
% yield = actual amount of desired chemical obtained x 100 / maximum theoretical amount
formed
Example 12.14.2a.1
o Magnesium metal dissolves in hydrochloric acid to form magnesium chloride.
o Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g)
o Atomic masses : Mg = 24 and Cl = 35.5, and formula mass MgCl 2 = 24 + (2 x 35.5) = 95
o (a) What is the maximum theoretical mass of magnesium chloride which can be made
from 12g of magnesium?
 Reacting mass ratio calculation from the balanced equation:
 1 Mg ==> 1 MgCl2, so 24g ==> 95g or 12g ==> 47.5g MgCl2
o (b) If only 47.0g of purified magnesium chloride was obtained after crystallising the salt
from the solution, what is the % yield of the salt preparation?
 % yield = actual amount obtained x 100 / maximum theoretical amount possible
 % yield = 47.0 x 100 / 47.5 = 98.9% (to 1dp)

Example 12.14.2a.2
o 2.8g of iron was heated with excess sulphur to form iron sulphide.
o Fe + S ==> FeS
o The excess sulphur was dissolved in a solvent and the iron sulphide filtered off, washed
with clean solvent and dried.
o If 4.1g of purified iron sulphide was finally obtained, what was the % yield of the
reaction?
o 1st a reacting mass calculation of the maximum amount of FeS that can be formed:
 Relative atomic/formula masses: Fe =56, FeS = 56 + 32 = 88
 This means 56g Fe ==> 88g FeS, or by ratio, 2.8g Fe ==> 4.4g FeS
 because 2.8 is 1/20th of 56, so theoretically you can get 1/20th of 88g of FeS.
o 2nd the % yield calculation itself.
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 % yield = actual amount obtained x 100 / maximum theoretical amount possible
 % yield = 4.1 x 100 / 4.4 = 93.2% (to 1dp)

Example 12.14.2a.3
o (a) Theoretically how much iron can be obtained from 1000 tonne of pure haematite
ore, formula Fe2O3 in a blast furnace?
o If we assume the iron(III) oxide ore (haematite) is reduced by carbon monoxide, the
equation is:
o Fe2O3(s) + 3CO(g) ==> 2Fe(l) + 3CO2(g)
o (atomic masses: Fe = 56, O = 16)
o For every Fe2O3 ==> 2Fe can be extracted, formula mass of ore = (2 x 56) + (3 x 56) =
160
o Therefore reacting mass ratio is: 160 ==> 112 (from 2 x 56)
o so, solving the ratio, 1000 ==> 112 / 160 = 700 tonne copper = max. can be extracted
o (b) If in reality, only 670 tonne of iron is produced what is the % yield of the
overall blast furnace process?
o % yield = actual yield x 100 / theoretical yield
o % yield = 670 x 100 / 700 = 95.7%
In other words, 4.3% of the iron is lost in waste e.g. in the slag.
12.14.2b Atom economy
The atom economy of a reaction is a theoretical measure of the amount of starting materials that
ends up as 'desired' reaction product. The greater the atom economy of a reaction, the more
'efficient' or 'economic' it is likely to be, though this is a gross simplification when complex and costly
chemical synthesis are looked at. It can be defined numerically in words in several ways, all of which
amount to the same theoretical % number!
mass of atoms desired product
mass of useful product
total Mr of useful product
Atom economy ------------------------------------ ---------------------------- ------------------------------=
- x 100 =
x 100 =
- x 100
total mass of reactant atoms
mass of all products
total Mr of all products
Example 12.14.2b.1
This is illustrated by using the blast furnace reaction from example 14.2a.3 above.
Fe2O3(s) + 3CO(g) ===> 2Fe(l) + 3CO2(g)
Using the atomic masses of Fe = 56, C = 12, O = 16, we can calculate the atom economy for extracting
iron.
the reaction equation can be expressed in terms of theoretical reacting mass units
[(2 x 56) + (3 x 16)] + [3 x (12 + 16)] ===> [2 x 56] + [3 x (12 + 16 + 16)]
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[160 of Fe2O3] + [84 of CO] ===> [112 of Fe] + [132 of CO2]
so there are a total of 112 mass units of the useful/desired product iron, Fe
out of a total mass of reactants or products of 112 + 132 = 244.
Therefore the atom economy = 112 x 100 / 244 = 45.9%
Note: It doesn't matter whether you use the total mass of reactants or the total mass products in the
calculations, they are the same from the law of conservation of mass.
12.14.3 Dilution calculations
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It is important to know how to accurately dilute a more concentrated solution to a specified
solution of lower concentration. It involves a bit of logic using ratios of volumes. The
diagram above illustrates some of the apparatus that might be used when dealing with
solutions.
Example 12.14.3.1
A purchased standard solution of sodium hydroxide had a concentration of 1.0 mol/dm 3.
How would you prepare 100 cm3 of a 0.1 mol/dm3 solution to do a titration of an acid?
o The required concentration is 1/10th of the original solution.
o To make 1dm3 (1000 cm3) of the diluted solution you would take 100 cm3 of the original
solution and mix with 900 cm3 of water.
o The total volume is 1dm3 but only 1/10th as much sodium hydroxide in this diluted
solution, so the concentration is 1/10th, 0.1 mol/dm3.
o To make only 100 cm3 of the diluted solution you would dilute 10cm3 by mixing it with 90
cm3 of water.
o How to do this in practice is described at the end of Example 14.3.2 below and a variety
of accurate/inaccurate apparatus is illustrated above.
o
Example 12.14.3.2
o Given a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare
250cm3 of a 0.5 mol/dm3 solution?
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o
o
o
o
o
o
The required 0.5 mol/dm3 concentration is 1/4 of the original concentration of 2.0
mol/dm3.
To make 1dm3 (1000 cm3) of a 0.5 mol/dm3 solution you would take 250 cm3 of the stock
solution and add 750 cm3 of water.
Therefore to make only 250 cm3 of solution you would mix 1/4 of the above quantities i.e.
mix 62.5 cm3 of the stock solution plus 187.5 cm3 of pure water.
This can be done, but rather inaccurately, using measuring cylinders and stirring to mix
the two liquids in a beaker.
It can be done much more accurately by using a burette or a pipette to measure out the
stock solution directly into a 250 cm3 graduated-volumetric flask.
Topping up the flask to the calibration mark (meniscus should rest on it). Then putting on
the stopper and thoroughly mixing it by carefully shaking the flask holding the stopper
on at the same time!
12.14.4 Water of crystallisation
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Example 12.14.4.1: Calculate the % of water in hydrated magnesium sulphate MgSO 4.7H2O
o Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1
o relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246
o 7 x 18 = 126 is the mass of water
o so % water = 126 x 100 / 246 = 51.2%
o
Example 12.14.4.2 The % water of crystallisation and the formula of the salt are
calculated as follows:
o Suppose 6.25g of blue hydrated copper(II) sulphate, CuSO4.xH2O, (x unknown) was
gently heated in a crucible until the mass remaining was 4.00g. This is the white
anhydrous copper(II) sulphate.
o The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.254.00 = 2.25g
o The % water of crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%
o [ Ar's Cu=64, S=32, O=16, H=1 ]
o The mass ratio of CuSO4 : H2O is 4.00 : 2.25
o To convert from mass ratio to mole ratio, you divide by the molecular mass of each
'species'
o CuSO4 = 64 + 32 + (4x18) = 160 and H2O = 1+1+16 = 18
o The mole ratio of CuSO4 : H2O is 4.00/160 : 2.25/18
o which is 0.025 : 0.125 or 1 : 5, so the formula of the hydrated salt is CuSO4.5H2O
12.14.5 Calculation of quantities required for a reaction and %
yield (atom economy)
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Suppose you want to make 25g of the blue hydrated copper(II) sulphate crystals.
o Your reactants are dilute sulphuric acid and the black solid copper(II) oxide.
o copper(II) oxide + sulphuric acid ==> copper(II) sulphate + water
o CuO(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l)
o on crystallisation you get the blue hydrated crystals
How much copper(II) oxide is needed?
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A 'non-moles' calculation first of all, involving a reacting mass calculation.
The change overall is CuO ==> CuSO4.5H2O
Atomic masses: Cu = 64, S = 32, H = 1, O = 16
Formula masses are for: CuO = 64+16 = 80 and
CuSO4.5H2O = 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250
The reacting mass ratio is: 80 ==> 250 since formula ratio is 1:1
Therefore, theoretically to make 50g of the crystals (1/5th of 250)
So you need 1/5th of 80g of copper(II) oxide,
and 80/5 = 16g of copper(II) oxide is required.
However, in reality, things are not so simple because the method involves adding excess
copper(II) oxide to the dilute sulphuric acid.
o So in practice you would need to take more of the CuO to get 25g of the salt crystals.
o However, there is another way to calculate the quantities required.
How much dilute sulphuric acid (of concentration 1 mol dm-3) is required?
o Mol = mass in g / formula mass,
o so moles of CuSO4.5H2O required = 25/250 = 0.1 mol (see basics of moles)
o Therefore 0.1 mol of H2SO4 is required.
o 1dm3 (1000 cm3) of a 1 mol dm-3 solution of contains 1 mole (by definition o Therefore 1/10th of 1dm3 is required, so 100 cm3 of 1 mol dm-3 dilute sulphuric acid
is required.
o You would then add copper(II) oxide until no more dissolves and the excess is filtered
off.
Suppose after carrying out the preparation you finally crystallise 17g of pure the blue
crystals of CuSO4.5H2O and weigh the product when dry.
What is the 'atom economy' of the preparation?
That is, what is the % yield? i.e. comparing what you actually get with the maximum possible.
o % yield = mass of product obtained x 100 / theoretical mass from the equation
o % yield = 17 x 100 / 25 = 68%
o
o
o
o
o
o
o
o
o
o
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12.15 Calculation of heat transfer using bond energies (bond enthalpies)
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Atoms in molecules are held together by chemical bonds which are the electrical attractive
forces between the atoms.
The bond energy is the energy involved in making or breaking bonds and is usually quoted in kJ
per mole of the particular bond involved.
To break a chemical bond requires the molecule to take in energy to pull atoms apart, which is an
endothermic change. The atoms of the bond vibrate more until they spring apart.
To make a chemical bond, the atoms must give out energy to become combined and electronically
more stable in the molecule, this is an exothermic change.
The energy to make or break a chemical bond is called the bond energy and is quoted in kJ/mol
of bonds.
o Each bond has a typical value e.g. to break 1 mole of C-H bonds is on average about
413kJ,
o the C=O takes an average 743 kJ/mol, and note the stronger double bond, so more
energy is needed,
o and not surprisingly, a typical double bond needs more energy to break than a typical
single bond.
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During a chemical reaction, energy must be supplied to break chemical bonds in the molecules,
this the endothermic 'upward' slope on the reaction profile on diagrams below.
When the new molecules are formed, new bonds must be made in the process, this is the
exothermic 'downward' slope on the reaction profile on diagrams above.
If we know all the bond energies of the molecules involved in a reaction, we can theoretically
calculate what the net energy change is for that reaction and determine whether the reaction is
exothermic or endothermic.
We do this by calculating the energy taken in to break the bonds in the reactant molecules. We
then calculate the energy given out when the new bonds are formed. The difference between
these two gives us the net energy change.
o In a reaction energy must be supplied to break bonds (energy absorbed, taken in,
endothermic).
o Energy is released when new bonds are formed (energy given out, releases, exothermic).
o If more energy is needed to break the original existing bonds of the reactant molecules,
than is given out when the new bonds are formed in the product molecules, the reaction
is endothermic.
o If less energy is needed to break the original existing bonds of the reactant molecules,
than is given out when the new bonds are formed in the product molecules, the reaction
is exothermic.
o So the overall energy change for a reaction (ΔH) is the overall energy net change from
the bond making and bond forming processes. These ideas are illustrated in the
calculations below.
o
Example 12.15.1 Hydrogen + Chlorine ==> Hydrogen Chloride
o The symbol equation is: H2(g) + Cl2(g) ==> 2HCl(g)
o but think of it as: H-H + Cl-Cl ==> H-Cl + H-Cl
o (where - represents the chemical bonds to be broken or formed)
o the bond energies in kJ/mol are: H-H 436; Cl-Cl 242; H-Cl 431
o Energy needed to break bonds = 436 + 242 = 678 kJ taken in
o Energy released on bond formation = 431 + 431 = 862 kJ given out
o The net difference between them = 862-678 = 184 kJ given out (92 kJ per mole of
HCl formed)
o More energy is given out than taken in, so the reaction is exothermic.
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Example 12.15.2 Hydrogen Bromide ==> Hydrogen + Bromine
o The symbol equation is: 2HBr(g) ==> H2(g) + Br2(g)
o but think of it as: H-Br + H-Br ==> H-H + Br-Br
o (where - represents the chemical bonds to be broken or formed)
o the bond energies in kJ/mol are: H-Br 366; H-H 436; Br-Br 193
o Energy needed to break bonds = 366 + 366 = 732 kJ taken in
o Energy released on bond formation = 436 + 193 = 629 kJ given out
o The net difference between them = 732-629 = 103 kJ taken in (51.5kJ per mole
of HBr decomposed)
o More energy is taken in than given out, so the reaction is endothermic
o
Example 12.15.3 hydrogen + oxygen ==> water
o 2H2(g) + O2(g) ==> 2H2O(g)
o or 2 H-H + O=O ==> 2 H-O-H (where - or = represent the covalent bonds)
o bond energies in kJ/mol: H-H is 436, O=O is 496 and O-H is 463
o bonds broken and energy absorbed (taken in):
o (2 x H-H) + (1 x O=O) = (2 x 436) + (1 x 496) = 1368 kJ
o bonds made and energy released (given out):
o (4 x O-H) = (4 x 463) = 1852 kJ
o overall energy change is:
o 1852 - 1368 = 484 kJ given out (242 kJ per mole hydrogen burned or water
formed)
o since more energy is given out than taken in, the reaction is exothermic.
o NOTE: Hydrogen gas can be used as fuel and a long-term possible alternative to
fossil fuels (see methane combustion below in example 12.15.5..
 It burns with a pale blue flame in air reacting with oxygen to be oxidised to
form water.
 hydrogen + oxygen ==> water
 2H2(g) + O2(g) ==> 2H2O(l)
 It is a non-polluting clean fuel since the only combustion product is water
and so its use would not lead to all environmental problems associated with
burning fossil fuels.
 It would be ideal if it could be manufactured by electrolysis of water e.g.
using solar cells.
 Hydrogen can be used to power fuel cells.

Example 12.15.4 nitrogen + hydrogen ==> ammonia
o N2(g) + 3H2(g) ==> 2NH3(g)
o
o
o
o
o
o
or N N + 3 H-H ==> 2
bond energies in kJ/mol: N N is 944, H-H is 436 and N-H is 388
bonds broken and energy absorbed (taken in):
 (1 x N N) + (3 x H-H) = (1 x 944) + (3 x 436) = 2252 kJ
bonds made and energy released (given out):
 2 x (3 x N-H) = 2 x 3 x 388 = 2328 kJ
overall energy change is:
 2328 - 2252 = 76 kJ given out (38 kJ per mole of ammonia formed)
since more energy is given out than taken in, the reaction is exothermic.
CHAPTER 13: CHEMICAL CALCULATIONS
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CHEMISTRY NOTES

o
Example 12.15.5 methane + oxygen ==> carbon dioxide + water
o CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(g)
or
bond energies in kJ/mol:
 C-H single bond is 412, O=O double bond is 496, C=O double bond is 743,
H-O single bond is 463
o bonds broken and heat energy absorbed from surroundings, endothermic change
 (4 x C-H) + 2 x (1 x O=O) = (4 x 412) + 2 x (1 x 496) = 1648 + 992 =
2640 kJ taken in
o bonds formed and heat energy released and given out to surroundings, exothermic
change
 (2 x C=O) + 2 x (2 x O-H) = (2 x 743) + 2 x (2 x 463) = 1486 + 1852 =
3338 given out
o overall energy change is:
 3338 - 2640 = 698 kJ/mol given out per mole methane burned,
 since more energy is given out than taken in, the reaction is exothermic.

Example 12.15.6 analysing the bonds in more complex molecules
o
o





2
8
2
1
x
x
x
x
ethyl ethanoate
C-C single covalent bonds
C-H single covalent bonds
C-O single covalent bonds
C=O double covalent bond




1
5
1
1
x
x
x
x
ethanol
C-C single covalent bond
C-H single covalent bonds
C-O single covalent bond
O-H single covalent bond
o
o
CHAPTER 13: CHEMICAL CALCULATIONS
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