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12ChF321 Moles and Equations Quantitative Chemistry - AS Calculations Relative Masses Atoms of different elements weigh different amounts because of the different numbers of protons and neutrons (and electrons – although they don't weigh much) that form them. Atoms of the same element can also weigh different amounts because although they must all have the same number of protons, they can have different numbers of neutrons. We call these isotopes. The chemical properties of an element depend on the electron configuration. Isotopes have the same number of electrons (because they have the same number of protons) and therefore have the same chemical properties . We need a scale to be able to compare masses, so we need a STANDARD – a fixed value to compare to. The standard is chosen to be an atom of carbon with 6 protons and 6 neutrons (mass number = 12). We call this isotope carbon-12. We define this as having a mass of exactly 12, then compare other elements against this – so we call the masses we work out RELATIVE masses. Page 1 Candidates should be able to: (a) explain the terms: (i) amount of substance, (ii) mole (symbol ‘mol’), as the unit for amount of substance, (iii) the Avogadro constant, N A , as the number of particles per mole (6.02 × 10 23 mol –1 ); (b) define and use the term molar mass (units g mol –1 ) as the mass per mole of a substance; (d) calculate empirical and molecular formulae, using composition by mass and percentage compositions; (f) carry out calculations, using amount of substance in mol, involving: (i) mass, (ii) gas volume, (iii) solution volume and concentration; (g) deduce stoichiometric relationships from calculations; (h) use the terms concentrated and dilute as qualitative descriptions for the concentration of a solution. Definition: Isotopes of an element have the same number of protons (same atomic number) but different numbers of neutrons.

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Page 1: F321 Calculations

12ChF321 Moles and Equations

Quantitative Chemistry - AS Calculations

Relative MassesAtoms of different elements weigh different amounts because of the different numbers of protons and neutrons (and electrons – although they don't weigh much) that form them.

Atoms of the same element can also weigh different amounts because although they must all have the same number of protons, they can have different numbers of neutrons. We call these isotopes.

The chemical properties of an element depend on the electron configuration. Isotopes have the same number of electrons (because they have the same number of protons) and therefore have the same chemical properties.

We need a scale to be able to compare masses, so we need a STANDARD – a fixed value to compare to. The standard is chosen to be an atom of carbon with 6 protons and 6 neutrons (mass number = 12). We call this isotope carbon-12. We define this as having a mass of exactly 12, then compare other elements against this – so we call the masses we work out RELATIVE masses.

The ‘compared with…’ bit defines the standard against which the mass is compared – the carbon-12 isotope.

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Candidates should be able to:(a) explain the terms:

(i) amount of substance,(ii) mole (symbol ‘mol’), as the unit for amount of substance,(iii) the Avogadro constant, NA, as the number of particles per mole (6.02 × 1023 mol–1);

(b) define and use the term molar mass (units g mol–1) as the mass per mole of a substance; (d) calculate empirical and molecular formulae, using composition by mass and percentage compositions; (f) carry out calculations, using amount of substance in mol, involving:

(i) mass,(ii) gas volume,(iii) solution volume and concentration;

(g) deduce stoichiometric relationships from calculations;(h) use the terms concentrated and dilute as qualitative descriptions for the concentration of a solution.

Definition: Isotopes of an element have the same number of protons (same atomic number) but different numbers of neutrons.

Relative Atomic Mass Symbol: Ar (unofficially RAM)Definition: The weighted mean mass of an atom of an element, compared with 1/12

the mass of an atom of carbon-12, which is taken as 12 exactly.

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12ChF321 Moles and Equations

The ‘weighted mean mass of an atom of an element’ means the average mass of an atom. Individual atoms don’t vary in mass, but not every atom of an element weighs the same, because not all atoms of an element contain the same number of neutrons. This is an average over all the naturally occurring isotopes.

The mass of an individual atom is given by its RELATIVE ISOTOPIC MASS, using the same scale as before:

e.g. 1 in 4 of all chlorine atoms have a relative isotopic mass of 37, and the rest have a relative isotopic mass of 35. That’s because some contain 20 and some 18 neutrons respectively. This means that the average mass of a chlorine atom is (25% x 37) + (75% x 35) = 35.5, which is chlorine's relative atomic mass.

We measure the mass of atoms accurately using a mass spectrometer. The output from a mass spectrometer tells us the masses of each isotope present (on the same scale compared to carbon-12), and the abundance (%) of that isotope. We can therefore calculate the Ar:

Ar = Σ (fractional abundance for each isotope x relative isotopic mass)

N.B fractional abundance = percentage abundance / 100

Worked example:Mass spectrometry shows that a sample of Br has 2 isotopes; 79Br and 81Br. Their abundances are 50.52% and 49.48% respectively. Show that the RAM of bromine in this sample is 79.98.

RAM = (0.5052 x 79)+(0.4948 x 81) = 79.98

Check your understanding: Calculate the relative atomic masses of these samples. Give answers to 2dp.xix) A sample of Li containing 8% of 6Li and 92% of 7Li

xx) A sample containing 78.5% of 24Mg, 10.11% of 25Mg, and 11.29% of 26Mg

Now we know the masses of atoms, we can use them to calculate masses of molecules (relative molecular mass, Mr) by adding the masses of the atoms given in the formula.

Not everything is formed of simple molecules (e.g. giant structures) but even these have a formula, so more generally the relative formula mass (use this one if in doubt) is the sum of the relative atomic masses in the formula.

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Definition: The relative isotopic mass is the mass of an atom of an isotope of the element compared with 1/12 the mass of an atom of carbon-12, which is taken as 12 exactly.

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Units:In all these cases, there are NO UNITS – we’re comparing to a unitless scale.Examples: relative formula mass sodium chloride NaCl 58.5 sodium hydroxide NaOH 40 nickel II sulphate NiSO4 154.8 aluminium hydroxide Al(OH)3 78

Check your understanding:Calculate relative molecular/formula masses for:xxi) propanol (C3H7OH)xxii) sulphur dioxidexxiii) nitrogen gasxxiv) trichloromethane (CHCl3)xxv) ammonium nitrate

Moles

Weighing out the RFM in grams is called measuring out a mole of that substance. The mole (symbol ‘mol’) is the official measure of ‘amount of substance’.

A mole of any substance contains the same number of particles of that substance – the number is 6.02 x 1023 (Avagadro’s number, symbol NA). e.g. 24g of magnesium contains 6.02 x 1023 Mg atoms

39.9g of argon contains 6.02 x 1023 Ar atoms18g of water contains 6.02 x 1023 H2O molecules 2g of hydrogen contains 6.02 x 1023 H2 molecules (12.04 x 1023 H atoms)

Units:A mole of something has units of grams (g)We sometimes refer to the molar mass of a substance. The units are then grams per mole (g mol-1)

e.g. a mole of magnesium is 24g both of these meanthe molar mass of magnesium is 24 g mol-1 the same thing

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Definition: The mole is the mass of an element or compound that contains exactly the same number of particles as there are atoms in 12g of carbon-12 (i.e. 6.02 x 1023 particles)

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Empirical and Molecular formulaeThe molecular formula tells us the actual numbers of atoms of each element present in a molecule, e.g. C6H12O6 tells us exactly which atoms are in a glucose molecule.

Not every substance is comprised of simple molecules like this, though (e.g. giant ionic structures). For any compound we can always write down ratios of different atoms present. The empirical formula is “the simplest whole-number ratio of atoms of each element present in a compound”.

e.g. for glucose the empirical formula is CH2OThe empirical formula MAY be the same as the molecular formula, or may not.

Example: elemental analysis of CO2

e.g. "44g of a gas is analysed and found to contain 18g of carbon and 48g of oxygen. Calculate the empirical formula."

C OMass (g) 18 48Ar 12 16Moles 1.5 3.0Ratio 1 2 Empirical formula: CO2

Sometimes we get fractions in the ratio to deal with: e.g. A 14.2g sample containing P and O is found to contain 6.2g of P.

P OMass (g) 6.2 14.2 – 6.2 = 8.0Ar 31.0 16.0Moles 0.2 0.5Ratio 1 2.5 x 2 to clear fractions…

2 5 Empirical formula: P2O5

Check your understanding:xxvi) 8.00g of Na is burnt in excess oxygen to produce 10.78g of a yellow solid.

Find its empirical formula.

xxvii) 10.00g of an oxide of lead, when reduced, produced 9.07g of lead. What was the empirical formula of the oxide ?

Working out % by massThis allows us to calculate what contribution to the total mass of a compound any one of its component elements makes:

e.g. For CO2: C OMass of element 1 x 12 2 x 16RFM of CO2 44 44Fraction 0.2727 0.7273% 27.3% 72.3%

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Check your understanding:xxviii) Calculate the % by mass of P, O and Cl in POCl2

xxix) Calculate the % by mass of C, O and H in glucose C6H12O6

xxx) What % chlorine is there in aluminium chloride ?

So the elemental analysis of CO2 example could have been written as:"A gas was analysed and found to contain 27.27% by mass of carbon and 72.73% of oxygen… calculate its empirical formula."

We can use the same layout, with the % in place of the mass (this works because we can pretend the total mass of sample is 100g so each % is also a mass in g)

C OMass 27.27 72.73RAM 12 16Moles 2.2725 4.5456Ratio 1 2.00027 Empirical formula: CO2

Check your understanding:xxxi) Compound F contains 39.74% C, 7.28% H and 52.98% Br.

Show that its empirical formula is C5H11Br.

xxxii) A compound contains 29.1% Na, 40.5% S and 30.4% OCalculate its empirical formula.

Working out molecular formulaeThe additional information we would need, in addition to the empirical formula, is the relative molecular mass, Mr.

We can work out the "empirical formula mass" (EFM) and use it to work out how to “scale up” the empirical formula. Calculate it by adding up the RAMs in the empirical formula.

Calculate (Mr / EFM) and then multiply all the numbers of atoms in the empirical formula by this amount.

e.g. A hydrocarbon is found to have empirical formula CH and Mr = 78. What is its molecular formula ?

Mr 78Empirical formula mass = 12.0 + 1.0 13.0Mr/EFM 6Molecular formula = 6 x CH = C6H6

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Check your understanding:xxxiii) An alkene has empirical formula CH2 and Mr = 42

What's its molecular formula ?

xxxiv) A salt of silver comprises silver, carbon and oxygen only. It contains 7.89% C and 21.06% O by mass and has Mr = 304. What's its molecular formula ?

Mole calculations

Part 1 – coverting mass to/from molesRecall “molar mass” = mass of a mole of a substance, units: g mol-1. This is simply Mr (the RMM or RFM, in grammes per mole)

Practice calculation examples:How many moles of water do we have if we have 2.7g of water ? (Mr of H2O = 18.0)Use Moles = Mass ÷ Mr Moles = 2.7 ÷ 18 = 0.15 moles

What is the mass of 2.5 moles of S atoms ?Here we use RAM not RFM – its atoms we’re being asked about.. Ar of S = 32.1

Mass = Moles x Ar = 2.5 x 32.1 = 80.25g

0.3 moles of a gas A has a mass of 8.4g. What is the molar mass of the gas ?Use Mr = mass ÷ moles = 8.4 ÷ 0.3 = 28 So the molar mass is 28 g mol-1

Check your understanding:Work out the molar masses (to 1 decimal place) of the following substances:xxxv) HNO3

xxxvi) FeCl3xxxvii) FeCl3.6H2O

xxxvii) Na2CO3.10H2Oxxxviii)

(NH4)2SO4.Fe2(SO4)3.24H2O

Work out the mass (to 2 sig figs) of:xxxix) 0.1 mole of zincxl) 1 mole of potassium hydrogencarbonatexli) 0.1 moles of lead(II) nitratexlii) 0.05 moles of aluminium oxide

How many moles are there (to 2 sig figs) in:xliii) 80g of Br2

xliv) 3.0g Mgxlv) 8.5g AgNO3

xlvi) 2.0g of NaOHxlvii) 2.0kg of CaCO3

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Our key equation here is MOLES = MASS ÷ Mr

Or if we want to go the other way MASS = MOLES x Mr

Finally we might want to work out Mr or molar mass Mr = MASS ÷ MOLES

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We also need to be able to use Avagadro’s number to work out numbers of atoms (don’t panic – the numbers will be huge !) It is just a matter of working out a number of moles, then multiplying by NA:

Example: How many atoms are there in 0.0005g of He ?Step 1: convert mass to moles moles = mass ÷ RAM = 0.0005 ÷ 4

= 0.000125

Step 2: convert moles to atoms atoms = moles x NA = 0.000125 x 6.02x1023

= 7.525x1019

Check your understanding:xlviii) How many molecules are there in 80g of Br2 ?xlix) How many Br atoms are there in 80g of Br2 ?l) How many atoms in 3.0g of Mg ?li) How many oxygen atoms in 8.5g AgNO3 ?

Part 2 – Reacting MassesBalanced symbol equations show us the ratios in which reactants react and in which products are formed. The numbers in front are numbers of moles.

e.g. N2(g) + 3H2(g) 2 NH3(g) one mole of nitrogen molecules react with three molesof hydrogen molecules to make two moles of ammonia.

We can use the mole ratios to calculate masses of product formed, or masses of reactants needed. We’ll always have enough information given to us in the question to be able to work out the number of moles of one substance in the equation, and we can then use the mole ratios to work out the number of moles of all the others.

Example:30.0g of hydrogen react with nitrogen to form ammonia. What mass of ammonia will be formed ?

A tabular approach starting with the balanced equation can help collect all the relevant information in the right places. Use a ? to indicate what you are trying to find out.

N2(g) + 3H2(g) 2NH3(g)

mass (g) 30.0 ?

Now add in the RFMs, working these out and remembering NOT to include the numbers in front !

N2(g) + 3H2(g) 2NH3(g)

mass (g) 30.0 ?RFM 28 2 17

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Now calculate number of moles for the one substance where you have both bits of information that you need – in this case the hydrogen:

N2(g) + 3H2(g) 2NH3(g)

mass (g) 30.0 ?RFM 28 2 17moles = 30.0 ÷ 2 = 15.0

Then write the mole ratios down, and use these to work out the number of moles of each other substance:

N2(g) + 3H2(g) 2NH3(g)

mass (g) 30.0 ?RFM 28 2 17moles 5.0 15.0 10.0mole ratio 1 : 3 : 2

Finally you can use the moles and RFM to work out the mass corresponding to your '?'. It may be worth working out the other masses too – they may be asked about in subsequent questions:

N2(g) + 3H2(g) 2NH3(g)

mass (g) 140.0 30.0 170.0RFM 28 2 17moles 5.0 15.0 10.0mole ratio 1 : 3 : 2

The mass of ammonia formed would be 170g (and 140g of nitrogen would be needed).

Mole calculations involving gasesIt is not very convenient to measure the mass of gases. We normally measure the amount of a gas by volume.

Avagadro’s work in the 19th century led to the surprising conclusion that the same volumes of gases contain the same number of moles – in other words the particles in gases are all equally spread out, regardless of how heavy they are. (This result only holds if the gases are at the same temperature and pressure, of course).

We use two standard measures:At room temperature and pressure (denoted r.t.p.) the volume of one mole of any gas is taken to be 24.0 dm3 (which is 24,000 cm3). This figure is given on your exam data sheet.

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Avagadro’s Law: Equal volumes of all gases, measured at the same temperature and pressure contain the same number of molecules (and hence moles).

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Under standard conditions (0°C (273 Kelvin) and 101.3 kPa) the molar volume of any gas is 22.4dm3 mol-1 (which is 22,400 cm3 mol-1). If you are asked to do a calculation involving moles of gas, you will be GIVEN the molar volume of the gas at the temperature and pressure concerned.

This makes calculations involving moles of gas very straightforward. Normally, for a gas:

Moles = volume (dm3) ÷ 24.0 Moles = volume (cm3) ÷ 24000

Volume (dm3) = moles x 24.0 Volume (cm3) = moles x 24000

So now we have the tools to do mole calculations involving gases as well as solids.

Example 1:Calculate the volume of hydrogen produced when excess dilute sulphuric acid is added to 5.00g of zinc. Assume 1 mole of gas has a volume of 24.0 dm3.

We can use a modification of our tabular approach:

Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)

mass (g) 5.00 ? volume (dm3)RFM 64.5 24.0 molar vol (dm3)moles 0.0765 0.0765mole ratio 1 : 1 : 1 : 1

So volume of hydrogen = 0.0765 x 24.0 = 1.835dm3 (1,835cm3)

(Think about this for a moment, are you surprised that 5g of zinc, only a little, can produce nearly 2 litres of gas ? – It shows how dense solids are compared to gases)You could also work out the mass of zinc sulphate produced, and the mass of sulphuric acid used, if you want.

Example 2:Hydrogen peroxide (H2O2) is decomposed by heating to give water and oxygen. Construct a balanced equation, then calculate the mass of hydrogen peroxide required to produce 100cm3 of oxygen. Under standard conditions the molar volume of gases is 22,400cm3.

2 H2O2(aq) 2 H2O(l) + O2(g)

mass (g) ? 100 volume (cm3)RFM 34 22,400 molar volume (cm3)moles 0.00834 0.00417 molesmole ratio 2 : 2 : 1 mole ratio

So mass of H2O2 = 0.00834 x 34.0 = 0.284g H2O2

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We can write this as: Concentration (mol dm-3) = moles ÷ volume (dm3)(remember that 1 dm3 = 1000 dm3= 1 litre)

Conversely: Moles = concentration (mol dm-3) x volume (dm3)(usual mistake, forgetting to convert volume to dm3)

12ChF321 Moles and Equations

Check your understanding:lii) Calculate the volume of CO2 produced at r.t.p. when 5.005g of calcium carbonate

is heated until it decomposes into calcium oxide and carbon dioxide

liii) Calculate the mass of lithium required when reacting with excess HCl to produce 10dm3 of hydrogen gas at r.t.p.

liv) 5.58g of a metal M reacts fully with excess nitric acid producing 2.4dm3 of hydrogen at r.t.p. Calculate the Ar of the metal, and hence identify the metal. The equation for the reaction is M(s) + 2HNO3(aq) M(NO3)2(aq) + H2(g)

Mole calculations involving pure liquids (not solutions – look for the (l) state symbol) For pure liquids we can measure the mass of the liquid, and hence the number of moles just as we did for solids. If we know the volume of a pure liquid we can convert that to mass easily using the density of the liquid. We can then use the mass in a mole calculation.

Example:The density of ethanol is 0.79gcm-1. What will 50cm3 of ethanol weigh ?

Mass of ethanol = 50 x 0.79 = 39.5g

The density of glycerol is 1.26g. What volume will 10g of glycerol occupy ?Volume = 10 / 1.26 = 7.94cm3

Mole calculations involving solutions (look for the (aq) state symbol)The majority of liquids we work with will be solutions. The terms ‘concentrated’ and ‘dilute’ are used to describe how much of the solute there is present in a known volume of the solvent.

We measure concentrations in moles per dm3 (mol dm-3). The old term ‘Molar’ is sometimes seen on solutions – a 1 Molar solution (1M) contains 1 mole of solute per dm3 of solvent.

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Mass = volume x density (units of density gcm-1)

Volume = mass ÷ density

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12ChF321 Moles and Equations

Examples:How many moles of HCl in 100cm3 of 0.2 mol dm-3 HCl ?

moles = 0.2 x (100/1000) = 0.02 mol

What is the concentration of a solution containing 0.3 mol CaCl2 in 500cm3 of the solution ?Conc = moles / volume = 0.3 / 0.5 = 0.6 mol dm-3

What is the concentration of a solution containing 5.61g potassium hydroxide in 100cm3 of the solution ?

First convert mass to moles of KOH Mr = 56.1 Moles = 5.61/56.1 = 0.1Conc = moles / vol = 0.1 / 0.1 = 1 mol dm-3

Calculate the mass of NaCl required to produce 250cm3 of 0.1M solution ?First work out moles in the solution moles = conc x vol

= 0.1 x 0.25 = 0.025 molesNow work out mass of NaCl requiredMr NaCl = 58.5 mass = moles x Mr

= 0.025 x 58.5 = 1.4625g

Check your understanding:Calculate the number of moles of:lv) Sodium hydroxide in 25 cm3 of 1.5 mol dm-3 sodium hydroxide solutionlvi) Sodium carbonate in 10 cm3 of 3.0 mol dm-3 sodium carbonate solution

Calculate the mass of:lvii) Hydrogen chloride in 500 cm3 of 2M hydrochloric acidlviii) Ammonia in 20 cm3 of a solution of 8M ammonia

Calculate the concentration in mol dm-3 of:lix) A solution made by dissolving 2.92g of NaCl in 400cm3 of solution

Hydrated and anhydrous crystalsWhen hydrated crystals are heated the water is driven off and they become anhydrous. This obviously affects the relative formula mass- water of crystallization is part of the formula, and hence part of the RFM. Look at the bottles when you are weighing out salts, and check the formula and Mr given on the bottle !

Check your understanding:lx) Calculate the concentration in mol dm-3 of a solution which contains 6.95g of

hydrated iron II sulphate (FeSO4.7H2O) in 250 cm3 of solution

We may be asked to work out the formula for hydrated crystals given % composition data. We do this by converting the % to a mole ratio (divide by the Ar or Mr for each % given), just like we did with percentage composition calculations already:

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e.g. A sample of sodium sulphate contains 14.28% sodium ions, 29.84% sulphate ions and 55.88% of water, by mass. Determine the formula of the crystals.

Na+ SO42- H2O

% 14.28 29.84 55.88Mr 23.0 96.1 18moles 0.62 0.31 3.10ratio 2 1 10 (dividing all by the smallest moles, 0.31)

So the formula of the crystals will be Na2SO4.10H2O

We also need to be able to work out how much water crystals contain. We can do this using the mass of hydrated and anhydrous crystals (i.e. the mass loss on heating to dehydrate the crystals):

e.g. The mass of a sample of cobalt chloride CoCl2.nH2O is 5.00g. After heating to remove the water of crystallization the mass of the sample was 2.73g. Work out how many water molecules are represented by n.

CoCl2 H2OMass(g) 2.73 5.00 – 2.73 = 2.27Mr 129.9 18Moles 0.021 0.126Ratio 1 6

n = 0.126/0.021 = 6 so the formula is CoCl2.6H2O

Identifying the salt from dehydration dataA different form of this type of question involves a salt of an unknown metal, which is to be identified from the mass change on dehydration.

Using the same example as before:The mass of a sample of a chloride salt XCl2.6H2O is 5.00g. After heating to remove the water of crystallization the mass of the sample was 2.73g.

i) Write the formulae of the positive and negative ions present:X2+ and Cl- (valency of X can be worked out from knowledge of Cl- valency)

ii) Write an equation for the change which took place on heating (including state symbols):XCl2.6H2O(s) XCl2(s) + 6H2O(l)

iii) From the mass change, calculate the number of moles of water removedmoles of H2O = mass of H2O/Mr of H2O = (5.00 – 2.73)/18 = 0.1261

iv) Using the equation in ii) calculate moles of XCl2 produced after heatingmole ratio = 1:6 so moles XCl2 = 0.1261/6 = 0.02102

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v) Calculate the Mr of the anhydrous salt to one decimal placeMr of XCl2 = mass of XCl2 / moles of XCl2 = 2.73/0.02102 = 129.9

vi) Identify XIf Mr of XCl2 is 129.9 then Ar of X = 129.9 – (2 x 35.5) = 58.9Therefore X = Cobalt

This calculation can be arranged in a similar way to previously if not all these steps are required:

XCl2 H2Omass (g) 2.73 5.00 – 2.73 = 2.27Mr ? 18moles 0.1261/6 = 0.02102 0.1261 moles = mass ÷ Mr

mole ratio 1 : 6 from formula of hydrated salt

Mr of XCl2 = mass of XCl2 ÷ moles of XCl2 = 2.73 ÷ 0.02102 = 129.9Ar of X = 129.9 – (2 x 35.5) = 58.9X = Cobalt

Diluting solutionsSolutions are sold with specific concentrations. If we want to use different concentrations in our reactions we need to be able to make up solutions of different concentration by dilution with distilled water. REMEMBER ! Never add water to acids – always add the acid, slowly, to water. This minimizes the risk due to the heat produced causing spitting/boiling.

Firstly we need to know how to work out the concentration of a solution which has been diluted. The number of moles of solute remains the same, just contained in a bigger volume of solution. Since moles = conc x volume we can work out that:

e.g. if I dilute 50cm3 of a 2 mol dm-3 solution until I have 250cm3 of solution, its concentration will now be:

conc of diluted solution = (50 / 250) x 2 = 0.4 mol dm-3

Note: while there’s nothing wrong with converting to dm3 we don’t have to if both our volumes are in cm3 !

Secondly, we need to be able to work out how make up a certain volume of a solution with a certain concentration, using dilution. For example, hydrochloric acid is often sold as a solution of 10 mol dm-3 concentration. This is concentrated hydrochloric acid. In the laboratory we often use 2 mol dm-3 hydrochloric acid.

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conc of diluted solution = volume of original solution x conc. of original solution volume of diluted solution

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So for making lab hydrochloric acid from concentrated, the dilution factor = 10/2 = 5

The dilution factor tells us by how much the volume of the solution will change as well. In the case of our hydrochloric acid;

volume of diluted HCl = dilution factor x volume of concentrated HCl

So if we took 1.0dm3 of 10M HCl and diluted it to make 2M HCl for lab use;

volume of 2M HCl = (10/2) x 1.0 = 5.0dm3

We can use this in two different ways:1) To work out how much solution we’ll end up with:e.g. We dilute 100cm3 of HCl which has 10mol dm-3 concentration until we get a solution with 2 mol dm-3 concentration. What is the final volume of the solution ?

volume after dilution (cm3) = 100 x (10/2) = 500cm3

Note: we started with 100cm3 of acid and finished up with 500cm3 so we must have added the acid to 400cm3 of water.

2) More typically we are asked to work out what volume of original solution and what volume of water to mix in order to end up with a given volume and concentration:e.g. we want to dilute our 10 mol dm-3 HCl to make up 1dm3 of acid with 2 mol dm-3 concentration. What should we do ?

dilution factor = 10/2 = 5

1 dm3 (volume after dilution) = volume before dilution x 5 => volume before dilution = 1/ 5 = 0.2dm3

So we are going to need to add 0.2 dm3 of concentrated acid to water until the final volume is 1dm3 (therefore we’re going to need 0.8dm3 of water).

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The dilution factor is given by dilution factor = conc of solution before dilution conc of solution after dilution

or dilution factor = volume of diluted solution volume of original solution

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Check your understanding:lxi) Concentrated nitric acid is sold as 16 mol dm-3. How would I prepare 4dm3 of

1mol dm-3 nitric acid solution ?

lxii) What is the concentration (in mol dm-3) of the sodium hydroxide solution prepared by diluting 75cm3 of 0.3 mol dm-3 sodium hydroxide solution with water until its final volume was 450cm3 ?

Using concentrations in mole calculationsWe use the same steps are used as we’ve used, again using a modification of our tabular approach:

e.g.Excess magnesium oxide (MgO) was added to 10cm3 of 0.5 mol dm-3 sulphuric acid and the solution warmed until no more magnesium oxide reacted. The remaining unreacted magnesium oxide was filtered off, and the colourless solution obtained was allowed to crystallize. What is the maximum mass of magnesium sulphate, MgSO4.7H2O, that could be obtained by this method ? [MgSO4.7H2O has an Mr of 246.4]

MgO(s) + H2SO4(aq) MgSO4(aq) + H2O(l)

conc (moldm-3) 0.5 ?vol (dm3) 0.01 246.4moles 0.005 0.005mole ratio 1 : 1 : 1 : 1

Mass of MgSO4.7H2O = 0.005 x 246.4 = 1.232g

Calculation of results from titrationsTitrations are one of the more common examples where mole calculations involving concentrations are involved. Generally a solution of known concentration is used to determine experimentally the concentration of an unknown solution with which it reacts.

e.g. A solution of HCl of unknown concentration was titrated against 25cm3 of KOH solution of concentration 0.50 mol dm-3. The volume of acid needed for the indicator to show the endpoint was 18.5cm3. Calculate the concentration of the hydrochloric acid.

HCl(aq) + KOH(aq) KCl(aq) + H2O(l)

conc (moldm-3) ? 0.50vol (dm3) 0.0185 0.025moles 0.0125 0.0125mole ratio 1 : 1 : 1 : 1

Conc of HCl = 0.0125 ÷ 0.0185 = 0.68 mol dm-3 (to 2 sf)

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Check your understanding:lxiii) 10cm3 of a solution of sodium hydroxide was neutralized by 12.5cm3 of 0.1M HCl.

Calculate the concentration in mol dm-3 of the alkali.

lxiv) 25cm3 of a 0.050M solution of sodium hydroxide was neutralized by the addition 20.0cm3 of sulphuric acid. Calculate the concentration of the acid.

Answers to "Check your understanding" questions:

Calculating relative atomic masses (answers to 2dp)xix) A sample of Li contains 8% of 6Li and 92% of 7LiAns: RAM = 6.92

xx) A sample containing 78.5% of 24Mg, 10.11% 25Mg, 11.29% 26MgAns: RAM = 24.33

Calculating relative molecular massesxxi) propanol 60xxii) sulphur dioxide 64.1xxiii) nitrogen gas 28xxiv) trichloromethane 119.5xxv) ammonium nitrate 80

Empirical formula calculations8.00g of Na is burnt in excess oxygen to produce 10.78g of a yellow solid. xxvi) Find the empirical formula:

Na OMass (g) 8.00 10.78 – 8.00 = 2.78RAM 23.0 16.0Moles 0.3478 0.l7375 ** watch rounding !Ratio 2 : 1 Empirical Na2O

10.00g of an oxide of lead, when reduced, produced 9.07g of lead. xxvi) What was the empirical formula of the oxide ?

Pb OMass(g) 9.07 10.00 – 9.07 = 0.93RAM 207.2 16Moles 0.04377 0.058125Ratio 1.0 1.33 Recognise 1/3 so x all by 3Ratio 3 4 Empirical Pb3O4

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Percentage composition calculationsxxviii) Calculate the % by mass of P, O and Cl in POCl2

P O ClMass of element 1 x 31.0 1 x 16.0 2 x 35.5RFM 118 118 118Fraction 0.2627 0.1356 0.6017% 26.27% 13.56% 60.17%

xxix) Calculate the % by mass of C, O and H in glucose C6H12O6

C H OMass of element 6 x 12.0 12 x 1.0 6 x 16.0RFM 180 180 180Fraction 0.4 0.0667 0.5333% 40.0% 6.7% 53.3%

xxx) What % chlorine is there in aluminium chloride ?Al Cl

Mass of element 1 x 27.0 3 x 35.5RFM 133.5 133.5Fraction 0.79775% 79.8%

Empirical formulae from percentage composition dataxxxi) Compound F contains 39.74% C, 7.28%H and 52.98%Br.

Show its empirical formula is C5H11Br.C H Br

% 39.74 7.28 52.98RAM 12.0 1.0 79.9Moles 3.31167 7.28 0.66308Ratio 4.994 10.979 1

5 11 1 C5H11Br

xxxii) A compound contains 29.1% Na, 40.5% S and 30.4% OCalculate its empirical formula.

Na S O% 29.1 40.5 30.4RAM 23.0 32.1 16.0Moles 1.2652 1.2616 1.9Ratio 1 1 1.5 recognise ½ so x2Ratio 2 2 3 Empirical: Na2S2O3

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Molecular formulae from composition data and Mr

xxxiii) An alkene has empirical formula CH2 and Mr = 42 What's its molecular formula ?

Mr 42EFM = 12 + 2 14Mr/EFM 3Molecular formula = 3 x CH2 = C3H6

xxxiv) A salt of silver comprises silver, carbon and oxygen only. It contains 7.89% C and 21.06% O by mass and has Mr = 304. What's its molecular formula ?

Ag C O% 71.05 (calc) 7.89 21.06RAM 107.9 12.0 16.0Moles 0.6585 0.6575 1.316Ratio 1 1 2 AgCO2 empirical

Mr 304Empirical formula mass 151.9 (107.9 + 12 + 16 + 16)Mr/EFM 2.00Molecular formula: 2 x AgCO2 Molecular formula: Ag2C2O4

Molar masses and molar quantitiesWork out the molar masses (to 1 decimal place) of the following substances:xxxv) HNO3 63.0 g mol-1

xxxvi) FeCl3 162.3xxxvii) FeCl3.6H2O 270.3xxxvii) Na2CO3.10H2O 286.0xxxviii) (NH4)2SO4.Fe2(SO4)3.24H2O 964.0

Work out the mass (to 2 sig figs) of:xxxix) 0.1 mole of zinc

Zn 0.10 x 65.4 6.5g to 2sf

xxxx) 1 mole of potassium hydrogencarbonateKHCO3 1 x 100.1g (100.1g) 1.0 x 102 to 2sf

xxxxi) 0.1 moles of lead II nitratePb(NO3)2 0.1 x 331.2 (33.12g) 33g to 2sf

xxxxii) 0.05 moles of aluminium oxideAl2O3 0.05 x 102 5.1g to 2sf

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How many moles are there in:xxxxiii) 80g of Br2 80 / 79.9 = 1.001 1.0 to 2sfxxxxiv) 3.0g Mg 3.0 / 24.3 = 0.12345 0.12 to 2sfxxxxv) 8.5g AgNO3 8.5 / 169.9 = 0.05003 0.050 to 2sfxxxxvi) 2.0g of NaOH 2.0 / 40 = 0.05 0.05 to 2sfxxxxvii) 2.0kg of CaCO3 2000 / 100 = 20 to 2sf

Using Avogadro's number:xlviii) How many molecules are there in 80g of Br2 ? 6.02 x 1023

xlix) How many Br atoms are there in 80g of Br2 ? 1.204 x 1024

l) How many atoms in 3.0g of Mg ? 7.224 x 1022

li) How many oxygen atoms in 8.5g AgNO3 ? 3 x 0.05003 x NA = 9.035 x 1022

Mole calculations involving gaseslii) Calculate the volume of CO2 produced at r.t.p. when 5.005g of calcium carbonate

is heated until it decomposes into calcium oxide and carbon dioxide

Moles of CaCO3 = mass of CaCO3 / RFM of CaCO3 = 5.005 / (40.1 + 12 + 48) = 0.05Moles of CO2 produced = 0.05 (1:1 mole ratio since CaCO3 CaO + CO2)Volume of CO2 = moles x molar volume = 0.05 x 24.0 = 1.2dm3

liii) Calculate the mass of lithium required when reacting with excess HCl to produce 10dm3 of hydrogen gas at r.t.p.

Moles of H2 = volume of H2 / molar volume = 10 / 24.0 = 0.4166r

Moles of Li = 2 x 0.4166r = 0.833r (2:1 mole ratio since 2Li + 2HCl 2LiCl + H2)Mass of Li = moles of Li x Ar = 0.833r x 6.9 = 5.75g

liv) 5.58g of a metal M reacts fully with excess nitric acid producing 2.4dm3 of hydrogen at r.t.p. Calculate the Ar of the metal, and hence identify the metal. The equation for the reaction is: M(s) + 2HNO3(aq) M(NO3)2(aq) + H2(g)

Moles of H2 = volume of H2 / molar volume = 2.4 / 24.0 = 0.1 molesMoles of M = 0.1 since 1:1 molar ratio in equationMass of M = moles of M x Ar => Ar = mass of M / moles of M = 5.58/0.1 = 55.8The metal M has atomic mass 55.8 and is therefore Iron.

Calculating moles in solutionCalculate the number of moles of:lv) Sodium hydroxide in 25 cm3 of 1.5 mol dm-3 sodium hydroxide solution

moles of NaOH = conc x vol (in dm3) = 1.5 x (25/1000) = 0.0375 moles

lvi) Sodium carbonate in 10 cm3 of 3.0 mol dm-3 sodium carbonate solutionmoles of Na2CO3 = conc x vol(in dm3) = 3.0 x (10/1000) = 0.03 moles

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Calculate the mass of:lvii) Hydrogen chloride in 500 cm3 of 2M hydrochloric acid

moles of HCl = conc x vol = 2 x (500/1000) = 1 molemass of HCl = moles x RFM = 1 x 36.5 = 36.5g

lviii) Ammonia in 20 cm3 of a solution of 8M ammoniamoles of NH3 = conc x vol = 8 x (20/1000) = 0.16mass of NH3 = moles x RFM = 0.16 x 17 = 2.72g

Calculate the concentration in mol dm-3 of:lix) A solution made by dissolving 2.92g of NaCl in 400cm3 of solution

moles of NaCl = mass of NaCl / RFM = 2.92 / 58.5 = 0.05 molesconc = moles of NaCl / volume of solution (in dm3) = 0.05 / (400/1000)

= 0.125 mol dm-3

Solutions where the solute crystals are hydrated:lx) Calculate the concentration in mol dm-3 of a solution which contains 6.95g of

hydrated iron II sulphate (FeSO4.7H2O) in 250 cm3 of solution

RFM of FeSO4.7H2O = 55.8 + 32.1 + (16 x 4) + (7 x 18) = 277.9moles of iron sulphate = mass / RFM = 6.95 / 277.9 = 0.025conc = moles of iron sulphate / volume (in dm3) = 0.025 / (250/1000) = 0.1 mol dm-3

Dilution calculations:lxi) Concentrated nitric acid is sold as 16 mol dm-3. How would I prepare 4dm3 of

1mol dm-3 nitric acid solution ?

Dilution factor = 16 / 1 = 16Volume of nitric acid before dilution = 4 dm3 / 16 = 0.25dm3

The final diluted volume is to be 4 dm3 so I’ll need to add 3.75dm3 water

lxii) What is the concentration (in mol dm-3) of the sodium hydroxide solution prepared by diluting 75cm3 of 0.3 mol dm-3 sodium hydroxide solution with water until its final volume was 450cm3 ?

dilution factor = volume after dilution / volume before dilution = 450/75 = 6Dilution factor = conc before dilution / conc after dilutionSo conc after dilution = conc before dilution / dilution factor = 0.3 / 6 = 0.05 mol dm-3

Titration calculations:lxiii) 10cm3 of a solution of sodium hydroxide was neutralized by 12.5cm3 of 0.1M

HCl. Calculate the concentration in mol dm-3 of the alkali.

Moles of HCl = conc of HCl x vol of HCl (in dm3) = 0.1 x (12.5/1000) = 0.00125Moles of NaOH = 0.00125 (1:1 mole ratio: NaOH + HCl NaCl + H2O)Conc of NaOH = moles of NaOH / vol of NaOH (in dm3) = 0.00125 / (10 / 1000)

= 0.125 mol dm-3

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lxiv) 25cm3 of a 0.050M solution of sodium hydroxide was neutralized by the addition 20.0cm3 of sulphuric acid. Calculate the concentration of the acid.

Moles of NaOH = conc of NaOH x vol of NaOH (in dm3) = 0.050 x (25/1000) = 0.00125

Moles of H2SO4 = 0.00125 /2 = 0.000625 (1:2 mole ratio: H2SO4 + 2NaOH Na2SO4 + 2H2O)

Conc of H2SO4 = moles of H2SO4 / volume of H2SO4 (in dm3) = 0.000625 / (20/1000) = 0.03125 mol dm-3

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