1.1.2 Moles and equations 2012

Introduction Within this topic you will learn how to quantify the outcomes of chemical reactions; this will involve writing chemical equations, using ratios, converting units, deducing formulae, using gas volumes and understanding the mole concept.

1. Conversion of masses into different unitsWe need to convert g into mg, kg or tonnes (and the other way).The conversions are:1tonne = 1000kg1tonne = 1,000,000g1kg = 1000g1g = 1000mg

Convert the following masses as specified.a) 5 kg into g; ____

b) 45800 kg into g; ____

c) 6.34 g into kg; ____

d) 1.34 x 109 g into kg; ____

e) 450mg into g ____

2. Conversion of volumes into different unitsWe need to convert m3 into dm3 or cm3 (or the other way).

The conversions are:1 m3 = 1000dm3

1 m3 = 1,000,000cm3

1 dm3 = 1000cm3

Convert the following volumes as specified.a) 2 dm3 into cm3;

b) 81400 dm3 into cm3;

c) 9.85 cm3 into dm3;

d) 4.94 x 109 cm3 into dm3;

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3. Using significant figures

Often, in chemistry, we will work to a certain number of significant figures; this implies a certain level of accuracy. We often use 3 significant figures, although there are exceptions; Reading a burette always uses 2dp with the second digit a 0 or a 5 . When a number less than 1 has zeros directly after the decimal point these are not significant, i.e. 0.04 (is 1 sf), 0.0256 ( is 3sf). Note: 0.105( is 3 sf). Standard index form is also useful 0.04 is 4x10-2, (1sf), 0.0256 is 2.56 x10-2(3sf) and 0.105 is 1.05x10-1 (3sf)

Write the following numbers to 3 significant figures

a) 0.09936 ____

b) 234 ____

c) 1369.185 ____

d) 0.101 ____

e) 2.8769 x 1016 ____

4. Answering to a set number of decimal placesAlternatively, on an exam paper, you may be asked to answer to a set number of decimal places.

Write the following to 2 decimal places

a) 0.09936 ____

b) 0.10752 ____

c) 58.942 ____

d) 1369.185 ____

e) 0.101 ____

5. Categorizing materials as substances or mixtures

Underline the materials from the list below (in pencil) which you would describe as substances (Hint; substances are those which are not mixtures).

Sugar water air seawater, wood sodium chloride sulphur dioxide chlorinecalcium lemon juice bromine plastic

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6. Converting numbers of particles into moles

The mole concept allows chemists to link the reaction equation to the actual quantities of reactants or products. The number of moles is also known as the amount of substance.

DefinitionThe mole is defined as the amount of substance which contains as many elementary particles as there are atoms in 12g of carbon-12.

DefinitionThe number of atoms, molecules, ions or subatomic particles in one mole is a constant known as Avogadro’s constant. One mole of substance contains 6.02 x 1023 atoms.

So, to convert numbers of particles into moles, we divide by Avogadro’s constant. Moles = number of particles / 6.02x1023

Convert the following numbers of particles into moles

a) 6.02 x 1023 atoms of carbon;

b) 3.01 x 1023 molecules of methane;

c) 5.78 x 1026 molecules of oxygen.

7. Converting moles into numbers of particles

To convert moles into numbers of particles we need to multiply the number of moles by Avogadro’s constant.

a) How many particles are present in 3 moles of sulphur atoms?____

b) How many sodium ions are in a solution containing 10.7 moles of sodium chloride? ___

c) How many oxygen molecules are in 2.5 moles of oxygen gas?

____

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8. Using the periodic table to find the molar mass of an element

DefinitionThe molar mass is the mass (in g) of one mole of substance. It has units of gmol-1

The elements which exist as diatomic molecules are: H2, N2, O2, Cl2, Br2, I2 and At2. For these, we must multiply the relative atomic mass by 2.

DefinitionThe relative atomic mass (Ar) for an element is the average mass of the naturally occurring isotopes of the element relative to the mass of an atom of carbon-12, which has a mass of 12. Write the molar mass for the following elements.

a) carbon ___ b) sulphur ___ c) magnesium ___

d) hydrogen (H2) ___ e) potassium ___ f) iodine (I2) ___

Examination Question 1c Jan 2009

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Examination Question 1c Jun 2010

9. Finding the molar masses of simple compounds

To find the relative molecular mass (Mr) of a compound, add up the individual relative atomic masses of the elements (multiplied by the number of atoms of that element) which make up the compound. The molar mass is the relative molecular mass in g.

DefinitionThe relative molecular mass (Mr) for a compound, is the weighted average mass of a molecule of a compound relative to the mass of an atom of carbon-12, which has a mass of 12.

Calculate the molar mass to one decimal place of the following compounds from their formulae?

a) CH4 ___

b) (NH4)2SO4 ___

c) FeSO4.7H2O ___

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10. Calculating the percentage by mass of an element in a compound

To find the percentage by mass of an element within a compound, divide the relative mass of the atoms of the element in the compound by the total mass of the compound (Mr), and then multiply by 100.

Calculate the percentage by mass to three sig figs of the named element in each of the compounds listed.

a) Copper in CuSO4 ____

63.5/(63.5 + 32.1 + 4 x 16) x 100 = 39.8% ____b) Calcium in Ca(CN)2

c) Nitrogen in (NH4)2CO3 ____

11. Deducing empirical formulae from given molecular formulae

DefinitionThe empirical formula is the simplest whole number ratio of atoms of each element present in a compound.

What are the empirical formulae of the following compounds?

a) benzene C6H6; _____

b) heptene C7H14; _____

c) cyclopentane C5H10; _____

d) dinitrogen tetroxide N2O4; _____

e) ethane C2H6. _____

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12. Deducing the molecular formula from empirical formula and molar mass

The molar mass or Mr of a compound might be known (from the mass spectrum). If the empirical formula is also known, then we can deduce the molecular formula.a) A hydrocarbon of molar mass 142 gmol-1 has an empirical formula C5H11.

What is the molecular formula?

_____

b) A hydrocarbon of molar mass 56 gmol -1 has an empirical formula CH2. What is the molecular formula?

_____

13. Calculating the empirical formula from percentage composition

If the percentage of each element is given, then this will be equal to the number of grams of that element, if the mass of the compound is 100g. Division of this mass by the Ar will give the number of moles of each element. The ratio can then be calculated by dividing each number of moles by the smallest number of moles. Remember that the numbers may not give exact whole number ratios; a ratio of 1 to 1.48, is actually 2 to 3. In other words you treat % exactly the same as mass.

Work out the empirical formulae for the following compounds

a) Fe 77% O 23%;

b) C 92.3 % H 7.6%

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c) K 26.6% Cr 35.4% O 38%;

14. Calculating the empirical formula from the percentage of one reactant

If the percentage of only one element is given, then subtract this number from 100 to find the percentage of the other element; then complete the calculation as for Exercise 13 above.

Work out the empirical formulae for the following compoundsa) A chloride of cobalt which contains 54.7% chlorine;

b) A chloride of copper which contains 64.1% copper;

Examination Question 1d Jun 2010

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15. Calculating the empirical formula from composition by mass of all atoms present

If the mass of each element present is given, these can be converted into moles and the ratio deduced.

a) A compound contains 5.58g of iron and 3.21g of sulphur. The molar mass is 87.9. Find the molecular formula.

b) A compound contains 0.200g of C, 0.033g of H and 0.268g of O. The molar mass is 60. Find the molecular formula.

16. Producing formulae for compounds using a valency tableThe valency is the combining power of an atom or a group of atoms. Chemical formulae (of ionic compounds) are produced by combining ions which may be positively or negatively charged.

ExampleCompound name zinc chlorideIons present Zn2+ Cl-Balance the charges 1 zinc and 2 chlorides are neededFormula ZnCl2

For transition elements, the valency (of the positive ion) is given as Roman numerals in brackets after the metal.Use your valency table to construct formulae for the following compounds.

a) copper (II) oxide; _____ f) magnesium chloride; _____

b) potassium oxide; _____ g) aluminium sulphate; _____

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c) sodium bromide; _____ h) potassium dichromate; _____

d) copper (II) sulphate; _____ i) calcium hydroxide. _____

e) tin (II) nitrate; _____ j) ammonium carbonate; _____

17. Writing symbol equations for reactions we have studied

There are three steps in writing equations:1. Write (or think of) the word equation;2. Write the correct symbols and formulae (with state symbols);3. Balance the equation.

ExampleDuring the reaction of sodium with water; sodium hydroxide and hydrogen are produced.Word equation sodium + water → hydrogen + sodium hydroxide

Na(s) + H2O(l) → H2(g) + NaOH(aq)

2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)

Write symbol equations for the following.

a) hydrogen + oxygen → water

______________________________________________________________

b) lithium + sulphuric acid → lithium sulphate + hydrogen

______________________________________________________________

c) ammonium + calcium → ammonia + calcium chloride + water chloride hydroxide

______________________________________________________________

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18. Writing symbol equations for reactions with reactants or products stated

You may have to write equations for totally unfamiliar reactions, in which case, you will be provided with names of some reactants and products. Use the information given to write in formulae for some of the compounds. If you know the reaction then fill in the other formulae needed. If you don’t know the other products then make up a suitable compound, or compounds, with the atoms which are available; always follow valency rules.

Write symbol equations for the following reactions .

a) The formation of magnesium oxide from burning magnesium.

___________________________________________________________

b) The reaction of sulphur trioxide with water to produce sulphuric acid (H2SO4).

___________________________________________________________

c) The decomposition of hydrogen peroxide (H2O2) to produce oxygen and another product.

___________________________________________________________

19. Reacting masses: using moles=mass/Mr , n=m/Mr

You will often need to work out the number of moles from the mass of a reactant. This is often the first step towards working out the mass of a product formed. In order to calculate the number of moles (n), we can use the conversions below.

For elementsMoles = mass (g)

relative atomic mass (Ar)

For compoundsMoles = mass (g)

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relative molecular mass (Mr)

How many moles are in the following substances? (answer to 3 sf) a) 24g of carbon; ____

b) 1.7mg of ammonium oxide; ____

c) 40 kg of iron; ____

d) 2 tonnes of water; ____

e) 0.000257g of sodium hydroxide; ____

20. Calculating the mass from the number of moles

We can transform the equations above to make mass the subject of the equations. For elementsmass (g) = moles x relative atomic mass (Ar)

For compoundsmass (g) = moles x relative molecular mass (Mr)

Work out the masses of the following quantities.

a) 3 moles of carbon;

b) 2.4 moles of oxygen molecules;

c) 10.9 moles of carbon dioxide;

d) 1.45 x 10-4 moles of sulphur dioxide;

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21. Working out how many moles of each substance are used up or produced: using stoichiometric numbers

The reaction equation shows the ratio of product to reactant. The stoichiometric numbers are the big numbers in front of the formulae or symbols; these are the numbers which balance the equation. In the equation below the stoichiometric numbers for X, Y and Z are 4, 1 and 2 respectively.

4X + Y → 2Z

If we react 4 moles of X with one mole of Y we will produce 2 moles of Z. In order to produce 1 mole of Z, we would need to react 2 moles of X with 0.5 moles of Y. If we reacted 1 mole of X with excess Y, the maximum amount of Z we could produce would be 0.5 moles.

a) How many moles of carbon dioxide will be produced when 0.5 moles of carbon is burned?C(s) + O2(g) → CO2(g)

b) How many moles of lead (II) nitrate are needed to produce 2 moles of lead (II) oxide in the reaction shown below?2Pb(NO3)2 (s) → 2PbO (s) + 4NO2(g) + O2(g)

c) How many moles of sodium chloride can be produced by reacting 1.4 moles of chlorine?2Na(s) + Cl2(g) → 2NaCl(s)

d) How many moles of iron oxide can be made from 1 mole of iron (answer to 3 sf)?3Fe(s) + 2O2(g) → Fe3O4(s)

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22. Calculating the mass of a product from the mass of a reactant (for a known compound formula)

a) Calculate the mass of carbon needed to make 10g of carbon dioxide in the reaction below.

C(s) + O2(g) → CO2(g)

b) Iron will react with chlorine to form iron (III) chloride. Find the mass of iron (III) chloride that can be obtained from 4g of iron.

2Fe(s) + 3Cl2(g) → 2FeCl3(s)

c) Calculate the maximum mass of aluminium chloride which can be made from 213g of chlorine.

2Al(s) + 3Cl2(g) → 2AlCl3(s)

d) Calculate the mass of calcium phosphate needed to produce 15.0g of P4 by the reaction below.

2Ca3(PO4)2 + 6SiO2 + 5C → P4 + 6CaSiO3 +5CO2

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23. Reacting volumes: using moles=volume(in cm3)/24000 , n=v/24000Calculating the number of moles from gas volume

How many moles are in the following volumes of gases?a) 24000 cm3 of oxygen;

b) 12000 cm3 of hydrogen;

c) 72dm3 of argon;

d) 12.6 cm3 of ethane.

24. Calculating the gas volume from the number of moles,v=nx24000 (in cm3)

What volume would the following quantities of gas occupy (use whichever unit is most appropriate) ?

a) 5 moles of oxygen;

b) 0.25 moles of nitrogen

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c) 3.2g of oxygen;

d) 22 tonnes of carbon dioxide;

25a. Calculating the mass or volume of a product from the mass or volume of a reactant

a) What mass of phosphorus is required for the preparation of 200cm3 of phosphine (PH3) by the reaction shown below?P4(s) + 3NaOH(l) + 3H2O(l) → 3NaH2PO4(s) + PH3(g)

b) What mass of PbO2 is required to produce 40dm3 of oxygen by the equation shown below?2PbO2(s)) → 2PbO(s) + O2(g)

c) What volume of oxygen would be produced by completely decomposing 4.25g of sodium nitrate?2NaNO3(s)) → 2NaNO2(s) + O2(g)

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d)What volume of hydrogen (at room temperature and pressure) is evolved when 0.325g of zinc react with excess dilute hydrochloric acid?

Examination Question 5dii) Jan 2010

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25b. Using Gas Volumes to Calculate Mr and Ar.

a. 0.08g of element M is reacted with excess water to produce 138cm3 of hydrogen gas. Element M reacts with water according to the following equation:

2M (s) + 2H2O (l) → 2MOH (aq) + H2 (g)

Use the above information to work out the Ar of element M and therefore identify the element.

b. 0.10g of a hydrocarbon was vaporised in a syringe. Its volume, measure in the gas syringe, was 33.5cm3 when corrected to room temperature and pressure. Calculate the relative molecular mass of the hydrocarbon.

26. Using n=cv/1000Calculating the number of moles of dissolved substance from concentration and solution volume

Calculate the number of moles of dissolved substance in the following quantities of solutionsa) 1000cm3 of 1moldm-3 HCl;

b) 25cm3 of 0.05moldm-3 HCl;

c) 25cm3 of 0.025moldm-3 NaOH;

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27. Calculating the concentration from number of moles of dissolved substance and solution volume

Calculate the concentration when the following quantities of solute have been dissolved to produce the following volumes of solution.

a) 0.1 moles of HCl in 1000cm3 of solution;

b) 0.1 moles of HCl in 35cm3 of solution;

c) 0.245 moles of NaOH in 19.35cm3 of solution.

d) 0.5 moles of H2SO4 in 50cm3 of solution ( give your answer in gdm-3)

28. Calculating the solution volume from number of moles of dissolved substance and concentration, v=nx1000/c

Calculate the volume of solution required to provide the stated number of moles when using the solution of given concentration.

a) 2 moles of HCl; the concentration of HCl is 1moldm-3;

b) 3 moles of NaOH; the concentration of NaOH is 1moldm-3;

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c) 0.00845 moles of NaOH; the concentration of NaOH is 0.01moldm-3;

29. Calculating the concentration of a reactant from the concentration and volume of the other reactant in the titration, c=nx1000/v

a) What is the concentration of a solution of sodium hydroxide, 25cm3 of which required 20cm3 of hydrochloric acid of concentration 0.100 moldm-3 for neutralisation?NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

b) If 20.0cm3 of a solution of sodium hydroxide are neutralised by 24.25 cm3 of sulphuric acid of concentration 0.5 moldm-3, what is the concentration of the sodium hydroxide?2NaOH(aq) + H2SO4(aq) → Na2SO4 (aq) + 2H2O(l)

c) Calculate the concentration of sodium carbonate; 25.0cm3 of the carbonate required 21.30cm3 of 0.200 moldm-3 of acid, for complete neutralisation.Na2CO3 (aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

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Examination Question 2b Jan 2010

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30. Calculating reacting volumes, given the reaction equation and n=cv/1000 data

a) Predict the volume of hydrochloric acid (0.1moldm-3) needed to neutralise 25cm3 0.1moldm-3 sodium carbonate.Na2CO3 (aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

b) Predict the volume of sulphuric acid (0.25moldm-3) needed to neutralise 37.60cm3 0.20moldm-3 sodium hydroxide.2NaOH(aq) + H2SO4(aq) → Na2SO4 (aq) + 2H2O(l)

c) Predict the volume of phosphoric acid (0.1moldm-3) needed to neutralise 21cm3 0.1moldm-3 sodium hydroxide.3NaOH(aq) + H3PO4(aq) → Na3PO4 (aq) + 3H2O(l)

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Examination Question 3a Jun 2009

31. Dilution Effects for Moles Calculations

a) 4.0g of solid sodium hydroxide was dissolved in 10cm3 and the solution was made up to 250cm3, A 25cm3 portion was used in a titration reaction. i) How many moles of sodium hydroxide were present in the

original 250cm3 solution?

ii) How many moles of sodium hydroxide were present in the solution used in the titration?

iii) What was the concentration of the sodium hydroxide solution used in the titration?

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b)Lithium can be used to make Lithium Hydroxide 2Li + 2H2O 2LiOH + H2

Lithium Hydroxide can be neutralised by Sulfuric Acid

2LiOH + H2SO4 Li2SO4 + H2O

A student dissolved Lithium in water and made up the solution to 100cm3 in a volumetric flask. The student forgot to make a note of the mass of lithium used. The student then decided to do a titration to work out how much lithium they used.

10cm3 of LiOH needed 12.5cm3 of H2SO4 (0.0200moldm-3) for complete neutralisation.

i) Calculate the amount, in moles of LiOH that reacted?

ii) Calculate the amount, in moles of LiOH that was present originally in the volumetric flask.

iii) Use the equations to help calculate the mass of Lithium used; give your answer to 3 sf.

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Examination Question 2b Jun 2010

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32. Deducing the stoichiometric relationship from given masses and gas volumes

a) A piece of magnesium (0.243g) reacted with 120 cm3 of oxygen when heated. Deduce the empirical formula of magnesium oxide.

b) When heated, 1.117g of iron reacted with 720cm3 chlorine to produce a chloride of iron. Use calculations to help construct the reaction equation.

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33. Trickier Moles Questions

Examination Question 2 Jan 2011

AQA Jan 2011 (continues on the next page)

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34.Concentration of solutions

Complete the missing words activity using the terms concentrated and dilute, as appropriate.

a) A solution of 0.1 mol dm-3 HCl is more ______________ than 0.2 moldm-3 HCl.b) A solution of 0.5 mol dm-3 HCl is more ______________ than 0.2 moldm-3 HCl.c) A solution which contains 0.1g of dissolved NaOH in 100 cm3 is more ___________ than a solution which contains 0.2g of dissolved NaOH in 250 cm3.d) A solution which contains 0.5g of dissolved NaOH in 100 cm3 is more ___________ than a solution which contains 0.2g of dissolved NaOH in 250 cm3.e) A more ________________ acid will react less quickly with a piece of magnesium ribbon.f) A more _________________ salt solution will contain more sodium ions per unit volume.

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