1.1.2 Moles and equations This Powerpoint contains the questions and answers for the activities 1-34

1. Convert the following masses as specified. 5 kg into g; 45800 kg into g; 6.34 g into kg; 1.34 x 109 g into kg; 450mg into g.

1. Convert the following masses as specified- Answers 5 kg into g; 5000g 45800 kg into g; 4.58 x 107g 6.34 g into kg; 0.00634kg 1.34 x 109 g into kg; 1340000kg 450mg into g 0.450g

2. Conversion of volumes into different units 2 dm3 into cm3; b)81400 dm3 into cm3; c)9.85 cm3 into dm3; d)4.94 x 109 cm3 into dm3;

2. Conversion of volumes into different units 2 dm3 into cm3; 2000cm3 81400 dm3 into cm3; 81400000cm3 9.85 cm3 into dm3; 0.00985dm3 4.94 x 109 cm3 into dm3; 4940000dm3

3. Using significant figures Answering to 3 significant figures Write the following numbers to 3 significant figures 0.09936 234 1369.185 0.101 2.8769 x 1016

3. Using significant figures 0.09936 0.0994 234 234 1369.185 1370 0.101 0.101 2.8769 x 1016 2.88 x 1016

4. Answering 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

4. Answering to a set number of decimal places Write the following to 2 decimal places a) 0.09936 0.10 b) 0.10752 0.11 c) 58.942 58.94 d) 1369.185 1369.19 e) 0.101 0.10

5. Categorizing materials as substances or mixtures Underline the materials from the list below which you would describe as substances (Hint; substances are those which are not mixtures). Sugar, water, air, seawater , wood, sodium chloride, sulphur dioxide, chlorine, calcium, lemon juice, bromine, plastic

5. Categorizing materials as substances or mixtures Sugar, water, air, seawater, wood, sodium chloride , sulphur dioxide, chlorine, calcium, lemon juice, bromine, plastic

6. Converting numbers of particles into moles Convert the following numbers of particles into moles 6.02 x 1023 atoms of carbon; 3.01 x 1023 molecules of methane; 5.78 x 1026 molecules of oxygen.

6. Converting numbers of particles into moles 6.02 x 1023 atoms of carbon; 1 mole 3.01 x 1023 molecules of methane; 0.5 moles 5.78 x 1026 molecules of oxygen. 960 moles

7. Converting moles into numbers of particles 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?

7. Converting moles into numbers of particles a) How many particles are present in 3 moles of sulphur atoms? 1.81x1024 b) How many sodium ions are in a solution containing 10.7 moles of sodium chloride? 6.44 x1024 c) How many oxygen molecules are in 2.5 moles of oxygen gas? 1.51 x1024

8. Using the periodic table to find the molar mass of an element Write the molar mass for the following elements. a) carbon ___ b) sulphur ___ c) magnesium ___ d) hydrogen (H2) ___ e) potassium ___ f) iodine (I2) ___

8. Using the periodic table to find the molar mass of an element carbon 12.0g b) sulphur 32.1g c) magnesium 24.3g d) hydrogen (H2) 2.0g e) potassium 39.1g f) iodine (I2) 254.0g

Exam Q 1c Jan09

Exam Q 1c Jun 10

9. Finding the molar masses of simple compounds Calculate the molar mass of the following compounds from their formulae CH4 (NH4)2SO4 c) FeSO4.7H2O

9. Finding the molar masses of simple compounds a) CH4 16g b) (NH4)2SO4 132.1g c) FeSO4.7H2O 277.9g

10. Calculating the percentage by mass of an element in a compound Calculate the percentage by mass of the named element in each of the compounds listed. a) Copper in CuSO4 b) Calcium in Ca(CN)2 c) Nitrogen in (NH4)2CO3

10. Calculating the percentage by mass of an element in a compound a) Copper in CuSO4 39.8% Mr 159.5 b) Calcium in Ca(CN)2 43.5% Mr 92.1 c) Nitrogen in (NH4)2CO3 29.2% Mr 96

11. Deducing empirical formulae from given molecular formulae a) benzene C6H6; b) heptene C7H14; c) cyclopentane C5H10; d) dinitrogen tetroxide N2O4; e) ethane C2H6.

11. Deducing empirical formulae from given molecular formulae a) benzene C6H6; CH b) heptene C7H14; CH2 c) cyclopentane C5H10; CH2 d) dinitrogen tetroxide N2O4; NO2 e) ethane C2H6. CH3

12. Deducing the molecular formula from empirical formula and molar mass a) A hydrocarbon of molar mass 142 has an empirical formula C5H11. What is the molecular formula? b) A hydrocarbon of molar mass 56 has an empirical formula CH2. What is the molecular formula?

12. Deducing the molecular formula from empirical formula and molar mass a) A hydrocarbon of molar mass 142 has an empirical formula C5H11. What is the molecular formula? C10H22 b) A hydrocarbon of molar mass 56 has an empirical formula CH2. What is the molecular formula? C4H8

13. Calculating the empirical formula from percentage composition Work out the empirical formulae for the following compounds a) Fe 78% O 22%; b) C 92.3 % H 7.6% c) K 26.6% Cr 35.4% O 38%;

13. Calculating the empirical formula from percentage composition a) Fe 77% O 22%; FeO b) C 92.3 % H 7.6% CH c) K 26.6% Cr 35.4% O 38%; K2Cr2O7

14. Calculating the empirical formula from the percentage of one reactant Work out the empirical formulae for the following compounds A chloride of cobalt which contains 54.7% chlorine; A chloride of copper which contains 64.1% copper;

14. Calculating the empirical formula from the percentage of one reactant a) A chloride of cobalt which contains 54.7% chlorine; CoCl2 b) A chloride of copper which contains 64.1% copper; CuCl

Exam Qu 1d Jun 2010

15. Calculating the empirical formula from composition by mass of all atoms present A compound contains 5.58g of iron and 3.21g of sulphur. The molar mass is 87.9. Find the molecular formula. 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.

15. Calculating the empirical formula from composition by mass of all atoms present A compound contains 5.58g of iron and 3.21g of sulphur. The molar mass is 87.9. Find the molecular formula. Answer: FeS 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. Answer: C2H4O2 Mass 0.200g of C 0.033g of H 0.268g of O Divide by Ar 0.0167 0.033 0.0168 Ratio of atoms 1 2 1

16. Producing formulae for compounds using a valency table Use your valency table to construct formulae for the following compounds. a) copper (II) oxide; b) potassium oxide; c) sodium bromide; d) copper (II) sulphate; e) tin (II) nitrate; f) magnesium chloride; g) aluminium sulphate; h) potassium dichromate; i) calcium hydroxide. j) ammonium carbonate;

16. Producing formulae for compounds using a valency table a) copper (II) oxide; CuO b) potassium oxide; K2O c) sodium bromide; NaBr d) copper (II) sulphate; CuSO4 e) tin (II) nitrate; Sn(NO3)2 f) magnesium chloride; MgCl2 g) aluminium sulphate; Al2(SO4)3 h) potassium dichromate; K2Cr2O7 i) calcium hydroxide. Ca(OH)2 j) ammonium carbonate; (NH4)2CO3

17. Writing symbol equations for reactions we have studied 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

17. Writing symbol equations for reactions we have studied a) hydrogen + oxygen → water 2H2 + O2 → 2H2O b) Lithium + sulphuric acid → Lithium sulphate + hydrogen Li + H2SO4 → Li2SO4 + H2 c) ammonium + calcium → ammonia + calcium chloride + water chloride hydroxide 2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O

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

18. Writing symbol equations for reactions with reactants or products stated The formation of magnesium oxide from burning magnesium. 2Mg + O2 → 2MgO b) The reaction of sulphur trioxide with water to produce sulphuric acid (H2SO4). SO3 + H2O → H2SO4 c) The decomposition of hydrogen peroxide (H2O2) to produce oxygen and another product. 2H2O2 → 2H2O + O2

19. Reacting masses: using moles=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; ____

19. Reacting masses: using moles=mass/Mr a) 24.0g of carbon; 2.00 b) 1.7mg of ammonium oxide; 3.27x10-5 c) 40 kg of iron; 717 d) 2 tonnes of water; 111000 or 1.11x105 e) 2.57 x10-4 g of sodium hydroxide; 6.43 x10-6

20. Calculating the mass from the number of moles 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;

20. Calculating the mass from the number of moles a) 3 moles of carbon; 36.0g b) 2.4 moles of oxygen molecules; 76.8g c) 10.9 moles of carbon dioxide; 479.6g d) 1.45 x 10-4 moles of sulphur dioxide; 0.00928g

21. 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)

Answers a) How many moles of carbon dioxide will be produced when 0.5 moles of carbon is burned? 0.5 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? 2 2Pb(NO3)2 (s) → 2PbO (s) + 4NO2(g) + O2(g)

21. 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)

Answers c) How many moles of sodium chloride can be produced by reacting 1.4 moles of chlorine? 2.8 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)? 0.333 3Fe(s) + 2O2(g) → Fe3O4(s)

22. 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)

22. Answers a) Calculate the mass of carbon needed to make 10g of carbon dioxide in the reaction below. 2.73g 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. 11.63g 2Fe(s) + 3Cl2(g) → 2FeCl3(s)

22. 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

22. Answers c) Calculate the maximum mass of aluminium chloride which can be made from 213g of chlorine. 267g 2Al(s) + 3Cl2(g) → 2AlCl3(s) d) Calculate the mass of calcium phosphate needed to produce 15.0g of P4 by the reaction below. 75.1g (Ca3(PO4)2) 2Ca3(PO4)2 + 6SiO2 + 5C → P4 + 6CaSiO3 +5CO2

23. Calculating the number of moles from gas volume a) 24000 cm3 of oxygen; b) 12000 cm3 of hydrogen; c) 72dm3 of argon; d) 12.6 cm3 of ethane.

Answers a) 24000 cm3 of oxygen; 1 moles b) 12000 cm3 of hydrogen; 0.5 moles c) 72dm3 of argon; 3 moles d) 12.6 cm3 of ethane. 5.25 x 10-4moles

24. a) 5 moles of oxygen; b) 0.25 moles of nitrogen; c) 3.2g of oxygen; d) 22 tonnes of carbon dioxide;

Answers a) 5 moles of oxygen; 120dm3 b) 0.25 moles of nitrogen; 6 dm3 c) 3.2g of oxygen; 2.4dm3 d) 22 tonnes of carbon dioxide; 1.2 x 107dm3

Activity 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)

2NaNO3(s)) → 2NaNO2(s) + O2(g) Activity 25a Continued Calculating the mass or volume of a product from the mass or volume of a reactant c) What volume of oxygen would be produced by completely decomposing 4.25g of sodium nitrate? 2NaNO3(s)) → 2NaNO2(s) + O2(g) d) What volume of hydrogen (at room temperature and pressure) is evolved when 0.325 g of zinc react with excess dilute hydrochloric acid?

Activity 25a ANSWERS 1.03g 797.7g 600cm3 (0.6dm3) 119cm3 (0.119dm3)

Exam Qu 5dii) Jan 2010

Activity 25b Using Gas Volumes to Calculate Mr and Ar. 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, measured in the gas syringe, was 33.5cm3, when corrected to room temperature and pressure. Calculate the relative molecular mass of the hydrocarbon.

Activity 25b Answers a) n(H2) = 138/24,000 = 0.00575 n(M)=2x0.00575=0.0115 n=m/Ar so Ar =m/n Ar= 0.08/0.0115 = 6.96 (3sf) M is Lithium

Activity 25b Answers 0.10g is 33.5cm3 n(gas) = 33.5/24,000=1.395x10-3 Mr=m/n Mr=0.1/1.395x10-4 Mr=71.68 Mr= 72 MF = C5H12 (Pentane)

Activity 26 Calculate the number of moles of dissolved substance in the following quantities of solutions a) 1000cm3 of 1moldm-3 HCl; b) 25cm3 of 0.05moldm-3 HCl; c) 25cm3 of 0.025moldm-3 NaOH;

Exercise 26: Answers 1 moles 0.00125 moles (1.25x10-3)

Activity 27 Calculating the concentration from number of moles of dissolved substance and solution volume 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) 0.5moles of H2SO4 in 50cm3 of solution ( give your answer in gdm-3)

Exercise 27: Answers 0.1 mol dm-3 2.86 mol dm-3 12.66 mol dm-3 981gdm-3

Activity 28 Calculating the solution volume from number of moles of dissolved substance and 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; c) 0.00845 moles of NaOH; the concentration of NaOH is 0.01moldm-3;

Exercise 28: Answers 2000 cm3 3000 cm3 845 cm3

Activity 29 Calculating the concentration of a reactant from the concentration and volume of the other reactant 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)

Activity 29 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)  

Exercise 29: Answers 0.08 mol dm-3 1.21 mol dm-3 0.0852 mol dm-3

Exam Qu 2b Jan2010

Activity 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)

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

Exercise 30: Answers 50 cm3 15 cm3 7 cm3

Exam Qu 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. How many moles of sodium hydroxide were present in the original 250cm3 solution? How many moles of sodium hydroxide were present in the solution used in the titration?   What was the concentration of the sodium hydroxide solution used in the titration?

31. Dilution Effects for Moles Calculations i) n(NaOH) in 250cm3 =4/40 = 0.1 ii) n(NaOH) in 25cm3 = 0.1/10 = 0.01 iii) c(NaOH) in titration = 0.01x1000/25 = 0.4

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

31. Dilution Effects for Moles Calculations(Answers) i) n(H2SO4)= 0.02x12.5/1000= 2.5x10-4 n(LiOH) in 10cm3 = 2x2.5x10-4 = 5x10-4 ii) n(LiOH) in 100cm3 = 5x10-3 iii) n(Li)= n (LiOH) n=m/Ar so, m (Li) = 6.9x5x10-3= 3.45x10-2g of Lithium or 0.0345g

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

Exercise 32: Answers MgO FeCl3

Exam Qu 2 Jan 2011

AQA Qu 3 Jan 2011

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.

Exercise 34: Answers Dilute Concentrated