The Masses of chemicals

Transcription

1 The Masses of chemicals Boardworks Ltd 2003 WILF To give a definition of relative formula mass M r. To calculate relative formula mass if its formula and the relative atomic mass are given. To give a full definition of relative atomic mass A r. To explain what a mole is.

2 What s in a Symbol? Relative atomic mass (Mass number) This is equal to the number of protons and neutrons in the nucleus Proton number (Atomic number) 12 6 C is the symbol for carbon This is equal to the number of protons in the nucleus Its also equal to the number of electrons. Why?

3 Relative Atomic Mass of Elements Definiton The relative atomic mass of an element (A r ) compares the mass of atoms of the element with the carbon-12 isotope. It is an average value for the isotope of the element. Carbon is given a relative atomic mass (A r ) of 12. The A r of other atoms compares them with carbon. Eg. Hydrogen has a mass of only one twelfth that of carbon and so has a A r of 1. Element Symbol Times as heavy as carbon A r Helium He one third Beryllium Be three quarters Molybdenum Mo Eight Krypton Kr Seven

4 Relative Formula Mass (Molecular Mass) For a number of reasons it is useful to use something called the formula mass sometimes called the relative molecular mass. M r To calculate this we simply add together the atomic masses of all the atoms shown in the formula. (N=14; H=1; Na=23; O=16; Mg=24; Ca=40) Substance Formula Relative Formula Mass Ammonia NH (3x1)=17 Sodium oxide Na 2 O (2x23) + 16 =62 Magnesium hydroxide Mg(OH) 2 Calcium nitrate Ca(NO 3 ) (16+1)= (14+(3x16))=164

5 ISOTOPES Isotopes are virtually identical in their chemical reactions. (There may be slight differences in speeds of reaction). This is because they have the same number of protons and the same number of electrons. The uncharged neutrons make no difference to chemical properties but do affect physical properties such as melting point and density. Definition An isotope is an atom of the same element but with a different number of neutrons

6 Isotopes: Carbon Natural samples of elements are often a mixture of isotopes. About 1% of natural carbon is carbon C 13 99% 1% C Protons Electrons Neutrons Boardworks Ltd 2003

7 Isotopes: Chlorine About 75% of natural chlorine is 35 Cl the rest is 37 Cl. 35 Cl 17 75% 25% 37 Cl 17 Protons Electrons Neutrons Protons Electrons Neutrons Boardworks Ltd 2003

8 Isotopes and Relative Atomic Mass Many natural elements are a mixture of isotopes. This means that when we react atoms of an element we are using a mixture of atoms with different mass numbers. The relative atomic mass given in the periodic table takes account of this. E.g.. For 100 atoms of chlorine: Mass of 75 atoms of Chlorine 35: 75 x 35 =2625 Mass of 25 atoms of Chlorine 37: 25 x 37 =925 Total = 3550 Average (divide by 100) = 35.5

9 The Mole An Italian count called Amadeo Avogadro worked out that a mole contained 6.02 x10 23 particles. This is called the Avogadro Number He worked out that one mole of any substance is the relative atomic or formula mass expressed in grams. The relative atomic mass A R in grams is known as a MOLE. So a mole of carbon 12 atoms has a mass 12g. The relative formula mass M R of a compound in grams is also one mole. 1 mole of CO 2 =44g. C+0+0 = Amadeo Avogadro

10 How big is a mole? 1 mole of marshmallows would be enough marshmallows to make a 12 mile thick layer of marshmallows covering the entire face of the Earth. 1 mole of popcorn kernels could be spread uniformly over the USA if the thickness of the layer was about 9 miles. 1 mole of donut holes would cover the earth and be 5 miles deep. 1 mole of sheets of paper could form a million stacks from the surface of the earth, all that would pass the sun.

11 The Mole (cont d) 1 mole of any substance always contains the same number of particles 6 x mole of carbon weighs differently to 1 mole of oxygen 1 mole of carbon, C, atoms weighs 12g 1 mole of oxygen, O, atoms weighs 16g. Use a periodic able and work out the mass of each of the following: 1 mole of sodium Na 23g 2 moles carbon C 24g 1 mole of nitrogen N 14g 10 moles chlorine Cl 355g 1 mole of calcium Ca 40g 0.5 moles oxygen O 8g 1 mole of hydrogen H 1g 0.1 moles sulphur S 3.2g Worksheet

12 The Mole (cont d) Changing grams to moles No of moles= mass of substance (g) relative atomic (A r ) or formula mass (M r ) e.g. How many moles in 12g of magnesium? No moles = mass of substance (g) relative atomic mass (A r ) = 12 = 0.5 moles (A r Mg = 24) 24 e.g. How many moles in 4g of oxygen? Trick oxygen exists as molecules = O 2 No moles = mass of substance (g) relative formula mass (M r) = 4 32 (2x16) M r O 2 = 1 8

13 Check Up How many moles in the following: 1. 46g Sodium Na g lithium Li g oxygen O g Chlorine gas Cl g carbon dioxide CO g water H 2 O? ANSWERS

14 1. 46g sodium Na. No moles = mass A r = g Lithium Li = 2 moles No moles = mass A r = g oxygen O 2 = 0.5 moles No moles = mass M r = 3.2 (2x16) = = 0.1 moles 4. 71g chlorine gas Cl 2. No moles = mass M r = 71 (2 x 35.5) = = 1 mole g carbon dioxide CO 2 No moles = mass M r = g water H 2 O = 0.1 moles No moles = mass M r = = 0.2 moles

15 Changing Moles to grams: Mass of substance = no of moles x A r or M r E.G What is the mass of 4 moles of magnesium oxide MgO? Mass = no. moles x M r = 4 x ( ) MgO = 4 x 40 = 160g

16 Check Up What is the mass of the following? moles of carbon C 2. 5 moles of beryllium Be moles oxygen O moles magnesium chloride MgCl moles hydrogen H moles aluminium oxide Al 2 O 3 ANSWERS

17 1. 10 moles of carbon C mass = No moles x A r mass = 10 x 12 = 120g 2. 5 moles Be mass = No moles x A r mass = 5 x 9 = 45g moles H 2 mass = No moles x M r Mass = 0.25 x 2 = 0.5g moles A1 2 O moles O 2 mass = No moles x M r Mass = 0.2 x (2 x 16) = 0.2 x 32 = 6.4g moles MgCl 2 mass = No moles x M r Mass = 0.1 x (24 +(2 x 35.5) = 0.1 x ( ) = 9.5g mass = No moles x M r Mass = 0.2 x ((2x27) + (3x16)) = 0.2 x ( ) =0.2 x 102 = 20.4g

18 Percentage Composition by Mass It is sometimes useful to know how much of a compound is made up of some particular element. This is called the percentage composition by mass the percentage of an element in a compound. % mass of an element in a compound = Relative atomic mass of the element A r X No. of atoms of the element in the formula Relative formula mass of the compound M r %

19 Percentage Composition by Mass cont d E.g. % of oxygen in carbon dioxide (Atomic Masses: C=12. O=16) Formula = Number oxygen atoms = Rel. Atomic Mass of O = Rel. Formula Mass CO 2 = % oxygen = CO (2x16)=44 (2 x 16 ) x 100 = 72.7% Carbon % Oxygen Now work out the percentage composition of potassium K in Potassium nitrate, - KNO 3 K= 39, N = 14, O=16 16

20 Activity Calculate the percentage of oxygen in the compounds shown below % Z = (Number of atoms of Z) x (atomic Mass of Z) Formula Mass of the compound Formula Atoms of O MgO 1 Mass of O Formula Mass %age Oxygen =40 16x100/40=40% K 2 O 1 16 (2x39)+16 =94 16x100/94=17% NaOH =40 16x100/40=40% SO (2x16)= 64 32x100/64=50%

21 Activity Boardworks Ltd 2003 Nitrogen is a vital ingredient of fertiliser that is needed for healthy leaf growth. But which of the two fertilisers ammonium nitrate or urea contains most nitrogen? To answer this we need to calculate what percentage of nitrogen is in each compound

22 Activity Boardworks Ltd 2003 Formulae: Ammonium Nitrate NH 4 NO 3 : Urea CON 2 H 4 Formula Atoms of N Rel. At Mass of N NH 4 NO CON 2 H Relative Formula Mass 14+(1x4)+14+(3x16)= (2x14)+(4x1)= 60 %age Nitrogen 28x100 /80 = 35% 28x100 /60 = 46.7% Atomic masses H=1: C=12: N=14: O=16 Amm.Nitrate Urea And so, in terms of % nitrogen urea is a better fertiliser than ammonium nitrate st Qtr

23 Working out a formula from the percentage composition. This is just working backwards from the percentage composition. If we know the %age composition we can work out the ratio of the atoms of the elements in the compound. This is called the empirical formula. This is the simplest whole number ratio. Sometime it is the same as the actual number this is called the molecular formula. e.g. Empirical formula water H 2 O Molecular formula water H 2 O Empirical formula hydrogen peroxide HO Molecular formula hydrogen peroxide H 2 O 2

24 Working out a formula from the percentage composition. Example 9g of aluminium react with 35.5g chlorine gas. What is the empirical formula of the compound formed? Important bit you have to convert these masses into moles (number of atoms). How do we do that? Divide mass by relative atomic mass. No of moles in 9g of Aluminium = 9 27 = 1/3 No moles in 35.5g chlorine = = 1 Ratio = 1Al : 3Cl Empirical Formula = AlCl 3 1 mole Al react with 3 moles Cl So 1 atom Al reacts with 3 atoms Cl

25 Working out a formula from the percentage composition. Problems g of carbon react with 3.2 g of oxygen. What is the empirical formula of Carbon Dioxide? 2. 80g of calcium react with 32g of oxygen. What is the empirical formula of Calcium oxide?

26 Representing Chemical reactions: Equations. WILF 1. To balance symbol equations. 2. To interpret how many moles of reactants/products are shown in balanced equation. 3. To use a balanced symbol equation to calculate the mass of reactants or products.

27 Word Equations All equations take the general form: Reactants Products Word equations simply replace reactants and products with the names of the actual reactants and products. E.g Reactants Products Magnesium + oxygen Sodium + water Magnesium + lead nitrate Nitric acid + calcium hydroxide Magnesium oxide Sodium hydroxide + hydrogen Magnesium nitrate + lead Water + calcium nitrate

28 Activity Boardworks Ltd 2003 Write the word equations for the descriptions below. 1. The copper oxide was added to hot sulphuric acid and it reacted to give a blue solution of copper sulphate and water. Copper oxide + sulphuric acid copper sulphate + water 2. The magnesium was added to hot sulphuric acid and it reacted to give colourless magnesium sulphate solution plus hydrogen Magnesium + sulphuric acid Magnesium sulphate + hydrogen

29 Activity Boardworks Ltd 2003 Write the word equations for the descriptions below. 3. The methane burned in oxygen and it reacted to give carbon dioxide and water. methane + oxygen Carbon dioxide + water 4. The copper metal was placed in the silver nitrate solution. The copper slowly disappeared forming blue copper nitrate solution and needles of silver metal seemed to grow from the surface of the copper copper + Silver nitrate Copper nitrate + silver

30 Chemical Equations Step 1: Write down the word equation. Step 2: Replace words with the chemical formula. Step 3: Check that there are equal numbers of each type of atom on both sides of the equation. If not, then balance the equation by using more than one. Step 4: Write in the state symbols (s), (l), (g), (aq). Reactants Products magnesium + oxygen magnesium oxide Mg + O 2 MgO Oxygen doesn t balance.need 2 MgO and so need 2 Mg 2Mg + O 2 2Mg(s) +O 2 (g) 2MgO 2MgO(s)

31 Chemical Equations Step 1: Write down the word equation. Step 2: Replace words with the chemical formula. Step 3: Check that there are equal numbers of each type of atom on both sides of the equation. If not, then balance the equation by using more than one. Step 4: Write in the state symbols (s), (l), (g), (aq). Reactants Products sodium + water hydrogen + sodium hydroxide Na + H 2 O H 2 + NaOH Hydrogen doesn t balance. Use 2 H 2 O, NaOH, 2Na + + 2Na 2H 2 O 2NaOH 2Na(s) + + 2H 2 O(l) H 2 (g) 2NaOH(aq) H 2

32 Chemical Equations Step 1: Write down the word equation. Step 2: Replace words with the chemical formula. Step 3: Check that there are equal numbers of each type of atom on both sides of the equation. If not, then balance the equation by using more than one. Step 4: Write in the state symbols (s), (l), (g), (aq). Reactants Products magnesium + lead nitrate magnesium nitrate + lead Mg + Pb(NO 3 ) 2 Mg(NO 3 ) 2 + Pb Already balances. Just add state symbols Mg(s) + Pb(NO + 3 ) 2 (aq) Mg(NO 3 ) 2 (aq) Pb(s)

33 Activity Boardworks Ltd 2003 Below are some chemical equations where the formulae are correct but the balancing step has not been done. Write in appropriate coefficients (numbers) to make them balance. 2 Reactants Products AgNO 3 (aq) + CaCl 2 (aq) Ca(NO 3 ) 2 (aq) + AgCl(s) 2 CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) 2 2 Mg(s) + Ag 2 O(s) MgO(s) + Ag(s) NaOH + H 2 SO 4 (aq) Na 2 SO 4 (aq) + H 2 O(l) 2 2 2

34 Reacting Masses Boardworks Ltd 2003

35 Conservation of Mass New substances are made during chemical reactions However, the same atoms are present before and after reaction. They have just joined up in different ways. Because of this the total mass of reactants is always equal to the total mass of products. This idea is known as the Law of Conservation of Mass. Reaction but no mass change

36 Conservation of Mass There are examples where the mass may seem to change during a reaction. Eg. In reactions where a gas is given off the mass of the chemicals in the flask will decrease because gas atoms will leave the flask. If we carry the same reaction in a strong sealed container the mass is unchanged. Gas given off HCl Mg Mass of chemicals in flask decreases Same reaction in sealed container: No change in mass

37 Reacting Mass and formula mass The formula mass in grams of any substance contains the same number of particles. We call this amount of substance 1 mole. Symbol Atomic Masses: H=1; Mg=24; O=16; C=12; N=14 Formula Mass Contains H 2 MgO CH 4 HNO 3 1x (1x4) 1+14+(3x16) 1 mole of hydrogen molecules 1 mole of magnesium oxide 1 mole of methane molecules 1 mole of nitric acid

38 Reacting Mass and Equations By using the formula masses in grams ( moles) we can deduce what masses of reactants to use and what mass of products will be formed. Atomic masses: C=12; O=16 carbon + oxygen carbon dioxide C + O 2 CO x (2x16) 12g 32g 44g So we need 32g of oxygen to react with 12g of carbon and 44g of carbon dioxide is formed in the reaction.

39 Activity Boardworks Ltd 2003 What mass of aluminium and chlorine react together? Atomic masses: Cl=35.5; Al=27 aluminium + chlorine aluminium chloride 2Al + 3Cl 2 2AlCl 3 2 x x x (27+(3x35.5) 54g 213g 267g So 54g of aluminium react with 213g of chlorine to give 267g of aluminium chloride.

40 Activity Boardworks Ltd 2003 What mass of magnesium and oxygen react together? Atomic masses: Mg=24; O=16 magnesium + oxygen 2 + Magnesium oxide Mg O 2 2 MgO 2 x x16 2x(24+16) 48g 32g 80g So 48g of magnesium react with 32g of oxygen to give 80g of magnesium oxide.

41 Activity What mass of sodium chloride is formed when sodium hydroxide and hydrochloric acid react together? Atomic masses: Na = 23 O = 16 H = 1 Cl = 35.5 Sodium + hydrochloric Sodium + hydroxide + acid chloride NaOH + HCl NaCl + water H 2 O g 36.5g 58.5g (2x1)+16 18g So 40g of sodium hydroxide react with 36.5g of hydrochloric acid to give 58.5g of sodium chloride.

42 It is important to go through the process in the correct order to avoid mistakes. Step 1 Step 2 Step 3 Word Equation Replace words with correct formula. Balance the equation. Step 4 Write in formula masses. Remember: where the equation shows more than 1 molecule to include this in the calculation. Step 5 Add grams to the numbers.

43 Activity Reacting Mass and Scale Factors We may be able to calculate that 48g of magnesium gives 80g of magnesium oxide but can we calculate what mass of magnesium oxide we would get from burning 1000g of magnesium? There are 3 extra steps: Step 1 Will 1000g of Mg give more or less MgO than 48g? Step 2 I need to scale up? the 48g to 1000g. What scale factor does this give? Step 3 If 48g Mg gives 80g of MgO What mass does 1000g give? Answer more 1000 = x g

44 Activity Boardworks Ltd 2003 Mg + CuSO 4 MgSO 4 + Cu (4x16) (4x16) 64 24g 160g 120g 64g What mass of copper will I get when 2 grams of magnesium is added to excess (more than enough) copper sulphate? Step 1 Will 2g of Mg give more or less Cu than 24g? down Step 2 I need to scale? the 24g to 2g. What scale factor does this give? less 2 = Step 3 If 24g Mg gives 64g of Cu What mass does 2g give? Answer x

45 Activity Boardworks Ltd 2003 CaCO 3 CaO + CO (3x16) (2x16) 100g 56g 44g What mass of calcium oxide will I get when 20 grams of limestone is decomposed? Step 1 Will 20g of CaCO 3 give more or less CaO than 100g? Step 2 I need to scale down? the 100g to 20g. What scale factor does this give? less 20 = Step 3 If 100g CaCo 3 gives 56g of CaO What mass does 20g give? Answer 0.20 x g

46 Calculation of Empirical Formulae. Example 1.A compound contains 75% carbon and 25% hydrogen. What is its empirical formula? Elements Carbon Hydrogen Amount in question (% or mass) Atomic mass (periodic table) 12 1 Number of moles = Amount in question Atomic mass Mole ratio (divide each number of moles by the smallest number of moles) = = = = 4 Empirical formula C H 4 Empirical formula CH 4

47 Calculation of Empirical Formulae. Example 2. An oxide of carbon contains 27% carbon. What is its empirical formula? Boardworks Ltd 2003 Elements Carbon Oxygen Amount in question (% or mass) 27 Rel. Atomic mass (periodic table) 12 Number of moles = Amount in question Rel. Atomic mass = =4.56 Mole ratio (divide each number of moles by the smallest number of moles) = = 1 = = 2 Empirical formula C O 2

48 Moles and solutions When 1 mole of solute is dissolved in 1 litre of water we say it is a 1 Molar or 1M solution. A 1M solution of sodium hydroxide (NaOH) solution would contain 1 mole ( = 40g) of sodium hydroxide solid dissolved in 1 litre of water. Number of moles = volume(l) x Moles per litre (M) e.g. How many moles in 250cm 3 of a 2M solution? No of moles = 0.25 l x 2m = 0.5 moles.

49 Moles and gases Gases are measured by volume not by mass. At room temperature and pressure 1 mole of any gas occupies 24 litres. e.g. 0.2 moles of hydrogen H2 occupies what volume? Volume = 0.2 x 24 = 4.8 litres.

50 Formula from Composition by mass. WILF To calculate the formula of a compound from the mass of the reactants.

51 GCSE Question An exothermic reaction takes place when nitrogen reacts with hydrogen to make ammonia. The reaction can be represented by this equation. N 2 (g) + 3H 2 (g) 2NH 3 (g) (a) Calculate the maximum mass of ammonia that could be made from 1000 g of nitrogen. Relative atomic masses: H = 1; N = 14

52 Relative Formula Mass M r When a new compound is discovered we have to deduce its formula. This always involves getting data about the masses of elements that are combined together. What we have to do is work back from this data to calculate the number of atoms of each element and then calculate the ratio. In order to do this we divide the mass of each atom by its atomic mass. The calculation is best done in 5 stages: Boardworks Ltd 2003

53 Calculating Empirical Formula Boardworks Ltd 2003 We found 3.2g of copper reacted with 0.8g of oxygen. What is the formula of the oxide of copper that was formed? (At. Mass Cu=64: O=16) Substance Copper oxide 1. Elements Cu O 2. Mass of each element in question (g) 3. Mass Atomic Mass 4. Ratio (No. of moles) 5. Formula = =0.05 1:1 CuO

54 We found 5.5g of manganese reacted with 3.2g of oxygen. What is the formula of the oxide of manganese formed? (Atomic. Mass Mn=55: O=16) Substance Manganese oxide 1. Elements Mn O 2. Mass of each element (g) 3. Mass / Atomic Mass 4. Ratio (No. of moles) 5. Formula /55 = /16 =0.20 1:2 MnO 2

55 Activity A chloride of silicon was found to have the following % composition by mass: Silicon 16.5%: Chlorine 83.5% (Atomic. Mass Si=28: Cl=35.5) Substance Silicon Chloride 1. Elements Si Cl 2. Mass of each element (g per 100g) 3. Mass / Atomic Mass /28 = /35.5 = Ratio Divide biggest by smallest Cl Si = ( ) = (3.98) Ratio of Cl:Si =4:1 5. Formula SiCl 4

56 Activity Boardworks Ltd 2003 Calculate the formula of the compounds formed when the following masses of elements react completely: (Atomic. Mass Si=28: Cl=35.5) Element 1 Element 2 Atomic Masses Formula Fe = 5.6g Cl=106.5g Fe=56 Cl=35.5 K = 0.78g Br=1.6g K=39: Br=80 P=1.55g Cl=8.8g P=31: Cl=35.5 C=0.6g H=0.2g C=12: H=1 Mg=4.8g O=3.2g Mg=24: O=16 FeCl 3 KBr PCl 5 CH 4 MgO masses

57 Formula from Charges on ions WILF To calculate the formula of a compound from the charge(s) on the ions

58 Charges on ions. Many elements form ions with some definite charge (E.g. Na +, Mg 2+ and O 2-). It is often possible to work out the charge using the Periodic Table. If we know the charges on the ions that make up the compound then we can work out its formula. This topic is covered in more detail in the Topic on Bonding but a few slides are included here on how to work out the charges on ions and use these to deduce the formula of simple ionic compounds.

59 Charges and Metal ions Metals usually lose electrons to empty this outer shell. The number of electrons in the outer shell is usually equal to the group number in the Periodic Table. Eg. Li =Group 1 Mg=Group2 Al=Group3 Li Mg Al 2.1 Li Mg Al 3+

60 Charges and non-metal ions Elements in Groups 4 onwards generally gain electrons and the number of electrons they gain is equal to the Group Number. Oxygen (Group 6) gains (8-6) =2 electrons to form O 2- Chlorine (Group 7) gains (8-7)=1 electron to form Cl - O Cl O O Cl Cl -

61 Activity Boardworks Ltd Copy out and fill in the Table below showing what charge ions will be formed from the elements listed. H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Symbol Li N Cl Ca K Al O Br Na Group No Charge

62 The formulae of ionic compounds This is most quickly done in 5 stages. Remember the total + and charges must =zero Eg. The formula of calcium bromide. 1. Symbols: Ca Br 2. Charge on ions Need more of Br 4. Ratio of ions Formula CaBr 2 Ca 2 electrons Br Br Ca 2+ Br - Br - Boardworks Ltd 2003

63 The formulae of ionic compounds Eg. The formula of aluminium bromide. 1. Symbols: Al Br 2. Charge on ions Need more of Br 4. Ratio of ions Formula AlBr 3 Al Br Br Br - Al 3+ Br - 3 electrons Br Br -

64 The formulae of ionic compounds Eg. The formula of aluminium oxide. 1. Symbols: Al O 2. Charge on ions Need more of O 4. Ratio of ions 2 3 (to give 6 e - ) 5. Formula Al 2 O 3 Al Al 2e - 2e - 2e - O O O Al 3+ Al3+ O 2- O 2- O 2-

65 Activity The formulae of ionic compounds Eg. The formula of magnesium chloride. 1. Symbols: Mg Cl 2. Charge on ions 3. Need more of Cl 4. Ratio of ions 1:2 5. Formula MgCl 2 Mg 1e - 1e - Cl Cl Mg 2+ Cl - Cl -

66 Activity The formulae of ionic compounds Eg. The formula of sodium oxide. 1. Symbols: Na O 2. Charge on ions 3. Need more of 4. Ratio of ions 5. Formula Na 2 : 1 Na 2 O 1e Na - Na + Na O O 2-1e - Na+

67 Activity Boardworks Ltd 2003 Using the method shown on the last few slides, work out the formula of all the ionic compounds that you can make from combinations of the metals and non-metals shown below: Metals: Li Ca Na Mg Al K Non-Metals: F O N Br S Cl

68 Reacting Mass Industrial Processes Industrial processes use tonnes of reactants not grams. We can still use equation and formula masses to calculate masses of reactants and products. We simply swap grams for tonnes. E.g. What mass of CaO does 200 tonnes of CaCO 3 give? CaCO 3 CaO + CO So 100 tonnes would give 56? tonnes And 200 tonnes will give more Scale factor = 200/100 =2 So mass of CaO formed = 2 x? 56 tonnes = 112 tonnes

69 Activity Boardworks Ltd 2003 Iron is extracted from iron oxide Fe 2 O 3 E.g. What mass of Fe does 100 tonnes of Fe 2 O 3 give? Fe 2 O 3 + 3CO 2Fe + 3CO So 160 tonnes would give 112? tonnes And 100 tonnes will give less Scale factor = 100/160 =0.625 So mass of Fe formed = 0.625? x 112 = 70 tonnes

70 Activity Boardworks Ltd 2003 Ammonia is made from nitrogen and hydrogen E.g. What mass of NH 3 is formed when 50 tonnes of N 2 is completely converted to ammonia? N 2 + 3H 2 2NH So 28 tonnes would give? 34 tonnes And 50 tonnes will give Scale factor = more 50/28 =1.786 than 28 tonnes So mass of NH 3 formed = 1.786? x 34 = 60.7 tonnes

71 Na is the symbol for? 1. Nitrogen 2. Nickel 3. Neodynium 4. Sodium

72 Which of these does NOT exist as a diatomic molecule (2 bonded atoms)? 1. Nitrogen 2. Oxygen 3. Calcium 4. Chlorine

73 How many oxygen atoms are represented in the formula Pb(NO 3 ) 2? 1. One 2.Two 3.Three 4.Six

74 What is the formula mass of MgCl 2? Mg=24 Cl=

75 What is the formula mass of Mg(OH) 2? Mg=24 O=16 H =

76 What is the percentage nitrogen in ammonium sulphate (NH 4 ) 2 SO 4? 1. 21% 2. 42% 3. 63% 4. 84%

77 What is the formula of a compound containing 1.4g nitrogen and 3.2g of oxygen? (N=14 O=16) 1. N 2 O 2. NO 3. NO 2 4. N 2 O 3

78 What is the formula of a compound containing 6.5g zinc and 1.6g oxygen? (Zn=65 O=16) 1. ZnO 2. Zn 2 O 3 3. ZnO 2 4. Zn 2 O

79 What is the formula of a compound formed between Cr 3+ ions and O 2- ions? 1. CrO 2. Cr 2 O 3 3. CrO 2 4. Cr 3 O 2

80 What is the formula of a compound formed between Cr 3+ ions and OH - ions? 1.CrOH 3 2.Cr 3 OH 3.Cr(OH) 3 4.Cr 2 OH 3

81 What is the word equation for the reaction described below? A small piece of strontium metal was added to water. It fizzed giving off hydrogen gas leaving an alkaline solution of strontium hydroxide. 1.Strontium + water hydrogen + strontium hydride 2.Strontium + water oxygen + strontium hydroxide 3.Strontium + water hydrogen + strontium hydrate 4.Strontium + water hydrogen + strontium hydroxide

82 What numbers a - d are needed to balance the equation? Strontium + water hydrogen + strontium hydroxide a Sr + b H 2 O c H 2 + d Sr(OH) 2 1 a=1 b=1 c=1 d=1 2 a=1 b=2 c=1 d=1 3 a=1 b=1 c=2 d=1 4 a=1 b=1 c=1 d=2

83 What is the mass of 2 moles of magnesium nitrate Mg(NO 3 ) 2? 1. 86g g g g

84 How many moles of iron atoms is 280g of iron? (Fe=56) 1. One mole 2. Two moles 3. Four moles 4. Five moles

85 When iron rusts it forms the iron oxide Fe 2 O 3. What mass of oxygen reacts with 112g of iron? (Fe=56 O=16) 1. 1g 2. 16g 3. 48g g

86 Hydrogen reacts with chlorine to form hydrogen chloride HCl. H 2 + Cl 2 2HCl What mass of hydrogen chloride will be obtained from 4g of hydrogen gas? (H=1 Cl=35.5) g 2 73g g 4 146g