# Chapter 1: Moles and equations. Learning outcomes. you should be able to:

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1 Chapter 1: Moles and equations 1 Learning outcomes you should be able to: define and use the terms: relative atomic mass, isotopic mass and formula mass based on the 12 C scale perform calculations, including use of the mole concept involving: reacting masses (from formulae and equations) empirical formula and molecular formula the mole in terms of the Avogadro constant analyse and use mass spectra to calculate the relative atomic mass of an element calculate empirical and molecular formulae using combustion data or composition by mass volumes of gases (eg in the burning of hydrocarbons) volumes and concentrations of solutions deduce stoichiometric relationships from calculations involving reacting masses, volumes of gases and volumes and concentrations of solutions write and construct balanced equations

2 Cambridge International AS Level Chemistry Introduction For thousands of years, people have heated rocks and distilled plant juices to extract materials Over the past two centuries, chemists have learnt more and more about how to get materials from rocks, from the air and the sea, and from plants They have also found out the right conditions to allow these materials to react together to make new substances, such as dyes, plastics and medicines When we make a new substance it is important to mix the reactants in the correct proportions to ensure that none is wasted In order to do this we need to know about the relative masses of atoms and molecules and how these are used in chemical calculations 2 Figure 11 A titration is a method used to find the amount of a particular substance in a solution Masses of atoms and molecules Relative atomic mass, A r Atoms of different elements have different masses When we perform chemical calculations, we need to know how heavy one atom is compared with another The mass of a single atom is so small that it is impossible to weigh it directly To overcome this problem, we have to weigh a lot of atoms We then compare this mass with the mass of the same number of standard atoms Scientists have chosen to use the isotope carbon-12 as the standard This has been given a mass of exactly 12 units The mass of other atoms is found by comparing their mass with the mass of carbon-12 atoms This is called the relative atomic mass, A r The relative atomic mass is the weighted average mass of naturally occurring atoms of an element on a scale where an atom of carbon-12 has a mass of exactly 12 units From this it follows that: A r [element Y] average mass of one atom of element Y 12 = mass of one atom of carbon-12 We use the average mass of the atom of a particular element because most elements are mixtures of isotopes For example, the exact A r of hydrogen is This is very close to 1 and most periodic tables give the A r of hydrogen as 10 However, some elements in the Periodic Table have values that are not whole numbers For example, the A r for chlorine is 355 This is because chlorine has two isotopes In a sample of chlorine, chlorine-35 makes up about three-quarters of the chlorine atoms and chlorine-37 makes up about a quarter Relative isotopic mass Isotopes are atoms that have the same number of protons but different numbers of neutrons (see page 28) We represent the nucleon number (the total number of neutrons plus protons in an atom) by a number written at the top left-hand corner of the atom s symbol, eg 20 Ne, or by a number written after the atom s name or symbol, eg neon-20 or Ne-20 We use the term relative isotopic mass for the mass of a particular isotope of an element on a scale where an atom of carbon-12 has a mass of exactly 12 units For example, the relative isotopic mass of carbon-13 is 1300 If we know both the natural abundance of every isotope of an element and their isotopic masses, we can calculate

3 Chapter 1: Moles and equations the relative atomic mass of the element very accurately To find the necessary data we use an instrument called a mass spectrometer (see box on mass spectrometry) Relative molecular mass, M r The relative molecular mass of a compound (M r ) is the relative mass of one molecule of the compound on a scale where the carbon-12 isotope has a mass of exactly 12 units We find the relative molecular mass by adding up the relative atomic masses of all the atoms present in the molecule For example, for methane: formula CH 4 atoms present 1 C; 4 H add A r values (1 A r [C]) + (4 A r [H]) M r of methane = (1 120) + (4 10) = 160 Relative formula mass For compounds containing ions we use the term relative formula mass This is calculated in the same way as for relative molecular mass It is also given the same symbol, M r For example, for magnesium hydroxide: formula Mg(OH) 2 ions present 1 Mg 2+ ; 2 (OH ) add A r values (1 A r [Mg]) + (2 [O] + A r [H])) M r of magnesium hydroxide = (1 243) + (2 ( )) = 583 Accurate relative atomic masses MAss spectrometry box 11: biological drawing A mass spectrometer (Figure 12) can be used to measure the mass of each isotope present in an element It also compares how much of each isotope is present the relative abundance (isotopic abundance) A simplified diagram of a mass spectrometer is shown in Figure 13 You will not be expected to know the details of how a mass spectrometer works, but it is useful to understand how the results are obtained Figure 12 A mass spectrometer is a large and complex instrument 3 1 Use the Periodic Table on page 473 to calculate the relative formula masses of the following: a calcium chloride, CaCl 2 b copper(ii) sulfate, CuSO 4 vaporised sample positively charged electrodes accelerate positive ions magnetic field c ammonium sulfate, (NH 4 ) 2 SO 4 d magnesium nitrate-6-water, Mg(N ) 2 6H 2 O Hint: for part d you need to calculate the mass of water separately and then add it to the M r of Mg(N ) 2 heated filament produces high-energy electrons ionisation chamber flight tube ion detector recorder computer Figure 13 Simplified diagram of a mass spectrometer

4 Cambridge International AS Level Chemistry MASS SpeCTrOMeTry (COnTInued) The atoms of the element in the vaporised sample are converted into ions The stream of ions is brought to a detector after being deflected (bent) by a strong magnetic field As the magnetic field is increased, the ions of heavier and heavier isotopes are brought to the detector The detector is connected to a computer, which displays the mass spectrum The mass spectrum produced shows the relative abundance (isotopic abundance) on the vertical axis and the mass to ion charge ratio (m/e) on the horizontal axis Figure 14 shows a typical mass spectrum for a sample of lead Table 11 shows how the data is interpreted 3 Determination of A r from mass spectra We can use the data obtained from a mass spectrometer to calculate the relative atomic mass of an element very accurately To calculate the relative atomic mass we follow this method: multiply each isotopic mass by its percentage abundance add the figures together divide by 100 We can use this method to calculate the relative atomic mass of neon from its mass spectrum, shown in Figure 15 The mass spectrum of neon has three peaks: 20 Ne (909%), 21 Ne (03%) and 22 Ne (88%) A r of neon (20 909) + (210 03) + (22 88) = = Note that this answer is given to 3 significant figures, which is consistent with the data given 4 Detector current /ma Mass/charge (m/e) ratio Figure 14 The mass spectrum of a sample of lead For singly positively charged ions the m/e values give the nucleon number of the isotopes detected In the case of lead, Table 11 shows that 52% of the lead is the isotope with an isotopic mass of 208 The rest is lead-204 (2%), lead-206 (24%) and lead- 207 (22%) Isotopic mass Relative abundance / % total 100 Table 11 The data from Figure 14 Relative abundance/% % 03 % 88 % Mass/charge (m/e) ratio Figure 15 The mass spectrum of neon, Ne A high-resolution mass spectrometer can give very accurate relative isotopic masses For example 16 O = and 32 S = Because of this, chemists can distinguish between molecules such as SO 2 and S 2, which appear to have the same relative molecular mass

5 Chapter 1: Moles and equations 2 Look at the mass spectrum of germanium, Ge Abundance/% % 274 % 77 % 367 % 76% Mass/charge (m/e) ratio Figure 16 The mass spectrum of germanium a b Write the isotopic formula for the heaviest isotope of germanium Use the % abundance of each isotope to calculate the relative atomic mass of germanium 80 We often refer to the mass of a mole of substance as its molar mass (abbreviation M) The units of molar mass are g mol 1 The number of atoms in a mole of atoms is very large: atoms This number is called the Avogadro constant (or Avogadro number) The symbol for the Avogadro constant is L (the symbol N A may also be used) The Avogadro constant applies to atoms, molecules, ions and electrons So in 1 mole of sodium there are sodium atoms and in 1 mole of sodium chloride (NaCl) there are sodium ions and chloride ions It is important to make clear what type of particles we are referring to If we just state moles of chlorine, it is not clear whether we are thinking about chlorine atoms or chlorine molecules A mole of chlorine molecules, Cl 2, contains chlorine molecules but twice as many chlorine atoms, as there are two chlorine atoms in every chlorine molecule Amount of substance the mole and the Avogadro constant The formula of a compound shows us the number of atoms of each element present in one formula unit or one molecule of the compound In water we know that two atoms of hydrogen = 10) combine with one atom of oxygen = 160) So the ratio of mass of hydrogen atoms to oxygen atoms in a water molecule is 2 : 16 No matter how many molecules of water we have, this ratio will always be the same But the mass of even 1000 atoms is far too small to be weighed We have to scale up much more than this to get an amount of substance that is easy to weigh The relative atomic mass or relative molecular mass of a substance in grams is called a mole of the substance So a mole of sodium = 230) weighs 230 g The abbreviation for a mole is mol We define the mole in terms of the standard carbon-12 isotope (see page 28) One mole of a substance is the amount of that substance that has the same number of specific particles (atoms, molecules or ions) as there are atoms in exactly 12 g of the carbon-12 isotope Figure 17 Amedeo Avogadro ( ) was an Italian scientist who first deduced that equal volumes of gases contain equal numbers of molecules Although the Avogadro constant is named after him, it was left to other scientists to calculate the number of particles in a mole Moles and mass The Système International (SI) base unit for mass is the kilogram But this is a rather large mass to use for general laboratory work in chemistry So chemists prefer to use the relative molecular mass or formula mass in grams (1000 g = 1 kg) You can find the number of moles of a substance by using the mass of substance and the relative atomic mass ) or relative molecular mass (M r ) number of moles (mol) = mass of substance in grams (g) molar mass (g mol 1 ) 5

6 Cambridge International AS Level Chemistry 1 How many moles of sodium chloride are present in 1170 g of sodium chloride, NaCl? values: Na = 230, Cl = 355) molar mass of NaCl = = 585 g mol 1 mass number of moles = molar mass = = 20 mol 2 What mass of sodium hydroxide, NaOH, is present in 025 mol of sodium hydroxide? values: H = 10, Na = 230, O = 160) molar mass of NaOH = = 400 g mol 1 mass = number of moles molar mass = g = 100 g NaOH 4 Use these A r values: C = 120, Fe = 558, H = 10, O = 160, Na = 230 Calculate the mass of the following: a 020 moles of carbon dioxide, CO 2 b 0050 moles of sodium carbonate, Na 2 C c 500 moles of iron(ii) hydroxide, Fe(OH) 2 6 Figure 18 From left to right, one mole of each of copper, bromine, carbon, mercury and lead 3 a Use these A r values (Fe = 558, N = 140, O = 160, S = 321) to calculate the amount of substance in moles in each of the following: i 107 g of sulfur atoms ii 642 g of sulfur molecules (S 8 ) iii 6045 g of anhydrous iron(iii) nitrate, Fe(N ) 3 b Use the value of the Avogadro constant ( mol 1 ) to calculate the total number of atoms in 710 g of chlorine atoms value: Cl = 355) To find the mass of a substance present in a given number of moles, you need to rearrange the equation mass of substance in grams (g) number of moles (mol) = molar mass (g mol 1 ) mass of substance (g) = number of moles (mol) molar mass (g mol 1 ) Mole calculations Reacting masses When reacting chemicals together we may need to know what mass of each reactant to use so that they react exactly and there is no waste To calculate this we need to know the chemical equation This shows us the ratio of moles of the reactants and products the stoichiometry of the equation The balanced equation shows this stoichiometry For example, in the reaction Fe 2 + 3CO 2Fe + 3CO 2 1 mole of iron(iii) oxide reacts with 3 moles of carbon monoxide to form 2 moles of iron and 3 moles of carbon dioxide The stoichiometry of the equation is 1 : 3 : 2 : 3 The large numbers that are included in the equation (3, 2 and 3) are called stoichiometric numbers In order to find the mass of products formed in a chemical reaction we use: the mass of the reactants the molar mass of the reactants the balanced equation

7 Chapter 1: Moles and equations 4 Iron(III) oxide reacts with carbon monoxide to form iron and carbon dioxide Fe 2 + 3CO 2Fe + 3CO 2 Calculate the maximum mass of iron produced when 798 g of iron(iii) oxide is reduced by excess carbon monoxide values: Fe = 558, O = 160) step 1 Fe 2 + 3CO 2Fe + 3CO 2 step 2 1 mole iron(iii) oxide 2 moles iron (2 558) + (3 160) Figure 19 Iron reacting with sulfur to produce iron sulfide We can calculate exactly how much iron is needed to react with sulfur and the mass of the products formed by knowing the molar mass of each reactant and the balanced chemical equation 1596 g Fe g Fe 1116 step g = 558 g Fe You can see that in step 3, we have simply used ratios to calculate the amount of iron produced from 798 g of iron(iii) oxide 3 Magnesium burns in oxygen to form magnesium oxide 2Mg + O 2 2MgO We can calculate the mass of oxygen needed to react with 1 mole of magnesium We can calculate the mass of magnesium oxide formed step 1 Write the balanced equation step 2 Multiply each formula mass in g by the relevant stoichiometric number in the equation 2Mg + O 2 2MgO g g 2 (243 g g) 486 g 320 g 806 g From this calculation we can deduce that: 320 g of oxygen are needed to react exactly with 486 g of magnesium 806 g of magnesium oxide are formed If we burn 1215 g of magnesium (05 mol) we get 2015 g of magnesium oxide This is because the stoichiometry of the reaction shows us that for every mole of magnesium burnt we get the same number of moles of magnesium oxide In this type of calculation we do not always need to know the molar mass of each of the reactants If one or more of the reactants is in excess, we need only know the mass in grams and the molar mass of the reactant that is not in excess (the limiting reactant) 5 a Sodium reacts with excess oxygen to form sodium peroxide, Na 2 O 2 b 2Na + O 2 Na 2 O 2 Calculate the maximum mass of sodium peroxide formed when 460 g of sodium is burnt in excess oxygen values: Na = 230, O = 160) Tin(IV) oxide is reduced to tin by carbon Carbon monoxide is also formed SnO 2 + 2C Sn + 2CO Calculate the mass of carbon that exactly reacts with 140 g of tin(iv) oxide Give your answer to 3 significant figures values: C = 120, O = 160, Sn = 1187) the stoichiometry of a reaction We can find the stoichiometry of a reaction if we know the amounts of each reactant that exactly react together and the amounts of each product formed For example, if we react 40 g of hydrogen with 320 g of oxygen we get 360 g of water values: H = 10, O = 160) 7

8 Cambridge International AS Level Chemistry 8 hydrogen (H 2 ) + oxygen (O 2 ) water (H 2 O) (2 10) = 2 mol = 1 mol = 2 mol This ratio is the ratio of stoichiometric numbers in the equation So the equation is: 2H 2 + O 2 2H 2 O We can still deduce the stoichiometry of this reaction even if we do not know the mass of oxygen that reacted The ratio of hydrogen to water is 1 : 1 But there is only one atom of oxygen in a molecule of water half the amount in an oxygen molecule So the mole ratio of oxygen to water in the equation must be 1 : g of silicon, Si, reacts exactly with 2840 g of chlorine, Cl 2, to form 3402 g of silicon(iv) chloride, SiCl 4 Use this information to calculate the stoichiometry of the reaction values: Cl = 355, Si = 281) significant figures When we perform chemical calculations it is important that we give the answer to the number of significant figures that fits with the data provided The examples show the number rounded up to varying numbers of significant figures rounded to 4 significant figures = 5268 rounded to 3 significant figures = 527 rounded to 2 significant figures = 530 When you are writing an answer to a calculation, the answer should be to the same number of significant figures as the least number of significant figures in the data (COnTInued) Note 1 Zeros before a number are not significant figures For example, 0004 is only to 1 significant figure Note 2 After the decimal point, zeros after a number are significant figures has 2 significant figures and has 3 significant figures Note 3 If you are performing a calculation with several steps, do not round up in between steps Round up at the end percentage composition by mass We can use the formula of a compound and relative atomic masses to calculate the percentage by mass of a particular element in a compound % by mass atomic mass number of moles of particular element in a compound = 100 molar mass of compound 6 Calculate the percentage by mass of iron in iron(iii) oxide, Fe 2 values: Fe = 558, O = 160) % mass of iron = (2 558) + (3 160) 100 = 699 % 5 How many moles of calcium oxide are there in 29 g of calcium oxide? values: Ca = 401, O = 160) If you divide 29 by 561, your calculator shows The least number of significant figures in the data, however, is 2 (the mass is 29 g) So your answer should be expressed to 2 significant figures, as 0052 mol Figure 110 This iron ore is impure Fe 2 We can calculate the mass of iron that can be obtained from Fe 2 by using molar masses

9 Chapter 1: Moles and equations empirical formulae The empirical formula of a compound is the simplest whole number ratio of the elements present in one molecule or formula unit of the compound The molecular formula of a compound shows the total number of atoms of each element present in a molecule Table 12 shows the empirical and molecular formulae for a number of compounds 7 Calculate the percentage by mass of carbon in ethanol, C 2 H 5 OH values: C = 120, H = 10, O = 160) The formula for an ionic compound is always its empirical formula The empirical formula and molecular formula for simple inorganic molecules are often the same Organic molecules often have different empirical and molecular formulae compound empirical formula Molecular formula water H 2 O H 2 O hydrogen peroxide HO H 2 O 2 sulfur dioxide SO 2 SO 2 butane C 2 H 5 C 4 H 10 cyclohexane CH 2 C 6 H 12 Table 12 Some empirical and molecular formulae 8 Write the empirical formula for: a hydrazine, N 2 H 4 b octane, C 8 H 18 c benzene, C 6 H 6 d ammonia, NH 3 S 7 Deduce the formula of magnesium oxide This can be found as follows: burn a known mass of magnesium (0486 g) in excess oxygen record the mass of magnesium oxide formed (0806 g) calculate the mass of oxygen that has combined with the magnesium ( g) = 0320 g calculate the mole ratio of magnesium to oxygen values: Mg = 243, O = 160) 0486 g moles of Mg = = mol 243 g mol g moles of oxygen = = mol 160 g mol 1 The simplest ratio of magnesium : oxygen is 1 : 1 So the empirical formula of magnesium oxide is MgO 8 When 155 g of phosphorus is completely combusted 355 g of an oxide of phosphorus is produced Deduce the empirical formula of this oxide of phosphorus values: O = 160, P = 310) p o step 1 note the mass 155 g of each element = 200 g step 2 divide by atomic masses step 3 divide by the lowest figure 155 g 200 g 310 g mol g mol 11 = 005 mol = 0125 mol = = 25 step 4 if needed, obtain P 2 O 5 the lowest whole number ratio to get empirical formula An empirical formula can also be deduced from data that give the percentage composition by mass of the elements in a compound 9 The empirical formula can be found by determining the mass of each element present in a sample of the compound For some compounds this can be done by combustion An organic compound must be very pure in order to calculate its empirical formula Chemists often use gas chromatography to purify compounds before carrying out formula analysis

10 Cambridge International AS Level Chemistry 10 Molecular formulae The molecular formula shows the actual number of each of the different atoms present in a molecule The molecular formula is more useful than the empirical formula We use the molecular formula to write balanced equations and to calculate molar masses The molecular formula is always a multiple of the empirical formula For example, the molecular formula of ethane, C 2 H 6, is two times the empirical formula, CH 3 In order to deduce the molecular formula we need to know: 9 A compound of carbon and hydrogen contains 857% carbon and 143% hydrogen by mass Deduce the empirical formula of this hydrocarbon values: C = 120, O = 160) c H step 1 note the % by mass step 2 divide by A r values = = 143 step 3 divide by the lowest = figure 7142 = 2 Empirical formula is CH 2 9 The composition by mass of a hydrocarbon is 10% hydrogen and 90% carbon Deduce the empirical formula of this hydrocarbon values: C = 120, H = 10) the relative formula mass of the compound the empirical formula 10 A compound has the empirical formula CH 2 Br Its relative molecular mass is 1878 Deduce the molecular formula of this compound values: Br = 799, C = 120, H = 10) step 1 find the empirical formula mass: (2 10) = 939 (COnTInued) step 2 divide the relative molecular mass by the empirical formula mass: = 2 step 3 multiply the number of atoms in the empirical formula by the number in step 2: 2 CH 2 Br, so molecular formula is C 2 H 4 Br 2 10 The empirical formulae and molar masses of three compounds, A, B and C, are shown in the table below Calculate the molecular formula of each of these compounds values: C = 120, Cl = 355, H = 10) compound empirical formula M r A C 3 H 5 82 B CCl C CH chemical formulae and chemical equations Deducing the formula The electronic structure of the individual elements in a compound determines the formula of a compound (see page 33) The formula of an ionic compound is determined by the charges on each of the ions present The number of positive charges is balanced by the number of negative charges so that the total charge on the compound is zero We can work out the formula for a compound if we know the charges on the ions Figure 111 shows the charges on some simple ions related to the position of the elements in the Periodic Table The form of the Periodic Table that we shall be using has 18 groups because the transition elements are numbered as Groups 3 to 12 So, aluminium is in Group 13 and chlorine is in Group 17 For simple metal ions in Groups 1 and 2, the value of the positive charge is the same as the group number For a simple metal ion in Group 13, the value of the positive charge is 3+ For a simple non-metal ion in Groups 15 to 17, the value of the negative charge is 18 minus the group

Page 1 of 14 Amount of Substance Key terms in this chapter are: Element Compound Mixture Atom Molecule Ion Relative Atomic Mass Avogadro constant Mole Isotope Relative Isotopic Mass Relative Molecular

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