Chemistry Past Paper - Mass Spectrometry PDF

Tip: for part d you need to calculate the mass of water separately and then add it to the Mr of Mg(NO3)2. 各种国际课程资料都有， 全网最全 请加微信：CLASS0608 3.3 Accurate relative atomic masses Mass spectrometry A mass spectrometer (Figure 3.2) can be used to measure the mass of each isotope pres...

Tip: for part d you need to calculate the mass of water separately and then add it to the Mr of Mg(NO3)2. 各种国际课程资料都有， 全网最全 请加微信：CLASS0608 3.3 Accurate relative atomic masses Mass spectrometry A mass spectrometer (Figure 3.2) 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 (relative isotopic abundance). A simplified diagram of a mass spectrometer is shown in Figure 3.3. 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 3.2: A mass spectrometer is a large and complex instrument. Figure 3.3: Simplified diagram of a mass spectrometer. The atoms of the element in the vaporised sample are converted into ions. The stream of ions travels to a detector after being deflected by a strong magnetic field. As the magnetic field is increased, the ions of heavier and heavier isotopes move towards 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 3.4 shows a typical mass spectrum for a sample of lead. Table 3.1 shows how the data is interpreted. Figure 3.4: 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 3.1 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 / % 204 2 206 24 207 22 208 52 total 100 Table 3.1: The data from Figure 3.4. Determination of Ar from mass spectra IMPORTANT You do not need to know the exact working of the mass spectrometer. 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 3.5. Figure 3.5: The mass spectrum of neon, Ne. The mass spectrum of neon has three peaks: 20Ne (90.9%), 21Ne (0.3%) and 22Ne (8.8%). Ar of neon = = 20.2 Note that this answer is given to 3 significant figures, which is consistent with the data given. A high-resolution mass spectrometer can give very accurate relative isotopic masses. For example, 16O = 15.995 and 32S = 31.972. Because of this high level of precision, chemists can distinguish between molecules such as SO2 and S2, which appear to have the same relative molecular mass. Question 2 Look at the mass spectrum of germanium, Ge. a Write isotopic formula for the heaviest isotope of germanium. b Use the % abundance of each isotope to calculate the relative atomic mass of germanium. Identification of an organic compound using mass spectrometry The main use of mass spectrometry is in the identification of organic compounds. As in other forms of spectroscopy, a substance can be identified by matching its spectrum against the spectra of known substances stored in a database. This technique is known as ‘fingerprinting’. The high energy electrons knock electrons from the molecules and break covalent bonds, fragmenting the molecule. Figure 3.6 shows the mass spectrum produced by propanone, CH3COCH3. Figure 3.6: The mass spectrum of propanone, CH3COCH3. Figure 3.7: The mass spectrum of germanium, Ge. IMPORTANT A molecular ion, M+, is formed when one electron is removed from a molecule to form an ion with a single positive charge e.g. CH4 → CH4+ + e−. Remember that no atoms are removed. The peak at the highest mass-to-charge ratio is caused by the molecular ion (M+). This ion is formed by the sample molecule with one electron knocked out. It gives us the relative molecular mass of the sample. We can assume the ions detected carry a single positive charge, so the reading on the horizontal axis gives us the mass. In the case of propanone, CH3COCH3, the molecular ion has a relative mass of 58.0. This corresponds to CH3COCH3+, with a mass of (3 × 12.0) + (1 × 16.0) + (6 × 1.0). We also get large peaks at 15 and 43 on the mass spectrum. These peaks are due to fragments that are produced when propanone molecules are broken apart by the electron bombardment. Knowing the structure of propanone, we should be able to identify the fragment responsible for each peak (Figure 3.8). Figure 3.8: The fragmentation of propanone: +CH3 causes the peak at 15 and CH3CO+ causes the peak at 43. IMPORTANT Remember that fragmentation (breaking apart) of a compound in a mass spectrometer causes certain bonds to break. You can deduce what the fragment is by adding up the atomic masses of carbon, hydrogen and / or other atoms. So a fragment of m/e 15 is C + 3H = 12 + (3 × 1) which is +CH3 and a fragment of m/e 43 could be +C3H7 or CH3CO+. The electron bombardment has caused the C─C single bonds in the propanone molecules to break. This has resulted in the fragments at m/e 15 and 43 that are observed in Figure 3.6. The breaking of single bonds, such as C─C, C─O or C─N, is the most common cause of fragmentation. IMPORTANT These fragments are very common in mass spectra of organic compounds. Mass Fragment 15 +CH 3  28 +CO or C2H4+ 29 CH3CH2+ 43 C3H7+ or CH3CO+ Table 3.2: Common fragments in the mass spectra of organic compounds. Question 3 Look at Figure 3.10, which shows the mass spectrum of ethanol, C2H5OH. A structural isomer of ethanol is methoxymethane, an ether with the formula CH3OCH3. a Predict the mass-to-charge ratio of a fragment that would appear on the mass spectrum of methoxymethane but does not appear on ethanol’s mass spectrum. b Give the formula of the ion responsible for the peak in your answer to part a. c Look at the mass spectrum of ethanoic acid: Figure 3.9: Mass spectrum of ethanoic acid. Identify the fragments with mass-to-charge ratios of: i 15 ii 43 iii 45 iv 60. High-resolution mass spectra High-resolution mass spectrometers can distinguish between ions that appear to have the same mass on a low-resolution mass spectrum. Table 3.3 shows the accurate relative isotopic masses of the most common atoms found in organic molecules. Isotope Relative isotopic mass 1H 1.007 824 6 12C 12.000 000 0 14N 14.003 073 8 16O 15.994 914 1 Table 3.3: Accurate masses of isotopes. These accurate isotopic masses allow us to measure the mass of the molecular ion so accurately that it can only correspond to one possible molecular formula. For example, a molecular ion peak at 45 could be caused by C2H7N or CH3NO. However, a high-resolution mass spectrum would show the C2H7N+ peak at 45.057846 and the CH3NO+ peak at 45.021462. We can, therefore, be sure which molecule is being analysed. Using the [M + 1] peak There will always be a very small peak just beyond the molecular ion peak at a mass of [M + 1]. This is caused by molecules in which one of the carbon atoms is the 13C isotope. This is shown in the mass spectrum of ethanol in Figure 3.10. Figure 3.10: The mass spectrum of ethanol, showing the [M + 1] peak. In any organic compound there will be 1.10% carbon-13. We can use this fact to work out the number of carbon atoms (n) in a molecule. We apply the equation: WORKED EXAMPLE 1 An unknown compound has a molecular ion peak, M+, with a relative abundance of 54.5% and has an [M + 1]+ peak with a relative abundance of 3.6%. How many carbon atoms does the unknown compound contain? Solution Substituting the values of relative abundance into the equation: we get: n= × = 6.0 There are 6 carbon atoms in each molecule. Question 4 A hydrocarbon has a molecular ion peak at a mass-to-charge ratio of 84 (relative abundance of 62.0%) and an [M + 1] peak with a relative abundance of 4.1%. How many carbon atoms are in the hydrocarbon? Using [M + 2] and [M + 4] peaks Note: Material relating to [M + 4] peaks is extension content. It is not part of the syllabus. IMPORTANT  We can tell whether there is chlorine or bromine in an organic compound by comparing the relative heights of the M and [M + 2] peaks. If the peak heights are equal, there is one atom of bromine per molecule. If the peak heights are in the ratio 3 [M] to 1 [M + 2], there is one atom of chlorine per molecule. If the sample compound contains chlorine or bromine atoms, we also get peaks beyond the molecular ion peak because of isotopes of chlorine and bromine. Chlorine has two isotopes, 35Cl and 37Cl, as does bromine, 79Br and 81Br. Table 3.4 shows the approximate percentage of each isotope in naturally occurring samples. Isotope Approximate % 35Cl 75 37Cl 25 79Br 50 81Br 50 Table 3.4: Naturally occurring isotopes of chlorine and bromine. One Cl or Br atom per molecule Imagine a sample of chloromethane, CH3Cl. We will have molecules of CH335Cl (75%) and molecules of CH337Cl (25%). The molecular ion will be CH335Cl+, and two units beyond that on the mass spectrum will be the peak for CH337Cl+. The peak for CH337Cl+ will be one-third the height of the molecular ion. This is the [M + 2] peak. In the mass spectrum of bromomethane, CH3Br, we will have two molecular ion peaks of approximately the same height: one for CH379Br+ and the other for CH381Br+ (the [M + 2] peak). You should look out for the relative heights mentioned here when interpreting mass spectra. If the [M + 2] peak is one-third the height of the M peak, this suggests the presence of one chlorine atom per molecule. If the [M + 2] peak is the same as the height of the M peak, this suggests the presence of one bromine atom per molecule. An example of the [M + 2] peak is shown on the mass spectrum of chlorobenzene (Figure 3.11). Two Cl or Br atoms per molecule The situation is a little more complex with two chlorine atoms in a molecule, as there are three possibilities. Considering dichloromethane, CH2Cl2, we have: 35ClCH 35Cl+ the M peak 2 35ClCH 37Cl+ the [M + 2] peak 2 37ClCH 35Cl+ the [M + 2] peak 2 37 ClCH237Cl+ the [M + 4] peak The relative heights of the peaks must take into account the natural abundances: it works out as 9 : 6 : 1 for molecules with two Cl atoms. Figure 3.11: The mass spectrum of chlorobenzene, showing the [M + 2] peak. (Note that there are also tiny [M + 1] and [M + 3] peaks corresponding to 13C in the molecule.) The M, [M + 2] and [M + 4] peaks also occur in dibromomethane but the relative heights of peaks are easier to work out. Because the ratio 79Br : 81Br is 1 : 1, the M : [M + 2] : [M + 4] height ratio is 1 : 2 : 1. Question Note: Question 5 contains extension content as [M + 4] peaks are not included in the syllabus. 5 a List the ions responsible for the M, [M + 2] and [M + 4] peaks in a mass spectrum of dibromomethane. b What would be the mass-to-charge ratio and relative abundances of the major peaks with the highest charge-to-mass ratios in the mass spectrum of chloroethane? c How many peaks would you see beyond the molecular ion peak in 1,1-dibromoethane? What would be their mass-to-charge ratios and abundances relative to the molecular ion? (Ignore peaks due to 13C.) 3.4 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. We know that two atoms of hydrogen (Ar = 1.0) combine with one atom of oxygen (Ar = 16.0) in water, 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 number of particles equivalent to the relative atomic mass or relative molecular mass of a substance in grams is called the Avogadro constant (or Avogadro number). The symbol for the Avogadro constant is L (the symbol NA may also be used).The numerical value of the Avogadro constant is 6.02 × 1023. The mass of substance with this number of particles is called a mole. The Avogadro constant is chosen so that the mass of one mole in grams equals the average mass of an atom of an element in unified atomic mass units. So a mole of sodium (Ar = 23.0) contains 6.02 × 1023 atoms and has a mass of 23.0 g. The abbreviation for a mole is mol. We can define the mole in terms of the Avogadro constant: A mole is the amount of substance which contains 6.02 × 1023 specified particles. The particles can be atoms, molecules, ions or electrons. IMPORTANT In many books you will see an alternative definition of the mole based on the carbon-12 isotope: 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. This definition is now out of date. The Avogadro constant applies to atoms, molecules, ions and electrons. So in 1 mole of sodium there are 6.02 × 1023 sodium atoms and in 1 mole of sodium chloride (NaCl) there are 6.02 × 1023 sodium ions and 6.02 × 1023 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, Cl2, contains 6.02 × 1023 chlorine molecules but twice as many chlorine atoms, as there are two chlorine atoms in every chlorine molecule. IMPORTANT Molar mass is a general term used for the mass in grams of 1 mole of a compound, whether ionic, simple molecules or giant covalent structures. ‘Relative molecular mass’ refers only to molecules. Figure 3.12: Amedeo Avogadro (1776–1856) 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. 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. 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 mass of one mole of that substance (the molar mass) number of moles (mol) = IMPORTANT Some students find it helpful to use a triangle such as that shown to work out mass from molar mass and number of moles. By covering the quantity you want to find, you will see the correct form of the equation to use. Figure 3.13: Triangle diagram for mole calculations. However, it is far better to learn how to cross multiply because such triangles will not work with more complex equations: ask another learner or a teacher to help you. WORKED EXAMPLES 2 How many moles of sodium chloride are present in 117.0 g of sodium chloride, NaCl?  (Ar values: Na = 23.0, Cl = 35.5) Solution Step 1: molar mass of NaCl = 23.0 + 35.5 = 58.5 g mol−1 Step 2: number of moles = = 2.0 mol 3 What mass of sodium hydroxide, NaOH, is present in 0.25 mol of sodium hydroxide? (Ar values: H = 1.0, Na = 23.0, O = 16.0) Solution Step 1: molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g mol−1 Step 2: mass = number of moles × molar mass = 0.25 × 40.0 g = 10.0 g NaOH IMPORTANT When dealing with moles, it is important to be clear about the type of particle referred to. For example: 32.0 grams of oxygen O2 contains 1 mole of oxygen molecules but 2 moles of oxygen atoms and 1 mole of MgCl2 contains 1 mol of Mg2+ ions but 2 moles of Cl− ions. Question 6 a Use these Ar values (Fe = 55.8, N = 14.0, O = 16.0, S = 32.1) to calculate the amount of substance in moles in each of the following: i 10.7 g of sulfur atoms ii 64.2 g of sulfur molecules (S8) iii 60.45 g of anhydrous iron(III) nitrate, Fe(NO3)3 b Use the value of the Avogadro constant (6.02 × 1023 mol−1) to calculate the total number of atoms in 7.10 g of chlorine atoms. (Ar value: Cl = 35.5) To find the mass of a substance present in a given number of moles, you need to rearrange the equation number of moles (mol) = mass of substance (g) = number of moles (mol) × molar mass (g mol−1) Figure 3.14 shows what the molar mass of six elements looks like. Figure 3.14: One mole quantities of common elements. The cylinders from left to right hold mercury, lead and copper. Sulfur is in the left flask, magnesium in the right and chromium on the watch-glass. Question 7 Use these Ar values: C = 12.0, Fe = 55.8, H = 1.0, O = 16.0, Na = 23.0. Calculate the mass of the following: a 20 moles of carbon dioxide, CO2 b 0.050 moles of sodium carbonate, Na2CO3 c 5.00 moles of iron(II) hydroxide, Fe(OH)2 3.5 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 (Figure 3.15). 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 Fe2O3 + 3CO → 2Fe + 3CO2 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. Figure 3.15: 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. WORKED EXAMPLE 4 Magnesium burns in oxygen to form magnesium oxide. 2Mg + O2 → 2MgO Calculate the mass of oxygen needed to react with 1 mole of magnesium. What is the mass of the magnesium oxide formed?. Solution Step 1: Write the balanced equation. 2Mg + O2 → 2MgO Step 2: Multiply each formula mass in g by the relevant stoichiometric number in the equation. 2 × 24.3 g 1 × 32.0 g 2 × (24.3 g + 16.0 g) 48.6 g 32.0 g 80.6 g From this calculation we can deduce that: 32.0 g of oxygen are needed to react exactly with 48.6 g of magnesium  80.6 g of magnesium oxide are formed. If we burn 12.15 g of magnesium (0.5 mol) we get 20.15 g of magnesium oxide. This is because the stoichiometry of the reaction shows us that for each mole (or part of mole) of magnesium burnt, we get the same number of moles of magnesium oxide. This means that for every 0.5 mole burnt, we get 0.5 mole of product. 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 reactant which has the number of moles in excess is called the excess reagent. The reactant which is not in excess is called the limiting reagent. IMPORTANT Remember that in calculating which reactant is limiting, you must: work out the number of moles of the reactant take into account the ratio of the reactants shown in the equation (the stoichiometry). WORKED EXAMPLE 5 A sample of 79.8 g of iron(III) oxide is mixed with 9.36 g of carbon and heated. A reaction occurs. 2Fe2O3 + 3C → 4Fe + 3CO2 Show by calculation that iron(III) oxide is the limiting reactant. (Ar values: Fe = 55.8, O = 16.0, C = 12.0) Solution Step 1: Calculate the moles of each reagent. moles of Fe2O3 = = 0.50 mol moles of C = = 0.78 mol Step 2: Refer to the stoichiometry of the equation: For every 2 Fe2O3 which react, 3 C are needed So for 0.5 mol Fe2O3 we need 0.5 × = 0.75 mol C Step 3: Determine which is in excess. number of moles C required = 0.75 number of moles C from step 1 = 0.78 So C is in excess by 0.78 − 0.75 = 0.03 mol and iron(III) oxide is limiting. WORKED EXAMPLE 6 Iron(III) oxide reacts with carbon monoxide to form iron and carbon dioxide. Fe2O3 + 3CO → 2Fe + 3CO2 Calculate the maximum mass of iron produced when 798 g of iron(III) oxide is reduced by excess carbon monoxide. (Ar values: Fe = 55.8, O = 16.0) Solution Step 1: Write the balanced equation.  Fe2O3 + 3CO → 2Fe + 3CO2 Step 2: Calculate molar masses taking into account the number of moles in the equation. 1 mole iron(III) oxide → 2 moles iron (2 × 55.8) + (3 × 16.0) → 2 × 55.8 159.6 g Fe2O3 → 111.6 g Fe Step 3: Calculate mass of Fe in 798 g Fe2O3 × 798 = 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. Question 8 a Sodium reacts with excess oxygen to form sodium peroxide, Na2O2. 2Na + O2 → Na2O2 Calculate the maximum mass of sodium peroxide formed when 4.60 g of sodium is burnt in excess oxygen. (Ar values: Na = 23.0, O = 16.0) b Tin(IV) oxide is reduced to tin by carbon. Carbon monoxide is also formed. SnO2 + 2C → Sn + 2CO Calculate the mass of carbon that exactly reacts with 14.0 g of tin(IV) oxide. Give your answer to 3 significant figures. (Ar values: C = 12.0, O = 16.0, Sn = 118.7) 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 4.0 g of hydrogen with 32.0 g of oxygen we get 36.0 g of water. (Ar values: H = 1.0, O = 16.0) This ratio is the ratio of stoichiometric numbers in the equation. So the equation is: 2H2 + O2 → 2H2O 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 : 2. Question 9 56.2 g of silicon, Si, reacts exactly with 284.0 g of chlorine, Cl2, to form 340.2 g of silicon(IV) chloride, SiCl4. Use this information to calculate the stoichiometry of the reaction. (Ar values: Cl = 35.5, Si = 28.1) 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 here show the number 526.84 rounded up to varying numbers of significant figures. rounded to 4 significant figures = 526.8 rounded to 3 significant figures = 527 rounded to 2 significant figures = 530 IMPORTANT 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. Do not round numbers to the correct number of significant figures until the end of a calculation or you risk introducing errors. 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. WORKED EXAMPLES 7 How many moles of calcium oxide are there in 2.9 g of calcium oxide? (Ar values: Ca = 40.1, O = 16.0) Solution If you divide 2.9 by 56.1, your calculator shows 0.051 693 … The least number of significant figures in the data, however, is 2 (the mass is 2.9 g). So your answer should be expressed to 2 significant figures, as 0.052 mol. Note: 1 Zeros before a number are not significant figures. For example, 0.004 is only to 1 significant figure. 2 After the decimal point, zeros after a number are significant figures. 0.0040 has 2 significant figures and 0.004 00 has 3 significant figures. 3 If you are performing a calculation with several steps, do not round up in between steps. Round up at the end. 8 Calculate the percentage by mass of iron in iron(III) oxide, Fe2O3. (Ar values: Fe = 55.8, O = 16.0) Solution % mass of iron = × 100 = 69.9 %  Figure 3.16: This iron ore is impure Fe2O3. We can calculate the mass of iron that can be obtained from Fe2O3 by using molar masses. Question 10 Calculate the percentage by mass of carbon in ethanol, C2H5OH. (Ar values: C = 12.0, H = 1.0, O = 16.0) Percentage yield In many chemical reactions, especially organic reactions, not all the reactants are changed to the products you want. This is because there are other reactions going on at the same time; reactants or products are lost to the atmosphere; or the reaction does not go to completion. The percentage yield tells you how much of a particular product you get from the reactants compared with the maximum theoretical amount that you can get. percentage yield = × 100 The actual yield is the moles or mass of product obtained by experiment. The predicted yield is the moles or mass of product obtained by calculation if no side products are formed and all of a specific reactant is converted to a specific product. WORKED EXAMPLE 9 A sample of aluminium chloride, AlCl3, is made by reacting 18 g of aluminium powder with excess chlorine. The mass of aluminium chloride produced is 71.0 g. Calculate the percentage yield of aluminium oxide. (Ar values: Al = 27.0, Cl = 35.5) 2Al + 3Cl2 → 2AlCl3 Solution Step 1: Calculate the predicted mass from the stoichiometry of the equation if all the aluminium is converted to aluminium chloride. 2 × 27 g Al produces 2 × (27 + (3 × 35.5)) g AlCl3 54 g Al produces 267 g AlCl3 Step 2: Calculate the mass of aluminium chloride formed from the given amount of aluminium using simple proportion. 18 g Al produces 267 × = 89.0 g AlCl3 Step 3: Calculate the percentage yield. × 100 = 79.8% 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 number of atoms of each element present in a molecule. Table 3.5 shows the empirical and molecular formulae for a number of compounds. 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 2O H 2O hydrogen peroxide HO H 2O 2 sulfur dioxide SO2 SO2 butane C 2H 5 C4H10 cyclohexane CH2 C6H12 Table 3.5: Some empirical and molecular formulae. Question 11 Deduce the empirical formula of: a hydrazine, N2H4 b octane, C8H18 c benzene, C6H6 d ammonia, NH3 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. WORKED EXAMPLES 10 Deduce the formula of magnesium oxide. Solution This can be found as follows: burn a known mass of magnesium (0.486 g) in excess oxygen record the mass of magnesium oxide formed (0.806 g) calculate the mass of oxygen that has combined with the magnesium (0.806 − 0.486 g) = 0.320 g calculate the mole ratio of magnesium to oxygen (Ar values: Mg = 24.3, O = 16.0) moles of Mg = = 0.0200 mol moles of oxygen = = 0.0200 mol  The simplest ratio of magnesium : oxygen is 1 : 1. So the empirical formula of magnesium oxide is MgO. 11 When 1.55 g of phosphorus is completely combusted 3.55 g of an oxide of phosphorus is produced. Deduce the empirical formula of this oxide of phosphorus. (Ar values: O = 16.0, P = 31.0) Solution P O Step 1: note the mass of each element 1.55 g 3.55 − 1.55 = 2.00 g Step 2: divide by atomic masses = 0.05 mol = 0.125 mol Step 3: divide by the lowest figure =1 = 2.5 Step 4: if needed, obtain the lowest whole number ratio to get empirical P 2O 5 formula Table 3.6: Steps to deduce an empirical formula. An empirical formula can also be deduced from data that give the percentage composition by mass of the elements in a compound. WORKED EXAMPLE 12 A compound of carbon and hydrogen contains 85.7% carbon and 14.3% hydrogen by mass. Deduce the empirical formula of this hydrocarbon. (Ar values: C = 12.0, H = 1.0) Solution C H Step 1: note the % by mass 85.7 14.3 Step 2: divide by Ar values = 7.142 = 14.3 Step 3: divide by the lowest figure =1 =2 Empirical formula is CH2. Question 12 The composition by mass of a hydrocarbon is 10% hydrogen and 90% carbon. Deduce the empirical formula of this hydrocarbon. (Ar values: C = 12.0, H = 1.0) Molecular formulae The molecular formula shows the actual number of each of the different atoms present in a molecule. 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, C2H6, is twice the empirical formula, CH3. In order to deduce the molecular formula we need to know: the relative formula mass of the compound the empirical formula. WORKED EXAMPLE 13 A compound has the empirical formula CH2Br. Its relative molecular mass is 187.8. Deduce the molecular formula of this compound. (Ar values: Br = 79.9, C = 12.0, H = 1.0) Solution Step 1: Find the empirical formula mass: 12.0 + (2 × 1.0) + 79.9 = 93.9 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 × CH2Br, so molecular formula is C2H4Br2. Question 13 The empirical formulae and molar masses of three compounds, A, B and C, are shown in Table 3.7. Calculate the molecular formula of each of these compounds. (Ar values: C = 12.0, Cl = 35.5, H = 1.0) Compound Empirical formula Mr A C 3H 5 82 B CCl3 237 C CH2 112 Table 3.7: Information table for Question 13. 3.6 Chemical formulae and chemical equations Deducing the formula for ionic and covalent compounds Ionic compounds The electronic structure of the individual elements in a compound determines the formula of a compound (see Section 2.1). 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 3.17 shows the charges on some 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 number. The charge on the ions of transition elements can vary. For example, iron forms two types of ions, Fe2+ and Fe3+ (Figure 3.18). Figure 3.17: The charge on some simple ions is related to their position in the Periodic Table. Figure 3.18: Iron(II) chloride (left) and iron(III) chloride (right). These two chlorides of iron both contain iron and chlorine, but they have different formulae. The roman numerals after the ions are called oxidation numbers. There are rules for deducing oxidation numbers (see Section 7.2): For most metal ions in Groups 1 to 3, the oxidation numbers are equal to the positive charge on the ion, e.g. an Al3+ ion has an oxidation number of +3. When naming compounds containing these ions, we do not include the oxidation number of the positive ion, e.g. magnesium oxide (not magnesium(II) oxide). Simple negative ions have oxidation numbers which are the same as the charge on the ion, e.g. O2− has an oxidation number of −2 and Cl− has an oxidation number of −1. When naming compounds containing these ions we do not include the oxidation number of the negative ion as these do not generally vary. For transition elements, the oxidation numbers do vary. They show the charge on the ion, e.g. chromium(II) chloride contains a Cr2+ ion and chromium(III) chloride contains a Cr3+ ion. When naming  these compounds we must include the oxidation number to avoid confusion. Ions that contain more than one type of atom are called compound ions. Some common compound ions and two metal ions that you should learn are listed in Table 3.8. The formula for an ionic compound is obtained by balancing the charges of the ions. Ion Formula ammonium NH4+ carbonate CO32− hydrogencarbonate HCO3− hydroxide OH− nitrate NO3− phosphate PO43− sulfate SO42− zinc Zn2+ silver Ag+ Table 3.8: The formulae of some common ions. WORKED EXAMPLES 14 Deduce the formula of magnesium chloride. Solution Ions present: Mg2+ and Cl−. For electrical neutrality, we need two Cl− ions for every Mg 2+ ion. (2 × 1−) + (1 × 2+) = 0 So the formula is MgCl2. 15 Deduce the formula of aluminium oxide. Ions present: Al3+ and O2−. For electrical neutrality, we need three O2− ions for every two Al3+ ions. (3 × 2−) + (2 × 3+) = 0 So the formula is Al2O3. Covalent compounds You can work out the formula of a covalent compound from the number of electrons needed to achieve the stable electronic configuration of a noble gas (see Section 4.1). In general, carbon atoms form four bonds with other atoms, hydrogen and halogen atoms form one bond and oxygen atoms form two bonds. So the formula of water, H2O, follows these rules. The formula for methane is CH4, with each carbon atom bonding with four hydrogen atoms. However, there are many exceptions to these rules. Compounds containing a simple metal ion and non-metal ion are named by changing the end of the name of the non-metal element to ‘-ide’. sodium + chlorine → sodium chloride zinc + sulfur → zinc sulfide The names of compound ions containing oxygen often end in ‘-ate’. Sometimes they end in ‘-ite’. For example, the sulfate ion contains sulfur and four oxygen atoms, the phosphate ion contains phosphorus and oxygen. Question 14 a Write down the formula of each of the following compounds: i magnesium nitrate ii calcium sulfate iii sodium iodide iv hydrogen bromide v sodium sulfide b Name each of the following compounds: i Na3PO4 ii (NH4)2SO4 iii AlCl3 iv Ca(NO3)2 Balancing chemical equations When chemicals react, atoms cannot be either created or destroyed. So there must be the same number of each type of atom on the reactants side of a chemical equation as there are on the products side. A symbol equation is a shorthand way of describing a chemical reaction. It shows the number and type of the atoms in the reactants and the number and type of atoms in the products. If this number of each type of atom is the same on both sides, we say the equation is balanced. Follow these examples to see how we balance an equation. IMPORTANT When balancing chemical equations remember: Do not change any of the formulae. The numbers used to balance are put in front of the formulae. In equations involving combustion reactions of organic compounds, balance the carbon first and then the hydrogen. Balance the oxygen last. WORKED EXAMPLES 16 Write a balanced equation for the reaction of hydrogen and oxygen to form water. Solution Step 1: Write down the formulae of all the reactants and products. For example: H 2 + O 2 → H 2O Step 2: Count the number of atoms of each reactant and product. H2 + O2 → H 2O 2[H] 2[O] 2[H] + 1[O] Step 3: Balance one of the atoms by placing a number in front of one of the reactants or products. In this case the oxygen atoms on the right-hand side need to be balanced, so that they are equal in number to those on the left-hand side. Remember that the number in front multiplies everything in the formula. For example, 2H2O has 4 hydrogen atoms and 2 oxygen atoms. H2 + O2 → 2H2O  2[H] 2[O] 4[H] + 2[O] Step 4: Keep balancing in this way, one type of atom at a time until all the atoms are balanced. 2H2 + O2 → 2H2O 4[H] 2[O] 4[H] + 2[O] Note that when you balance an equation you must not change the formulae of any of the reactants or products. 17 Write a balanced equation for the reaction of iron(III) oxide with carbon monoxide to form iron and carbon dioxide. Solution Step 1: Formulae: Fe2O3 + CO → Fe + CO2 Step 2: Count the number of atoms. Fe2O3 + CO → Fe + CO2 2[Fe] 1[C] 1[Fe] 1[C] 3[O] 1[O] 2[O] Step 3: Balance the iron. Fe2O3 + CO → 2Fe + CO2 2[Fe] 1[C] 2[Fe] + 1[C] 3[O] 1[O] 2[O] Step 4: Balance the oxygen. Fe2O3 + 3CO → 2Fe + 3CO2 2[Fe] 3[C] 2[Fe] 3[C] 3[O] 3[O] 6[O] In step 4, the oxygen in the CO2 comes from two places, the Fe2O3 and the CO. Note that trial and error is involved here as the carbon also needs to be balanced. In this particular case, in order to balance the equation, three oxygen atoms come from the iron oxide and three oxygen atoms come from the carbon monoxide. IMPORTANT When balancing chemical equations follow these steps: Write the formulae of the reactants and products Count the numbers of atoms in each reactant and product Balance the atoms one at a time until all the atoms are balanced. Question 15 Write balanced equations for the following reactions. a Iron reacts with hydrochloric acid to form iron(II) chloride, FeCl2, and hydrogen. b Aluminium hydroxide, Al(OH)3, breaks down on heating to form aluminium oxide, Al2O3, and water. c Hexane, C6H14, burns in oxygen to form carbon dioxide and water. Using state symbols We sometimes find it useful to specify the physical states of the reactants and products in a chemical reaction. This is especially important where chemical equilibria and rates of reaction are being discussed (see Chapter 8 and Chapter 9). We use the following state symbols: (s) solid (l) liquid (g) gas (aq) aqueous (a solution in water). State symbols are written after the formula of each reactant and product. For example: ZnCO3(s) + H2SO4(aq) → ZnSO4(aq) + H2O(l) + CO2(g) Question 16 Write balanced equations, including state symbols, for the following reactions. a Solid calcium carbonate reacts with aqueous hydrochloric acid to form water, carbon dioxide and an aqueous solution of calcium chloride. b An aqueous solution of zinc sulfate, ZnSO4, reacts with an aqueous solution of sodium hydroxide. The products are a precipitate of zinc hydroxide, Zn(OH)2, and an aqueous solution of sodium sulfate. Figure 3.19: The reaction between calcium carbonate and hydrochloric acid. The equation for the reaction shown in Figure 3.19, with all the state symbols, is: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) Question 17 Which is the correct equation for the reaction between iron(III) oxide and carbon monoxide? A Fe2O3(g) + 3CO(g) → 2Fe(s) + 3CO2(l) B 2FeO(s) + 2CO(g) → 2Fe(s) + 2CO2(l) C Fe2O3(g) + 3CO(g) → 2Fe(s) + 3CO2(g) D 2Fe2O3(g) + 3CO(g) → 4Fe(s) + 3CO2(g) Balancing ionic equations When ionic compounds dissolve in water, the ions separate from each other. For example: NaCl(s) + aq → Na+(aq) + Cl−(aq) Ionic compounds include salts such as sodium bromide, magnesium sulfate and ammonium nitrate. Acids and alkalis also contain ions. For example, H+(aq) and Cl−(aq) ions are present in hydrochloric acid and Na+ (aq) and OH−(aq) ions are present in sodium hydroxide. Many chemical reactions in aqueous solution involve ionic compounds. Only some of the ions in solution take part in these reactions. The ions that play no part in the reaction are called spectator ions. An ionic equation is simpler than a full chemical equation. It shows only the ions or other particles taking part in a reaction. Spectator ions are omitted (left out). Ionic equations are often written for reactions involving a change in oxidation state. Compare the full equation for the reaction of zinc with aqueous copper(II) sulfate with the ionic equation. full chemical equation: Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) with charges: Zn(s) + Cu2+SO42−(aq) → Zn2+SO42−(aq) + Cu(s) cancelling spectator ions: Zn(s) + Cu2+ → Zn2+ + Cu(s) ionic equation: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) In the ionic equation you will notice that: there are no sulfate ions: these are the spectator ions as they have not changed both the charges and the atoms are balanced. IMPORTANT When writing an ionic equation remember: The product(s) will often include a precipitate or simple molecules such as iodine or water e.g. Cl2(aq) + 2I−(aq) → I2(aq) + 2Cl−(aq) Acids are represented by H+ ions, e.g. ZnO(s) + 2H+(aq) → Zn2+(aq) + H2O(l) The next examples show how we can change a full equation into an ionic equation. WORKED EXAMPLE 18 Writing an ionic equation: Solution Step 1: Write down the full balanced equation. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Step 2: Write down all the ions present. Any reactant or product that has a state symbol (s), (l) or (g) or is a covalent molecule in solution such as chlorine, Cl2(aq), does not split into ions. Mg(s) + 2H+(aq) + 2Cl−(aq) → Mg2+(aq) + 2Cl−(aq) + H2(g)  Step 3: Cancel the ions that appear on both sides of the equation (the spectator ions). Mg(s) + 2H+(aq) + → Mg2+(aq) + + H2(g) Step 4: Write down the equation omitting the spectator ions. Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) 19 Write the ionic equation for the reaction of aqueous chlorine with aqueous potassium bromide. The products are aqueous bromine and aqueous potassium chloride. Solution Step 1: The full balanced equation is: Cl2(aq) + 2KBr(aq) → Br2(aq) + 2KCl(aq) Step 2: The ions present are: Cl2(aq) + 2K+(aq) + 2Br−(aq) → Br2(aq) + 2K+(aq) + 2Cl−(aq) Step 3: Cancel the spectator ions: Cl2(aq) + + 2Br−(aq) → Br2(aq) + + 2Cl−(aq) Step 4: Write the final ionic equation: Cl2(aq) + 2Br−(aq) → Br2(aq) + 2Cl−(aq) Questions 18 Change these full equations to ionic equations. a H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(I) b Br2(aq) + KI(aq) → 2KBr(aq)+I2 19 Write ionic equations for these precipitation reactions. a CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq) b Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Chemists usually prefer to write ionic equations for precipitation reactions. A precipitation reaction is a reaction where two aqueous solutions react to form a solid: the precipitate. For these reactions the method of writing the ionic equation can be simplified. All you have to do is: write the formula of the precipitate as the product write the ions that go to make up the precipitate as the reactants. WORKED EXAMPLE 20 An aqueous solution of iron(II) sulfate reacts with an aqueous solution of sodium hydroxide. A precipitate of iron(II) hydroxide is formed, together with an aqueous solution of sodium sulfate. Solution  Step 1: Write the full balanced equation FeSO4(aq) + 2NaOH(aq) → Fe(OH)2(s) + Na2SO4(aq) Step 2: Identify the precipitate formed: Fe(OH)2(s) Step 3: Identify the ions which form the precipitate: Fe2+(aq) and OH−(aq) Step 4: Write the ionic equation: Fe2+(aq) + 2OH−(aq) → Fe(OH)2(s) REFLECTION 1 Work with another learner to discuss these points: a What are the stages in balancing a chemical equation? b What sort of problems have you found when trying to write ionic equations? c How do you know which species to include in the equation? 2 Make a list of the key words in this chapter so far, e.g. mole, and see if you can explain their meaning to another learner. 3 Work with another learner to discuss an activity (it could be memory aid or a game) to help you learn how to write formulae or do mole calculations. Can you teach this to someone else in your class? 3.7 Solutions and concentration Calculating the concentration of a solution The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 (one cubic decimetre) of solution. The solvent is usually water. There are 1000 cm3 in a cubic decimetre. When 1 mole of a compound is dissolved to make 1 dm3 of solution the concentration is 1 mol dm−3. concentration (mol dm−3) = We use the terms ‘concentrated’ and ‘dilute’ to refer to the relative amount of solute in the solution. A solution with a low concentration of solute is a dilute solution. If there is a high concentration of solute, the solution is concentrated. When performing calculations involving concentrations in mol dm−3 you need to: change mass in grams to moles change cm3 to dm3 (by dividing the number of cm3 by 1000). IMPORTANT When making solutions of known concentration, you dissolve a weighed amount of solute in a small amount of solvent. Then you make it up to the required volume by adding more solvent. You do not add the solute to the required final volume of solvent because the total volume of the solution may then change. WORKED EXAMPLE 21 Calculate the concentration in mol dm−3 of sodium hydroxide, NaOH, if 250 cm3 of a solution contains 2.0 g of sodium hydroxide. Solution (Mr value: NaOH = 40.0) Step 1: Change grams to moles. = 0.050 mol NaOH Step 2: Change cm3 to dm3. 250 cm3 = 0.25 dm3 Step 3: Calculate concentration. = 0.20 mol dm−3 Figure 3.20: The concentration of chlorine in the water in a swimming pool must be carefully controlled. We often need to calculate the mass of a substance present in a solution of known concentration and volume. To do this we: rearrange the concentration equation to: number of moles (mol) = concentration (mol dm−3) × volume (dm3) multiply the moles of solute by its molar mass mass of solute (g) = number of moles (mol) × molar mass (g mol−1) WORKED EXAMPLE 22 Calculate the mass of anhydrous copper(II) sulfate in 55 cm3 of a 0.20 mol dm−3 solution of copper(II) sulfate. (Ar values: Cu = 63.5, O = 16.0, S = 32.1) Solution Step 1: Change cm3 to dm3. = 0.055 dm3 Step 2: moles = concentration (mol dm−3) × volume of solution (dm3) = 0.20 × 0.055 = 0.011 mol Step 3: mass (g) = moles × M = 0.011 × (63.5 + 32.1 + (4 × 16.0)) = 1.8 g (to 2 significant figures) Question 20 a Calculate the concentration, in mol dm−3, of the following solutions: (Ar values: C = 12.0, H = 1.0, Na = 23.0, O = 16.0) i a solution of sodium hydroxide, NaOH, containing 2.0 g of sodium hydroxide in 50 cm3 of solution ii a solution of ethanoic acid, CH3CO2H, containing 12.0 g of ethanoic acid in 250 cm3 of solution. b Calculate the number of moles of solute dissolved in each of the following: i 40 cm3 of aqueous nitric acid of concentration 0.2 mol dm−3 ii 50 cm3 of calcium hydroxide solution of concentration 0.01 mol dm−3. PRACTICAL ACTIVITY 3.2 Carrying out a titration A procedure called a titration is used to determine the amount of substance present in a solution of unknown concentration. There are several different kinds of titration. One of the commonest involves the exact neutralisation of an alkali by an acid (Figure 3.21). Figure 3.21: a A funnel is used to fill the burette with hydrochloric acid. b A graduated pipette is used to measure 25.0 cm3 of sodium hydroxide solution into a conical flask. c An indicator called litmus is added to the sodium hydroxide solution, which turns blue. d 12.5 cm3 of hydrochloric acid from the burette have been added to the 25.0 cm3 of alkali in the conical flask. The litmus has gone red, showing that this volume of acid was just enough to neutralise the alkali. If we want to determine the concentration of a solution of sodium hydroxide of unknown concentration we use the following procedure. Use a volumetric pipette to place a given volume (usually 10 cm3 or 25 cm3) of sodium hydroxide of unknown concentration into a flask. Get some of acid of known concentration, e.g. hydrochloric acid of concentration 1.00 mol dm−3 Fill a clean burette with the acid (after having washed the burette with a little of the acid). Record the initial burette reading. Add a few drops of an indicator solution to the alkali in the flask, e.g. methyl orange. Slowly add the acid from the burette to the flask, swirling the flask all the time until the indicator changes colour (the end-point). Record the final burette reading. The final reading minus the initial reading is called the titre. This first titre is normally known as a ‘rough’ value. Repeat this process, adding the acid drop by drop near the end-point. Repeat again, until you have two titres that are no more than 0.10 cm3 apart. Take the average of these two titre values. Your results should be recorded in a table, looking like this: Rough 1 2 3 Final burette reading / cm3 37.60 38.65 36.40 34.75 Initial burette reading / cm3 2.40 4.00 1.40 0.00 Titre / cm3 35.20 34.65 35.00 34.75  Table 3.9: Example results table. You should note: all burette readings are given to an accuracy of 0.05 cm3 the units are shown like this: / cm3 the two titres that are no more than 0.10 cm3 apart are 1 and 3, so they would be averaged the average titre is 34.70 cm3. In every titration there are five important things you need to know or work out: 1 the balanced equation for the reaction 2 the volume of the solution from the burette (in this example this is hydrochloric acid) 3 the concentration of the solution in the burette 4 the volume of the solution in the titration flask (in this example this is sodium hydroxide) 5 the concentration of the solution in the titration flask. If we know four of these five things, we can calculate the fifth. So, in order to calculate the concentration of sodium hydroxide in the flask, we need to know the first four of these points. Calculating solution concentration by titration A titration is often used to find the exact concentration of a solution. Worked example 23 shows the steps used to calculate the concentration of a solution of sodium hydroxide when it is neutralised by aqueous sulfuric acid of known concentration and volume. WORKED EXAMPLE 23 25.0 cm3 of a solution of sodium hydroxide is exactly neutralised by 15.10 cm3 of sulfuric acid of concentration 0.200 mol dm−3. 2NaOH + H2SO4 → Na2SO4 + 2H2O Calculate the concentration, in mol dm−3, of the sodium hydroxide solution. Solution Step 1: Calculate how many moles of acid. moles = concentration (mol dm−3) × volume of solution (dm3) 0.200 × = 0.00302 mol H2SO4 Step 2: Use the stoichiometry of the balanced equation to calculate how many moles of NaOH. moles of NaOH = moles of acid (from step 1) × 2 = 0.00604 mol Step 3: Calculate the concentration of NaOH.  Note: In the first step we use the reagent for which the concentration and volume are both known. In step 2, we multiply by 2 because the balanced equation shows that 2 mol of NaOH reacts with every 1 mol of H2SO4. In step 3, we divide by 0.0250 because we have changed cm3 to dm3 0.0250 = The answer is given to 3 significant figures because the smallest number of significant figures in the data is 3. IMPORTANT When doing calculations from the results of titrations, remember: average only those titres which are very close to each other, e.g. within 0.1–0.2 cm3 of each other, and ignore the rough (rangefinder) titration. keep the units the same. It is often easier to convert volumes in cm3 to dm3 because solution concentration is usually given in mol dm−3. Note that in some older books M is sometimes used in place of mol dm−3. Question 21 a The equation for the reaction of strontium hydroxide with hydrochloric acid is shown below. Sr(OH)2 + 2HCl → SrCl2 + 2H2O 25.0 cm3 of a solution of strontium hydroxide was exactly neutralised by 15.00 cm3 of 0.100 mol dm−3 hydrochloric acid. Calculate the concentration, in mol dm−3, of the strontium hydroxide solution. b 20.0 cm3 of a 0.400 mol dm−3 solution of sodium hydroxide was exactly neutralised by 25.25 cm3 of sulfuric acid. Calculate the concentration, in mol dm−3, of the sulfuric acid. The equation for the reaction is: H2SO4 + 2NaOH → Na2SO4 + 2H2O Deducing stoichiometry by titration We can use titration results to find the stoichiometry of a reaction. In order to do this, we need to know the concentrations and the volumes of both the reactants. Worked example 24 shows how to determine the stoichiometry of the reaction between a metal hydroxide and an acid. WORKED EXAMPLE 24 25.0 cm3 of a 0.0500 mol dm−3 solution of a metal hydroxide was titrated against a solution of 0.200 mol dm−3 hydrochloric acid. It required 12.50 cm3 of hydrochloric acid to exactly neutralise the metal hydroxide. Deduce the stoichiometry of this reaction. Solution Step 1: Calculate the number of moles of each reagent.  Step 2: Deduce the simplest mole ratio of metal hydroxide to hydrochloric acid. 1.25 × 10−3 moles of hydroxide : 2.50 × 10−3 moles of acid = 1 hydroxide : 2 acid Step 3: Write the equation. M(OH)2 + 2HCl → MCl2 + 2H2O One mole of hydroxide ions neutralises one mole of hydrogen ions. As one mole of the metal hydroxide neutralises two moles of hydrochloric acid, the metal hydroxide must contain two hydroxide ions in each formula unit. Questions 22 20.0 cm3 of a metal hydroxide of concentration 0.0600 mol dm−3 was titrated with 0.100 mol dm−3 hydrochloric acid. It required 24.00 cm3 of the hydrochloric acid to exactly neutralise the metal hydroxide. a Calculate the number of moles of metal hydroxide used. b Calculate the number of moles of hydrochloric acid used. c What is the simplest mole ratio of metal hydroxide to hydrochloric acid? d Write a balanced equation for this reaction using your answers to parts a, b and c to help you. Use the symbol M for the metal. 23 0.4 moles of aluminium sulfate, Al2(SO4)3, are dissolved in water and the volume of the solution was made up to 500 cm3. Which one of these statements is correct? A The solution contains a total of 2 moles of ions. B The concentration of aluminium ions is 0.8 mol dm−3. C The concentration of sulfate ions in solution is 1.2 mol dm−3. D The solution contains a total of 4 moles of ions. 3.8 Calculations involving gas volumes Using the molar gas volume In 1811 the Italian scientist Amedeo Avogadro suggested that equal volumes of all gases contain the same number of molecules. This is called Avogadro’s hypothesis. This idea is approximately true as long as the pressure is not too high or the temperature too low. It is convenient to measure volumes of gases at room temperature (20 °C) and pressure (1 atmosphere). At room temperature and pressure (r.t.p.) one mole of any gas has a volume of 24.0 dm3. So, 24.0 dm3 of carbon dioxide and 24.0 dm3 of hydrogen both contain one mole of gas molecules. We can use the molar gas volume of 24.0 dm3 at r.t.p. to find: the volume of a given mass or number of moles of gas the mass or number of moles of a given volume of gas. WORKED EXAMPLES 25 Calculate the volume of 0.40 mol of nitrogen at r.t.p. Solution 26 Calculate the mass of methane, CH4, present in 120 cm3 of methane. (Mr value: methane = 16.0) Solution 120 cm3 is 0.120 dm3 Question 24 a Calculate the volume, in dm3, occupied by 26.4 g of carbon dioxide at r.t.p. (Ar values: C = 12.0, O = 16.0) b A flask of volume 120 cm3 is filled with helium gas at r.t.p. Calculate the mass of helium present in the flask. (Ar value: He = 4.0) Figure 3.22: Anaesthetists have to know about gas volumes so that patients remain unconscious during major operations. Gas volumes and stoichiometry We can use the ratio of reacting volumes of gases to deduce the stoichiometry of a reaction. If we mix 20 cm3 of hydrogen with 10 cm3 of oxygen and explode the mixture, we will find that the gases have exactly reacted together and no hydrogen or oxygen remains. According to Avogadro’s hypothesis, equal volumes of gases contain equal numbers of molecules and therefore equal numbers of moles of gases. So the mole ratio of hydrogen to oxygen is 2 : 1. We can summarise this as: We can extend this idea to experiments involving combustion data of hydrocarbons. Worked example 27 shows how the formula of propane and the stoichiometry of the equation can be deduced. Propane is a hydrocarbon: a compound of carbon and hydrogen only. WORKED EXAMPLE 27 When 50 cm3 of propane reacts exactly with 250 cm3 of oxygen, 150 cm3 of carbon dioxide is formed. propane + oxygen → carbon dioxide + water C xH y (O2) (CO2) (H2O) 3 3 3 50 cm 250 cm 150 cm Solution ratio of moles: 1 5 3 As 1 mole of propane produces 3 moles of carbon dioxide, there must be 3 moles of carbon atoms in one mole of propane, x = 3. C3Hy + 5O2 → 3CO2 + zH2O The 5 moles of oxygen molecules are used to react with both the carbon and the hydrogen in the propane. 3 moles of these oxygen molecules have been used in forming carbon dioxide. So 5 − 3 = 2 moles of oxygen molecules must be used in reacting with the hydrogen to form water. There are 4 moles of atoms in 2 moles of oxygen molecules, so there must be 4 moles of water formed. C3Hy + 5O2 → 3CO2 + 4H2O  So there must be 8 hydrogen atoms in 1 molecule of propane. C3H8 + 5O2 → 3CO2 + 4H2O Questions 25 50 cm3 of a gaseous hydride of phosphorus, PHn reacts with exactly 150 cm3 of chlorine, Cl2, to form liquid phosphorus trichloride and 150 cm3 of hydrogen chloride gas, HCl. a How many moles of chlorine react with 1 mole of the gaseous hydride? b Deduce the formula of the phosphorus hydride. c Write a balanced equation for the reaction. 26 What is the volume of carbon dioxide produced when 14 g of ethene, C2H4, reacts with excess oxygen? (Mr ethene = 28) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) A 96.0 dm3 B 24.0 dm3 C 0.0467 dm3 D 12.0 dm3 REFLECTION 1. Make a list of the chemical expressions that you have learned during this chapter, e.g. the relationship between moles, mass and relative molecular mass. 2. Compare your list with another learner then write the expressions out in as many different ways as you can, e.g. the expression density = could be rewritten as mass = density × volume What problems did you come across? How did you solve them? SUMMARY Unified atomic mass unit is one twelfth of the mass of a carbon-12 atom. Relative atomic mass is the ratio of the weighted average mass of atoms in a given sample of an element to the unified atomic mass unit. The mass spectrometer can be used to determine atomic masses and isotopic abundances, to deduce the molecular mass of an organic compound and to deduce the formula of an organic molecule from its fragmentation pattern and isotopic ratios. One mole of a substance is the amount of substance that has the Avogadro number of specified particles (6.02 × 1023). The particles can be atoms, molecules, ions or electrons. Molecular formulae show the total number of atoms of each element present in one molecule or one formula unit of the compound. Empirical formulae show the simplest whole number ratio of atoms in a compound. The mole concept can be used to calculate: reacting masses volumes of gases volumes and concentrations of solutions. The stoichiometry of a reaction can be obtained from calculations involving reacting masses, gas volumes, and volumes and concentrations of solutions. EXAM-STYLE QUESTIONS 1 a i State the meaning of the term relative atomic mass. 10 ii A sample of boron was found to have the following % composition by mass: 5 B (18.7%), 115B (81.3%) Calculate a value for the relative atomic mass of boron. Give your answer to 3 significant figures. b Boron ions, B3+, can be formed by bombarding gaseous boron with high-energy electrons. Deduce the number of electrons in one B3+ ion. c Boron is present in compounds called borates. i Use the Ar values below to calculate the relative molecular mass of iron(III) borate, Fe(BO2)3. (Ar values: Fe = 55.8, B = 10.8, O = 16.0) ii The accurate relative atomic mass of iron, Fe, is 55.8. Explain why the accurate relative atomic mass is not a whole number. [Total: 6] 2 This question is about some metals and metal compounds. a i Hafnium, Hf, forms a hydrated peroxide whose formula can be written as HfO3·2H2O. Use the Ar values below to calculate the relative molecular mass of hydrated hafnium peroxide. (Ar values: Hf = 178.5, H = 1.0, O = 16.0) ii What is meant by the term hydrated salt? b A particular isotope of hafnium has 72 protons and a nucleon number of 180. Write the isotopic symbol for this isotope, showing this information. c The mass spectrum of zirconium is shown below. i Give the isotopic symbol for the most abundant isotope of zirconium. ii Use the information from this mass spectrum to calculate the relative atomic mass of zirconium. Give your answer to 3 significant figures. iii High-resolution mass spectra show accurate relative isotopic masses. State the meaning of the term relative isotopic mass? d A sample of 15.2 g of tin(IV) oxide is mixed with 2.41 g of carbon and heated. A reaction occurs. SnO2 + 2C → Sn + 2CO Show by calculation that the reagent in excess is tin(IV) oxide. (Ar values: Sn = 118.7, C  = 12.0, O = 16.0). e A sample of zirconium is made by reacting 58.30 g of zirconium tetrachloride, ZrCl4, with excess magnesium. ZrCl4 + 2Mg → Zr + 2MgCl2 The mass of zirconium produced is 20.52 g. Calculate the percentage yield of zirconium. (Ar values: Zr = 91.2, Cl = 35.5, Mg = 24.3) [Total: 11] 3 Solid sodium carbonate reacts with aqueous hydrochloric acid to form aqueous sodium chloride, carbon dioxide and water. Na2CO3 + 2HCl → 2NaCl + CO2 + H2O a Rewrite this equation to include state symbols. b Calculate the number of moles of hydrochloric acid required to react exactly with 4.15 g of sodium carbonate. (Ar values: C = 12.0, Na = 23.0, O = 16.0, H = 1.0, Cl = 35.5) c Define the term mole. d An aqueous solution of 25.0 cm3 sodium carbonate of concentration 0.0200 mol dm−3 is titrated with hydrochloric acid. The volume of hydrochloric acid required to exactly react with the sodium carbonate is 12.50 cm3. i Calculate the number of moles of sodium carbonate present in the solution of sodium carbonate. ii Calculate the concentration of the hydrochloric acid. e How many moles of carbon dioxide are produced when 0.2 mol of sodium carbonate reacts with excess hydrochloric acid? f Calculate the volume of this number of moles of carbon dioxide at r.t.p. (1 mol of gas occupies 24 dm3 at r.t.p.) [Total: 10] 4 Hydrocarbons are compounds of carbon and hydrogen only. Hydrocarbon Z is composed of 80% carbon and 20% hydrogen. a Calculate the empirical formula of hydrocarbon Z. (Ar values: C = 12.0, H = 1.0) b The molar mass of hydrocarbon Z is 30.0 g mol−1. Deduce the molecular formula of this hydrocarbon. c When 50 cm3 of hydrocarbon Y is burnt, it reacts with exactly 300 cm3 of oxygen to form 200 cm3 of carbon dioxide. Water is also formed in the reaction. Construct the equation for this reaction. Explain your reasoning. d Propane has the molecular formula C3H8. Calculate the mass of 600 cm3 of propane at r.t.p. (1 mol of gas occupies 24 dm3 at r.t.p.) (Ar values: C = 12.0, H = 1.0) [Total: 10] 5 When sodium reacts with titanium chloride (TiCl4), sodium chloride (NaCl) and titanium (Ti) are produced. a Construct the balanced symbol equation for the reaction. b Calculate the mass of titanium produced from 380 g of titanium chloride. Give your answer to 3 significant figures. (Ar values: Ti = 47.9, Cl = 35.5) c Calculate the mass of titanium produced using 46.0 g of sodium. Give your answer to 3 significant figures. (Ar value: Na = 23.0) [Total: 6] 6 In this question give all answers to 3 significant figures. The reaction between NaOH and HCl can be written as: HCl + NaOH → NaCl + H2O In this reaction, 15.0 cm3 of hydrochloric acid was neutralised by 20.0 cm3 of 0.0500 mol dm−3 sodium hydroxide.  a Calculate the volume in dm3 of: i the acid ii the alkali b Calculate the number of moles of alkali. c Calculate the number of moles of acid and then its concentration. [Total: 5] 7 Give all answers to 3 significant figures. Ammonium nitrate decomposes on heating to give nitrogen(II) oxide and water as follows: NH4NO3(s) → N2O(g) + 2H2O(l) (Ar values: N = 14.0; H = 1.0; O = 16.0) a Deduce the formula mass of ammonium nitrate. b How many moles of ammonium nitrate are present in 0.800 g of the solid? c Calculate the volume of N2O gas that would be produced from this mass of ammonium nitrate? [Total: 5] 8 Give all answers to 3 significant figures. a 1.20 dm3 of hydrogen chloride gas was dissolved in 100 cm3 of water. i Calculate the number of moles of hydrogen chloride gas present. ii Calculate the concentration of the hydrochloric acid formed. b 25.0 cm3 of the acid was then titrated against sodium hydroxide of concentration 0.200 mol dm−3 to form NaCl and water: NaOH + HCl → H2O + NaCl i Calculate the number of moles of acid used. ii Calculate the volume of sodium hydroxide used. [Total: 7] 9 Give all answers to 3 significant figures. 4.80 dm3 of chlorine gas was reacted with sodium hydroxide solution. The reaction taking place was as follows: Cl2(g) + 2NaOH(aq) → NaCl(aq) + NaOCl(aq) + H2O(l) (Ar values: Na = 23.0; Cl = 35.5; O = 16.0) a Calculate the number of moles of Cl2 reacted. b Calculate the mass of NaOCl formed. c The concentration of the NaOH was 2.00 mol dm−3. Calculate the volume of sodium hydroxide solution required. d Write an ionic equation for this reaction. [Total: 6] 10 Calcium oxide reacts with hydrochloric acid according to the equation: CaO + 2HCl → CaCl2 + H2O Ar values: H = 1.0; O = 16.0; Cl = 35.5; Ca = 40.1 a Calculate the mass of calcium chloride formed when 28.05 g of calcium oxide reacts with excess hydrochloric acid. b Calculate the mass of hydrochloric acid that reacts with 28.05 g of calcium oxide. c Calculate the mass of water produced. 11 When ammonia gas and hydrogen chloride gas mix together, they react to form a solid [Total: 5] called ammonium chloride. Ar values: H = 1.0; N = 14.0; Cl = 35.5 a Write a balanced equation for this reaction, including state symbols. b Calculate the molar masses of ammonia, hydrogen chloride and ammonium chloride. c Calculate the volumes of ammonia and hydrogen chloride gases that must react at r.t.p. in order to produce 10.7 g of ammonium chloride? (1 mol of gas occupies 24 dm3 at r.t.p.) [Total: 8] 12 The mass spectrum of an organic compound with the structure CH3CH2CH2CH3 is shown. a Identify the fragments with mass/charge ratios of i 15 ii 29 iii 43 iv 58 b An organic compound, Z, has a molecular ion peak, M, with a relative abundance of 36.4% and an [M + 1] peak with a relative abundance of 4.8 %. Deduce the number of carbon atoms this compound contains. c The mass spectrum of chlorobutane shows an [M + 2] peak beyond the Mr peak. Explain why there is an [M + 2] peak. d The relative height of the [M + 2] peak in organic compound T is the same as that of the Mr peak. Explain how this shows that the compound contains one atom of bromine and not one atom of chlorine. [Total: 10] SELF-EVALUATION After studying this chapter, complete a table like this: See Needs Almost Ready to I can section... more work there move on define and use the terms unified atomic mass unit, relative 3.1 atomic mass, relative isotopic mass and relative formula mass analyse mass spectra in terms of isotopic abundances 3.3 deduce the accurate relative atomic mass of an element using 3.3 data from a mass spectrum deduce the molecular mass of an organic molecule from the 3.3 molecular ion peak in a mass spectrum suggest the identity of molecules formed by simple 3.3 fragmentation in a given mass spectrum deduce the number of carbon atoms in a compound using the [M + 1] peak and the relevant formula: 3.3 deduce the presence of chlorine and bromine atoms in a 3.3 compound using the [M + 2] peak define and use the terms mole and Avogadro constant 3.4 understand the meaning of the terms anhydrous, hydrated 3.2 and water of crystallisation write formulae of ionic compounds from ionic charges and oxidation numbers including: a the prediction of ionic charge from the position of an element in the Periodic Table 3.6 b recall of the names and formulae for the ions NO3−, CO32−, SO42−, OH−, NH4+, Zn2+, Ag+, HCO3− and PO43− write and construct balanced equations, including ionic 3.6 equations use the correct state symbols in equations 3.6 define and use the terms empirical formula and molecular 3.5 formula calculate empirical and molecular formulae using given data 3.5 perform calculations using the mole concept involving reacting 3.5 masses understand the terms excess reactant and limiting reactant 3.5 perform calculations using mole concept involving a percentage yield 3.5, 3.7, b volumes of gases 3.8 c concentrations of solutions d limiting reagents and excess reagent deduce stoichiometry of a reaction using reacting masses, volumes of gases and volumes and concentrations of 3.5 solutions. images Chapter 4 Chemical bonding LEARNING INTENTIONS In this chapter you will learn how to: define electronegativity and explain the factors influencing the electronegativity values of the elements explain the trends in electronegativity across a period and down a group in the Periodic Table use differences in the Pauling electronegativity values to predict if a compound has ionic or covalent bonds define ionic bonding and describe ionic bonding in compounds such as sodium chloride, magnesium oxide and calcium fluoride define covalent bonding and describe covalent bonding in molecules such as hydrogen, oxygen, nitrogen, chlorine, hydrogen chloride, carbon dioxide, ammonia, methane, ethane and ethene describe how some atoms in Period 3 can expand their octet of electrons to form compounds such as sulfur dioxide, phosphorus pentachloride and sulfur hexafluoride describe co-ordinate bonding (dative covalent bonding) in ions such as NH4+ and molecules such as Al2Cl6 use dot-and-cross diagrams to show the arrangement of electrons in compounds with ionic, covalent and co- ordinate bonding describe covalent bonding in terms of orbital overlap giving sigma (σ) and pi (π) bonds  describe how sigma and pi bonds form in molecules such as H2, C2H6, C2H4, HCN and N2 describe hybridisation of atomic orbitals to form sp, sp2 and sp3 orbitals define the terms bond energy and bond length and use these to compare the reactions of covalent molecules describe and explain the shapes and bond angles in simple molecules (such as BF3, CO2, CH4, NH3, H2O, SF6 and PF5) using ‘valence shell electron pair repulsion’ theory (VSEPR) predict the shapes and bond angles in other molecules and ions similar to those above describe hydrogen bonding and explain, in terms of hydrogen bonding, why some physical properties of water are unusual for a molecular compound use electronegativity values to explain bond polarity and dipole moments in molecules describe and understand the different types of intermolecular forces (van der Waals’ forces) as either instantaneous dipoles or permanent dipoles describe metallic bonding describe the relative bond strengths of ionic, covalent and metallic bonds compared with intermolecular forces. BEFORE YOU START 1 Take it in turns to explain to another learner what is meant by each of the following: a covalent bonding b ionic bonding c metallic bonding d intermolecular forces Mention the types of particles that are responsible for the attractive forces between the atoms or ions. 2 Draw a labelled diagram of an ionic compound and a simple covalent compound. Compare your diagrams with those of another learner. 3 Make a list of the typical physical properties of metals such as iron and non-meals such as sulfur. Compare your list with the list made by others in the class. 4 Ask another learner to select one of the first 20 elements in the Periodic Table. Can you deduce the simple electronic configuration for an atom of this element? Take turns in doing this until you are sure that you can write simple electronic configurations. 5 Explain to another learner how to work out the number of electrons in the outer principal energy level of atoms in Groups 1 and 2 and in Groups 13–18 (you may know Groups 13 to 18 as Groups 3 to 8) 6 Ask another learner to select one of the positive ions: sodium, magnesium or calcium, and one of the negative ions: oxygen, chlorine or bromine. Draw the simple electronic configuration for the positive and negative ions selected and then draw a dot-and-cross diagram for the compound formed from these ions. 7 How confident are you in drawing dot-and-cross diagrams for molecules? Without looking at a textbook, work with a partner to draw dot-and-cross diagrams for the following molecules: a hydrogen b chlorine c hydrogen chloride d methane (CH4) e water f ammonia g oxygen Show only the outer shell electrons. 8 Try this problem by yourself and then discuss your ideas with a partner. Lone pairs of electrons repel each other more than a pair of electrons in a bond. Draw the electronic structure of a methane molecule and a water molecule and suggest why the methane molecule has an H─C─H bond angle of about 109.5° but a water molecule has an H─O─H bond angle of 104.5°. SKATING ON WATER Figure 4.1: The intermolecular forces in water allow some insects to skate over its surface. Some insects can skate across the surface of a pond (Figure 4.1). How do they do this? Part of the answer is that water has unusual physical properties due to the strong forces of attraction (intermolecular forces) between the water molecules. The unusually strong intermolecular forces in water are due to hydrogen bonding. These attractive forces make it more difficult for water molecules to slide over each other compared with most other liquids. A water molecule below the surface is attracted equally in all directions to other water molecules. At the surface, where water and air meet, there is hardly any attraction between the molecules in the air and the water molecules. But the water molecules at the surface are strongly attracted to each other. So there is a downward force at the surface of the liquid which pulls the surface inwards. This surface tension causes the surface of the water to appear to have an elastic skin. You can float a paper clip on water to demonstrate surface tension: Fill a glass right up to the top with water. Take a paper clip and lower it gently onto the surface of the water using a fork. Skating insects take advantage of the high surface tension of water by having special features such as wide feet. These feet are also covered in a hard layer which repels water. Questions for discussion Discuss with another learner or group of learners: Suggest why skating insects have wide feet and not narrow feet. Can you think of any liquids that do not mix with water and are lighter than water? Suggest why a soft paper-like layer on insect’s feet would not allow them to stay on the surface of the water. Imagine that you are designing an insect that can skate perfectly on water. What other features would be useful? 4.1 Types of chemical bonding IMPORTANT When elements form compounds, they either gain, lose or share electrons to get to the nearest stable noble gas electronic configuration. There are exceptions to this rule for some Group 15, 16 and 17 elements. Ionic bonding is the electrostatic attraction between positive ions (cations) and negative ions (anions) in an ionic crystal lattice. Covalent bonds are formed when the outer electrons of two atoms are shared. The ionic or covalent bonds formed are usually very strong: it takes a lot of energy to break them. There is also a third form of strong bonding: metallic bonding. Although the atoms within molecules are kept together by strong covalent bonds, the forces between molecules are weak. We call these weak forces between molecules van der Waals’ forces. The term van der Waals’ forces is a general term used to describe all intermolecular forces. There are several types of van der Waals forces: dipole (instantaneous dipole–induced dipole (id–id) forces. These are also called London dispersion forces. permanent dipole–permanent dipole (pd–pd) forces. hydrogen bonding, which is a stronger form of permanent dipole–permanent dipole force. If you understand these different types of chemical bonding and understand intermolecular forces, this will help you to explain the structure and some physical properties of elements and compounds. 4.2 Ionic bonding How are ions formed? One way of forming ions is for atoms to gain or lose one or more electrons. Positive ions are formed when an atom loses one or more electrons. Metal atoms usually lose electrons and form positive ions. Negative ions are formed when an atom gains one or more electrons. Non-metal atoms usually gain electrons and form negative ions. The charge on the ion depends on the number of electrons lost or gained (see Section 2.4). When metals combine with non-metals, the electrons in the outer shell of the metal atoms are transferred to the non-metal atoms. Each non-metal atom usually gains enough electrons to fill its outer shell. As a result of this, the metal atoms and non-metal atoms usually end up with outer electron shells that are complete: they have the electronic configuration of a noble gas. In Figure 4.2 we can see that: the sodium ion has the same electronic structure as neon: [2,8]+ the chloride ion has the same electronic structure as argon: [2,8,8]−. The strong force of attraction between the positive ions and negative ions in the ionic crystal lattice results in an ionic bond. An ionic bond is sometimes called an electrovalent bond. In an ionic structure such as sodium chloride crystals (Figure 4.3), the ions are arranged in a regular repeating pattern (see Chapter 5). As a result of this, the force between one ion and the ions of opposite charge that surround it is very great. In other words, ionic bonding is very strong. Figure 4.2: The formation of a sodium ion and chloride ion by electron transfer. Figure 4.3: These crystals of salt are made up of millions of sodium ions and chloride ions. Dot-and-cross diagrams You will notice that in Figure 4.2 we use dots and crosses to show the electronic configuration of the chloride and sodium ions. This helps us keep track of where the electrons have come from. It does not mean that the electron transferred is any different from the others. Diagrams like this are called dot-and- cross diagrams. When drawing a dot-and-cross diagram for an ionic compound it is usually acceptable to draw the outer electron shell of the metal ion without any electrons. This is because it has transferred these electrons to the negative ion. Figure 4.4 shows the outer shell dot-and-cross diagram for sodium chloride. A dot-and-cross diagram shows: the outer electron shells only the charge of the ion is spread evenly, by using square brackets the charge on each ion, written at the top right-hand corner. Figure 4.4: Dot-and-cross diagram for sodium chloride. Some examples of dot-and-cross diagrams Magnesium oxide When magnesium reacts with oxygen to form magnesium oxide, the two electrons in the outer shell of each magnesium atom are transferred to the incompletely filled orbitals of an oxygen atom. By losing two electrons, each magnesium atom achieves the electronic configuration [2,8] (Figure 4.5). By gaining two electrons, each oxygen atom achieves the electronic configuration [2,8]. [2,8] is the electronic configuration of neon. It is sometimes called a noble-gas configuration. When ions or atoms have 8 electrons in their outer shell like this it is called an octet of electrons (the prefix oct means 8). Figure 4.5: Dot-and-cross diagram for magnesium oxide. Calcium fluoride Each calcium atom has two electrons in its outer shell, and these can be transferred to two fluorine atoms. By losing two electrons, each calcium atom achieves the electronic configuration [2,8,8] (Figure 4.6). The two fluorine atoms each gain one electron to achieve the electronic configuration [2,8]. [2,8] is the electronic configuration of neon; it is a ‘noble-gas configuration’. Figure 4.6: Dot-and-cross diagram for calcium fluoride. Question 1 Draw dot-and-cross diagrams for the ions in the following ionic compounds. Show only the outer electron shells. a Potassium chloride, KCl b Sodium oxide, Na2O c Calcium oxide, CaO d Magnesium chloride, MgCl2 4.3 Covalent bonding Single covalent bonds When two non-metal atoms combine, they share one or more pairs of electrons. A shared pair of electrons is called a single covalent bond, or a bond pair. A single covalent bond is represented by a single line between the atoms: for example, Cl─Cl. You can see that when chlorine atoms combine not all the electrons are used in bonding. The pairs of outer- shell electrons not used in bonding are called lone pairs. Figure 4.7: a Bromine and b iodine are elements. They both have simple covalent molecules. Each atom in a chlorine molecule has three lone pairs of electrons and shares one bonding pair of electrons (Figure 4.8). Figure 4.8: Atoms of chlorine share electrons to form a single covalent bond. IMPORTANT A dot-and-cross diagram shows the arrangement of the electrons in the main energy levels. We generally only show the outer energy level when we draw dot-and-cross diagrams. Remember that we only draw the electrons as dots and crosses to show us which atoms the electrons come from. When drawing the arrangement of electrons in a molecule: use a ‘dot’ for electrons from one of the atoms and a ‘cross’ for the electrons from the other atom if there are more than two types of atom, use additional symbols such as a small circle or a small triangle draw the outer electrons in pairs, to show the number of bond pairs and the number of lone pairs. Some examples of dot-and-cross diagrams for simple covalently bonded molecules are shown in Figure 4.9. Figure 4.9: Dot-and-cross diagrams for some covalent compounds: a hydrogen, H2, b methane, CH4, c water, H2O, d ammonia, NH3, e hydrogen chloride, HCl and f ethane, C2H6. There are some cases in which the electrons around a central atom may not have a noble gas configuration (an octet of electrons). For example: boron trifluoride, BF3, has only six electrons around the boron atom. We say that the boron atom is ‘electron deficient’. sulfur hexafluoride, SF6, has twelve electrons around the central sulfur atom. We say that the sulfur atom has an ‘expanded octet’ (Figure 4.10). phosphorus(V) chloride, PCl5, has 10 electrons around the central phosphorus atom. Phosphorus has expanded its octet (Figure 4.10). sulfur dioxide, SO2, the sulfur atom forms a double bond with each oxygen atom. This leaves a pair of non-bonding electr