Quantitative chemistry 1
Downloaded from www.heinemann.co.uk/ib 1 Quantitative chemistry es Chemistry was a late developer as a physical science. Newton was working on the laws of physics more than a century before the work of the French chemist Antoine Lavoisier (1743–1794) brought chemistry into the modern age. Chemical reactions involve changes in smell, colour and texture and these are difficult to quantify. Lavoisier appreciated the importance of attaching numbers to properties and recognized the need for precise measurement. His use of the balance allowed changes in mass to be used to analyse chemical reactions. There are practical problems with this approach as powders scatter, liquids splash and gases disperse. It is essential to keep track of all products and it is perhaps significant that Lavoisier was a tax collector by profession. A quantitative approach to the subject helped chemistry to develop beyond the pseudoscience of alchemy. Mole calculations are used to work out the relative amounts of hydrogen and oxygen needed to launch the space shuttle. Sa m pl e pa g This chapter is central to the practice of chemistry as it builds a foundation for most of the numerical work in the course. The two threads to this chapter, a description of the states of the matter and its measurement, are both based on a particulate model of matter. The unit of amount, the mole, and the universal language of chemistry, chemical equations, are introduced. UNCORRECTED PROOF COPY 1 1 Downloaded from www.heinemann.co.uk/ib Quantitative chemistry Assessment statements 1.1 The mole concept and Avogadro’s constant 1.1.1 Apply the mole concept to substances. 1.1.2 Determine the number of particles and the amount of substance (in moles). 1.2 Formulas 1.2.1 Define the terms relative atomic mass (Ar) and relative molecular mass (Mr). 1.2.2 Calculate the mass of one mole of a species from its formula. 1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass and molar mass. 1.2.4 Distinguish between the terms empirical formula and molecular formula. 1.2.5 Determine the empirical formula from the percentage composition or from other experimental data. 1.2.6 Determine the molecular formula when given both the empirical formula and experimental data. pa g es 1.3 Chemical equations 1.3.1 Deduce chemical equations when all reactants and products are given. 1.3.2 Identify the mole ratio of any two species in a chemical equation. 1.3.3 Apply the state symbols (s), (l), (g) and (aq). Sa m pl e 1.4 Mass and gaseous volume relationships in chemical reactions 1.4.1 Calculate theoretical yields from chemical equations. 1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given. 1.4.3 Solve problems involving theoretical, experimental and percentage yield. 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases. 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations. 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. 1.4.7 Solve problems using the ideal gas equation, PV nRT . 1.4.8 Analyse graphs relating to the ideal gas equation. 1.5 Solutions 1.5.1 Distinguish between the terms solute, solvent, solution and concentration (g dm–3 and mol dm–3). 1.5.2 Solve problems involving concentration, amount of solute and volume of solution. 1.1 The mole concept and Avogadro’s constant Measurement and units Scientists have developed the SI system – from the French Système International, to allow the scientific community to communicate effectively both across disciplines and across borders. 2 Scientists search for order in their observations of the world. Measurement is a vital tool in this search. It makes our observations more objective and helps us find relationships between different properties. The standardization of measurement of mass and length began thousands of years ago when kings and emperors used units of length based on the length of their arms or feet. Because modern science is an international endeavour, a more reliable system of standards, the Système International is needed. UNCORRECTED PROOF COPY Downloaded from www.heinemann.co.uk/ib A lot of experimental chemistry relies on the accurate measurement and recording of the physical quantities of mass, time, temperature and volume. The SI units for these are given below. Property Unit Symbol for unit mass kilogram kg time second s temperature kelvin K volume cubic metre m3 pressure pascal Base SI units of some physical quantities used in chemistry; volume is a derived unit as it depends on length. Pa or N m2 1 m3 Figure 1.1 A cube of 1 m3. This has a volume of 100 cm 100 cm 100 cm 1000 000 cm3. A typical gas syringe has a volume of 100 cm3 so ten thousand of these would be needed to measure a gas with this volume. e 1m pa g es These units are however not always convenient for the quantities typically used in the laboratory. Volumes of liquids and gases, for example, are measured in cubic centimetres. pl 1 cm 1m mass time Unit gram Sa Property m 1m Symbol for unit g minute min degree celsius °C volume cubic centimetre cm3 pressure atmosphere atm temperature Other units used in chemistry. Amounts of substance Chemists need to measure quantities of substances for many purposes. Pharmaceutical companies need to check that a tablet contains the correct amount of the drug. Food manufacturers check levels of purity. In the laboratory, reactants need to be mixed in the correct ratios to prepare the desired product. We measure mass and volume routinely in the lab but they are not direct measures of amount. Equal quantities of apples and oranges do not have equal masses or equal volumes but equal numbers. The chemist adopts the same approach. As all matter is made up from small particles (see Chapter 2), we measure amount by counting particles. If the substance is an element we usually count atoms, if it is a compound we count molecules or ions. UNCORRECTED PROOF COPY A chemical species may be an atom, a molecule or an ion. 3 1 Quantitative chemistry Downloaded from www.heinemann.co.uk/ib A standard unit of amount can be defined in terms of a sample amount of any substance. Shoes and socks are counted in pairs, eggs in dozens and atoms in moles. A mole is the amount of a substance which contains the same number of chemical species as there are atoms in exactly 12 grams of the isotope carbon-12 (see Chapter 2). The mole is a SI unit with the symbol mol. The word derives from the Latin for heap or pile. A mole is the amount of a substance which contains the same number of chemical species as there are atoms in exactly 12 g of the isotope carbon-12. As the average relative mass of a carbon atom is actually greater than 12 owing to the presence of heavier isotopes (see Chapter 2), one mole of the element carbon has a mass of 12.01 g. The mass of one mole of atoms of an element is simply the relative atomic mass (see page 41 in Chapter 2) expressed in grams. One mole of hydrogen atoms has a mass of 1.01 g, one mole of helium 4.00 g and so on. e pl m Some elements exist as molecules not as individual atoms. The composition of a molecule is given by its molecular formula. Hydrogen gas, for example, is made from diatomic molecules and so has the molecular formula H2.Water molecules are made from two hydrogen atoms and one oxygen atom and have the molecular formula H2O. The relative molecular mass (Mr ) is calculated by adding the relative atomic masses of the atoms making up the molecule. Sa • The mass of one mole of a species is called the molar mass. It is the relative mass expressed in g and has units of g mol1. • The molar mass of an element which exists as atoms is the relative atomic mass expressed in g. • The relative molecular mass (Mr) is defined as the sum of the relative atomic masses of the atoms in the molecular formula. The molar mass (M) of a compound is the relative molecular mass expressed in g. pa g es One mole of carbon-12. Worked example Calculate the relative molecular mass of ethanol C2H5OH. Solution The compound is made from three elements, carbon, hydrogen and oxygen. Find the number of atoms of each element and their relative atomic masses from the Periodic Table. relative atomic mass C H O 12.01 1.01 16.00 2 516 1 number of atoms in one molecule of compound Calculate the relative molecular mass of the molecule. Relative molecular mass (2 12.01) (6 1.01) 16.00 46.08 4 UNCORRECTED PROOF COPY Downloaded from www.heinemann.co.uk/ib The molar mass of ethanol is 46.08 g mol1. The molar mass of a compound is calculated in the same way as that of the elements. It is the relative molecular mass in grams. It is incorrect to use the term relative molecular mass for ionic compounds as they are made from ions not molecules. The term relative formula mass is used. It is calculated in the same way. Once the molar mass is calculated and the mass is measured, the number of moles can be determined. Worked example Calculate the amount (in mol) in 4.00 g of sodium hydroxide, NaOH. es pa g M 40.00 g mol1 4.00 0.100 mol m _____ n __ M 40.00 m pl e It is important to be precise when calculating amounts. One mole of hydrogen atoms has a molar mass of 1.01 g mol1 but one mole of hydrogen molecules, H2, has a mass of 2 1.01 2.02 g mol1. Counting particles Examiner’s hint: Pay attention to decimal places. When adding or subtracting, the number of decimal figures in the result should be the same as the least precise value given in the data. Number of moles (n) mass (m) ______________ molar mass (M) Solution The relative atomic masses are Na: 22.99, O: 16.00 and H: 1.01. The relative formula mass 22.99 16.00 1.01 40.00 The relative formula mass of an ionic compound is the sum of the relative atomic masses of the atoms in the formula. The molar mass of an ionic compound is the relative formula mass expressed in g. Sa As the mass of an individual atom can be measured using a mass spectrometer (see Chapter 2), the mole is a counting unit. Examiner’s hint: Pay attention to significant figures. When multiplying or dividing, the number of significant figures in the result should be the same as the least precise value in the data. Examiner’s hint: Use the accepted shorthand to solve problems more quickly in exams, for example n for moles, m for mass and M for molar mass. Avogadro’s constant (L) has the value 6.02 1023 mol1. It has units as it is the number of particles per mole. From mass spectrometer measurements: mass of 1 atom of 12C 1.99252 1023 g 12 Number of atoms in one mole (12 g of 12C) ______________ 1.99252 1023 602 000 000 000 000 000 000 000. This is a big number. It is called Avogadro’s number (L) and is more compactly written in scientific notation as 6.02 1023. It is the number of atoms in one mole of an element and the number of molecules in one mole of a covalent compound. Examiner’s hint: Although Avogadro’s constant is given in the IB Data booklet you need to know its value for Paper 1. • If 6.02 1023 pennies were distributed to everyone currently alive they could all spend two million pounds every hour, day and night of their lives. • 6.02 1023 grains of sand would cover a city the size of Los Angeles to a height of 600 m. • 6.02 1023 soft drink cans would cover the surface of the earth to a height of over 300 km. This is the number of carbon atoms in a tablespoon of soot! UNCORRECTED PROOF COPY 5 1 Quantitative chemistry Downloaded from www.heinemann.co.uk/ib Imagine you emptied a glass of labelled water molecules into the sea and then allowed sufficient time for the molecules to disperse throughout the oceans. What is the probability that you could catch one of the original molecules if you placed the same glass into the sea? (see the Answer section). The magnitude of Avogadro’s constant is beyond the scale of our everyday experience. This is one reason why ‘moles’ is a challenging subject and why natural sciences don’t always come naturally! Number of particles (N) number of moles (n) Avogadro’s constant (L) N nL Although we could never count to L, even with the most powerful computer, we can prepare samples with this number of atoms. The atoms are counted in the same way as coins are counted in a bank; we use a balance: 3.01 1023 atoms of C _12 mol 0.5 12.01 g 6.005 g. 3.01 1023 carbon atoms are ‘counted out’ when we prepare a sample of 6.005 g. Worked example 2.99 mol e pa g Solution Use the shorthand notation: N nL N n __ L 1.80 1024 __________ 6.02 1023 es Calculate the amount of water, H2O, that contains 1.80 1024 molecules. Sa m pl Note the answer should be given to 3 significant figures – the same precision as the data given in the question. If the amount given was 1.8 1024 the correct answer would be 3.0. Worked example Calculate how many hydrogen atoms are present in 3.0 moles of ethanol, C2H5OH. Solution In 1 molecule of ethanol there are 6 H atoms. In 1 mole of ethanol molecules there are 6 moles of H atoms. In 3 moles of ethanol there are 18 moles of H atoms. Number of H atoms 18L 6.01 1023 18 1.08 1025 Exercises 1 Calculate how many hydrogen atoms are present in 0.040 moles of C2H6. 2 Calculate the molar mass of magnesium nitrate, Mg(NO3)2. 3 Calculate how many hydrogen atoms are contained in 2.3 g of C2H5OH (Mr 46). 4 The relative molecular mass of a compound is 98.0. Calculate the number of molecules in a 4.90 g sample of the substance. 6 UNCORRECTED PROOF COPY Downloaded from www.heinemann.co.uk/ib 1.2 Formulas Finding chemical formulas in the laboratory When magnesium is burned in air its mass increases as it is combining with oxygen. The mass changes can be investigated experimentally. crucible magnesium ribbon pipe clay triangle heat tripod es Figure 1.2 This apparatus reduces the chance of the product escaping when magnesium is heated. Mass/g (0.001) pa g Item empty crucible 25.000 crucible with magnesium before heating 25.050 crucible with solid after heating Magnesium burns in air to produce a white residue of magnesium oxide. Examiner’s hint: The uncertainties in all the measurements should be included in all data tables. This is discussed in Chapter 11. 25.084 Mass/g (0.002) Moles pl Element e The masses of the magnesium and oxygen are then calculated. 25.050 25.000 0.050 oxygen 25.084 25.050 0.034 m magnesium 0.050 0.0021 ______ 24.34 0.034 0.0021 ______ 16.00 The precision of the calculated value is limited by the precision of the mass measurements to 2 significant figures. The empirical formula gives the ratio of the atoms of different elements in a compound. It is the molecular formula expressed as its simplest ratio. Sa The ratio of magnesium : oxygen atoms 0.0021:0.0021 1:1. This is expressed as an empirical formula: MgO. Worked example A 2.765 g sample of a lead oxide was heated in a stream of hydrogen gas and completely converted to elemental lead with a mass of 2.401 g. What is the empirical formula of the oxide? Solution The mass loss is caused by a loss of oxygen. Set out the calculation in a table. Pb O mass/g 2.401 2.765 2.401 0.364 moles 2.401 0.01159 _______ 207.19 0.01159 1 ________ 0.01159 0.364 0.0228 ______ 16.00 0.0228 1.97 2 ________ 0.01159 simplest ratio Empirical formula: PbO2 UNCORRECTED PROOF COPY 7 1 Quantitative chemistry Downloaded from www.heinemann.co.uk/ib Worked example A hydrocarbon contains 85.7% by mass of carbon. Deduce the empirical formula. Solution A hydrocarbon is a compound of carbon and hydrogen only (see Chapter 10). When data is given in percentages consider a 100 g sample. C Carry out your own carbon hydrogen analysis. Now go to www.heinemann.co.uk/ hotlinks, insert the express code 4259P and click on this activity. H mass/g 85.7 100 85.7 14.3 moles 85.7 7.14 ______ 12.01 14.3 14.16 _____ 1.01 simplest ratio 7.14 1 _____ 7.14 14.16 1.98 2 ______ 7.14 es Empirical formula: CH2 Exercises pa g 5 An oxide of sulfur contains 60% by mass of oxygen. Deduce the empirical formula. 6 Pure nickel was discovered in1751. It was named from the German word ‘kupfernickel’, meaning ‘devil’s copper’. A compound of nickel was analysed and shown to have the following composition by mass: Ni 37.9%, S 20.7 %, O 41.4 %. Deduce the empirical formula. e The molecular formula shows the actual number of atoms of each element present in a molecule. pl Molecular formula m The empirical formula does not give the actual number of atoms in the molecule. The hydrocarbon in the previous worked example had an empirical formula of CH2 but no stable molecule with this formula exists. The molecular formula, which is a multiple of the empirical formula, can only be determined once the relative molecular mass is known. This can either be measured by a mass spectrometer (Chapter 2.2, page 40) or calculated from the ideal gas equation (Chapter 1.4, page 24). Sa State symbols indicate the state of a substance: (s) is for solid, (l) is for liquid, (g) is for gas and (aq) is for aqueous – dissolved in water. The molecular formulas of some compounds. The state symbols identify the state at room temperature and atmospheric pressure. Substance Formula Substance hydrogen H2(g) carbon dioxide CO2(g) oxygen O2(g) ammonia NH3(g) nitrogen N2(g) methane CH4(g) water H2O(l) glucose C6H12O6(s) Worked example What is the empirical formula of glucose? Solution From the table above, the molecular formula C6H12O6 Express this as the simplest ratio: CH2O. 8 Formula UNCORRECTED PROOF COPY Downloaded from www.heinemann.co.uk/ib Worked example The compound with the empirical formula of CH2 is analysed by a mass spectrometer and its relative molecular mass found to be 42.09. Deduce its molecular formula. Solution Empirical formula CH2 Examiner’s hint: Practise empirical formula calculations. All steps in the calculation must be shown. ‘Keep going’ as errors are carried forward so that a correct method in a later part of the question is rewarded even if you have made earlier mistakes. One common problem is the use of too few significant figures in intermediate answers. Molecular formula CnH2n (where n is an integer) Mr 42.09 (12.01n) (2n 1.01) 14.03n 42.09 3 n _____ 14.03 Molecular formula: C3H6 Exercises I only II only Both I and II Neither I nor II pa g A B C D es 7 Which formula can be determined by only using the percent mass composition data of an unknown compound? I Molecular formula II Empirical (simplest) formula 8 CFCs are compounds of carbon, chlorine and fluorine which catalyse the depletion of the ozone layer. The composition of one CFC is shown below. chlorine 17.8% 1.5% 52.6% Sa States of matter 28.1% m The value of its Mr is 135. Determine the molecular formula of the CFC. fluorine e hydrogen pl carbon If you were hit with 180 g of solid water (ice) you could be seriously injured, but you would be only annoyed if it was 180 g of liquid water. 180 g of gaseous water (steam) could also be harmful. These three samples are all made from same particles – 10 moles of water molecules. The difference in physical properties is explained by kinetic theory. The basic ideas are: • All matter consists of particles (atoms or molecules) in motion. • As the temperature increases, the movement of the particles increases. The three states can be characterized in terms of the arrangement and movement of the particles and the forces between them. Most substances can exist in all three states. The state at a given temperature and pressure is determined by the strength of the interparticle forces. Solid: the particles are closely packed in fixed positions. The interparticle forces restrict the movement to vibration about a fixed position. Solids have a fixed shape. Figure 1.3 Comparison of the three states of matter. Liquid: the particles are still relatively close together. The interparticle forces are sufficiently weak to allow the particles to change places with each other, but their movement is constrained to a fixed volume. Liquids can change shape but not volume. UNCORRECTED PROOF COPY Gas: the interparticle forces between the particles are negligible; they are zero for an ideal gas (see Section 1.4). The particles move freely occupying all the space available to them. Gases have no fixed shape or volume. 9 1 Quantitative chemistry Downloaded from www.heinemann.co.uk/ib Changes of state The movement or kinetic energy of the particles depends on the temperature. When the temperature increases enough for the particles to have sufficient energy to overcome the interparticle forces, a change of state occurs. The heating curve below shows how the temperature changes as ice is heated from 40°C to steam at 140°C. Figure 1.4 Heating curve for water. The phase change (l) →(g) needs more energy than (s)→(l) as all the interparticle bonds are broken during this process. 60 40 solid 20 and liquid 0 �20 �40 solid liquid melting heat To understand these changes it is more helpful to use the absolute or Kelvin scale of temperature. Consider a sample of ice at 40°C 233 K. The water molecules vibrate at this temperature about their fixed positions. • As the ice is heated the vibrational energy of its particles increases and so the temperature increases. • At the melting point of 273 K the vibrations are sufficiently energetic for the molecules to move away from their fixed positions and liquid water starts to form. The added energy is needed to break the bonds between the molecules – the intermolecular bonds. There is no increase in kinetic energy so there is no increase in temperature. • As the water is heated, the particles move faster and so the temperature increases. • Some molecules will have sufficient energy to break away from the surface of the liquid so some water evaporates. • At the boiling point of water there is sufficient energy to break all the intermolecular bonds. The added energy is used for this process, not to increase the kinetic energy, and so the temperature remains constant. • As steam is heated the average kinetic energy of the molecules increases and so the temperature increases. pl Ice cubes melting over a period of 4 hours. The water is absorbing heat from the surroundings to break some of the intermolecular bonds. boiling es temperature/°C melting point liquid and gas 80 Sa The absolute temperature of a substance is proportional to the average kinetic energy of its particles. 10 100 m Temperature differences measured on either the Celsius or Kelvin scale are the same. Absolute zero 0 K. This is the temperature of minimum kinetic energy. boiling point e The kelvin is the SI unit of temperature. Temperature in kelvin temperature in °C 273 gas 120 pa g The Celsius scale of temperature is defined relative to the boiling and freezing point of water. The original scale, developed by the Swedish astronomer Anders Celsius, made the boiling point of water zero and the freezing point 100. This may now seem absurd but the modern scale is just as arbitrary. 140 UNCORRECTED PROOF COPY Downloaded from www.heinemann.co.uk/ib Worked example In which sample do molecules have the greatest average kinetic energy? A He at 100 K B H2 at 200 K C O2 at 300 K D H2O at 400 K Solution Answer D The sample at the highest temperature has the greatest kinetic energy. The kinetic energy of a particle depends on its mass (m) and speed (v). All gases have the same kinetic energy at the same temperature, so particles with smaller mass move at faster speeds. Kinetic energy _12 mv2. Some substances change directly from a solid to gas at atmospheric pressure. This change is called sublimation. Figure 1.5 The different phase changes. melting freezing es liquid Sublimation is the conversion of a solid directly to a vapour state. Dry ice, CO2(s), sublimes when it is mixed with water producing thick clouds of fog. Sa m pl e solid pa g lim sub g/ ilin ng bo ati r po eva ng nsi de con ati on rev sub er lim se ati on gas The coldest place in nature is in the depths of outer space at a temperature of 3 K. The 2001 Nobel Prize in Physics was awarded to a team who cooled a sample of helium atoms down to only a few billionths (0.000 000 001) of a degree above absolute zero. Under these conditions helium atoms crawl along at a speed of only about 3 mm s1! Exercise 9 When a small quantity of perfume is released into the air, it can be detected several metres away in a short time. Use the kinetic theory to explain why this happens. UNCORRECTED PROOF COPY 11 1 Quantitative chemistry Downloaded from www.heinemann.co.uk/ib We can smell perfumes because they evaporate at body temperature. In a mixture of molecules the most volatile evaporate first and the least volatile last. The skill of the perfumer is to use the laws of chemistry to make sure that chemicals are released steadily in the same proportions. m pl e pa g es The Celsius scale gives an artificial description of temperature and the Kelvin scale a natural description. Do the units we use help or hinder our understanding of the natural world? Sa Worked example A flask contains water and steam at boiling point. Distinguish between the two states on a molecular level by referring to the average speed of the molecules and the relative intermolecular distances. Solution As the two phases are at the same temperature they have the same average kinetic energy and are moving at the same speed. The separation between the particles in a gas is significantly larger than that in a liquid. Exercise 10 Which of the following occur when a solid sublimes? I The molecules increase in size. II The distances between the molecules increase. A B C D 12 I only II only Both I and II Neither I nor II UNCORRECTED PROOF COPY
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