@ Oxford Fajar Sdn. Bhd. ( T) Matter. 1.1 Atoms and Molecules 1.2 Mole Concept 1.3 Stoichiometry

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1 1 Matter 1.1 Atoms and Molecules 1.2 Mole Concept 1.3 Stoichiometry

2 2 Chemistry for Matriculation Semester Atoms and Molecules LEARNING OUTCOMES Describe proton, electron and neutron in terms of the relative mass and relative charge Define proton number, Z, nucleon number, A and isotope Write isotope notation Define relative atomic mass, A r and relative molecular mass, M r based on the C-12 scale Calculate the average atomic mass of an element given the relative abundances of isotopes or a mass spectrum Subatomic Particles of an Atom 1 An atom consists of a positively charged nucleus surrounded by a negatively charged electron cloud. 2 The particles in the nucleus of an atom are called nucleons. Nucleons consist of positively charged protons and neutrons which are neutral. 3 Protons, neutrons and electrons are called fundamental or subatomic particles. The properties of subatomic particles are given in Table 1.1. Table 1.1 Properties (mass and charge) of subatomic particles Particle Symbol Charge Relative mass (a.m.u.) Approximate relative mass (a.m.u.) Proton p Neutron n Electron e or 1/ [1 a.m.u. (atomic mass unit) = kg] 4 The radius of an atomic nucleus is approximately m. However, a large volume of the atom is made up of empty space which is occupied by the electron cloud. Almost all (99.98%) the mass of the atom is concentrated in the nucleus. An atom is the smallest particle in an element that can participate in a chemical reaction. Proton Number, Nucleon Number and Isotopes Proton number 1 The proton number is the number of protons in the nucleus of an atom. 2 The symbol for proton number is Z. 3 Each element has a different proton number.

3 4 Atoms are neutral particles because the number of protons equals the number of electrons. 5 In the modern periodic table (Appendix 1), the elements are arranged in ascending order of proton number. Matter 3 Nucleon number 1 The nucleon number (A) is the total number of protons and neutrons in the nucleus. Number of neutrons = Nucleon number Proton number = A Z 2 The structure of an atom can be written in a number of ways. For example, the carbon atom, which contains six protons and six neutrons can be written in the following ways. (a) carbon-12: the number 12 denotes the nucleon number (b) 12 C or 12 C: the nucleon number 12 is indicated as a superscript 6 and the proton number 6 indicated as a subscript on the left of the carbon atom symbol (C) 3 In general, the structure of an atom, X, can be written in symbol form as a b X. nucleon number proton number a X b symbol of the element 4 In the case of an anion or a cation, the charge of the ion is shown as a superscript on the right of the symbol for the ion. For example, the symbol Cl denotes a chloride ion which consists of the atom chlorine-35 with a negative charge of 1. Similarly, 16 O22 denotes a 8 peroxide ion which contains two bonded oxygen atoms with a negative charge of 2. Proton number is also known as atomic number. Nucleon number is also known as mass number. t EXAMPLE 1.1 Plutonium is a radioactive substance. The plutonium atom has 94 protons and 150 neutrons. Write the symbol for the plutonium atom in three different ways. Proton number = 94 Nucleon number = = 244 The plutonium atom can be written as plutonium-244, 244 Pu and 244Pu. 94 Isotopes 1 Isotopes are atoms of the same element having the same number of protons but different number of neutrons. For example, the isotopes of hydrogen are called hydrogen, deuterium and tritium. 1 1H hydrogen deuterium 2 H 3 H 1 1 tritium Deuterium can also be written as 2 D 1.

4 4 Chemistry for Matriculation Semester 1 Isotope Fluorine-19 Aluminium-27 Chromium-50 Chromium-52 Chromium-53 Chromium-54 Table 1.2 Atomic structures and isotopic abundances of some elements Relative abundance Atomic structure (%) Number of protons Number of neutrons Number of electrons Isotopes have the same chemical properties because they have the same electronic configuration. 3 Isotopes have different physical properties such as melting point, boiling point, density and rate of diffusion. Isotopic abundance 1 A few elements, such as fluorine and aluminium, have been found to consist of only one isotope. 2 However, most elements exist as mixtures of two or more naturally occurring isotopes. For example, tin (Sn) has 10 isotopes. 3 The abundance of each isotope in the mixture is called its isotopic abundance. 4 Isotopic abundances can be expressed in terms of relative or percentage abundances (Table 1.2). Isotopic abundance (%) Number of atoms for a given isotope = Total number of atoms for all isotopes of the element The isotopic abundance can also be expressed as fractional abundance or isotopic ratio. For example: (a) The fractional abundances of the isotopes of chromium (Table 1.2) are 0.04, 0.84, 0.1 and (b) Chlorine is a mixture of 75.5% chlorine-35 and 24.5% chlorine-37. The isotopic ratio of 35 Cl : 37 Cl is 75.5 : 24.5, usually expressed as 3 : 1. Cations and anions 1 Atoms and molecules do not carry any charge whereas ions are charged particles. 2 An ion is an atom or a group of atoms that has lost one or more electrons, making it positively charged, or gained one or more electrons, making it negatively charged. 3 Positive ions are called cations. Negative ions are called anions. 4 Tables 1.3 and 1.4 show the number of protons, neutrons and electrons in some atoms and their ions.

5 Matter 5 Table 1.3 Numbers of protons, electrons and neutrons in atoms and their positive ions Atom or positive ion protons Number of electrons neutrons Na Na Mg Mg Al Al Remarks Loss of 1e Loss of 2e Loss of 3e Table 1.4 Numbers of protons, electrons and neutrons in atoms and their negative ions Atom or Number of Remarks negative ion protons electrons neutrons Cl Gain of 1e Cl O Gain of 2e O Polyatomic ions are ions which contain two or more covalently bonded atoms. 6 Examples of polyatomic ions are the hydroxide ion (OH ), the nitrate ion (NO 3 ), the carbonate ion (CO 3 2 ) and the ammonium ion (NH 4 + ). EXAMPLE 1.2 State the numbers of protons, neutrons and electrons present in the following sulphur species. 32 S 33 S S2 Species Number of protons Number of neutrons Number of electrons 32S S S QUICK CHECK A beam of particles containing protons, electrons and neutrons is passed between charged plates. The diagram shows the effect of the charged plates on these particles. (a) Which beam (X, Y or Z) contains (i) protons, (ii) electrons, and (iii) neutrons? Explain your answers. (b) Explain why beam X is deflected more than beam Z. beam of p, e, n positive plate + negative plate X Z Y

6 6 Chemistry for Matriculation Semester 1 2. With reference to the periodic table (Appendix 1) at the end of this book, state the numbers of subatomic particles in the following atoms. silicon-28; chlorine-37; bromine Determine the numbers of protons, neutrons and electrons in the following species O2 ; Al3+ ; 7 3 Li+ 4. With the help of the periodic table, identify the species with the compositions shown on the right. Species A B C D Number of protons neutrons electrons Relative Atomic Mass and Relative Molecular Mass Relative atomic mass 1 In 1961, the International Union of Pure and Applied Chemistry (IUPAC) adopted a scale based on the isotope carbon-12 ( 12 C). On 6 this scale (the carbon-12 scale), one carbon-12 atom is assigned a mass of exactly atomic mass units (a.m.u.). 2 Thus, one atomic mass unit is defined as 1 the mass of one atom 12 of the carbon-12 isotope. The mass of one carbon-12 atom is g. Therefore, 1 a.m.u. = g = g = kg 3 The relative atomic masses of all elements are found by comparing the mass of one atom of the element with the mass of a carbon-12 atom. 4 The relative atomic mass (A r ) of an element is defined as the ratio of the mass of one atom of the element to 1 12 of the mass of a carbon-12 atom. Mass of one atom of the element Relative atomic mass = 1 of the mass of one atom of carbon = Mass of one atom of the element 12 Mass of one atom of carbon-12 5 If the mass of one atom of an element X is one third of the mass of one carbon-12 atom, then the relative atomic mass of the element is 1 12 = 4 a.m.u. Similarly, if the mass of one atom of another element 3 is 2 times the mass of a carbon-12 atom, then its relative atomic mass is 2 12 = 24 a.m.u. 6 The mass of a proton ( a.m.u.) is almost the same as the mass of a neutron ( a.m.u.) while the mass of an electron is very small.

7 Hence, the relative atomic mass of an element can be considered to be the same as its nucleon number. For example, the relative atomic mass of fluorine-19 = The relative isotopic mass is the ratio of the mass of one atom of an isotope to 1 12 of the mass of one atom of carbon-12. Matter 7 Mass of one atom of the isotope Relative isotopic mass = 1 of the mass of one carbon-12 atom 12 Mass of one atom of the isotope 12 = Mass of one carbon-12 atom Determining relative atomic mass from isotopic abundance 1 For elements that do not have isotopes (for example, fluorine-19), the relative atomic mass is the same as the relative isotopic mass. 2 However, for elements that have isotopes, the relative atomic mass is calculated by multiplying the relative isotopic mass of each isotope by its relative abundance and adding all these values together. For example, for an element with two isotopes: mx + ny Relative atomic mass = 100 where m, n = relative isotopic mass of each isotope x, y = relative abundance of isotopes m and n respectively EXAMPLE 1.3 Sulphur is a non-metallic element found in many minerals. Calculate the relative atomic mass of naturally occurring sulphur from the following data: Isotope Relative abundance (%) Sulphur Sulphur Sulphur Relative atomic ( ) + ( ) + ( ) = mass of sulphur 100 = 32.1 EXAMPLE 1.4 An element Y has two isotopes, P and Q. The ratio of the abundances of the isotopes relative to each other is: P Q = 0.32

8 8 Chemistry for Matriculation Semester 1 What is the relative atomic mass of Y? (Based on 12 C, the relative isotopic mass of P is and the relative isotopic mass of Q is ) P = 0.32 = 0.32 Q % of P = 100 = % of Q = = 75.8 Relative atomic mass of Y = = = Many compounds do not exist as molecules. When considering ionic compounds such as sodium chloride, Na + Cl, the term relative formula mass is used in place of relative molecular mass. Relative molecular mass 1 The relative molecular mass of an element or a compound is the ratio of the mass of one molecule of the substance to 1 of the mass 12 of an atom of carbon-12. Relative molecular mass = Mass of one molecule of a substance 1 of the mass of one atom 12 of carbon-12 2 Relative molecular mass, M r, has no units and is calculated by adding up the relative atomic masses of all the atoms present in one molecule of the substance. For example, the relative molecular mass (M r ) of ethanoic acid, CH 3 COOH, is ( ) = 60. QUICK CHECK (a) By using chlorine as an example, explain the meaning of isotope. (b) Define relative atomic mass. (c) What is the relative atomic mass of an element X if the atomic mass ratio of X to 12 C is 2.25? 2. The relative atomic mass of a sample of bromine which contains two isotopes, Br and Br, is Calculate the relative abundance of 79 Br in the sample of bromine Naturally occurring copper is a mixture of 69.09% Cu isotope and 30.9% Cu isotope. If the masses of the isotopes Cu and Cu are and respectively, what is the relative atomic mass of copper? 4. The relative molecular mass of a hydrated salt, (NH 4 ) 2 X(SO 4 ) 2.6H 2 O is 392. X is a metallic element. What is (a) the relative atomic mass of X, (b) the percentage of water of crystallisation in the hydrated salt? 5. The three stable isotopes of magnesium are magnesium-24, magnesium-25 and magnesium-26. The relative atomic mass of magnesium is Write down the proton number and the number of neutrons for the most abundant magnesium isotope.

9 Mass spectrometer 1 Mass spectrometry is a modern technique that uses a mass spectrometer to determine (a) the relative isotopic mass of an atom, (b) the relative abundance of an isotope in a sample of the element, (c) the relative atomic mass of an element, (d) the relative molecular mass of a compound. 2 A mass spectrometer (Figure 1.1) is an instrument that (a) separates positive ions by their mass-to-charge (m/e) ratios, (b) shows their masses and abundance. 3 There are many types of mass spectrometers, but all mass spectrometers have five basic parts. 2 ionisation chamber 3 acceleration chamber P1 P2 positive ions 4 magnetic field 1 2 The sample is vaporised Positive ions are produced 3 The positive ions are accelerated Matter 9 Photo 1.1 The mass spectrometer is a very expensive instrument. It is found only in some modern chemistry laboratories Figure 1.1 The schematic representation of a mass spectrometer 1 vaporisation chamber electron gun 5 ion detector recorder 4 The positive ions are deflected 5 The positive ions are detected 1 1 Vaporisation chamber The sample to be analysed is first vaporised by heating (unless the sample is already a gas). 2 2 Ionisation chamber The sample in the gaseous state is ionised by electron bombardment. The heated filament releases high-energy electrons which knock out electrons from the molecules or atoms in the sample to form positive ions. X(g) + e X + (g) + 2e sample high-energy electron 3 Acceleration chamber The positive ions are then accelerated by applying a high negative potential to the plates P1 and P2. The slits in P2 produce a fine beam of ions moving at high speeds. 4 Magnetic field The accelerated ions then enter a magnetic field where they are deflected into the arc of a circle according to their mass charge ratio (m/e). Positive ions with lower mass-to-charge ratio will be deflected more than those with higher mass-to-charge ratio. Thus, 79 Br + is deflected more than 81 Br + because it has a smaller mass-to-charge ratio (Figure 1.2).

10 10 Chemistry for Matriculation Semester 1 Figure 1.2 Deflection of positive ions in a mass spectrometer 79 Br + and 81 Br + magnetic field 79 Br + 81 Br + Mass spectrometers can be used to confirm the presence of forbidden drugs in urine. Figure 1.3 The mass spectrum of naturally occurring boron (a) Ion detector By varying the strength of the magnetic field, ions with different masses can be deflected to the ion detector. This produces a flow of current which is amplified and recorded as a peak on a chart called the mass spectrum. The relative heights of the peaks give the relative proportions of the ions present. 4 The mass spectrum of naturally occurring boron is shown in Figure 1.3(a). This mass spectrum can be changed into a bar chart (Figure 1.3(b)). The mass spectrum of boron contains two peaks at m/e 10 and m/e 11. This shows that (a) naturally occurring boron is made up of two isotopes, 10 B and 11 B, (b) boron-11 is four times more abundant than boron-10, that is, the relative abundances of boron-10 and boron-11 are 20% and 80% respectively. (b) 100 relative abundance relative abundance m/e m/e Determining Relative Atomic Mass from Mass Spectra 1 If an element has n isotopes with m/e ratios of m 1, m 2,..., m n, then: Average atomic mass = (m Q ) i i = m Q + m Q m n Q n Q i Q 1 + Q Q n where m i = m/e ratio Q i = relative height of peak or relative abundance Examples: (a) An element A has three isotopes X, Y and Z with m/e ratios of a, b and c, and relative heights of h 1, h 2 and h 3. Average atomic mass of A = (m Q ) i i = ah + bh + ch Q i h 1 + h 2 + h 3 (b) An element A has three isotopes X, Y and Z with m/e ratios of a, b and c, and relative abundances of x%, y% and z% respectively. Average atomic mass of A = (m Q ) i i ax + by + cz ax + by + cz = = Q i x + y + z Take note that the unit for average atomic mass is u (atomic mass unit). In contrast, the relative atomic mass (A r ) has no unit (see Example 1.5).

11 Matter 11 EXAMPLE 1.5 Calculate the relative atomic mass of silicon from the following data. Isotope Relative abundance (%) Silicon Silicon Silicon Average atomic mass of silicon = (m Q ) i i Q i ( ) + ( ) + ( ) = = 28.1 u Average atomic mass of Si Relative atomic mass (A r ) of Si = 1/12 mass of one C-12 atom = 28.1 EXAMPLE 1.6 The mass spectrum of an element X is given in the figure below. What information can be deduced with regard to the element X? relative abundance m/e There are four peaks in the mass spectrum at m/e values of 64, 66, 67 and 68. Therefore, the naturally occurring element, X, has four isotopes with relative isotopic masses of 64, 66, 67 and 68. The most abundant isotope is X-64. Applying formula 1.1, (64 100) + (66 56) + (67 8) + (68 38) Average atomic mass of X = ( ) = Average atomic mass of X A r of X = 1/12 mass of one C-12 atom = With reference to the periodic table (Appendix 1), the element X is zinc.

12 12 Chemistry for Matriculation Semester 1 EXAMPLE 1.7 Figure (a) shows the passage of 16 O + and 17 O + through a magnetic field and Figure (b) shows the passage of 18 O + and 18 O 2+ through a magnetic field. 16 O + and magnetic field 18 O + and magnetic field 17 O + 18 O 2+ In the mass spectrometer, the heavier the particle, the smaller the deflection. The higher the charge of the particle, the bigger the deflection. A (a) Identify the particles A, B, C and D. STRATEGY B In Figure (a), 16 O + and 17 O + have the same charge but different mass. 16 O + is the lighter particle. It will be deflected more than 17 O +. Hence, line A is caused by 16 O +. In Figure (b), 18 O + and 18 O 2+ have the same mass but different charges. 18 O 2+ has the higher charge, that is, lower mass-to-charge ratio. Hence, 18 O 2+ is deflected more than 18 O +. (b) C D Particle A is 16 O + and particle B is 17 O +. Particle C is 18 O 2+ and particle D is 18 O +. QUICK CHECK (a) Name the components in a mass spectrometer where (i) positive ions are produced, (ii) positive ions are deflected. (b) Naturally occurring bromine consists of two isotopes, 79 Br and 81 Br. The following positive ions are produced in the ionisation chamber: 79 Br +, 81 Br +, ( 79 Br + 79 Br) +, ( 79 Br + 81 Br) + and ( 81 Br + 81 Br) + Which ion will be deflected the most and which ion the least? Explain your answer. 2. A sample of oxygen which contains isotopes with nucleon numbers 16 and 18 is analysed in a mass spectrometer. How many peaks are recorded in the mass spectrum obtained? Explain your answer. 3. (a) A sample of chlorine is analysed in a mass spectrometer. The mass spectrum contains five peaks at m/e 35, 37, 70, 72 and 74. Identify the ions that are responsible for these peaks. (b) The mass spectrum of neon contains three lines which correspond to the mass 100 ratios of 20, 21 charge and 22 with relative intensities of 0.91 : : Explain the meaning of these data and hence, calculate the relative atomic mass of neon. 4. The mass spectrum of an element is shown on the right. With the help of the periodic table, identify the element. 5. Boron consists of two isotopes, 10B and 11 B. The 5 5 relative atomic mass of boron is Calculate the relative abundances of 10B and 11B. 5 5 percentage abundance (%) m/e

13 Matter Mole Concept LEARNING OUTCOMES Define mole in terms of mass of carbon-12 Define mole in terms of Avogadro constant, N A Interconvert between moles, mass, number of particles, molar volume of gas at s.t.p. and room temperature Define empirical formula and molecular formula Determine empirical and molecular formulae from mass composition or combustion data Define and perform calculations for concentration measurements 1 In the field of science, the mass of a substance is usually measured in units of grams (g) or kilograms (kg). 2 Chemists use another SI unit to state the amount of a substance. This unit is called mole (mol). Mole Concept and the Avogadro Constant 1 One mole (1 mol) is defined as the quantity of a substance that contains the same number of particles (atoms, electrons, ions or molecules) as there are atoms in exactly 12 g of carbon The mass of a single carbon-12 atom has been found to be g using a mass spectrometer. The number of atoms in exactly one mole of 12 C is This can be shown by the following calculation. Atomic substances 1 mole of 12 C contains carbon atoms 1 mole of sodium contains sodium atoms (Na) Molecules 1 mole of chlorine molecules contains chlorine molecules (Cl 2 ) or chlorine atoms (Cl) Number of particles Ions 1 mole of calcium bromide (CaBr 2 ) contains calcium ions (Ca 2+ ) and bromide ions (Br ) Electrons 1 mole of electrons contains electrons Figure 1.4 Number of particles in one mole of a substance

14 14 Chemistry for Matriculation Semester 1 Mass per mole of 12 C N A = Mass of one atom of 12 C 12 g mol 1 = g = mol 1 3 Thus one mole of any substance contains particles (Figure 1.4). This number is called the Avogadro constant (N A ), and has the unit mol 1. 4 The relative atomic mass for all elements, stated in grams, contains the same number of atoms. 12 g of the element carbon, 1 g of the element hydrogen, and 32 g of the element sulphur contains atoms, even though the element carbon exists as atoms, and hydrogen (H 2 ) and sulphur (S 8 ) exist as molecules. Avogadro constant number of moles, n number of particles Avogadro constant Calculating the number of atoms and molecules 1 The mole concept enables chemists to count the number of atoms by weighing. For example, 12 g of carbon contains atoms. Therefore 1.0 g of carbon contains atoms and g of carbon contains atoms The relationship between the number of moles of a substance and the number of atoms or molecules in the substance is given by the formula: Number of moles of substance = Number of atoms or molecules Avogadro constant EXAMPLE 1.8 (a) How many atoms are found in 0.6 mole of copper? (b) How many moles of chlorine molecules are found in molecules of chlorine gas? (a) Number of copper atoms = Number of moles Avogadro constant = 0.6 mol ( mol 1 ) = atoms Number of molecules (b) Number of moles of chlorine molecules = Avogadro constant = mol 1 = mol

15 Mole Concept and Mass 1 The mass (in grams) of one mole of a substance is called molar mass (M). The unit for molar mass is g mol 1. For example: (a) The relative atomic mass of magnesium is So the molar mass of magnesium is 24.3 g mol 1. (b) The relative molecular mass of chlorine (Cl 2 ) is = 71.0 (A r of Cl = 35.5). Thus, the molar mass of chlorine molecules is 71.0 g mol 1. (c) The relative molecular mass of hydrogen chloride (HCl) is = The molar mass of hydrogen chloride is 36.5 g mol 1. 2 It is important to note that the type of particle (atom, molecule or ion) needs to be carefully specified. For example, a statement such as 1.0 mole of oxygen is ambiguous, because it can mean 1.0 mole of oxygen atoms (16.0 g) or 1.0 mole of oxygen molecules (32.0 g). 3 The relationship between the mass (m) and the number of moles (n) of a substance is given by the formula: Mass (m) = Number of moles (n) M where M = A r (relative atomic mass) or M r (relative molecular mass) Matter 15 Relative atomic mass and relative molecular mass have no units. In contrast, molar mass has the unit g mol 1. EXAMPLE 1.9 Calculate (a) the number of atoms in 192 g of ozone gas, (b) the mass of one aluminium atom, (c) the mass of sulphur atoms. (Relative atomic mass: O, 16.0; Al, 27.0; S, 32.1) 192 (a) Number of moles of ozone gas (O 3 ) = = 4.0 mol One molecule of ozone gas contains three oxygen atoms. Number of oxygen atoms = 3 ( ) = (b) Mass of aluminium atoms = 27.0 g 27.0 Mass of one aluminium atom = = g (c) Number of moles of sulphur atoms = = 0.5 Mass = = 16.1 g EXAMPLE 1.10 Calculate (a) the mass of oxygen contained in g of calcium hydroxide, (b) the number of moles of oxygen atoms in 25.5 g of hydrogen peroxide, H 2 O 2. molar mass (g mol 1 ) number of moles, n mass, m (g) molar mass (g mol 1 )

16 16 Chemistry for Matriculation Semester 1 (a) M r of calcium hydroxide, Ca(OH) 2 = (16 + 1) = 74.1 Number of moles of Ca(OH) 2 = = 2.0 Number of moles of oxygen atoms = = 4.0 Mass of oxygen atoms = = 64.0 g (b) Relative molecular mass of H 2 O 2, M r = (2 1) + (2 16) = 34 Number of moles of H 2 O 2 = Mass = 25.5 M r 34 = 0.75 Number of moles of oxygen atoms = = dm 3 mol 1 at s.t.p.* volume of gas (dm 3 ) number of moles, n * Use 24 dm 3 at room temperature dm 3 mol 1 at s.t.p.* Mole Concept and Molar Volume of Gases 1 For reactions that involve gases, the reacting volumes of gases (rather than their reacting masses) are often used. 2 The volume occupied by one mole of any gas is called the molar volume. (a) At s.t.p. (standard temperature and pressure), the molar gas volume is 22.4 dm 3. The conditions for s.t.p. are 0 C and 1 atm pressure. (b) At room temperature and pressure (20 C and 1 atm pressure), the molar gas volume is 24 dm 3. 3 The relationship between the number of moles of a gas and the volume of a gas is given by the formula: Number of moles, n = Volume of gas (dm 3 ) 22.4 (dm 3 mol 1 ) at s.t.p. = Volume of gas (dm 3 ) 24 (dm 3 mol 1 ) at r.t.p. The interconversion between moles, mass, number of particles and molar volume of gas is shown below. Number of particles (atom, molecule or ion) N A N A Number of moles 1 mole of any substance contains particles A r or M r A r or M r Mass (g) 22.4 dm 3 mol 1 at s.t.p. or 24 dm 3 mol 1 at room temperature 22.4 dm 3 mol 1 at s.t.p. or 24 dm 3 mol 1 at room temperature Volume of gas (dm 3 )

17 Matter 17 EXAMPLE 1.11 When potassium chlorate(v) solid is heated strongly, oxygen gas is liberated. 2KClO 3 (s) 2KCl(s) + 3O 2 (g) (a) A sample of KClO 3 released 48 dm 3 of oxygen gas at s.t.p. How many moles of oxygen gas are there? (b) How many molecules of oxygen are there in 48 dm 3 of oxygen gas? (c) Calculate the volume of oxygen gas evolved at s.t.p. when 0.5 g of potassium chlorate(v) is heated. (a) Number of moles of O 2 = 48.0 dm 3 = 2.14 mol 22.4 dm 3 1 mol (b) Number of molecules = 2.14 mol mol 1 = (c) 2 moles of KClO 3 produce 3 moles of oxygen gas (O 2 ). 2 moles of KClO 3 produce dm 3 of oxygen gas. M r for KClO 3 = (3 16) = Number of moles of KClO 3 used = 0.5 g 1 = mol g mol Volume of oxygen gas produced = mol dm 3 mol 1 = dm 3 Empirical and Molecular Formulae 1 The empirical formula of a compound is the formula which shows the simplest ratio of the atoms of the elements present in a compound. For example, an ethane (C 2 H 6 ) molecule contains 2 carbon atoms and 6 hydrogen atoms. The ratio of carbon : hydrogen = 2 : 6 = 1 : 3. Thus, its empirical formula is CH 3. EXAMPLE 1.12 When 2.67 g of copper reacts with excess sulphur, the mass of the compound obtained is 4.01 g. What is the empirical formula of the compound? (Relative atomic mass: Cu, 63.5; S, 32) Number of moles of copper = 2.67 = mol 63.5 Mass of sulphur in the compound = = 1.34 g Number of moles of sulphur = 1.34 = mol 32 Mole ratio of Cu : S = : = 1 : 1 The empirical formula of the copper compound is CuS. 2 The empirical formula of a compound is usually determined from the percentage composition (by mass) of the elements present in the molecule.

18 18 Chemistry for Matriculation Semester 1 EXAMPLE 1.13 A compound contains the elements potassium, chromium and oxygen. The composition by mass of the compound is 40.2% potassium, 32.9% oxygen and 26.9% chromium. Find the empirical formula of the compound. (Relative atomic mass: O, 16; K, 39; Cr, 52) Element Potassium Oxygen Chromium Percentage by mass Relative atomic mass Number of moles = = = 0.52 Atomic ratio = = = 1 Therefore, the empirical formula is K 2 CrO 4. 3 The molecular formula is the chemical formula which shows the exact number of atoms for each element in a molecule. 4 For many compounds, such as carbon dioxide and ammonia, the empirical formula and the molecular formula are the same. However, there are also many compounds (especially organic compounds) whose molecular formulae differ from their empirical formulae. Table 1.5 The molecular and empirical formulae of some compounds Compound Molecular formula Empirical formula Carbon dioxide CO 2 CO 2 Ammonia NH 3 NH 3 Nitrogen dioxide N 2 O 4 NO 2 Ethyne C 2 H 2 CH Benzene C 6 H 6 CH Glucose C 6 H 12 O 6 CH 2 O When the molecular formula of a compound is different from its empirical formula, the molecular formula is always a multiple of the empirical formula. For example, the empirical formula of glucose is CH 2 O. Its molecular formula is 6(CH 2 O) = C 6 H 12 O 6. 5 It is possible for different compounds to have the same empirical formula. For example, ethyne (C 2 H 2 ) and benzene (C 6 H 6 ) are two different compounds with the same empirical formula, CH. 6 To find the molecular formula of a substance, we need to know two things: (a) Its empirical formula (b) Its relative molecular mass 7 Empirical and molecular formulae are related as follows: Relative molecular mass Mass of empirical formula = n Molecular formula = n Empirical formula (where n = 1, 2, 3...)

19 Matter 19 EXAMPLE 1.14 Compound X contains the following composition by mass: 40.0% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative molecular mass is 180. What is the molecular formula of X? (Relative atomic mass: H, 1; C, 12; O, 16) Step 1 Find the empirical formula of compound X. Element Carbon Hydrogen Oxygen % composition by mass Relative atomic mass Number of moles = = = 3.33 Atomic ratio The empirical formula is CH 2 O = = = 1 Step 2 Find the molecular formula of X. Relative molecular mass Mass of empirical formula = n Mass of empirical formula = 12 + (2 1) + 16 = = n = 6 The molecular formula of X = 6 Empirical formula = 6 CH 2 O = C 6 H 12 O 6. To determine molecular formula from combustion data 1 Organic compounds are flammable. The empirical and molecular formulae of organic compounds can be obtained from combustion data. EXAMPLE g of an organic compound containing carbon, hydrogen and oxygen gives g of carbon dioxide and g of water on combustion. Find the empirical formula of the compound. (Relative atomic mass: H, 1; C, 12; O, 16) Step 1 Find the mass of carbon and hydrogen in 0.50 g of the organic compound from the combustion data. 1 mole (44 g) of carbon dioxide contains 1 mole (12 g) of carbon. Mass of carbon in g of CO 2 = = g

20 20 Chemistry for Matriculation Semester 1 1 mole (18 g) of H 2 O contains 2 g of hydrogen atoms. Mass of hydrogen (H) in g of H 2 O = = g Step 2 Find the composition by mass of carbon, hydrogen and oxygen. % of carbon (C) in the compound = = 37.5% % of hydrogen (H) in the compound = = 12.5% 2 % of oxygen (O) in the compound = = 50% Step 3 Calculate the empirical formula. Element C H O Percentage Number of moles = = = Atomic ratio = = = 1 The empirical formula of the compound is CH 4 O. 2 The molecular formula of a gaseous hydrocarbon can be determined by (a) measuring the volume of carbon dioxide produced, (b) measuring the volume of oxygen that reacts with the hydrocarbon, (c) using the general equation for the combustion of a gaseous hydrocarbon (shown below) to find the molecular formula of the hydrocarbon. C x H y (g) + x + y 4 O 2 (g) xco 2 (g) + y 2 H 2 O(l) EXAMPLE 1.16 When 8.0 cm 3 of a hydrocarbon, P, were burnt in 60 cm 3 of oxygen, the volume of the gaseous products occupied 44 cm 3 after being cooled to room temperature. After absorption by aqueous potassium hydroxide (KOH), the residual gases occupied 12.0 cm 3. What is the molecular formula of the hydrocarbon? (Relative atomic mass: H, 1; C, 12)

21 Matter 21 Step 1 Calculate the volume of CO 2 produced. Volume of gaseous products = Volume of unreacted O 2 + Volume of CO 2 produced = 44 cm 3... (1) KOH(aq) was used to absorb CO 2 so that only oxygen gas remained. Residual gas after absorption of CO 2 by KOH(aq) = Volume of unreacted oxygen = 12 cm 3 From equation (1), volume of CO 2 produced = (44 12) cm 3 = 32 cm 3 Step 2 Calculate the volume of oxygen that reacted. Volume of O 2 reacted = Initial volume before reaction Volume remaining after reaction = = 48 cm 3 Step 3 Use the general equation for the combustion of a gaseous hydrocarbon to find the molecular formula of the hydrocarbon. C x H y + x + 4 y O xco + y H 2O... (2) Volume of hydrocarbon used = 8.0 cm 3 From equation (2), volume of CO 2 produced = 8.0x = (3) volume of oxygen reacted = 8.0 x + 4 y = (4) From equation (3), x = 4 From equation (4), y 4 = 48 y = 8 The molecular formula of the hydrocarbon is C 4 H 8. Concentrations of Solutions The concentration of a solution can be expressed in several different quantitative ways: (a) Molarity (d) Percentage by mass, % w/w (b) Molality, m (e) Percentage by volume, % v/v (c) Mole fraction, X Molarity, M 1 The concentration of a solution can be measured as mass (g) of solute per volume (dm 3 ) of solution (g dm 3 ). Chemists usually measure the concentrations of solutions as moles of solute per cubic decimetre of solution (mol dm 3 ). 1 dm 3 = 1000 cm 3 = 1000 ml = 1 L

22 22 Chemistry for Matriculation Semester 1 2 The concentration of a solution stated in mol dm 3 is called the molarity of the solution, M. 3 The formulae below are used in calculations involving solutions. Concentration (mol dm 3 ) = Amount of solute (mol) Volume of solution (dm 3 ) Concentration (g dm 3 ) = Concentration (mol dm 3 ) Relative molecular (formula) mass Amount of solute (mol) = Concentration (mol dm 3 ) Volume (cm 3 ) 1000 EXAMPLE 1.17 (a) What is the molarity of a solution containing 14.8 g of KOH in 750 cm 3 aqueous solution? (b) 3.15 g of H 2 C 2 O 4.2H 2 O are dissolved in water and made up to a 1.0 dm 3 solution. Calculate the molarity of (i) hydrogen ions, (ii) C 2 O 4 2 ions, (iii) H 2 C 2 O 4 molecules. (a) M r of KOH = 56 Number of moles of KOH = = mol Volume of solution = 750 cm 3 = 0.75 dm 3 Molarity = = mol dm 3 (b) (i) M r of H 2 C 2 O 4.2H 2 O = 126 Number of moles of H 2 C 2 O 4.2H 2 O = = mol Volume of solution = 1.0 dm 3 Molarity = = mol dm Molarity of H + = = 0.05 mol dm 3 (ii) Molarity of C 2 O 2 4 = mol dm 3 (iii) Molarity of H 2 C 2 O 4 = mol dm 3 Relating density to concentration measurement EXAMPLE 1.18 A solution of hydrochloric acid, with 37.8% by mass of HCl, has a density of 1.19 g cm 3. Calculate the molarity of concentrated hydrochloric acid.

23 Matter 23 Step 1 Calculate the number of moles of HCl in 100 g of solution. Mass of HCl in 100 g solution = 37.8 g M r of HCl = 36.5 Number of moles of HCl in 100 g solution = 37.8 = mol 36.5 Step 2 Calculate the volume (in dm 3 ) of a 100 g solution of hydrochloric acid. Density = Mass Volume Mass Volume = Density Volume of hydrochloric acid = 100 g 1.19 g cm 3 = cm 3 = dm 3 Step 3 Calculate the molarity of hydrochloric acid. Molarity = Number of moles Volume (in dm 3 ) = = 12.3 mol dm Labels on concentrated acid bottles contain two important pieces of information: the percentage by mass and the density. Chemists make use of this information to calculate the molarity of the concentrated acid. Molality, m 1 Suppose we prepare a 1.0 mol dm 3 solution at 20 C. If the experiment is carried out at 30 C, the molarity of the solution will decrease slightly. This is because the amount of solute remains constant, but the volume of solution increases slightly due to thermal expansion. As a result, the number of moles of solute per dm 3 solution (molarity) decreases slightly. 2 Molarity, M is a useful concentration unit for calculation in most experiments, but it is temperature dependent. For experiments demanding a high precision, the concentration unit, molality is often used, because molality is independent of temperature. 3 The molality, m of a solution is defined as the number of moles of solute per kilogram of solvent (mol kg 1 ). Molality, m = Moles of solute (mol) Mass of solvent (kg) Since the masses of the solute and the solvent do not change when the solution is heated or cooled, molality is temperature independent. 4 To prepare a 1.0 molal (1.0 m) sodium hydroxide solution, we need to dissolve 1.0 mol (40 g) of NaOH in 1.0 kg (1000 g) of water. EXAMPLE 1.19 What is the molality of a glucose solution containing 9.5 g of glucose (C 6 H 12 O 6 ) in 185 g of water?

24 24 Chemistry for Matriculation Semester 1 M r of glucose (C 6 H 12 O 6 ) = 180 Number of moles of C 6 H 12 O 6 = = Mass of solvent = 185 g = kg Molality = = mol kg EXAMPLE 1.20 The density of 15.8 mol dm 3 nitric acid is 1.42 g cm 3. What is the molality of the nitric acid? Step 1 Calculate the mass of HNO 3 in 1.0 dm 3 solution Number of moles of HNO 3 in 1.0 dm 3 solution = 15.8 mol M r of HNO 3 = (16) = 63 Mass of HNO 3 in 1.0 dm 3 solution = Number of moles M r = = g Step 2 Calculate the mass of water in 1.0 dm 3 solution Density of nitric acid solution (H 2 O + HNO 3 ) = 1.42 g in 1 cm 3 solution Mass of H 2 O + HNO 3 in 1000 cm 3 (1 dm 3 ) = 1420 g Mass of H 2 O (solvent) = = g = kg Step 3 Calculate the molality of nitric acid Moles of solute (mol) Molality = Mass of solvent (kg) = 15.8 = 37.2 mol kg Mole fraction, X 1 Consider a solution containing n A moles of A and n B moles of B. The mole fraction of A in the solution is the number of moles of A divided by the total number of moles of A and B. Mole fraction of A, X A = n A + n B 2 Mole fractions are independent of temperature and have no units. The sum of the mole fractions of all the components in a solution is 1. For example, for a solution containing components A, B and C, X A + X B + X C = 1 where X A, X B and X C are the mole fractions of A, B and C respectively. EXAMPLE 1.21 A solution is prepared by dissolving 32.0 g of methanol (CH 3 OH) in 72.0 g of water. Calculate the mole fraction of methanol in the solution. Relative molecular mass of methanol (CH 3 OH) = 32.0 Number of moles of methanol = 32.0 = 1.0 mol 32.0 Relative molecular mass of water = 18.0 n A

25 Matter 25 Number of moles of water = 72.0 = 4.0 mol Mole fraction of methanol in the solution = 1 +4 = 0.2 Percentage by mass, % w/w 1 Percentage by mass is also called weight percent. The percentage by mass of any component in a solution is the mass of that component divided by the total mass of the solution multiplied by 100%. Percentage by Mass of component = mass (% w/w) Total mass of solution 100% 2 A 0.9% by mass sodium chloride solution is a solution in which the ratio of the solute (NaCl) to the solution (NaCl (aq)) is 0.9 g to 100 g. EXAMPLE 1.22 A 0.80 mol dm 3 solution of sulphuric acid has a density of 1.10 g cm 3 at room temperature. What is the concentration of this solution in weight percent? (Relative molecular mass of H 2 SO 4 = 98) Consider 1.0 dm 3 of H 2 SO 4 solution. Mass of H 2 SO 4 = = 78.4 g Density of H 2 SO 4 solution = 1.10 g cm 3 Mass of 1 dm 3 solution = = 1100 g Weight % of H 2 SO 4 = % = 7.13% 1100 Percentage by volume, % v/v 1 The percentage by volume of any component in a solution is the volume of that component divided by the total volume of the solution multiplied by 100%. Percentage by Volume of component = volume (% v/v) Total volume of solution 100% EXAMPLE 1.23 Calculate the volume of antifreeze required to make 10 dm 3 of a solution of antifreeze which is 40% by volume. Let volume of antifreeze = V dm 3 40 = V % V = 4 dm 3

26 26 Chemistry for Matriculation Semester 1 1. (a) Calculate the mass of hydrogen chloride in a sample containing molecules of HCl. (b) Calculate the relative atomic mass of argon if the mass of one atom of argon is kg. 2. (a) Calculate the number of ethane (C 2 H 6 ) molecules in 3.75 g of ethane. (b) Calculate the number of ions in kg of sodium oxide (Na 2 O). (c) What is the volume (in L) of 28.8 g of sulphur dioxide at s.t.p.? (Relative atomic mass: H, 1; C, 12; O, 16; Na, 23; S, 32) 3. Nitrogen monoxide reacts spontaneously with oxygen according to the equation 2NO(g) + O 2 (g) 2NO 2 (g) QUICK CHECK 1.4 In an experiment, 50 cm 3 of nitrogen monoxide was mixed with 100 cm 3 of oxygen. Calculate the total volume of the gaseous mixture produced after the reaction. Assume that the volumes of gases were measured at s.t.p. 4. (a) What is meant by empirical formula? (b) The composition by mass of an organic acid is shown below. H, 2.20%; C, 26.70%; O, 71.10% The molar mass of the acid is 90.0 g mol 1.. (i) What is the empirical formula of the organic acid? (ii) What is its molecular formula? 5. (a) What is relative molecular mass? (b) An organic compound P, with relative molecular mass of 46, contains only carbon, hydrogen and oxygen. When 1.38 g of P is burnt in excess oxygen, 2.64 g of carbon dioxide and 1.62 g of water are produced. (i) Calculate the mass of carbon, hydrogen and oxygen in 1.38 g of P. (ii) Hence, find the molecular formula of P cm 3 of a hydrocarbon (C x H y ) require 88 cm 3 of oxygen for complete combustion. The volume of carbon dioxide produced is 64 cm 3. Calculate the molecular formula of the hydrocarbon. 7. (a) Ammonia gas can be prepared by heating calcium hydroxide with ammonium chloride. Ca(OH) 2 + 2NH 4 Cl CaCl 2 + 2NH 3 + 2H 2 O A sample of ammonia gas is prepared by heating 5.56 g of calcium hydroxide with excess ammonium chloride. The ammonia gas produced is dissolved in 300 cm 3 of water. What is the molarity of the ammonia solution? (b) A solution of mol dm 3 sulphuric acid has a density of g cm 3. What is the molality of this solution? 8. Calculate (a) the mole fractions of C 2 H 5 OH and H 2 O in a solution prepared by dissolving 1.70 g of C 2 H 5 OH in 16.0 g of water, (b) the concentration (% w/w) of a solution formed by dissolving 15.0 g of ZnCl 2 in 45.0 g of water, (c) the percentage by volume (% v/v) of ethanol in a solution containing 18.6 cm 3 of ethanol in 120 cm 3 solution cm 3 of ethanol (C 2 H 5 OH) are added to distilled water in a volumetric flask and the resulting solution is made up to 100 cm 3. The density of ethanol is g cm 3. The density of the solution is g cm 3. What is the (a) molarity of the solution, (c) percentage by mass of ethanol in the solution, (b) molality of the solution, (d) mole fraction of (i) ethanol, (ii) water in the solution? 10. The density of an aqueous ammonia solution of concentration 15.2 mol dm 3 is 0.91 g cm 3. What is the mole fraction of (a) ammonia, (b) water in this solution? 11. A solution of potassium sulphate is labelled as 5.8% by mass. Calculate the mass of solution that contains 1.58 g of potassium sulphate.

27 Matter Stoichiometry LEARNING OUTCOMES Determine the oxidation number of an element in a chemical formula Write and balance (a) chemical equation by inspection method (b) redox equation by ion electron method Define limiting reactant and percentage yield Perform stoichiometric calculations using mole concept including limiting reactant and percentage yield Oxidation Numbers 1 The oxidation number of an atom is an arbitrary charge assigned to the atom according to a set of rules. 2 Oxidation numbers are also called oxidation states. Redox reactions (oxidation reduction reactions) are often considered in terms of changes in oxidation number or oxidation state for each reactant. 3 The rules for assigning oxidation numbers are given in Table 1.6. Table 1.6 Rules for assigning oxidation numbers Rule 1. The oxidation number of an element in its uncombined state (not combined with other elements) is zero. Example The oxidation number of each of the elements, chlorine (Cl 2 ), oxygen (O 2 ), sodium (Na) and sulphur (S 8 ) is zero. 2. (a) The more electronegative element in a binary compound (compound with only 2 elements) is given a negative oxidation number. (b) The less electronegative element is given a positive oxidation number. (c) The oxidation number of fluorine in its compounds is always 1 because fluorine is the most electronegative element. Fluorine acts as the reference when considering the oxidation numbers of other elements. 3. In a binary ionic compound, the oxidation number of the element is the same as the charge on the ion. In HCl, hydrogen is given the oxidation number +1 and chlorine 1 because chlorine is more electronegative than hydrogen. When hydrogen is bonded to a less electronegative element, for example, in metal hydrides (NaH, CaH 2 ), hydrogen has the oxidation number 1. The oxidation number of fluorine is 1 in the following compounds HF ClF ClF 3 CaF 2 The oxidation numbers of iron and chlorine in iron(ii) chloride, Fe 2+ (Cl ) 2, are Fe = +2 and Cl = The sum of the oxidation numbers of all the elements in a compound is zero. 5. The sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion. In FeCl 3, the sum of the oxidation numbers of Fe and Cl = ( 1) = 0. In SO 2 4, the sum of the oxidation numbers = ( 2) = 2 (the charge on the sulphate ion). 4 The oxidation numbers of some elements such as Group 1 and Group 2 elements, and aluminium are always the same in their compounds (Table 1.7). Interhalogen compounds are compounds formed between two halogens, for example, CIF 3 or ICI.

28 28 Chemistry for Matriculation Semester 1 Table 1.7 Oxidation numbers of some elements Element H Oxidation number +1 (in all covalent compounds except in metal hydrides) 1 (in metal hydrides) Group 1 elements in the periodic table +1 Group 2 elements in the periodic table +2 Al +3 Cl O 1 (except when combined with oxygen or in some interhalogen compounds) 2 (except in fluorides and peroxides) EXAMPLE 1.24 What are the oxidation numbers of (a) chromium in Cr 2 (SO 4 ) 3, (b) chlorine in Cl 2 O 7, (c) oxygen and fluorine in OF 2, (d) iodine in ICl? (a) Oxidation number of sulphate ion = 2 (rule 5) Let oxidation number of chromium = x 2x + 3( 2) = 0 (rule 4) x = +3 (b) The oxidation number of oxygen = 2 (rule 2) Let oxidation number of chlorine = x 2x + 7( 2) = 0 (rule 4) x = +7 (c) Oxidation number of fluorine = 1 (rule 2) Let oxidation number of oxygen = x x + 2( 1) = 0 (rule 4) Oxidation number of oxygen = +2 (d) Chlorine is more electronegative Oxidation number of chlorine = 1 (rule 2) Oxidation number of iodine = +1 (rule 4) EXAMPLE 1.25 What are the values of m and n (charges) in the following ions (a) AlO 2 m and (b) Pb(OH) 4 n? (a) Oxidation numbers: Al = +3, O = 2 Charge on the ion (m) = ( 2) = 1 that is, m = 1 and the formula of the ion is AlO 2 (rule 5) (b) Oxidation number of lead = +2, charge of hydroxide ion = 1 Charge on the ion (n) = ( 1) = 2 that is, n = 2 and the formula of the ion is Pb(OH) 4 2 (rule 5)

29 Chemical Equations Matter 29 1 The study of the relative proportions in which substances react is called stoichiometry. The chemical equations in which the number of moles of reactants (molecules or atoms) are in simple whole number ratios are called stoichiometric equations. 2 To represent a chemical equation correctly, the equation must be balanced. All balanced chemical equations have the following features. (a) The reactants must appear on the left, and the products appear on the right. The arrow joining them indicates the direction of reaction. (b) The number of each type of atom is the same on both sides of the equation. Writing chemical equations by inspection method 1 The equation given below is not balanced. The number of oxygen atoms in the reactants does not match the number of oxygen atoms in the product. H 2 (g) + O 2 (g) H 2 O(g)... (1) 2 oxygen atoms one oxygen atom 2 In order to balance equation (1), we need to put a 2 in front of H 2 as well as in front of H 2 O. 2H 2 (g) + O 2 (g) 2H 2 O(g) 4H atoms 2O atoms 4H atoms 2O atoms Oxidation and reduction 1 Oxidation is defined as the loss of electrons from a substance and reduction is the gain of electrons by a substance. 2 An oxidising agent is a substance which accepts electrons from another substance (brings about oxidation) and a reducing agent is a substance which donates electrons to another substance (brings about reduction). 3 Consider the following reaction: Sn 2+ (aq) + 2Fe 3+ (aq) Sn 4+ (aq) + 2Fe 2+ (aq)... (2) Reaction (2) can be considered as derived from two processes: Oxidation: Sn 2+ (aq) Sn 4+ (aq) + 2e... (3) Reduction: 2Fe 3+ (aq) + 2e 2Fe 2+ (aq)... (4) Equations (3) and (4) which show only oxidation or reduction in terms of loss and gain of electrons are called half-equations.