Chapter 1 Matter

LEARNING OUTCOMESAt the end of this topic, students should be able to:1.1 Atoms and Molecules a) Identify and describe protons, electrons and neutrons as sub-atomic particles. b) Define proton number, nucleon number and isotopes. Write isotope denotation. c) Define relative atomic mass, Ar and relative molecular mass, Mr based on the C-12 scale. d) Sketch and explain the function of the main components of a simple mass spectrometer: i. vaporization chamber, ii. ionization chamber, iii. acceleration chamber, iv. magnetic field, v. ion detector. e) Analyze mass spectrum of an element. Calculate the average atomic mass of an element given the relative abundance of its isotopes, or its mass spectrum.1.2 Mole Concept a) Define mole in terms of mass of carbon-12 and Avogadro constant, NA. b) Interconvert between moles, mass, number of particles, molar volume of gas at s.t.p and room temperature. c) Determine empirical and molecular formulae from mass composition or combustion data. d) Define and perform calculations for each the following concentration measurements: i. molarity, c ii. molality, m iii. mole fraction, X iv. percentage by mass, % w/w v. percentage by volume e) Determine the oxidation number of element in a chemical formula. f) Write and/or construct balance: i. chemical equation by inspection method. ii. redox equation by ion-electron method. g) Define limiting reagent and percentage yield. h) Perform stoichiometric calculations using mole concept including limiting reagent and percentage yield.

MATTER CHAPTER 11.1 ATOM AND MOLECULESAs soon as you are introduced to the subject of science, you will encounter the word atom.Atom is the smallest particle exists, discreet unit of every matter for non-living and livingthing (Campbell 1996). Believe it or not, atom is actually referring to the word zarah in Arabiclanguage. The word atom came from the word atomos of Greek language, introduced byDemocritus (460-370 SM) (Brown 2014). John Dalton (1766-1844) re-introduced the idea ofatom, while reading books written by Ibnu Sina (980 M). Ibnu Sina made a distinctiveconclusion that atom or zarah mentioned in Al-Quran (99: 7-8) is the smallest particlemaking up all matters.\"Then, whoever has done an atom's weight of good, shall see it; and whoeverhas done an atom's weight of evil, shall see it.\" (99: 7-8)The Surah Al-Zalzalah, chapter 99 of the Quran is talking about the Earthquake, which iswhen Doomsday (Qiamah) will happen. It has been said that men on that Day, rising fromtheir graves, will come out in their varied groups from all corners of the earth, to be showntheir deeds and works, and their presentation of the deeds will be so complete and detailedthat not an atom's weight of any good or evil act will be left unnoticed or hidden from Hiseyes.Later, in a different chapter of Al-Quran, Allah SWT explained that there are smaller particlethan atom. This findings and explanations is also written by Ibnu Sina (Al-Rawi 2002). Wetried to define the meaning of atom in general as mentioned in Surah Al-Zalzalah and weagreed that electron is actually the sub-particle mentioned in Surah As-Saba', chapter 34 ofAl-Quran (Mat 2016).Matter is anything that occupies space and possesses mass. The three states of matter aresolid, liquid and gas. Matter is made of atoms. An atom is comprised of a positively chargednucleus surrounded by a negatively charged electron cloud. The particles in the nucleus,called nucleons are consists of positively charged protons and neutrons. Protons, neutronsand electrons are the subatomic particles. Table 1.1 shows the properties of proton, neutronand electron while Table 1.1 and Figure 1.1 shows the properties of modern model of theatom. Table 1.1Particle Charge Relative Mass p +1 1.0 u n 0 1.0 u e -1  0.0 Nucleons Protons Electron waves Figure 1.1: Modern model of the atom 1

MATTER CHAPTER 1PROTON NUMBER, NUCLEON NUMBER AND ISOTOPESThe proton number, Z is the number of protons in the nucleus of an atom (also known asatomic number). In modern Periodic Table, proton number is used to arrange the elementsin ascending order. The nucleon number, A is the total number of protons and neutrons inthe nucleus, where we can say that the number of neutrons, n = A – Z.The isotopic notation or isotope symbol (Figure 1.2), shows the symbol of element,atomic number and nucleon number. In the case of an anion and cation, the charge of theion is shown as a superscript on the right of the symbol for the ion. X = element symbol A = Nucleon Number of the nuclide X =Z+n Z = Proton Number of the nuclide X =p Figure 1.2: Isotopic notation or isotope symbol.For example, the structure of an atom, plutonium which has 94 protons and 150 neutrons,can be expressed in the following ways; plutonium-244, 244Pu and 244 Pu . 94Atoms and molecules do not carry any charge but ions do. An ion is an atom or a group ofatoms that has lost one or more electrons making it positively charged (cation) or gained oneor more electrons making it negatively charged. Polyatomic ions are ions which contain twoor more covalently bonded atom, such as OH−.EXAMPLE 1.1Give the number of protons, neutrons and electrons in each of the following species in thetable below. Symbol Number of Charge Neutron Proton Electron200 Hg 8063 Cu2917 O 2- 859 Co 327Isotopes are atoms that have the same number of protons but different number of neutrons,for example the lithium atom has 3 isotopes; 6Li, 7Li and 8Li. They have the same chemicalproperties since they have the same electronic configuration, but different physicalproperties. 2

MATTER CHAPTER 1Most elements exist as mixtures of two or more naturally occurring isotopes and the relativeabundance is always expressed as percentage.Isotopes of an element have the same, number of protons (proton number) charge of nucleus of the atoms number of electrons in a neutral atom electronic configuration (the number of valence electrons) chemical properties (ionization energy; electron affinity; size of the atom; electronegativity are the same).Isotopes of an element have different, number of neutrons (nucleon number) in the nucleus of the atoms relative isotopic mass. physical properties (e.g boiling point/ melting point, density, effusion rate).Radioisotopes have a number of important applications beyond the production of energy orweapons of mass destruction. They are used in neutron activation analysis, geologicaldating, as tracers, in oncology research and more importantly in radiation detection.EXAMPLE 1.2Neon (proton number = 10) has three isotopes. Calculate the number of sub-atomic particlesin the three isotopes of Neon by completing the table below.Isotope 20Ne 21Ne 22NeNumber of protonsNumber of electronsNumber of neutronsRELATIVE ATOMIC MASS, Ar AND RELATIVE MOLECULAR MASS, MrRelative atomic mass, Ar is the ratio of the mass of one atom of that element to 1/12 of themass of a carbon-12 atom. The mass of a proton (1.0074 u) is almost the same as the massof a neutron (1.0089 u) while the mass of electron is very small. Therefore, the Ar of anelement is considered to be the same as its nucleon number.Relative molecular mass, Mr, has no unit and is calculated by adding up the relative atomicmasses of all the atoms present in one molecule of the substance.MASS SPECTROMETERMass spectrometer (Figure 1.3) is a modern technique to determine the Ar and relativeabundance of an isotope in a sample of an element. Mass spectrometer is also used todetermine relative atomic mass of an element, relative molecular mass of a compound, 3

MATTER CHAPTER 1types of isotopes that are found in the naturally occurring element including the abundanceof the isotopes and its relative isotopic mass and to recognize the structure of the compoundin an unknown sample.ionisation accelerationchamber chambervapourisationchamber magnetic field ion detector Figure 1.3: Mass spectrometerThere a five basic part in a mass spectrometer. (Refer Table 1.3) Vapourisation Chamber Ionisation Chamber Acceleration Chamber Magnetic Field Ion DetectorAt vapourisation chamber, the sample is vapourised and turned into gas. At ionizationchamber the gaseous particles is ionized by electron bombardment. Here, a gaseoussample (atom or molecule) is bombarded by a stream of high-energy electrons that areemitted from a hot filament. Collisions between the electrons and the gaseous atom (ormolecule) produce positive ions by dislodging an electron from each atom or molecule.The positive ions are then accelerated in acceleration chamber by an electric field towardsthe two oppositely charged plates. These ions are then deflected according to its mass/charge ratio at the magnetic field. Positive ions with lower mass to charge ratio will bedeflected more than those with higher mass to charge ratio. Finally, at ion detector this datawill be recorded as a peak in mass spectrum (Figure 1.4). 4

MATTER CHAPTER 1 Figure 1.4: Mass spectrum. (Source: http://earthandsolarsystem.wordpress.com/2011/05/18/xenon)Table 1.3: Summary of five main components of the mass spectrometer and their function.Component Function 5

MATTER CHAPTER 1MASS SPECTRUM ANALYSISThe relative atomic mass can be calculated from the mass spectrum. For example, therelative atomic mass, Ar for Xe in Figure 1.5 is calculated to be 14.0037. % abundance 99.63 0.37 m/e 14 15 Figure 1.5: Mass spectrum of element Xe.The calculation is shown below.Average atomic mass   fimi  (99.63x14)  (0.37x15)  14.0037u  fi (99.63  0.37)Ar  (Averageatomic mass) u  14.0037u  14.0037 1 x12.00 u 1 x12.00u 12 12 Whereby, the relative abundance / percentage abundance of an isotope f= of the element m= the relative isotopic mass of the elementEXAMPLE 1.3Figure below shows the mass spectrum of the element Rubidium, Rb. relative abundance 18 7 m/e 85 87 6

MATTER CHAPTER 1(a) What isotopes are present in Rb?(b) What is the percentage abundance of each isotope?(c) Calculate the relative atomic mass of Rb. PRACTICE 1.11. Calculate the relative atomic mass for Ne based on the. % abundan9c0.5 e 9.2 0.3 m/e 20 Ne 1201Ne 22 Ne 10 102. The relative atomic mass of 6 Li and 7 Li are 6.01 and 7.02 respectively. What is 3 3 the percentage abundance of each isotope if the relative atomic mass of Li is 6.94?3. The ratio of relative abundance of naturally occuring of chlorine isotopes is 35 Cl  3.127 based on the carbon-12 scale, the relative atomic mass of 37 Cl 35Cl= 34.9681 and 37Cl= 36.9659. Calculate the Ar of chlorine.4. Naturally occuring iridium, Ir is composed of 2 isotopes 191Ir and 193Ir in the ratio of 5:8. The relative mass of 191Ir and 193Ir is 191.021 and 193.025 respectively. Calculate the relative atomic mass of iridium. 7

MATTER CHAPTER 11.2 MOLE CONCEPTOne mole is defined as the quantity of a substance that contains the same number ofparticles (atoms, electrons, ions, molecules) as there are atoms in exactly 12 g of carbon-12.By using the mass spectrometer, the mass of one single atom carbon-12 is 1.992 x 10-23,therefore the number of atoms in exactly one mole of carbon-12 is 6.022 x 1023. This numberis called the Avogadro constant, NA with the unit mol-1. Hence we can say that for 0.6 mole ofcopper we have 3.6 x 1023 atoms. 1 mol = NA = 6.022 x 1023 atomsThe mass (in grams) of one mole of a substance is named molar mass , M with the unit gmol-1. For example the relative atomic mass, Ar of Ne is 20.18. Hence the molar mass, M is20.18 g mol-1.In short, for any element; atomic mass (u) = molar mass (grams)EXAMPLE 1.41. Calculate the number of atoms in 0.551 g of potassium, K.2. How many atoms of carbon are in 0.750 moles MSG, NaC5H8NO4?3. How many atoms of oxygen are in 1.5 mol of aluminum sulfate, Al2(SO4)3? 8

MATTER CHAPTER 1Molecular mass (or molecular weight) is the sum of the atomic masses (in u) in a molecule.Conveniently, 1 u = 1 g mol-1. The atomic masses found on the Periodic Table are now thenumber of grams of the element in one mole. The same relation can be made for formulaweights as well. Figure 1.6 shows the interconversion between moles, mass, number ofparticles, molar volume of gas at s.t.p and room temperature. Figure 1.6: Interconvertion between moles, mass, number of particles, molar volume of gas at s.t.p and room temperature.EXAMPLE 1.51. Determine the molar masses for the following pure substances. (a) Mo (b) SO2 (c) Cu2SO4 (d) C6H12O62. How many atoms of sodium are in 13.0 g of sodium metal? 9

MATTER CHAPTER 13. For an experiment that I am performing, I need to add 12.5 mol of iron(II) sulfate, FeSO4. How many grams must I weigh out?Molecular volume is the volume occupied by ONE (1) mole of gas at a particulartemperature and pressure. At standard temperature and pressure (s.t.p), the molar volume is22.4 dm3, and at room temperature, the molar volume is 24 dm3.EXAMPLE 1.61. A graduated cylinder contains 22.5 cm3 of mercury. The density of mercury is 13.53 g cm-3 and molecular weight is 200.59 g mol-1. Calculate the number of moles of mercury in the cylinder.2. Determine the density of 1 g of nitrogen gas, N2 at s.t.p. 10

MATTER CHAPTER 1EMPIRICAL AND MOLECULAR FORMULAE FROM MASS COMPOSITION ORCOMBUSTION DATAThe empirical formula is the chemical formula which shows the types of elements present inthe compound and the simplest ratio of the atoms. For instance, an ethane (C2H4) moleculehas the 2 carbon atoms and 4 hydrogen atoms. The ratio of carbon: hydrogen = 2 : 4 = 1 : 2.Thus the empirical formula for ethane is CH4. C2H4 for ethane is the molecular formula whichshows the exact number of atoms for each element in the compound.In short, we can say molecular formula = (empirical formula) n Table 1.4: Empirical formula, molecular formula and n value for some molecules. Molecules Empirical formula Molecular formula n Water H2O H2O 1Hydrogen peroxide HO H2O2 2 Benzene CH C6H6 6 Etyne CH C2H2 2 NH3 NH3 1 Ammonia CO2 CO2 1 Carbon dioxideWe can say that the empirical formula and the molecular formula of a compound could bethe same, for example H2O, NH3 and CO2. Two molecules might have different molecularformula but the same empirical formula as we can see for hydrogen peroxide, benzene andethyne in Table 1.4. In calculating the simplest ratio, we must never round off values close to whole number, but multiply the value by a factor until we get a whole number. However, if the value is very close to whole number (0.01), it is allowed to round off the value.EXAMPLE 1.61. 18.3 g sample of hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of water only. Calculate its empirical formula. (Answer: Empirical formula = CaCl2.4H2O) 11

MATTER CHAPTER 12. Ascorbic acid (vitamin C) cures scurvy and may help prevent the common cold. It is composed of 40.92% carbon, 4.58% hydrogen and 54.50% oxygen by mass. The molar mass of ascorbic acid is 176 g mol1. Determine its empirical formula and molecular formula. (Answer: Empirical formula = C3H4O3)3. 1.00 g sample of compound A was burnt in excess oxygen producing 2.52 g of CO2 and 0.443 g of H2O. Determine the empirical formula of the compound. (Answer: Empirical formula = C7H6O2) 12

MATTER CHAPTER 1 PRACTICE 1.2 1. The pre-hormone androstenedione (commonly called andro) has been in the sports news in recent years owing to its arguable contribution to the breaking of the homerun record. The product has a composition that is 79.78% carbon (C), 9.05% hydrogen (H), and 11.17% oxygen by weight. What is the empirical formula for this compound? 2. Determine the empirical formula for acetominophen, the active ingredient in Tylenol from the elemental analysis. C 63.56%, H 6.00%, N 9.27%, O 21.17% 3. Determine the empirical formula for glucose. C 40.00%, H 6.72%, O 53.29% 4. Adipic acid is used in the commercial manufacture of Nylon. The composition of the acid is 49.3% C, 6.9% H, and 43.8% O by mass. The molecular weight is 146 g mol-1. What is the molecular formula for Adipic Acid? (1. C19H26O2, 2. C8H9NO2 ,3. CH2O,4. C6H10O4)CONCENTRATION MEASUREMENTSBefore we are able to calculate the measurements of concentration of a substance, we haveto know what the meaning of solution, solute and solvent is.A solution is a homogenous mixture of TWO (2) or more substances or in other words asolution is a mixture of solute in solvent. The solute is (are) the substance(s) present in thesmaller amount(s) and the solvent is the substance present in the larger amount. Mass of solution = mass of solute + mass of solventMixtures consist of TWO (2) or more components. A homogenous mixture of a liquid with agas, liquid or solid (the solute) is called a solution. Solution can contain varying amounts ofsolute. The concentration of a solution is defined as the content of solute in a given quantityof solution or of solvent.(a) Molarity, c Molarity, c is defined as the number of moles of solute dissolved in ONE (1) liter or decimeter (dm3) of solution. The unit can be mol L-1 or mol dm-3 or M. This unit of concentration is also known as amount of concentration, c. It is important here to note that 1 dm3 = 1000 cm3 = 1 L. 13

MATTER CHAPTER 1EXAMPLE 1.71. Calculate the molarity of a solution of 1.71 g sucrose, C12H22O11 dissolved in a 0.5 L of water.2. How many grams of potassium dichromate, K2Cr2O7 is required to prepare a solution of 250 mL with 2.16 M?(b) Molality, m Molality, m is the number of solute per 1 kg of solvent in a solution. The unit is mol kg-1 or m. The molality of a solution is equal to the number of moles of solute contained in one kilogram of solvent. The unit for molality is mol kg-1 or molal or m.  Mass of solution = mass of solute + mass of solventEXAMPLE 1.81. Calculate the molality of sulphuric acid solution, H2SO4 containing 24.4 g of sulphuric acid in 198 g of water. 14

MATTER CHAPTER 12. What is the molal concentration of a solution prepared by dissolving 0.30 mol of CuCl2 in 40.0 mol of water?(c) Mole fraction, X Mole fraction, X can be defined as the number of moles of one component in a mixture divided by the total number of moles of all substances present in the mixture. In other words, the mole fraction is the concentration of a solution as the number of moles of the solute divided by the total number of the moles of all components in the solution. The sum of mole fraction for all the components is ONE (1) and it carries no unit. Mole fraction of component A = XA XA = moles of A sum of moles of all components  nA nA  nB  nC  ... where XA + XB + ... = 1EXAMPLE 1.91. The density of 10.5 molal NaOH solution is 1.33 g cm-3 at 20 C. Calculate a) the mole fraction of NaOH 15

MATTER CHAPTER 1b) the percentage by mass of NaOH c) the molarity of the solution(d) Percentage by Mass, % w/w Percentage by mass (% w/w) means that for a given weight of solution, a certain percent of that weight is solute. We can say that percent by mass, or weight percent or mass percent of a solute is defined as the number of grams of solute dissolved in 100 grams of a solution. For example, 20% w/w NaCl means that there are 20 g of NaCl in 100 g of solution. Mass percent of solute = % w/w %w/w = mass solute × 100% total mass solutionEXAMPLE 1.101. Given 24 g of NaCl dissolved in 152 g of water, calculate the mass percent. 16

MATTER CHAPTER 12. Calculate the percent by mass concentration of a solution prepared by dissolving 10 g of certain substance into 175 mL of butyl ethanoate, which has a density of 0.88 g cm-3.(e) Percentage by Volume, % v/v A volume percent (%V/V) gives the volume of solute divided by the volume of solution. The concentration of a solution is expressed as a percent by volume or volume percent is defined as the number of cubic centimeters of solute contained in 100 cm3 of the solution. For example, 14% V/V ascorbic acid means that there are 12 mL of ascorbic acid in 100 mL of solution. Percent by volume or volume percent of solute = % v/v %v/v = volume solute × 100% total volume solutionEXAMPLE 1.111. A 200 mL of perfume contains 28 mL of alcohol. What is the percent by volume of alcohol in this solution? 17

MATTER CHAPTER 1 Table 1.5 The summary of concentration measurements.Concentration Symbol Formula UnitmeasurementMolarity CMolality mMole fraction X %w/w Percent by %v/v mass Percent by volume PRACTICE 1.31. Tamhidi student prepared a solution by dissolving 0.586 g of sodium carbonate, Na2CO3 in 250 cm3 of water. Calculate its molarity.2. What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?3. What is the molality of a solution containing 7.78 g of urea [(NH2)2CO)] in 203 g of water?4. A solution containing 121.8 g of Zn(NO2)2 per liter has a density of 1.107 gmL-1. Calculate its molal concentration.DETERMINE THE OXIDATION NUMBER OF ELEMENT IN A CHEMICAL FORMULADifferent compounds of the same element may have different physical and chemicalproperties because the element can exist in different oxidation states. Oxidation numbers oroxidation state is the charge that the atom of an element would have if complete transferof electron takes place in a redox equation. Redox reactions are often considered in terms ofchanges in oxidation state for each reactant.The oxidation number of uncombined element is zero, for example Fe (s) and Cl2 (g). Thetotal oxidation number of all atoms in a neutral molecule is zero for example H2O and NaCl.For simple ion like S2-, the oxidation number is equal to the charge carried by it. Forpolyatomic ions such as NO3-, the sum of oxidation number of all the atoms in that ion equalto the magnitude and charges carried by that ion. When an element is oxidized, its oxidationnumber increased, but when an element is reduced, its oxidation number is reduced. 18

MATTER CHAPTER 1Specifically, there are certain rules for assigning oxidation number.1. The oxidation number of the uncombined element in its elemental state is ZERO: Na(s), O2(g), C(s), H2(g) have oxidation number of 0.2. The oxidation number of a monoatomic ion is equal to its charge. Table 1.6Ion K+ Zn2+ Al3+ Cl- S2- N3-Oxidation +1 +2 +3 -1 -2 -3 number3. In compounds, the more electronegative elements are given a negative oxidation number. The sequence of electronegativity of some elements is shown below. Fluorine, Oxygen, Nitrogen, Chlorine, Bromine, Iodine Increasing electronegativity (a) Fluorine has an oxidation number of –1 in all its compounds because it is very electronegative.(b) The oxidation number of other halogens (Cl, Br, and I) in their compounds are -1 except when they combine with more electronegative elements such as oxygen and nitrogen.(c) The oxidation number of hydrogen is always +1, except when it is bonded to reactive metals in metal hydrides. In these cases (for example: LiH, NaH, CaH2), its oxidation number is –1.(d) The oxidation number of oxygen in most compounds is always –2, except in peroxides and when oxygen combines with a more electronegative element such as fluorine. 19

MATTER CHAPTER 14. In a neutral molecule, the sum of the oxidation numbers of all the elements must be ZERO.5. In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must be equal to the net charge of the ion.PRACTICE 1.41. Assign the oxidation number of Cr in Cr2O72-.2. Determine the oxidation number for the underlined element. (a) IF7 (b) NaIO3 (c) K2Cr2O73. What is the oxidation number of(a) Fe atom in FeCl3(b) Cl atoms in Cl2O7(c) S atoms in H2SO4(d) I atoms in IO3-(e) N atom in NH4+4. Determine the oxidation number of Mn in the following chemical compounds.(a) MnO2 (b) MnO4-5. Determine the oxidation number of Cl in the following chemical compounds.(a) KClO3 (b) Cl2O72-6. Determine the oxidation number of followings:(a) U in UO22+ (b) C in C2O42- 20

MATTER CHAPTER 1BALANCING CHEMICAL EQUATIONChemical EquationA chemical equation shows a chemical reaction using symbols for the reactants and theproducts. The formulae of the reactants are written on the left side of the equation while theproducts are on the right. xA + yB zC + wDThe total number of atoms of each element is the same on both sides in a balancedequation. The numbers x, y, z and w showing the relative number of molecules reacting, arecalled the stoichiometric coefficients.Consider a chemical reaction; 2 Mg + O2 2 MgOThis means;and NOT 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO 2 grams Mg + 1 gram O2 makes 2 g MgOThe relationship between the amounts (mol or g) of reactants and products involved in achemical reaction is called the stoichiometry of the reaction. As we know, in an equation, thereactant will always appear on the left and the products appear on the right, joined by anarrow indicates the direction of the reaction. The most important condition for the equation tobe balanced is that the number of each atom is the same on both sides.(a) Inspection Method1. Write down the unbalanced equation.2. Write the correct formulae for the reactants and products.3. Balance the metallic element, followed by nonmetallic atoms.4. Balance the hydrogen and followed by oxygen atoms.5. Check to ensure that the total number of atoms of each element is the same on both sides of equation.EXAMPLE 1.121. Balance the chemical equation below step by step;NH3 + CuO → Cu + N2 + H2O 21

MATTER CHAPTER 1 PRACTICE 1.51. Write and balance the equations when (a) Ethane, C2H6 reacts with oxygen, O2 to form carbon dioxide, CO2 and water. (b) Heamatite, FeO3 is also known as iron ore. When reacts with carbon monoxide, CO in a blast furnace, iron, Fe and carbon dioxide CO2 is formed. (c) Consider the oxidation of propane, C3H8 below and balance the equation. C3H8 + O2  CO2 + H2O2. Balance the chemical equations below: (a) AgNO3 + Na2CrO4  Ag2CrO4 + NaNO3 (b) H2 + N2 → NH3 (c) Al2O3 → Al + O2 (d) KClO3 → KCl + O2 (e) C6H6 + O2 → CO2 + H2O(b) Ion Electron Method Redox Reaction Redox reaction is a reaction that involves both reduction and oxidation. Redox equation can be balance by using ion-electron method. Oxidation happens when the substance loses one or more electrons. The substance will have an increase in oxidation number and it acts as a reducing agent (reductant). Reduction happens when the substance gains one or more electrons. The substance will have a decrease in oxidation number and it acts as an oxidising agent (oxidant). 22

MATTER CHAPTER 1EXAMPLE 1.13Balance the equation of the oxidation of Fe2+ to Fe3+ by Cr2O72- in basic solution.(a) Write the unbalanced equation for the reaction ion ionic form.(b) Separate the equation into two half-reactions.(c) Balance the atoms other than O and H in each half-reaction.(d) For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.(e) Add electrons to one side of each half-reaction to balance the charges on the half- reaction.(f) If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 23

MATTER CHAPTER 1(g) Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.(h) Verify that the number of atoms and the charges are balanced.(i) For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. PRACTICE 1.6 1. Balance the equations below. (a) Cu + NO3 + H+→ Cu2+ + NO2 + H2O (b) MnO4- + H2SO3 → Mn2+ + SO42- (acidic) (c) Zn + SO42- → Zn2+ + SO2 (basic)LIMITING REAGENT AND PERCENTAGE YIELDMost of the time in chemistry, we have more than one reactant. So we need to completelyuse up one reactant and not the other reactant. This reactant is said to be in excess (there istoo much). The limiting reagent limits the amount of products that can be formed. Thelimiting reagent present is insufficient quantity to consume the other reactant(s). Thissituation arises when reactants are mixed in non-stoichiometric ratios. Therefore, a limitingreactant is the reactant that is completely consumed in a reaction and limits the amount ofproducts formed. An excess reactant is the reactant that is not completely consumed in areaction and remains at the end of the reaction.As an example, for a reaction S + 3F2 → SF6, if 4 mol of S react with 10 mol of F2, which ofthe TWO (2) reactants is the limiting reagent? From the equation, we know that for 1 mol ofS, 3 moles of F2 will be consumed. So if 4 moles of S is used, 12 moles of F2 is required.The number of mol for F2 supplied is only 10 moles, meaning that F2 is the limiting reactant. 24

MATTER CHAPTER 1The percentage yield is the ratio of the actual yield (obtained from experiment) to thetheoretical yield (obtained from stoichiometry equation) multiply by 100 %.Percentage yield = actual yield x 100 % theoretical yieldSTOICHIOMETRIC CALCULATIONS USING MOLE CONCEPT INCLUDING LIMITINGREAGENT AND PERCENTAGE YIELDEXAMPLE 1.141. C is prepared by reacting A and B: A + 5B → C In one process, 2 mol of A react with 9 mol of B. (a) Which is the limiting reactant? (b) Calculate the number of mole(s) of C? (c) How much of the excess reactant (in mol) is left at the end of the reaction?2. A student conducts a single displacement reaction that produces 2.755 grams of copper. Mathematically he determines that 3.150 grams of copper should have been produced. Calculate the student's percentage yield. 25

MATTER CHAPTER 1PRACTICE 1.71. Carbon tetrachloride was prepared by reacting 100 g of carbon disulfide and 100 g of chlorine. Calculate the percentage yield if 65.0 g of CCl4 was obtained from the example.CS2 + Cl2 CCl4 + S2Cl22. In a certain experiment, 14.6 g of SbF3 was allowed to react with excess CCl4. After the reaction was finished, 8.62 g of CCl2F2 was obtained. 3CCl4 + 2SbF3 → 3CCl2F2 + 2SbCl3(a) What was the theoretical yield of CCl2F2 in grams?(b) What was the percentage yield of CCl2F2? 26

MATTER CHAPTER 1 CHAPTER 1: MATTER1. (a) An atom P is three times heavier than one carbon-12 atom. Calculate the relative atomic mass of element P.(b) The relative atomic mass of Q is 25. Calculate the mass of one atom of Q in gram.2. (a) What is the mass of atom copper, Cu?(b) How many grams of Cu are there in 2.55 mol Cu?3. (a) Plutonium 239 (Pu-239) is one of the fissile isotopes used to make nuclear weapons. What is the atomic number of Pu-239? How many protons, neutrons and electrons in Pu-239. (For this question, students need to refer to the Periodic Table).(b) Define nucleon number and isotope. Give the number of protons, neutrons and electrons in each of the following species.i. 79 iii. 210 Br At 35 85ii. 130 iv. 69 Ba2+ Ga3+ 56 314. Figure 1 below shows the mass spectrometer in the determination of atomic mass of an element. Name and explain of each component (A-E) in the mass spectrometer below. AC D To E vacuum 27 B pump Figure 1

MATTER CHAPTER 15. Mass spectrum of copper is shown in Figure 2. Relative intensity 69 31 63 Figur6e52 mass/ chargeBased on Figure 2,(a) Write all the isotopes of copper.(b) Determine the percentage abundance of each isotope.(c) Calculate the relative atomic mass of copper. (63.62)6. Copper exists as a mixture of two isotopes which are 63Cu and 65Cu with the relative mass of 62.93 and 64.93 respectively. Suppose the ratio of relative abundance of copper isotopes is 65Cu = 0.447. Calculate the relative atomic mass of copper. 63Cu (63.55)7. Naturally occurring silver has two isotopes, one of which has a relative mass of 106.9 and 51.9% abundance. Given the relative atomic mass of silver is 108.07, calculate the relative mass of another isotope. (109.33 u)8. There are 26 isotopes of rubidium, Rb known with naturally occurring rubidium being composed of just two isotopes; 85Rb (72.2%) and the radioactive 87Rb (27.8%). What is the average atomic mass of Rb? (85.556 u)9. The isotopes of chromium are 50Cr, 52Cr, 53Cr and 54Cr with the percentage of abundance and relative masses.Isotope % Abundance Relative mass 50Cr 4.31 49.9461 52Cr 83.76 51.9405 53Cr 9.55 52.9407 54Cr 2.38 53.9389Calculate the relative atomic mass for Cr. (51.9976) 28

MATTER CHAPTER 110. (a) Ascorbic acid cures scurvy and may help prevent the common cold. It is composed of 40.92% carbon, C, 4.58% hydrogen, H and 54.5% oxygen, O by mass. Determine its empirical formula.(b) Phosphoric acid, H3PO4 is a colourless, syrupy liquid used in toothpastes, detergents, fertilizers and in carbonated beverages for a ‘tangy’ flavour. Calculate the percent composition by mass of H, P and O in the compound.(c) A sample of a compound of boron, B and hydrogen, H contains 6.444 g of B and 1.803 g of H. The molar mass of the compound is about 30 g. What is its molecular formula?11. A hydrate of potassium carbonate has the formula K2CO3  x H2O. From 10.00 g of the hydrate, 7.95 g of anhydrous salt was left after heating. Determine the value of x in the formula.12. A 1.50 g sample of a compound, consisting of elements chlorine and chromium, was dissolved completely in water. The solution was added to excess silver nitrate solution, AgNO3. All the chlorine from the sample was converted to 4.074 g of silver chloride precipitate, AgCl. Determine, (a) The numbers of moles of Ag+ and Cl- in the silver chloride precipitate. i. (0.0284 mol)(b) The mass of chlorine and chromium. (1.00678 g, 0.49322 g)(c) The empirical formula of the compound.13. Balance the equations below:(a) PCl3 + H2O H3PO3 + HCl(b) AgNO3 + Na2Cr2O7 Ag2Cr2O7 + NaNO3(c) KClO3 KCl + O2(d) BCl3 + P4 + H2 BP + HCl(e) SO2 + Br2 SO42- + Br-(f) Fe3+ + Sn2+ Fe2+ + Sn4+(g) CrO2- + ClO- + OH- CrO42- + Cl- + H2O(h) MnO4- + Fe2+ Mn2+ + Fe3+(i) K4Fe(CN)6 + H2SO4 + H2O K2SO4 + FeSO4+ (NH4)2SO4 + CO(j) Fe2+ + Cr2O72- Fe3+ + Cr3+ 29

MATTER CHAPTER 1(k) Fe + Cl2 FeCl3(l) NH4NO3 N2 + H2O + O2(m) C3H8 + O2 CO2 + H2O(n) H+ + CO32- CO2 + H2O(o) I2 + Na2S2O3 NaI + Na2S4O6(p) MO2 + HCl MCl2 + Cl2 + H2O(q) NaOH + FeCl3 Fe(OH)3 + NaCl(r) C4H10 + O2 CO2 + H2O(s) Cl2 ClO4- + Cl- (basic)(t) Fe + H2O Fe3O4 + H2(u) Al + Fe3O4 Al2O3 + Fe(v) Cr(OH)3 + IO 3 - CrO 3 2 - + I- (acidic)(w) Fe2O3 + HCl FeCl3 + H2O(x) C2O42- + MnO42- + H+ Mn2+ + H2O + CO214. (a) What is the meaning of limiting reagent? (b) Metal Al reacts with H2SO4 to produce aluminium sulphate, Al2(SO4)3 and hydrogen, H2. If 10.0 g of Al are reacted with 60.0 g of sulphuric acid, 57.0 g of solid Al2(SO4)3 are produced. Determine the limiting reagent. Calculate the theoretical yield of Al2(SO4)3 and the percentage yield.15. Iron, Fe reacts with steam, H2O to produce Fe3O4 and hydrogen gas. In an experiment, 24.5 g of iron were reacted with 12.6 g of steam.(a) Write a balance equation for the reaction above.(b) Determine the limiting reagent.(c) Calculate i. the mass of Fe3O4 produced. (33.8526 g) ii. the volume of hydrogen gas formed (measured at s.t.p). (13.1018 dm3) 30

MATTER CHAPTER 116. Hydrogen gas is produced when a certain mass of zinc reacts with an aqueous solution of hydrochloric acid. The concentration of hydrochloric acid supplied is 12.5 M.(a) Write a balance equation for the preparation of hydrogen gas.(b) Calculate the volume of hydrochloric acid needed to prepare 100 cm3 of 0.5 Mhydrochloric acid. (4 cm3)(c) If 2.356 g of zinc were reacted with 25 cm3 of 0.5 M hydrochloric acid, calculate the mass of hydrogen gas produced. (0.0125 g)17. 1.50 g oxide of metal M, MO2 reacts with excess hydrochloric acid solution, HCl to produce 386 cm3 chlorine gas at s.t.p as given by the following equation: MO2 + HCl → MCl2 + Cl2 + H2O(a) Balance the above equation.(b) Determine the relative molecular mass of MO2 and the relative atomic mass of M.(c) In a separate experiment, 0.20g of MO2 was added to 25 mL of 0.10 M HCl solution. Determine the limiting reagent in the reaction. Calculate the mass of MCl2 produced in the reaction. 31