CHAPTER Te -_) EAD Relative Atomic Mass and Relative Molecular Mass 1 Chemists determine how heavy an atom is by comparing the mass of an atom with another atom, which is taken as the standard. 2. The comparison of the mass of an atom to another atom is called the relative atomic mass, RAM. 3, The relative atomic mass ofan atom is determined using a machine called mass spectrometer. FIGURE 3.1 Mass spectrometer 4 Before 1961, scientists used hydrogen as the standard for comparison of mass. + However, the RAM of hydrogen was then found to be higher than 1. To correct the anomaly, scientists agreed to use oxygen as it is easily obtained. However, oxygen also has three isotopes; “0, "0 and "0, In 1961, chemists and physicists agreed to use carbon-I2 as the standard because carbon is a solid, unlike hydrogen or oxygen which are gases, and thus require covered containers to contain them so that they would not get mixed with other gases from the atmosphere. * Carbon is easily available. By heating petroleum, natural gas or wood in limited amount of air, carbon is obtained, + The percentage of carbon-12 is high compared to the other carbon isotopes. w The Mole Concept, Chemic, Formula and Equation, TABLE 3.1 Percentage abundances of carbon isotopes Isotope J Carbon-12 | Carbon-13 | Carbon-14 Percentage abundance 98.1 1 0.8 (%). Therefore, the average atomic mass of carbon in any sample of carbon is (98.1 x 12) + (1.1.x 13) + (0.8 14) 100 a = 12.027 Thus, we can assume the RAM of any sample of carbon as 12 unit. As carbon-12 is used as the standard, the relative atomic mass of an element is defined as the number of times one atom of the element is heavier than one-twelfth of the mass of a carbon-12 atom. Mass of one atom of the element a ic mass of an element 1 (qq x mass of one carbon-12 atom) a The relative molecular mass, RMM, ofa molecule ofa compound is defined as the number of times one molecule of the compound is heavier that one-twelfth of the mass of a carbon-12 atom. The relative atomic mass of a molecule _ Mass of one molecule of the compound x (Gq x Mass of one carbon-12 atom) CEVA For covalent compounds, the term relative molecular mass, RMM, is used whereas for ionic: the term relative formula mass, RFM, is used.| Two antimony, Sb, atoms have the same mass as the sum of five carbon, C, atoms and eight ‘sodium, Na, atoms. Determine the relative atomic mass of an atom of antimony, [Relative atomic mass: C = 12, Na = 23] Answer: Assuming the relative atomic mass of Sb =m 2x Sb = (6 x C) + (8x Na) 2xm = (6x 12)+ (8x23) fear. Bananas are rich in vitamin B6. The diagram below shows the molecular structure of vitamin B6. Oo—H (@)_ Name the elements that make up a vitamin B6 molecule. (b) Write the molecular formula of a vitamin B6 molecule. (©) Determine the relative molecular mass of a vitamin B6 molecule. [Relative atomic mass: H = 1, C= 12, O=16 Tips: + Spell the name of each element correctly. + Count the number of each element in the molecule. The number of each element is written as subscripts after the symbol of the element. When writing the molecular formula, arrange the alphabets in alphabetical order; C, H, N, O. + There are 8 carbon atoms, 11 hydrogen atoms, 1 nitrogen atom and 3 oxygen atoms. Solution: (@) Carbon, hydrogen, nitrogen, oxygen (b) Molecular formula of the molecule is C,H,,NO, (©) Relative molecular mass of a vitamin BG molecule is 8(12) + 11(1) + 14 + 3(16) = 169 N= 14, __RMM of C.H,,Cl,NO,PS = 350.5 985.5) + 14+ 9(16) +31 +92 = 350.5 106.5 +14 +48 +31 +32 = 350.5i: acne Determine the relative formula mass of the following ionic compounds. Solutio! {@)_ Relative formula mass of (NH,),CO, = 2[14 + 4(1)] + 12 + 3(16) {@) Ammonium carbonate, (NH,),CO, ate (0) Aluminium sulphate, Al,(S0,), — * © eae mgso, mo a (0) Relative formula mass of Al,(SO,), salt, MgSO, 7H, (@) Cobalt) chioride hexehycrate, COCl,6H,O 207+ te + 4(16)) [Relative atomic mass: H = a i e 7 A : eas eee "| (@). Relative formula mass of MgSO,.7H,O fea 24 + 32 + 4(16) + 7[2(1) + 16] Tips: 24 +32 +64 +126 {@) There are two ammonium ions, NH", and one oa carbonate ion, €O,",In ammonium carbonate. elative formula mass of CoCl,.6H,0 (&) Aluminium sulphate has two aluminium fons, Ae Sooo oe) tial ‘Al, and three sulphate ions, $0". Be ee (©) MgS0,,7H,0 has seven water molecules. he EXAMPLE JF} > Calculate the mass percentage of water from crystallisation in CuSO,.5H,0. [Relative atomic mass: H = 1, O = 16, S = 32, Cu=64] Solution: Relative formula mass of CuSO,.5H,O = 64 +32 +4(16) + 5(18) = 250 The mass of water from crystallisation is 5 x 18 = 90 Thus, mass percentage of water from crystallisation Mass of water = Relative molecular mass * 100% = 20 = 759 «100% = 36% Relative formula mass of X,(PO,), is 259. Determine the relative atomic mass of element X. Answer: Assume relative atomic mass of element X = m 3m + 2[81 + 4(16)] = 259 | 3m + 2(31 + 64) = 250 3m + 2(95) = 259 3m + 190 = 259 3m = 259-190 = ees =23 | [Relative atomic mass: O = 16, P = 31)Mole Concept t 1 The relative atomic mass of hydrogen is 1, whereas the relative atomic mass of carbon is 12. 2. Therefore, Da + the mass of carbon will always be twelve times heavier than the mass of hydrogen provided the number of carbon atoms and hydrogen atoms are the same. + Thus, a sample containing 12 g of carbon will contain the same number of atoms as 1g of hydrogen. FIGURE 3.2 The number of atoms is the same in 12g of carbon and 1 g of hydrogen 3 How many atoms are there in 12g of carbon? Using mathematical calculations, scientists have determined that there are 6.02 000 000 000 000 000 000 000 or 6.02 x 10 atoms in 12 g of carbon. Therefore, 6.02 x 10® = one mole (abbreviation: mol) 4 The number 6.02 x 10” is called the Avogadro number or Avogadro constant (N,). 5 Therefore, one mole is the amount of a substance that contains the same number of particles as there are in 12 grams of carbon-12. For example, +L mole of gold contains 6.02 x 10” gold, Au, atoms +L mole of copper(II) ion, Cu™, contains 6.02 x 10” Cu” ions. +L mole of water, H,O, contains 6.02 X 10 H,O molecules. eras aac ‘Amedeo Avogadro was an Italan physics professor. He proposed that equal volumes of different gases ‘contain equal number of molecules provided they are at the same temperature and pressure. He showed that 22,4 dmn® of any gas at temperature of 0°C and pressure of 1 atmosphere contains 6,02 x 10° ‘molecules. Therefore, the value 6.02 x 10**is called the Avogadro number or Avogadro constant in Fis honour. ~~ TeHoleConeptChemialFomulaandémuation (NASI!Number of moles and number of particles 1 Since 1 mole contains 6.02 x 10° particles, Hence, Number of particles = n x.N,, where N, = 6.02 x 1 2 1f6.02 x 10” particles are found in 1 mole, the number of moles contai FRESE ee then n moles of a substance contain 7 x 6.02 x 10 particles 0? Calculate the number of particles in @) 0.2 mol of iron. (0) 1.25 mol of chloride ions, Cr. (©) 0.05 mol of water molecules, [Avogadro constant, N, = 6.02 x 10%] Solution: (a) 1 mol of iron contains 6.02 x 10° Fe atoms. Therefore, 0.2 mol of iron contain = 1.204 x 10® Fe atoms (0) 1 mol of chioride ion contains 6.02 x 10% Cr ions. Therefore, 1.25 mol of Cr ions contain AB.et 6.09 10°F ions = 7.525 x 10 Cl ions, (©) 1 mol of H.0 contains 6.02 x 102 HO molecules. ‘Therefore, 0.05 mol of Ho contain 0.05.rA0t = oak * 6.02 x 10° H,0 molecules 301 x 10% H,0 molecules 3.01 x 10” H,O molecules HOTTIPS Ensure both the Numerator and denominator are of the same unit Calculate the total number of ions in (@) 1.2 mol of calcium chloride, CaCl, (©) 0.02 mol of aluminium sulphate, Al,(SO)),. [Avogadro constant, N, = 6.02 x 10%] Solution: (@ CaCl, > Ca + 2Cr s 1mol 1mol = 2mol er aes 3 mol 1 mole of calcium chloride dissociates to Produce a total of 3 moles of ions. 1.2 mol of CaCl, dissociate to produce 1.2 mer 2mer 3 mol ions = 3.6 mol of ions 1 mol has 6.02 x 10% of ions Therefore, 3.6 mol = 3.6 x 6.02 x 10” ions 2.1672 x 10% ions ©) AYO), > 2am 4 3g0% 1 mol 2mol__ 3 mol 5 mol 1 mole of Al(SO)), jonises to produce 2 moles of AP* ion and 3 moles of SO,* ions, a total of 5 moles of ions, * 0.02 mol of aluminium sulphate will produce 4 total of 0.02 x 5 mol = 0.1 mol of ions 1 mol has 6.02 x 10% of ions Therefore, 0.1 mol = 0.1 x 6.02 x 10° ions .02 x 10 ions cy,Calculate the number of moles of the following { (b) 1 mole of water contains 6.02 x 10% H,O substances. molecules * (@ 3.01 x 10? copper atoms Number of moles, n, containing 1.204 x 10* (b) 1.204 x 10 water molecules H,0 molecules fAvogadro constant, N, = 6.02 x 10% mot] Solution: (@) 1 mole of copper contains 6.02 x 10*Cu atoms pene Number of moles, n, containing 3.01 x 10 Cu atoms _ 3.01 X 10 atoms ="6.02 x 107 atoms * 1 ™l = 0.05 mol Number of moles and mass of substances 1 One mole of a substance contains 6.02 X 10” particles, Therefore, the mass of one mole of a substance is the mass of 6.02 X 10? particles. 2 The mass of one mole of atom of a substance is the relative atomic mass of the substance expressed in grams. For example, given that the relative atomic mass of C = 12, Zn = 65 and He = 4. Therefore, + 1 mole of carbon atom is 12 g + 1 mole of zinc atom is 65 g ee 71 stom + 1 mole of helium atom is 4 g 3. The mass of one mole of molecule of a compound is the relative molecular mass of the compound expressed in grams. Fir cal, given that the relative molecular mass of H,O = 18, CO, = 44 and NH, = 17. Therefore, +L mole of water molecule is 18 g | Each contains + 1 mole of carbon dioxide molecule is 44 gf 60 102 molecules + Lmole of ammonia molecule is 17 g 4 Since 1 mole of an element is its relative atomic mass (RAM) in grams, then the mass of y mole of an element = oer RAM Conversely, since 1 mole of an element is its relative atomic mass in grams, then the number of moles of m grams of the 5 Similarly, 1 mole of a compound is its relative molecular mass (RMM) in grams, then the mass of y mole of the compound =~ pol MM Tot Conversely, : since 1 mole of a compound is its relative molecular mass in grams, then the number of moles of m grams of (RAM) or (x RMM) (@ RAM) or (* RMM) | aThe figure shows the str compound found in coffee and tea. (@) Determine the relative molecut (0) Ifa caffeine pill contains 600 mg ¢ moles of the compound is in the pill? [Relative atomic mass: H | Comment: formula of the compound. Answers: (@) The molecule is made up of 8 carbon, 10 hydrogen, 4 nitrogen and 2 oxygen atoms. “The molecular formula of a caffeine molecule is C,H, .N,O,- (12) + 10(1) + 4(14) + 2(16) RMM of C,H,,N,O, 600 fe) 090.8 00 = 0.6 gram | Relative molecular mass of caffeine | 1 mol of caffeine 194 uctural formula of caffeine, 4 lar mass of the compound. 19 of caffeine, how many =1,C=12,N=14,0=16) | Count the number of atoms of each element then write the | | 0.6 . = x 0.6 g of caffeine = 25-9 x 1 mol = 0.0031 mol Quick Access “ ‘Alydrogen atom can form 1 covalent bond, ‘oxygen atom forms 2 covalent bonds, nitrogen atom 3 covalent bonds, while a carbon atom forms 4 covalent bonds. Calculate the mass of ammonia that contain the same number of molecules as 2.2 g carbon dioxide. [Relative atomic mass: H = 1, C = 12, N= 14, O= 16] Solution: 0.05 mol of ammonia, NH,, have the same number of molecules as 0.05 mol of carbon dioxide, CO, 1 mol NH, 0.05 mol NH, Quick Access “ ‘Substances with the same number of moles ‘contain the same number of particles. The following diagram shows the structural formula of halothane gas, which is used as a general anaesthesia during medical operations. It is also|, called the sleeping gas. Be F [Relative atomic mass H = 1, rb C=12,0=16,F=19, H-C-C-F = a Ll Cl = 35.5, Br = 80] bt (@) Calculate the number of moles of halothane in 39.5 g of the compound. (b) Calculate the mass of oxygen gas which Contains two times the number of molecules as in 39.5 g of halothane. Solution: (@) The RMM of halothane = 2(12) +1 +80 + 35.5 + 3(19) = 197.5 1 mol of halothane = 197.5 g 39.5 Therefore, 39.5 g of halothane = =o 7% = 0,2 mol (©) 0.4 mole of oxygen gas (O,) contain two times the number of molecules as in 0.2 mole 2 halothane, molNumber of particles and mass of substances PPA BIRR RPS: peveyrerenon: prow cap al “To convert Ni ‘ ie particles to mass, -Avogadro constant. ; - i lumber of particles is converted _ thenumber of HIE tornumber of mois by cividing > ‘the number of particles by the 8 “Orelatve molecular mass, mane iB Ot A ‘sce ati “rl ec ri Calculate the number of particles in (@) 13g of chromium, Cr. (0) 3.2.g of methane, CH,. (©) 10g of sodium hydroxide, NaOH. [Relative atomic mass H = 1, C N, = 6.02 x 10® mol] Solution: (@ 1 molofCr=529 Therefore, 1.3 g of Cr = 5 mol 025 mol 1 mol of chromium contains 6.02 x 10 atoms Therefore, 0.025 mol of chromium contain 0.025 mao Se ie = 1.505 x 10 atoms (0) 1 mole of CH, = 169 Hence, 3.2 g of CH, 1 mol of CH, contains 6.02 x 10 molecules Thus, 0.2 mol of CH, = 0.2 x 6.02 x 10% molecules = 1.204 x 10 molecules 12, O = 16, Na = 23, Cr = 52; Avogadro constant, (©). 1 mol of NaOH = 40g 10 Hence, 10 g of NaOH =—>- mol = 0.25 mol NaOH ——» Na* + OH 1 mol 4 mol 1 mol — 2mol 14 mole of NaOH dissociates to produce 2 moles of ions. Therefore, 0.25 mol of NaOH will produce 0.5 mol of ions. 1 mol of NaOH contains 6.02 x 10% of ions Thus, 0.5 mol of NaOH contains = 0.5 X 6.02 X 10 of ions 3.01 x 10® of ions—_e Calculate the mass of (a) 7.525 x 10 titanium, Tiatoms {p)3.01 x 10% glucose, CxH,,0, molecere {6) 1506 x 10% carbon dioxide, COp molecules. [Relative atomic mass: H = 1, C= 8A, O= “18, a yroey®. Avogadro constant, N, =6.02 x 10m Solution: {@)_ 1 mol of Ti contains 6.02 10 atoms 71505 x 10" T atoms = EAS = 0.125 mol qmol Ti =489 0.125 mol Ti = 0.125 x 489 269 (0) 1 molof C,H,0, contains 6.02 x 10® molecules 1x 10° mol A molof CO, = 44g 2.5 mol of CO, = 2.5 x 44g Number of moles, ey: The relative atomic mass of X and Y is 16 and 4 respectively. Which of the following statements jg true about atoms XandY? ‘A. The mass of one atom Y is 4 gram B Thenumber of protons in one atom X is 16 8 moles of ¥ has the same mass as 1 mole of D x The density of one atom X is 4 times the density of one atom Y Comments: : 4 The mass of 1 atom YiS = ops age % 49 (is false) The relative atomic mass is 16. This is the total number of protons and neutrons in the nucleus of ‘the atom. (Bis false) 1 mole of Xis 16 g Hence, 4 moles of Yis 4x 4g=169 (Cis true) ‘The density of each atom cannot be determined because the volumes of the atoms are not given. (Dis false) Answer: C - alice molar volume and volume of gases 1 The volume occuy ipied by a gas depends on the temperature and pressure, temperature 1) SBS FIGURE 3.3When pressure is decreased (by decompressing the gas), the volume of the gas increases. FIGURE 3.4 2. At standard temperature and pressure, STP, * one mole of gas occupies a volume of 22.4 dm’ or 22 400 cm? + The conditions at STP are 0°C and 1 atmospheric pressure. 3. Atroom temperature and pressure (room conditions), + which is at 25°C and 1 atmosphere pressure, + one mole of the gas occupies a volume of 24 dm or 24 000 cm’, 4 The volume occupied by one mole of gas is called the molar volume. Examples: 1 mole of N, gas 1 mole of CH, gas occupies 22.4 dm? at STP or 1 mole of CO, gas [24 dm’ at room conditions 1 mole of NH, gas Number of moles and volume of gases 1 Since 1 mole of gas occupies 22.4 dm? at STP (or 24 dm? at room conditions), then n moles of gas occupy a volume of tee X 22.4 dm? at STP. or TE X 24 dm’ at room conditions 2 ‘Since 1 mole of gas occupies 224d? at STP (or24 dn at room condition) then the number of moles of gas in V dm of gas is Viet V gat ., a da 1 mol at STP or 24 dnt X 1 mol at room conditions (% 22.4 dm’ at STP) or (X 24 dur’ at room conditions) (22.4 dm? at STP) or (24 dm’ at room conditions) 2Eo" Calculate the volume of (@) 0.75 mol oxygen gas at STP. {b) 3 mol carbon dioxide gas at STP. [Molar volume = 22.4 dm? at STP] Solution: @ © ‘4 mol of gas occupies 22.4 dm? at STP. Therefore, 0.75 mol of oxygen gas occupies =o ea 22.4dm* am x 22. = 16.8 dm?® 1 mol of gas occupies 22.4 dm? at STP. Therefore, 3 mol of carbon dioxide gas occupy _ 3 Bo, dm? Tor * 224 im‘ = 67.2 dm? (: Calculate the number of moles of the following gases at room conditions. (@) 6 dm’ of nitrogen gas (0) 360 cm? of chlorine gas (6) 4800 cm? of ammonia gas [Molar volume = 24 dm? at room conditions] Solution: (@ 1 mole of gas occupies a volume of 24 dm? at room conditions. Thus, the number of moles of gas in 6 dm? of oxygen gas = a x1 mol = 0.26 mol © © 4 mole of gas occupies 24 dm? or 24 000 cm? at room conditions. Thus, the number of moles of gas in 360 cm* of x1 mol ; __ 360 oar” chlorine 988 = 33 909 ca = 0.015 mol 1 mole of gas occupies 24 000 cm® at room conditions. Thus, the number of moles of gas in 4800 cm’ 4800 cer? of ammonia gas = 79-96 aaa 1 Mo! =0.2 mol Molar mass and volume of gasesNumber of particles and volume of gases a" Teconvert | ABU oF 8 IS converted e (988 is converted to Number of moles is conver Teer? ME number ofmoles by dhviging NN fonumber of paricie by. A the volume of gas by the molar multiplying the number of pagictes, volume. tmole withthe Aveoado (+224 dm?) or (+ 24 dm’) oun : (22.4 dm) or (X 24 dm’) Ai EE Se RGIS Number of moles of gas is then ‘Number of particles is piece vert tee Converted to volume of gas by converted to number of moles am MUMBer Of multiplying the number of moles by dividing the number of Peee sees oe with the molar volume. particles by the Avogadro of gas ‘constant, N, i 4 mod site Which of the following gases contain 6.02 x 10% 1 1 molofH, gas=2g9 | of molecules? eile ae / [Relative atomic mass of H =12,N=14, oe ee = 16; Avogacro constant, N, = 6.02 x 19° mot =05 mol Wis incorrect] 1.0 g of hydrogen gas 2.8 g of nitrogen gas 3.8 g of carbon dioxide gas 1.60 g of methane gas land i Wand iv andi 1 and 1 ‘ Mt jv A B c D Comment: 1 mol = 6.02 x 10 molecules 02 x 10% 02 x 10% £02 x 107 6.02 X 10% molecules = -205 * TPE x 1 mol = 0.1 mol I 1 mol of N, gas = 28 g is correct] {lis incorrect] 1.60 g of CH, 1 gas = +59 mot =O0.1mol [Vis correct] Answer: B Ties ee 1 Calculate the number of atoms in (@) 0.12 mol of carbon dioxide, CO,. (0). 0.05 mol of methane, CH,- 2 Calculate the volume of gas occupied by. (@ 336g of ethene gas, C,H, at STR. « (©) 1.149 of methyl isocyanate gas, ‘CH,NCO, at STP. [Avogadro constant, N, = 6.02 x4 Relative atomic mass: H = 1, C . N=14, 0 = 16; Molar volume = 22. “at STF] Te Pt tas eeChemical Formula 1 The chemical formula of a compound contains two information: + the elements present (denoted by their symbols) and » the number of atoms of each element that combined to form the compound. The number is indicated py subscripts written after the symbol of the element. TABLE 3.2 Examples of compounds and its chemical formula . Water 2 hydrogen atoms + {oxygen atom Dichloro-dipheny!- C,H,Cl, + 14 carbon atoms trichloroethane, DDT + 9 hydrogen atoms (an example of a pesticide) + 5 chlorine atoms Penicillin CygHyeN,0,S * 16 carbon atoms (an example of a antibiotic) + 18 hydrogen atoms + 2nitrogen atoms + 4 oxygen atoms + 1 sulphur atom Chemical formulae of ionic compounds 1. The chemical formula of an ionic compound can be written if we know the charges of its cation and anion are known. 2 In general, if cation is X" and an anion is Y*, then the formula of the ionic compound is X,Y,, so that the net charge is zero. If m =n, the chemical formula is XY, where it is written in the simplest ratio. 3 For example, + Ammonium phosphate : HOTTIPS '* For polyatomic ions (ions with more than one clement), they are writen in brackets, with the number of the ions in the ‘compound written afte the 2 e losing bracket. Formula: Cr80)), “Transition elements form | ions with more than one Check: Cr, Cr, SO, S07, SO? ‘@xidation ee 4 Ae numerals (|, Il Ill, IM, .--) (8) 8) are used to denote the ‘oxidation number of the ions.Empirical formula and molecular formula 1 The e Pound shows the simplest rato ofthe atoms ofthe elements that combine £0 tla of a compound shows the actual number of the atoms of the elements that combine TABLE 3.3 Examples of compounds with their molecular and empirical formulae Propene CH, C:H=14: cH, Ethane CH, C:H= CH, my Glucose G4H,,0, C:H:0=1:2:1 a0 Quinine H.-N, C:HiN = 5:7:1 CHIN 3 The following tes can be taken io deternine he epi formula of compound pl, to determine the empirical formula of a compound that was produced by the reaction between 1.40 g of iron and 2.67 g of chlorine, follow the steps shown below fy Beacon [Relative atomic mass: C1= 35.5, Fe = 56] dividing the mole of each iment with the smallest mole Quinine is a drug used to treat malaria. Chemical analysis show that it contains 74.10% carbon, 7.40% hydrogen, 8.64% nitrogen and the rest is oxygen according to mass. Determine the empirical formula of quinine. [Relative atomic mass: H = 1, C= 12, N= 14, O= 16] Solution: (% mass C) + (9% mass H) + (% mass N) + (9 mass 0) = 100% 74A0% + 7.40% + 8.64% + (% oxygen) = 100% (% oxygen) = 100% - 90.14% = 9.86%c H N 3 74.10% 7.40% 8.64% 9.86% 9.86 ¥ 7.40 8.64 986 T4419 x 400g | ~Aqq-* 109 | ~too “199 | “100 100 g 274.109 =7.409 =8.64g = 9.86 g 7.40 8.64 9.86 ae mol Tae 7450! 16 mol =6.175 mol =7.40 mol = 0.617 mol =0.616 mol 175 7.40. _ 0.616 _ SUB -10mol | Faig = 12mol O16. — + ml Empirical formula of quinine is C,,H,,NO <7 me Zinc phosphide is a rat poison. 2.57 g of phosphorus combines with a gram of zinc to form a compound with the formula Zn,P,. Determine | the value of a. [Relative atomic mass: P = 31, Zn = 65] Answer: | Element Zn || Mass agram ‘Number of moles a- 5 = mol ‘Simplest ratio 3 a Number of moles of Zn _ 3 Number of moles of P_ ~ 2 a | 65 3 0.083 “2 Paar | we 7 x 0.088 | a = 0.1245 x 65 a = 8.093 gOctyl ethanoate is an artificial flavouring that gives off orange flavour. A chemical analysis shows that it contains 69.8% carbon, 11.6% hydrogen and the remaining is oxygen according to mass. The relative molecular mass of octyl ethanoate is 172, Determine the (a) empirical formula and (b) molecular formula of octyl ethanoate. [Relative atomic mass: H = 1, C = 12, O = 16] Answers: Percentage mass of oxygen = 100% - 69.8% - 11.6% = 18.6% (@) [Element H Percentage 11.6% Number of moles a8 = 11.6 mol Simplest ratio 32 = 5 mol 18 - 10mol Empirical formula of octyl ethanoate is C,H,,0 (0) Assume the molecular formula of the compound is (C,H,,0), Relative molecular mass of (C,H,,0), = 172 | [5(12) + 10 + 16)n = 172 86n = 172 a2. 86 =2 Molecular formula of octyl ethanoate is (C,Hj,0), OF C,Ha,0,- ‘A gaseous hydrocarbon X contain 85.7% carbon according to mass. 5.6 g of gas X occupies a volume of 4.48 din? at STP. [Relative atomic mass: H = 1, C = 12; Molar volume = 22.4 dm® at STP] (@)_ Determine the empirical formula of X. () Determine the relative molecular mass of X. (©) Determine the molecular formula of X. Answers: (a) Percentage mass of hydrogen = ons 85.7% = 14. c H 85.7% 7.14 mol iT Oe +etioine to &. igang rv. 2eee 5.6 g of the hydrocarbon occupies 4.48 dm? at STP mole of gas occupies a volume of 22.4 dm® at STP me ey 2Adnf 56 Therefore, the mass of 1 mol of gas X = -7°75 oe 6g = 289 Relative molecular mass of the hydrocarbon is 28 (@) Assume the molecular formula is (CH,), Relative molecular mass of (CH,), = 28 (12 42)n =28 14n = 28 n ~~ LES Determining the empirical formula of magnesium oxide Materials and apparatus: Crucible with lid, tripod stand, sandpaper, Bunsen burner, clay pipe triangle, electronic balance, tongs, magnesium ribbon Procedure: 11 A10 cm magnesium ribbon is polished with sandpaper Until itis shiny, to remove the oxide layer on its surface. 2 The magnesium ribbon is then coiled around a pencil into a loose coil 3 Anempty crucible and its lid are weighed and the mass is recorded, 4 The coiled magnesium is then placed into the crucible. The crucible, its lid and the magnesium are weighed and the mass is recorded. 5 The crucible is placed on a clay pipe triangle supported by a tripod stand as shown in Figure 3.5. 6 The crucible is heated without its lid strongly until the magnesium starts to bun. When the magnesium ribbon started to burn, the crucible is closed with its lid. 7 Using a pair of tongs, the lid is opened slightly from time to time and quickly placed back to allow oxygen in the air to enter the crucible for the combustion of Same magnesium. 8 When the burning of the magnesium ribbon is complete, the lid is removed and the content in the Crucible is heated without the lid for 1 to 2 minutes to ensure all of the magnesium is oxidised. 9 Then, the lid is placed back, the crucible Is cooled and weighed together with its id, 10 The heating, cooling and weighing process is repeated until a constant mass is obtained, The constant mass is recorded. Results: @ gram b gram gramMass of magnesium used = (b - a) gram Mass of oxygen that combined with magnesium = (¢ - b) gram Calculation: Mg ° (6-2) gram (¢-b) gram b-a c-b ae mol Te mol x y Empirical formula of magnesium oxide is Mg,O, Safety precautions: + The magnesium ribbon must be polished with sandpaper to remove a layer of oxide on its surface as magnesium is a reactive metal and its surface has been oxidised by oxygen in the air. The heating and cooling process is repeated until a constant mass is obtained. This indicate all the magnesium had been oxidised. After lifting the crucible lid, it must be closed quickly again to avoid the white smoke of magnesium oxide from escaping. Materials and apparatus: Water, copper) oxide powder, zinc granules, 1.0 mol dr hydrochloric acid, wooden splinter, boiling tube, cotton buds, rubber stoppers, rubber tube, 12 cm glass tube, 10 cm glass tube, spirit lamp, retort stand with clamp, wooden block, electronic balance, spatula Procedure: glass tube (12cm) copper) oxide 4.0 mol der hydrochloric acid }— zinc granules ‘wooden block FIGURE 3.6 1 The mass of a 12 cm glass tube is weighed with an electronic balance and the mass is recorded. 2 Some copperill) oxide is put into the glass tube. The wooden splinter is used to move the copper) oxide powder to the middle of the glass tube. 3 The mass of the glass tube together with its contents and the mass is recorded. 4 Aboiling tube is filled until 2/3 full with water.ther boiling tube. 4.0 mol dir? hydrochloric acid is added into the anot i few zinc granules is inserted into i. has a 10 om glass tube. The boiling tube is clampeq boiling tube until it is 1/3 full. 6 The baling tube is closed with a rubber stopper that SE tt cota wader is connected as shown in Figure 3.6. ie Pr ey that contains oppo by alowng the air bubbles to be released in the water before The hydrogen g: ing the heating process. eet aide is heated with a spirit lamp. The hydrogen 9 ‘through the glass tube. 40 The eating is stopped when the colour of copper! ; 41. The hydrogen gas is kept flowing until the glass tube 's 42 The glass tube that contains the brown powder is remo tube is removed with a cotton bud. . oS f the glass tube together with its contents is weighed and rec . : is The heating, coating ‘and weighing process is repeated until a constant mass is obtained. This is to ensure that all of the copper(ll) oxide has been reduced. 45 The constant mass is recorded. Results: on as is allowed to continuously flow (ll oxide turned brown completely. cooled back to room temperature. ved, The water droplets at the end of the glass ‘The mass of copper produced = (43.40 - 35.38) g = 8.029 ‘The mass of oxygen that has combined with copper = (45.58 - 43.40) g 18g cu ° 8.02.9 2.189 02 8.02 mol = 0.125 mol 2.18 _ Er = 0-136 mol 0.125 _ Bage = 1-00 mol G588 = 1.09 mot Empirical formula of copperill) oxide is CuO Safety precautions: The chenioa une ws Gry hydrogen gas is anhydrous calcium chloride. glass tube is displaced by allowing a stream of hydro pass it gen gas to ause a mixture of hydrogen and oxygen can explode when lighted, Hydrogen gas is Saat ty i sting a mists af granulated zine metal and dilute hydrochloric acid aaa taim 1e heating, cooling and weighi ig ila ; re cee. cling ad Hohn Process is repeated until a constant mass is obtained to ensure all the ‘The determination of empirical formula of oxide of rt € an elemer hydrogen in the reactivity series, the metal is heated direct! For metals below hydrogen in the reactivity series, nt depends on its reactivity. For metals above ly in the air in a crucible (Activity 3.1). % louse : the empirical formula of its oxide is determined bY * Hydrogen gas which is more reactive than the metals, has a stronger a nd can oor ; rate. Is, iger attraction for the oxygen at K, Na, Mg, Al, Zn, Hydrogen, Fe, Sn, Po, Cu, Hg, Ag Reactivity decreases =a) ssn a eee eal‘The empirical formula of lead oxide is determined by reducing it with hydrogen gas. (2) () Name two chemicals that can be used to (i). Write a chemical equation for the reaction. (b) Name the chemical used to dry the hydrogen gas. (c) Why is hydrogen gas passed through the combustion tube for a period of time before lighting the hydrogen gas? (@ After reduction, a stream of hydrogen is allowed to flow through the combustion tube. | Explain why. | produce hydrogen gas. Chemical Equation Symbol of elements (@)_ The process of heating, cooling and weighing is repeated until a constant mass is obtained. Explain why. Answers: (@) (i) Zinc and dilute hydrochloric acid (i) HQ) + 2HCKaq) -» Zncl(aq) + H,9) (b) Anhydrous calcium chloride (©) To displace all of the air in the tube because a mixture of oxygen and hydrogen can explode iflighted. (@) To cool the lead metal so that it will not be oxidised by the oxygen from the ait (©) To obtain a constant mass which indicates all lead oxide has been reduced. 1 The symbols of most elements are derived from the first alphabet of the name of the element. 2 Some elements have the same first alphabet. Thus, a second alphabet is used. The second alphabet is written in small letters. Carbon: c Boron B Cobalt Co Beryllium Be Chromium Cr Barium Ba Cadmium Cd Bromine Br Caesium Cs ‘Sulphur s Calcium Ca Silicon Si Chlorine a Selenium Se Hydrogen H Scandium Sc Helium He ‘Strontium, sr 3. The symbols of some elements are derived from their Latin names. TABLE 3.5 Examples of symbols of elements derived from their Latin names Tin ‘Stannum Sn Lead Plumbum Pb Silver Argentum Ag Mercury Hydragyrum Hg Gold Aurum ‘Au Sodium Natrium Na Potassium Kalium K | 2How to write a balanced chemical equation? 7 ETE RATT 5058 Tere DUPE Write the correct formula of every reactant on the “left-hand side of the equation. For instance, e® OO Reactants (left-hand side) Hydrogen gas and iodine gas Balance the equation. ‘Write a number in front of the chemical formula of cub reactants or products. _Make sure that the number of atoms of each element are the same on either side of the arrow, ‘When balancing the equation, we must not "change the chemical formula of the reactants or Products. ue nae equation. Fea ha eo anna ete Products (right-hand side) Hydrogen iodide gas Dm. IPT RR \Wiite the physical states of the réactants and > products. (6) for solid cum to dedemeat : “+ (for tiquid | * (Q) for gas leat | + (@q) for aqueous state (when a substance is For example, the balanced equation between hydrogen gas and iodine gas that produces 5a Hg) * ue Ss > 2g) When potassium iodide aqueous solution is poured into lead(\l) nitrate aqueous solution, a yellow lead{\I) iodide precipitate (solid) and potassium(|l) nitrate aqueous solutions are produced. Write a balanced equation for this chemical reaction. Comment: Reactants: Potassium iodide - KI Leadi(|I) nitrate — Pb(NO,), Products: Lead{tl) iodide ~ Pol, Potassium nitrate - KNO, Equation: Ki + Pb(NO,), -> Pbl, + KNO, + This equation is not balanced as the number of iodide and nitrate ions are not the same on the left and right side of the equation. The equation is balanced by writing Kl and 2’ in front of KNO,. 2KI + Pb(NO,), — Pbl, + 2KNO, + Finally, write the physical states of the reactants and products. Solution: | 2Kilaq) + Pb(NO,),(aq) > PbI,(s) + 2KNO,(2q) | Ely)sure in srivay sLUICHOMetry problems 1 Stoichiometry is the study of quantitative changes during a chemical reaction. 2. Such knowledge is important for us to predict the quantity of yield of the products formed during a reaction, The coefficient or number written in front ofa chemical represents the number of moles ofthe reactants taking part in the reaction and the number of moles of products formed during a chemical reaction, 4 For example, from the equation 2SnO(s) + C(s) > 2Sn(s) + CO,(g), 2 moles of tin(II) oxide reacts with 1 mole of carbon to produce 2 moles of tin and 1 mole of carbon dioxide. Ethanol burns in air according to the following equation, C,H,OH() + 30,9) > 2C0,(q) + 3H,0() lf9.2 g of ethanol is burned completely in air, calculate (@)_ the volume of carbon dioxide produced. (0) the volume of oxygen gas needed for the reaction at STP. [Relative atomic mass: H = 1, C = 12, O = 16, Molar volume = 22.4 dm? at STP] Solution: RMM of ethanol, C,H,OH = 2(12) +6 + 16 = 46 (@_1 mole of C,H,OH produces 2 moles of CO, 46 g of C,H,OH produces 2 x 22.4 dm? of CO, at STP ‘Thus, 9.2 g of C,H,OH will produce = FERS x 2 x 22.4 de of CO, = 8.96 dm’ of CO, (©) 1 mole of C,H,OH reacts with 3 moles of O, 46 g of C,H,OH react with 3 x 22.4 dm? of O, Thus, 9.29 of C,H,OH need 22. x 3 x 22.4 dm’ of O, = 13.44 dm? £.0.9 of magnesium oxide powder is reacted with excess aqueous nic saltandwate. Ee @) Write a balanced chemical equation for the reaction. _ (2) MgO(s)+ 2HNO, aq) —> MaINO,, (4) +H,00) b) -1’mole of MgO produces 1 mole of Mg(NO, _ Therefore,