Moles And Equations
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore information1Chapter 1:Moles and equationsLearning outcomesyou should be able to: define and use the terms:– relative atomic mass, isotopic mass andformula mass based on the 12C scale– empirical formula and molecular formula– the mole in terms of the Avogadro constantanalyse and use mass spectra to calculate therelative atomic mass of an elementcalculate empirical and molecular formulae usingcombustion data or composition by masswrite and construct balanced equations in this web service Cambridge University Press perform calculations, including use of the moleconcept involving:– reacting masses (from formulae and equations)– volumes of gases (e.g. in the burning ofhydrocarbons)– volumes and concentrations of solutionsdeduce stoichiometric relationships fromcalculations involving reacting masses, volumes ofgases and volumes and concentrations of solutions.www.cambridge.org
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationCambridge International AS Level ChemistryIntroductionFor thousands of years, people have heated rocks anddistilled plant juices to extract materials. Over thepast two centuries, chemists have learnt more andmore about how to get materials from rocks, fromthe air and the sea, and from plants. They have alsofound out the right conditions to allow these materialsto react together to make new substances, such asdyes, plastics and medicines. When we make a newsubstance it is important to mix the reactants in thecorrect proportions to ensure that none is wasted. Inorder to do this we need to know about the relativemasses of atoms and molecules and how these areused in chemical calculations.Figure 1.1 A titration is a method used to find the amount ofa particular substance in a solution.2Masses of atoms and moleculesRelative atomic mass, ArAtoms of different elements have different masses. Whenwe perform chemical calculations, we need to know howheavy one atom is compared with another. The mass ofa single atom is so small that it is impossible to weigh itdirectly. To overcome this problem, we have to weigh a lotof atoms. We then compare this mass with the mass of thesame number of ‘standard’ atoms. Scientists have chosento use the isotope carbon-12 as the standard. This has beengiven a mass of exactly 12 units. The mass of other atoms isfound by comparing their mass with the mass of carbon-12atoms. This is called the relative atomic mass, Ar.The relative atomic mass is the weighted average mass ofnaturally occurring atoms of an element on a scale wherean atom of carbon-12 has a mass of exactly 12 units.From this it follows that:Ar [element Y]average mass of one atom of element Y 12 mass of one atom of carbon-12 in this web service Cambridge University PressWe use the average mass of the atom of a particular elementbecause most elements are mixtures of isotopes. For example,the exact Ar of hydrogen is 1.0079. This is very close to 1 andmost periodic tables give the Ar of hydrogen as 1.0. However,some elements in the Periodic Table have values that are notwhole numbers. For example, the Ar for chlorine is 35.5. Thisis because chlorine has two isotopes. In a sample of chlorine,chlorine-35 makes up about three-quarters of the chlorineatoms and chlorine-37 makes up about a quarter.Relative isotopic massIsotopes are atoms that have the same number of protonsbut different numbers of neutrons (see page 28). We representthe nucleon number (the total number of neutrons plusprotons in an atom) by a number written at the topleft-hand corner of the atom’s symbol, e.g. 20Ne, or bya number written after the atom’s name or symbol, e.g.neon-20 or Ne-20.We use the term relative isotopic mass for the massof a particular isotope of an element on a scale wherean atom of carbon-12 has a mass of exactly 12 units. Forexample, the relative isotopic mass of carbon-13 is 13.00.If we know both the natural abundance of every isotopeof an element and their isotopic masses, we can calculatewww.cambridge.org
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationChapter 1: Moles and equationsthe relative atomic mass of the element very accurately.To find the necessary data we use an instrument called amass spectrometer (see box on mass spectrometry).Relative molecular mass, MrThe relative molecular mass of a compound (Mr) is therelative mass of one molecule of the compound on ascale where the carbon-12 isotope has a mass of exactly12 units. We find the relative molecular mass by addingup the relative atomic masses of all the atoms present inthe molecule.For example, for methane:formulaCH4atoms present1 C; 4 H(1 Ar[C]) (4 Ar[H])add Ar values (1 12.0) (4 1.0)Mr of methane 16.0Accurate relative atomic massesMAss spectRoMetRybiologicaldrawing1.1: 1.2)A mass spectrometerbOx(Figurecan be usedto measure the mass of each isotope presentin an element. It also compares how much ofeach isotope is present – the relative abundance(isotopic abundance). A simplified diagram of amass spectrometer is shown in Figure 1.3. You willnot be expected to know the details of how a massspectrometer works, but it is useful to understandhow the results are obtained.Relative formula massFor compounds containing ions we use the term relativeformula mass. This is calculated in the same way as forrelative molecular mass. It is also given the same symbol,Mr. For example, for magnesium hydroxide:formulaions presentadd Ar valuesMr of magnesiumhydroxideMg(OH)21 Mg2 ; 2 (OH–)(1 Ar[Mg]) (2 (Ar[O] Ar[H])) (1 24.3) (2 (16.0 1.0)) 58.33Figure 1.2 A mass spectrometer is a large and complexinstrument.queSTIOnvaporised samplepositively chargedelectrodes accelerate positive ions1 Use the Periodic Table on page 473 to calculate therelative formula masses of the following:a calcium chloride, CaCl2magnetic fieldb copper(II) sulfate, CuSO4cammonium sulfate, (NH4)2SO4d magnesium nitrate-6-water, Mg(NO3)2.6H2OHint: for part d you need to calculate the mass ofwater separately and then add it to the Mr of sionisationchamber flight tubeiondetectorrecordercomputerFigure 1.3 Simplified diagram of a mass spectrometer. in this web service Cambridge University Presswww.cambridge.org
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationCambridge International AS Level ChemistryDetermination of Ar from mass spectraMASS SpeCTrOMeTry (COnTInued)The atoms of the element in the vaporised sampleare converted into ions. The stream of ions isbrought to a detector after being deflected (bent)by a strong magnetic field. As the magnetic field isincreased, the ions of heavier and heavier isotopesare brought to the detector. The detector isconnected to a computer, which displays themass spectrum.The mass spectrum produced shows the relativeabundance (isotopic abundance) on the verticalaxis and the mass to ion charge ratio (m/e) on thehorizontal axis. Figure 1.4 shows a typical massspectrum for a sample of lead. Table 1.1 showshow the data is interpreted. multiply each isotopic mass by its percentage abundanceadd the figures togetherdivide by 100.We can use this method to calculate the relative atomicmass of neon from its mass spectrum, shown in Figure 1.5.The mass spectrum of neon has three peaks:20Ne(90.9%), 21Ne (0.3%) and 22Ne (8.8%).Ar of neon(20 90.9) (21.0 0.3) (22 8.8) 20.2100204205 206 207 208Mass/charge (m/e) ratio209Figure 1.4 The mass spectrum of a sample of lead.For singly positively charged ions the m/e valuesgive the nucleon number of the isotopes detected.In the case of lead, Table 1.1 shows that 52% of thelead is the isotope with an isotopic mass of 208.The rest is lead-204 (2%), lead-206 (24%) and lead207 (22%).Isotopic massRelative abundance / %2042206242072220852total10060402008.8 %0800.3 %190.9 %1002Relative abundance / %Detector current / mA Note that this answer is given to 3 significant figures,which is consistent with the data given.34We can use the data obtained from a mass spectrometerto calculate the relative atomic mass of an element veryaccurately. To calculate the relative atomic mass we followthis method:19202122Mass/charge (m/e) ratio23Figure 1.5 The mass spectrum of neon, Ne.A high-resolution mass spectrometer can give veryaccurate relative isotopic masses. For example 16O 15.995and 32S 31.972. Because of this, chemists can distinguishbetween molecules such as SO2 and S2, which appear tohave the same relative molecular mass.Table 1.1 The data from Figure 1.4. in this web service Cambridge University Presswww.cambridge.org
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationChapter 1: Moles and equationsqueSTIOn36.7 %2 Look at the mass spectrum of germanium, Ge.10027.4 %7.6 %207.7 %3020.6 %Abundance / %407075Mass/charge (m/e) ratio80Figure 1.6 The mass spectrum of germanium.a Write the isotopic formula for the heaviest isotopeof germanium.We often refer to the mass of a mole of substance as itsmolar mass (abbreviation M). The units of molar mass–1are g mol .The number of atoms in a mole of atoms is very large:6.02 1023 atoms. This number is called the Avogadroconstant (or Avogadro number). The symbol for theAvogadro constant is L (the symbol NA may also be used).The Avogadro constant applies to atoms, molecules, ionsand electrons. So in 1 mole of sodium there are 6.02 1023sodium atoms and in 1 mole of sodium chloride (NaCl) thereare 6.02 1023 sodium ions and 6.02 1023 chloride ions.It is important to make clear what type of particleswe are referring to. If we just state ‘moles of chlorine’, it isnot clear whether we are thinking about chlorine atomsor chlorine molecules. A mole of chlorine molecules, Cl2,contains 6.02 1023 chlorine molecules but twice as manychlorine atoms, as there are two chlorine atoms in everychlorine molecule.b Use the % abundance of each isotope to calculatethe relative atomic mass of germanium.Amount of substance5the mole and the Avogadro constantThe formula of a compound shows us the number ofatoms of each element present in one formula unit or onemolecule of the compound. In water we know that twoatoms of hydrogen (Ar 1.0) combine with one atom ofoxygen (Ar 16.0). So the ratio of mass of hydrogen atomsto oxygen atoms in a water molecule is 2 : 16. No matterhow many molecules of water we have, this ratio willalways be the same. But the mass of even 1000 atoms isfar too small to be weighed. We have to scale up muchmore than this to get an amount of substance that is easyto weigh.The relative atomic mass or relative molecular mass ofa substance in grams is called a mole of the substance. So amole of sodium (Ar 23.0) weighs 23.0 g. The abbreviationfor a mole is mol. We define the mole in terms of thestandard carbon-12 isotope (see page 28).One mole of a substance is the amount of that substancethat has the same number of specific particles (atoms,molecules or ions) as there are atoms in exactly 12 g of thecarbon-12 isotope. in this web service Cambridge University PressFigure 1.7 Amedeo Avogadro (1776–1856) was an Italianscientist who first deduced that equal volumes of gasescontain equal numbers of molecules. Although the Avogadroconstant is named after him, it was left to other scientists tocalculate the number of particles in a mole.Moles and massThe Système International (SI) base unit for mass is thekilogram. But this is a rather large mass to use for generallaboratory work in chemistry. So chemists prefer to usethe relative molecular mass or formula mass in grams(1000 g 1 kg). You can find the number of moles of asubstance by using the mass of substance and the relativeatomic mass (Ar) or relative molecular mass (Mr).mass of substance in grams (g)number of moles (mol) molar mass (g mol–1)www.cambridge.org
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationCambridge International AS Level ChemistryWOrked exAMpLeWOrked exAMpLe1 How many moles of sodium chloride are present in117.0 g of sodium chloride, NaCl?2 What mass of sodium hydroxide, NaOH, is present in0.25 mol of sodium hydroxide?( Ar values: Na 23.0, Cl 35.5)( Ar values: H 1.0, Na 23.0, O 16.0)molar mass of NaCl 23.0 35.5molar mass of NaOH 23.0 16.0 1.0 number of moles58.5 g mol–1 40.0 g mol–1mass molar massmass number of moles molar mass 0.25 40.0 g117.0 58.5 10.0 g NaOH 2.0 molqueSTIOn4 Use these Ar values: C 12.0, Fe 55.8, H 1.0,O 16.0, Na 23.0.Calculate the mass of the following:a 0.20 moles of carbon dioxide, CO2b 0.050 moles of sodium carbonate, Na2CO3c5.00 moles of iron(II) hydroxide, Fe(OH)2Mole calculations6Reacting massesFigure 1.8 From left to right, one mole of each of copper,bromine, carbon, mercury and lead.queSTIOn3 a Use these Ar values (Fe 55.8, N 14.0, O 16.0,S 32.1) to calculate the amount of substance inmoles in each of the following:iWhen reacting chemicals together we may need to knowwhat mass of each reactant to use so that they react exactlyand there is no waste. To calculate this we need to knowthe chemical equation. This shows us the ratio of molesof the reactants and products – the stoichiometry of theequation. The balanced equation shows this stoichiometry.For example, in the reactionFe2O3 3CO10.7 g of sulfur atomsii 64.2 g of sulfur molecules (S8)iii 60.45 g of anhydrous iron(III) nitrate, Fe(NO3)3.b Use the value of the Avogadro constant (6.02 1023 mol–1) to calculate the total number of atoms in7.10 g of chlorine atoms. (Ar value: Cl 35.5)To find the mass of a substance present in a given numberof moles, you need to rearrange the equationmass of substance in grams (g)number of moles (mol) molar mass (g mol–1)2Fe 3CO21 mole of iron(III) oxide reacts with 3 moles of carbonmonoxide to form 2 moles of iron and 3 moles of carbondioxide. The stoichiometry of the equation is 1 : 3 : 2 : 3.The large numbers that are included in the equation(3, 2 and 3) are called stoichiometric numbers.In order to find the mass of products formed in achemical reaction we use: the mass of the reactantsthe molar mass of the reactantsthe balanced equation.mass of substance (g) number of moles (mol) molar mass (g mol–1) in this web service Cambridge University Presswww.cambridge.org
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationChapter 1: Moles and equationsWOrked exAMpLe4 Iron(III) oxide reacts with carbon monoxide to formiron and carbon dioxide.Fe2O3 3CO2Fe 3CO2Calculate the maximum mass of iron produced when798 g of iron(III) oxide is reduced by excess carbonmonoxide.(Ar values: Fe 55.8, O 16.0)step 1 Fe2O3 3CO2Fe 3CO2step 2 1 mole iron(III) oxideFigure 1.9 Iron reacting with sulfur to produce iron sulfide.We can calculate exactly how much iron is needed to reactwith sulfur and the mass of the products formed byknowing the molar mass of each reactant and the balancedchemical equation.WOrked exAMpLe3 Magnesium burns in oxygen to form magnesium oxide.2Mg O22MgOWe can calculate the mass of oxygen needed to reactwith 1 mole of magnesium. We can calculate the massof magnesium oxide formed.step 1 Write the balanced equation.step 2 Multiply each formula mass in g by therelevant stoichiometric number in the equation.2Mg2 24.3 g48.6 g O21 32.0 g32.0 g2MgO2 (24.3 g 16.0 g)80.6 gFrom this calculation we can deduce that: 32.0 g of oxygen are needed to react exactly with48.6 g of magnesium80.6 g of magnesium oxide are formed.If we burn 12.15 g of magnesium (0.5 mol) we get20.15 g of magnesium oxide. This is because thestoichiometry of the reaction shows us that for everymole of magnesium burnt we get the same number ofmoles of magnesium oxide.In this type of calculation we do not always need to knowthe molar mass of each of the reactants. If one or more ofthe reactants is in excess, we need only know the mass ingrams and the molar mass of the reactant that is not inexcess (the limiting reactant). in this web service Cambridge University Press2 moles iron(2 55.8) (3 16.0)2 55.8159.6 g Fe2O3111.6 g Fe111.6 798159.6 558 g Festep 3 798 gYou can see that in step 3, we have simply used ratiosto calculate the amount of iron produced from 798 g ofiron(III) oxide.queSTIOn5 a Sodium reacts with excess oxygen to form sodiumperoxide, Na2O2.2Na O2Na2O2Calculate the maximum mass of sodiumperoxide formed when 4.60 g of sodium is burntin excess oxygen.(Ar values: Na 23.0, O 16.0)b Tin(IV) oxide is reduced to tin by carbon. Carbonmonoxide is also formed.SnO2 2CSn 2COCalculate the mass of carbon that exactly reactswith 14.0 g of tin(IV) oxide. Give your answer to 3significant figures.(Ar values: C 12.0, O 16.0, Sn 118.7)the stoichiometry of a reactionWe can find the stoichiometry of a reaction if we know theamounts of each reactant that exactly react together andthe amounts of each product formed.For example, if we react 4.0 g of hydrogen with 32.0 g ofoxygen we get 36.0 g of water. (Ar values: H 1.0, O 16.0)www.cambridge.org7
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationCambridge International AS Level Chemistryhydrogen (H2) oxygen (O2)32.04.02 1.02 16.0 2 mol 1 molwater (H2O)36.0(2 1.0) 16.0 2 molThis ratio is the ratio of stoichiometric numbers in theequation. So the equation is:2H2 O22H2OWe can still deduce the stoichiometry of this reactioneven if we do not know the mass of oxygen that reacted.The ratio of hydrogen to water is 1 : 1. But there is only oneatom of oxygen in a molecule of water – half the amountin an oxygen molecule. So the mole ratio of oxygen towater in the equation must be 1 : 2.queSTIOn6 56.2 g of silicon, Si, reacts exactly with 284.0 g ofchlorine, Cl2, to form 340.2 g of silicon(IV) chloride,SiCl4. Use this information to calculate thestoichiometry of the reaction.(Ar values: Cl 35.5, Si 28.1)8significant figuresWhen we perform chemical calculations it is importantthat we give the answer to the number of significantfigures that fits with the data provided. The examples showthe number 526.84 rounded up to varying numbers ofsignificant figures.rounded to 4 significant figures 526.8rounded to 3 significant figures 527rounded to 2 significant figures 530When you are writing an answer to a calculation, theanswer should be to the same number of significant figuresas the least number of significant figures in the data.WOrked exAMpLe (COnTInued)Note 1 Zeros before a number are not significantfigures. For example, 0.004 is only to 1 significantfigure.Note 2 After the decimal point, zeros after a numberare significant figures. 0.0040 has 2 significant figuresand 0.004 00 has 3 significant figures.Note 3 If you are performing a calculation withseveral steps, do not round up in between steps.Round up at the end.percentage composition by massWe can use the formula of a compound and relative atomicmasses to calculate the percentage by mass of a particularelement in a compound.% by massatomic mass number of moles of particularelement in a compound 100molar mass of compoundWOrked exAMpLe6 Calculate the percentage by mass of iron in iron(III)oxide, Fe2O3.(Ar values: Fe 55.8, O 16.0)2 55.8% mass of iron 100(2 55.8) (3 16.0) 69.9 %WOrked exAMpLe5 How many moles of calcium oxide are there in 2.9 g ofcalcium oxide?(Ar values: Ca 40.1, O 16.0)If you divide 2.9 by 56.1, your calculator shows0.051 693 . The least number of significant figuresin the data, however, is 2 (the mass is 2.9 g). So youranswer should be expressed to 2 significant figures,as 0.052 mol. in this web service Cambridge University PressFigure 1.10 This iron ore is impure Fe2O3. We can calculatethe mass of iron that can be obtained from Fe2O3 by usingmolar masses.www.cambridge.org
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationChapter 1: Moles and equationsqueSTIOnWOrked exAMpLeS7 Calculate the percentage by mass of carbon inethanol, C2H5OH.7 Deduce the formula of magnesium oxide.(Ar values: C 12.0, H 1.0, O 16.0)empirical formulaeThe empirical formula of a compound is the simplest wholenumber ratio of the elements present in one molecule orformula unit of the compound. The molecular formula of acompound shows the total number of atoms of each elementpresent in a molecule.Table 1.2 shows the empirical and molecular formulaefor a number of compounds. The formula for an ionic compound is always itsempirical formula.The empirical formula and molecular formula for simpleinorganic molecules are often the same.Organic molecules often have different empirical andmolecular aterH2OH2Ohydrogen peroxideHOH2O2sulfur le 1.2 Some empirical and molecular formulae.queSTIOn8 Write the empirical formula for:This can be found as follows: burn a known mass of magnesium (0.486 g) inexcess oxygen record the mass of magnesium oxide formed(0.806 g) calculate the mass of oxygen that has combinedwith the magnesium (0.806 – 0.486 g) 0.320 g calculate the mole ratio of magnesium to oxygen ( Ar values: Mg 24.3, O 16.0)0.486 gmoles of Mg 0.0200 mol24.3 g mol–10.320 gmoles of oxygen 0.0200 mol16.0 g mol–1The simplest ratio of magnesium : oxygen is 1 : 1. So theempirical formula of magnesium oxide is MgO.8 When 1.55 g of phosphorus is completely combusted3.55 g of an oxide of phosphorus is produced. Deducethe empirical formula of this oxide of phosphorus.(Ar values: O 16.0, P 31.0)postep 1 note the massof each element1.55 g3.55 – 1.55 2.00 gstep 2 divide by atomicmasses1.55 g2.00 gstep 3 divide by thelowest figurestep 4 if needed, obtainthe lowest wholenumber ratioto get empiricalformula31.0 g mol–1 0.05 mol16.0 g mol–11 0.125 mol0.05 10.125 2.50.050.05P2O5a hydrazine, N2H4b octane, C8H18cbenzene, C6H6d ammonia, NH3An empirical formula can also be deduced from data thatgive the percentage composition by mass of the elementsin a compound.The empirical formula can be found by determiningthe mass of each element present in a sample of thecompound. For some compounds this can be done bycombustion. An organic compound must be very pure inorder to calculate its empirical formula. Chemists oftenuse gas chromatography to purify compounds beforecarrying out formula analysis. in this web service Cambridge University Presswww.cambridge.org9
Cambridge University Press978-1-107-63845-7 – Cambridge International AS and A Level ChemistryLawrie Ryan and Roger NorrisExcerptMore informationCambridge International AS Level ChemistryWOrked exAMpLe (COnTInued)WOrked exAMpLe9 A compound of carbon and hydrogen contains 85.7%carbon and 14.3% hydrogen by mass. Deduce theempirical formula of this hydrocarbon.(Ar values: C 12.0, O 16.0)cHstep 1 note the % by mass85.714.3step 2 divide by Ar values85.714.3 7.142 14.312.07.142step 3 divide by the lowest 1figure7.14214.3 27.142queSTIOn9 The composition by mass of a hydrocarbon is 10%hydrogen and 90% carbon. Deduce the empiricalformula of this hydrocarbon.(Ar values: C 12.0, H 1.0)Molecular formulaeThe molecular formula shows the actual number of eachof the different atoms present in a molecule. The molecularformula is more useful than the empirical formula. Weuse the molecular formula to write balanced equationsand to calculate molar masses. The molecular formula isalways a multiple of the empirical formula. For example,the molecular formula of ethane, C2H6, is two times theempirical formula, CH3.In order to deduce the molecular formula we needto know: the relative formula mass of the compoundthe empirical formula.WOrked exAMpLe10 A compound has the empirical formula CH2Br. Itsrelative molecular mass is 187.8. Deduce the molecularformula of this compound.(Ar values: Br 79.9, C 12.0, H 1.0)step 1 find the empirical formula mass:12.0 (2 1.0) 79.9 93.9 in this web service Cambridge University Pressstep 3 multiply the number of atoms in the empiricalformula by the number in step 2:2 CH2Br, so molecular formula is C2H4Br2.1.0Empirical formula is CH2.10step 2 divide the relative molecular mass by187.8the empirical formula mass: 293.9queSTIOn10 The empirical formulae and molar masses of threecompounds, A, B and C, are shown in the table below.Calculate the molecular formula of each of thesecompounds.(Ar values: C 12.0, Cl 35.5, H 1.0)compoundempirical formulaMrAC3H582BCCl3237CCH2112chemical formulae andchemical equationsDeducing the formulaThe electronic structure of the individual elements in acompound determines the formula of a compound (seepage 33). The formula of an ionic compound is determined bythe charges on each of the ions present. The number of positivecharges is balanced by the number of negative charges so thatthe total charge on the compound is zero. We can work out theformula for a compound if we know the charges on the ions.Figure 1.11 shows the charges on some simple ions related tothe position of the elements in the Periodic Table. The formof the Periodic Table that we shall be using has 18 groupsbecause the transition elements are numbered as Groups3 to 12. So, aluminium is in Group 13 and chlorine is inGroup 17.For simple metal ions in Groups 1 and 2, the value ofthe positive charge is the same as the group number. Fora simple metal ion in Group 13, the value of the positivecharge is 3 . For a simple non-metal ion in Groups 15 to17, the value of the negative charge is 18 minus the groupwww.cambridge.org
of atoms. We then compare this mass with the mass of the same number of ‘standard’ atoms. Scientists have chosen to use the isotope carbon-12 as the standard. This has been given a mass of exactly 12 units. The mass of other atoms is found by comparing their mass with the mass of carbon-12 atom
PRACTICE PACKET: Unit 3 Moles & Stoichiometry 3 www.mrpalermo.com Objective: Calculate Molar Mass (gram formula mass) LESSON 1: Moles and Molar Mass 1. Fill in the table below Formula Moles of each atom Total moles of atoms Formula Moles of each atom Total moles of atoms a. HClO 3 1 mol of H atoms 1 mol of Cl atoms 3 mol of O atoms 5 mol of
13 Molarity Moles of solute/Liters of Solution (M) Molality Moles of solute/Kg of Solvent (m) Mole Fraction Moles solute/total number of moles Mass % Mass solute/total mass x 100 Volume % volume solute/total volume x 100 ppm parts per million * ppb parts per billion * Chemical concentrations * mass for solutions, volume for gasses Molarity Moles of solute/Liters of Solution (M)
point at which an indicator changes color is called the end point of the titration. Phenolphthalein is an appropriate choice for this titration. In acidic solution, phenolphthalein is colorless, and in basic solution, it is pink. At the equivalence point, the number of moles of acid equals the number of moles of base. (1) moles of H 3O moles of OH
NEET (UG)-2019 (Code- 1)Q 6. The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is : (1) 20 (2) 30 (3) 40 (4) 10 Answer (2) Sol. Haber's process N22 3(g) 3H (g) 2NH (g) 20 moles need to be produced 2 moles of NH 3 3 mol
Total nurnber of moles of solution before adsorption, moles Change in mole fraction of surfactant resulting from adsorption Mass of insoluble adsorbent, kg Amount adsorbed, mol rnS2 Number of moles of surfactant adsorbed. moles Surface area of substrate, m' Zeta potential, V Fluid density, kg m" Particle density, kg m-' Fluid viscosity. N s rn"
EQUATIONS AND INEQUALITIES Golden Rule of Equations: "What you do to one side, you do to the other side too" Linear Equations Quadratic Equations Simultaneous Linear Equations Word Problems Literal Equations Linear Inequalities 1 LINEAR EQUATIONS E.g. Solve the following equations: (a) (b) 2s 3 11 4 2 8 2 11 3 s
1.2 First Order Equations 5 1.3 Direction Fields for First Order Equations 14 Chapter 2 First Order Equations 2.1 Linear First Order Equations 27 2.2 Separable Equations 39 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations 48 2.5 Exact Equations 55 2.6 Integrating Factors 63 Chapter 3 Numerical Methods 3.1 Euler’s Method 74
Chapter 1 Introduction 1 1.1 ApplicationsLeading to Differential Equations 1.2 First Order Equations 5 1.3 Direction Fields for First Order Equations 16 Chapter 2 First Order Equations 30 2.1 Linear First Order Equations 30 2.2 Separable Equations 45 2.3 Existence and Uniqueness of Solutionsof Nonlinear Equations 55
