Chemical Formulas

Main objectives for this chapter: Empirical and molecular formulas. Structural formulas. Simple examples Calculations of empirical formulas, given the percentage composition by mass. Calculation of empirical formulas, given the masses of reactants and products.HIGHER LEVEL only Calculation of molecular formulas, given the empirical formulas and the relative molecular masses (examples should include simple biological substances, such as glucose and urea). Percentage composition by mass. Calculations

Structural formulas, molecular formulas, empirical formulas

Structural Formulas Gives the arrangement of the atoms in a molecule of the compound The structural formula of ethane: HH HH HH The structural formula of ethene: H HH H

Molecular Formula Water Carbon dioxide Methane Molecular FormulaStructural FormulaName C H H H H C O O H H O Molecular formula tells you the number of atoms of each element present in a molecule of a compound CH 4 CO 2 H2OH2O

Formula of ionic compounds Ionic compounds are giant structures. butThere can be any number of ions in an ionic crystal - but always a definite ratio of ions. NameRatioFormula Sodium chloride1:1 Magnesium chloride1:2 Aluminium chloride1:3 Aluminium Oxide2: Sodium chloride A 1:1 ratio NaCl AlCl 3 Al 2 O 3 MgCl 2

Empirical Formula Gives only the ratios in which different atoms are present in a molecule of a compound For glucose C 6 H 12 O 6 is the molecular formula The empirical formula is CH 2 O

The structural formula of ethane is : The molecular formula is: C 2 H 6 The empirical formula is : CH 3 HH HH HH Once you know the structural formula you can work out the molecular formula and the empirical formula

The structural formula of ethene is: Find the (i) molecular formula (ii) empirical formula H HH H Find the (i) molecular formula= C 2 H 4 (ii) empirical formula= CH 2

Check your learning Define Structural formula Molecular formula Emperical formula

Objectives for today Calculations of empirical formulas: (i) Given the % mass of elements present in the compound (ii) given the masses of reactants and products.HIGHER LEVEL (ii) Calculation of molecular formulas

Calculating Empirical Formulas

A compound contains 40% sulfur and 60 % oxygen. What is its empirical formula? Therefore the empirical formula of this compound is SO 3. Element%molesRatio sulfur4040/ 32 = oxygen6060/ 16 = Method 1- when given % mass of element

A compound contains 48.8% carbon and 13.5% hydrogen and 37.7% nitrogen respectively by mass. Determine the empirical formula of the compound. Therefore the empirical formula of this compound is Element%/ ArRatio Carbon / 12 = Hydrogen / 1= Nitrogen37.737,5/14= Q282

A compound contains 64.9% carbon and 13.5% hydrogen and 21.6 % oxygen respectively by mass. Determine the empirical formula of the compound. Therefore the empirical formula of this compound is C 4 H 10 O Element%/ ArRatio Carbon / 12 = Hydrogen / 1= Oxygen /16= Q284 Finding empirical formula

Conservation of Mass in a reaction During chemical reactions the same atoms are present before and after reaction. They have just joined up in different ways. Because of this the total mass of reactants is always equal to the total mass of products. (Law of Conservation of Mass) Reaction but no mass change

Mg HCl Gas given off. Mass of chemicals in flask decreases Same reaction in sealed container: No change in mass Conservation of Mass

You must know the masses of all of the elements in the compound. You might have to work this out... REMEMBER SUM OF MASS OF REACTANTS = SUM OF MASS OF PRODUCTS When 3.175g of copper reacts with chlorine gas 6.725g of copper chloride is formed. Find the empirical formula of the copper chloride Copper + Chlorine gas Copper Chloride 3.175g ? 6.725g Example 3.55g ElementmassmolesRatio Copper3.175g0.051 Chlorine3.55g0.12 Method 2- when given mass of element in compound The empirical formula is CuCl 2

When`1.44g of Magnesium was completely burned in oxygen it resulted in the formation of 2.40g of magnesium oxide. Find the empirical formula of magnesium oxide Magnesium + Oxygen Magnesium oxide 1.44g ? 2.40g The empirical formula is MgO Example 0.96g ElementmassmolesRatio Magnesium1.44g0.061 Oxygen0.96g0.061 Method 2- when given mass of element in compound

When`2.07g of lead reacts with iodine, 4.61g of lead iodide was formed. Find the empirical formula of lead iodide Lead + Iodine Lead iodide 2.07g ? 4.61g The empirical formula is PbI 2 Q ElementmassmolesRatio Lead2.07g0.011 Iodine Method 2- when given mass of element in compound

When`3.94g of hydrated copper (II) sulfate was heated, 2.52g of anhydrous salt remained. Calculate the formula of the hydrated salt. Hydrated copper(II) sulfate Anhydrous copper sulphate + water 3.94g 2.52g The empirical formula is CuSO 4 (H 2 0) 5 Q289 Method 2- when given mass of element in compound 1.42g GroupmassmolesRatio Water Copper sulfate

9.76g of a metal forms 20.9g of its oxide whose formula is M 2 0. Calculate the relative molecular mass of the metal. Metal + Oxygen Metal oxide 9.76g 20.90g (1.3925/ 9.76= Mass of one mole of the metal is 7.009g. This is the relative molecular mass Q290 Method 2- when given mass of element in compound 11.14g GroupmassmolesRatio Metal9.76?2 oxygen

Calculating Molecular Formulas

Calculation of the molecular formula: You need: 1. The empirical formula 2. The molecular mass (can work out using the periodic table)

Calculating molecular mass Urea is used as a fertiliser and an animal feed. It has a relative molecular mass of 60 and is composed of 46.66% N, 26.66%O, 20%C and 6.66% H. Determine the molecular formula of urea You need: 1. The empirical formula 2. The relative molecular mass 1. The empirical formula Element%molesratio nitrogen oxygen Carbon Hydrogen N 2 OCH 4 2. Relative molecular mass of urea = Empirical formula = molecular formula Mass according to EF = 2(14) (1) = 60 Molecular formula N 2 OCH 4

284. Error – see 286 An alcohol was found on analysis to contain 64.9% carbon, 13.5% hydrogen and 21.6% oxygen. If the relative molecular mass of the alcohol is 74 show that the molecular formula is C 4 H 10 O You need: 1. The empirical formula 2. The relative molecular mass 1. The empirical formula Element%molesratio Carbon oxygen Hydrogen C 4 H 10 O 2. Relative molecular mass of urea = Empirical formula = molecular formula Mass according to EF = 4(12) (1) = 74 C 4 H 10 O

285. Calculating molecular mass An organic acid contain 27.6% carbon, 2.2% hydrogen and 71.1% oxygen. If the relative molecular mass of the alcohol is 90. Determine the molecular formula You need: 1. The empirical formula 2. The relative molecular mass 1. The empirical formula Element%molesratio Carbon oxygen Hydrogen2.2 1 CO 2 H 2. Relative molecular mass of the organic acid = Empirical formula x 2 = molecular formula Mass according to EF = 12 +2(16) +1 = 45 Molecular formula = C 2 O 4 H 2

286. Calculating molecular mass Determine the molecualr formula of a compound whose composition is carbon 64.8%, hydrogen 13.6%and oxygen 21.6% and whose relative molecular mass is 74 You need: 1. The empirical formula 2. The relative molecular mass 1. The empirical formula Element%molesratio Carbon oxygen Hydrogen C 4 H 10 O 2. Relative molecular mass of urea = Empirical formula = molecular formula Mass according to EF = 4(12) (1) = 74 Molecular formula = C 4 H 10 O

Calculating % compositions by mass

By the end of today’s class you should be able to: (iii)Calculate the Percentage composition by mass of an element in a compound

Percentage composition by mass Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

What is the percentage by mass of Fe in Fe 2 O 3 ? 2(56) x % Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

291b)What is the percentage by mass of nitrogen in NH 4 NO 3 ? 2(14) x % Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

291c) What is the percentage by mass of carbon in methylbenzene ? 7(12) x % Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

291(d)What is the percentage by mass of N in NO 2 ? 14 x % Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

291(e)What is the percentage by mass of water in Na 2 CO 3. 10H 2 0 ? 10(18) x % Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

291(f)What is the percentage by mass of Fe in Fe 3 O 4. ? 3(56) x % Mass % of A in compound = Mass of A in compoundx 100 Relative molecular mass of the compound 1

Extra questions not on worksheet

Urea has an empirical formula of CON 2 H 4 and a relative molecular mass of 60. Find its molecular formula. The empirical formula of urea is a simple whole number ratio of the atoms from each element that are present. If the atoms in each molecule actually present were CON 2 H 4 then the molecular mass of urea would be: = 60 We are told the molecular mass of urea is 60. molecular formula = empirical formula! Molecular formula = CON 2 H 4

Glucose has an empirical formula of CH 2 O and a relative molecular mass of 180. Find its molecular formula. If the atoms in each molecule actually present were CH2O then the molecular mass of glucose would be: = 30 But we are told the molecular mass of glucose is 180. So the molecular formula is not CH 2 O ! 30 x n = 180 n = 6 So the molecular formula = C 6 H 12 O 6

If the atoms in each molecule actually present were C 7 H 16 then the molecular mass of heptane would be: (12 X 7) + (1 X 16)= 100 We are told the molecular mass of Heptane is 100. Molecular formula = empirical formula Molecular formula of Heptane = C 7 H 16 Q1. Heptane has an empirical formula of C 7 H 16 and a relative molecular mass of 100. Find its molecular formula.

If the atoms in each molecule actually present were C 2 H 4 O 2 then the molecular mass of Butanoic acid would be: = 44 We are told the molecular mass of Butanoic acid is 88. Molecular formula x 2 = empirical formula Molecular formula of Butanoic acid = C 4 H 8 O 2 Q2. Butanoic acid has an empirical formula of C2H4O and a relative molecular mass of 88. Find its molecular formula.

Q3. Fructose has the following composition by mass – 40% carbon, 6.66% hydrogen, 53.33% oxygen. If the relative molecular mass of fructose is 180 find its molecular formula.

Element present Mass present in 100g of fructose Moles present in 100g of fructose Molar ratio Simplest ratio Simplest whole number ratio Carbon40g Hydrogen6.66g Oxygen53.33g First you need to find the empirical formula:

Element present Mass present in 100g of fructose Moles present in 100g of fructose Molar ratio Simplest ratio Simplest whole number ratio Carbon40g40 = Hydrogen6.66g6.66 = Oxygen53.33g53.33 = First you need to find the empirical formula: Moles present = mass in grams Ar

Element present Mass present in 100g of fructose Moles present in 100g of fructose Molar ratio Simplest ratio Simplest whole number ratio Carbon40g40 = = Hydrogen6.66g6.66 = = Oxygen53.33g53.33 = = First you need to find the empirical formula: To get a simple ratio divide each number by the smallest number present!

Element present Mass present in 100g of fructose Moles present in 100g of fructose Molar ratio Simplest ratio Simplest whole number ratio Carbon40g40 = = Hydrogen6.66g6.66 = = Oxygen53.33g53.33 = = First you need to find the empirical formula: empirical formula = CH 2 O

the relative molecular mass of fructose is 180, the empirical formula is CH 2 O - find its molecular formula. If the atoms in each molecule actually present were CH 2 O then the molecular mass of fructose would be: = 30 We are told the molecular mass of Fructose is 180. Molecular formula x 6 = empirical formula Molecular formula of Fructose = C 6 H 12 O 6

What is the percentage composition by mass of carbon present in ethanol ( C 2 H 5 OH )? Moles of carbon present : 2 Mass of carbon present : Ar x number of moles = mass present in grams 12 x 2 = 24g Total mass of C 2 H 5 OH = = 46 Percentage of carbon by mass in ethanol = 24 x 100 = 52.17% 46

What is the percentage composition by mass of nitrogen present in (NH 4 ) 2 HPO 4 ? (NH 4 ) 2 HPO 4 = N 2 H 9 PO 4 Moles of nitrogen present : 2 Mass of nitrogen present : Ar x number of moles = mass present in grams 14 x 2 = 28g Total mass of N 2 H 9 PO 4: (14 x 2)+ (1 x 9)+ 40 +(16 x4) = 109g Percentage of nitrogen by mass in N 2 H 9 PO 4 28 x 100 = 25.68% 109

What is the percentage composition by mass of sodium in sodium hydroxide NaOH? Moles of sodium present : 1 Mass of sodium present : Ar x number of moles = mass present in grams 23 x 1 = 23g Total mass of NaOH = = 40 Percentage of sodium by mass in sodium hydroxide= 23 x 100 = 57.5% 40

Chemical equations

Chemical Equations We can describe what happens in a chemical reaction using words : Carbon + oxygen Carbon dioxide “+” means “and” means “react to give”

Chemical Equations Another way in which we can we can describe a chemical reaction is using symbols to represent the reactants and products instead of words! C + O CO 2

Describe what the following chemical equations tell… Fe + HCl FeCl 2 + H 2 Al + O 2 Al 2 O 3 H 2 +Cl 2 AlCl 3 ZnS +O 2 ZnO + SO 2

Law of conservation of Mass The law of conservation of matter states that, in any chemical reaction, matter doesn’t get created or destroyed but it only changes from one form into another. C + O CO 2 In a chemical reaction: No. of atoms of an element present at the start of reaction = No. of atoms of the element present at the end of the reaction

Balanced Chemical Equations Chemical equations should always be balanced to describe accurately what happens during a chemical reaction C + O CO 2 For an equation to be balanced the total number of atoms of each element of reactant must equal the total number of atoms of that element in the product

Balancing chemical equations Rules: 1.Chemical formulas can’t be changed 2.Chemical formulas can be multiplied by a suitable number

Balance the following equation: Fe + O 2 Fe 3 O 4 Iron atoms: Reactant sideProduct side Oxygen atoms: To get 3 Fe atoms on the reactant side multiply it by 3! To get 4 oxygen atoms on the product side multiply it by 2! 3Fe + 2O 2 Fe 3 O 4 Step one – see what's present Step two – make changes

Step three – check its balanced 3Fe + 2O 2 Fe 3 O 4 Reactant sideProduct side Iron atoms: Oxygen atoms:

Try this one: N 2 +O 2 NO Nitrogen atoms: Reactant sideProduct side Oxygen atoms: Step one – see what's present Step two – make changes N 2 +O 2 2NO Step three – check its balanced Nitrogen atoms: Reactant sideProduct side Oxygen atoms:

Try this one: Al + Fe 2 O 3 Al 2 O 3 + Fe Aluminium atoms: Reactant sideProduct side Iron atoms: Step one – see what's present Step two – make changes Step three – check its balanced Oxygen atoms: 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Reactant sideProduct side Aluminium atoms: Iron atoms: Oxygen atoms:

Try now…

Balancing redox equations

MnO 4 ― + Cl -1 + H + Mn +2 + Cl 2 + H 2 O x + 4(-2) = -1 x – 8 = -1 x = x = 7 ( + 7 ) ( - 2) ( -1) ( +1) (+2) ( +1 ) ( -2) Each Mn goes down 5 in number (reduction) RIG – Each Mn is gaining 5 electrons. Each Cl goes up 1 in number ( oxidation) OIL – Each Cl is losing 1 electron (0) 5 Once the oxidation numbers are balanced, Make sure the overall equation still balances Ratio has to be 1 Mn : 5 Cl

Cr 2 O 7 ―2 + Fe +2 + H + Cr +3 + Fe +3 + H 2 O x + 7(-2) = -2 2(x) – 14 = -2 2x = 12 x = 6 ( + 6 ) ( - 2) ( +2) ( +1) (+3) ( +1 ) ( -2) Each Cr goes down 3 in number (reduction) RIG – Each Cr is gaining 3 electrons. Each Fe goes up 1 in number ( oxidation) OIL – Each Fe is losing 1 electron (+3) 6 Once the oxidation numbers are balanced, Make sure the overall equation still balances Ratio has to be 1 Cr : 3 Fe 7

Calculations based on balanced chemical equations A balanced equation tells you the relative amounts of each reactant and each product involved in the reaction. In this reaction: 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 2 moles of Aluminium 1 mole of Iron oxide 1 mole of Aluminium Oxide 2 moles of Iron

What do these chemical equations tell you? N 2 +O 2 2NO 3Fe + 2O 2 Fe 3 O 4 2H 2 O O 2 +H 2 O N 2 +3H 2 2NH 3

Calculations based on balanced chemical equations The reaction between oxygen and nitrogen is described by the balanced chemical equation: Question: If 2 moles of nitrogen were reacted: (i) How many moles of O 2 would it react with? (ii) How many moles of NO would be formed? N 2 +O 2 2NO

Answer… N 2 +O 2 NO Moles in B.E: therefore… Two moles of O 2 would react Four moles of NO would be produced

Answer: N 2 +3H 2 2NH 3 Moles in B.E: therefore… Six moles of H 2 would react Four moles of NH 3 would be produced The balanced equation: Question: If 2 moles of nitrogen reacted in this reaction, How many moles of hydrogen would react? How many moles of ammonia would be formed? N 2 +3H 2 2NH 3

Answer: 2H 2 O 2 O 2 +H 2 O Moles in B.E: therefore… Three moles of O 2 would be produced Three moles of NH 3 would be produced The balanced equation: Question: If 6 moles of H 2 O 2 reacted in this reaction, How many moles of oxygen would be formed? How many moles of water would be formed? 2H 2 O 2 O 2 +H 2 O

The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: C 6 H C 2 H 5 OH + 2CO 2 I f 126g of glucose are consumed.. Q293 (i) How many moles of glucose does this represent? 126g/ RMM = moles of glucose (126) /180 = 0.7 It represents 0.7 moles of glucose

The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: C 6 H C 2 H 5 OH + 2CO 2 I f 126g of glucose are consumed.. Q293 (ii) How many moles of ethanol are produced? Answer: C 6 H C 2 H 5 OH + 2CO 2 Moles in B.E: therefore… moles of ethanol would be produced

The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: C 6 H C 2 H 5 OH + 2CO 2 I f 126g of glucose are consumed.. Q293 (i) What volume of carbon dioxide, measured at s.t.p is produced? 1.4 moles of CO 2 x 22.4 = Volume of gas at stp (1.4)(22.4) = L of carbon dioxide will be formed

The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: C 6 H C 2 H 5 OH + 2CO 2 I f 126g of glucose are consumed.. Q293 (i) How many molecules of carbon dioxide does this volume contain? 1.4 moles of CO 2 x 6 x = molecules of carbon dioxide (1.4)(6 x ) = 8.4 x x molecules of carbon dioxide would be formed

Q10a Bottle of contents 2.5L of concentrated hydrochloric acid were spilled. It was neutralised with sodium carbonate. The spilled acid was 36%(w/v) (i) Calculate the number of moles of hydrochloric acid spilled. 2.5 L of a 36% (w/v) HCl solution is how many moles? 36/100 x 2500 = 900g There are 900 g of HCL in this much solution. 900g of HCl is how many moles? 900g / RMM = moles 900/ 36.5g = There are moles of HCL in this much solution.

. Bottle of contents 2.5L of concentrated hydrochloric acid were spilled. It was neutralised with sodium carbonate. The spilled acid was 36%(w/v) (i) What is the minimum mass of anhydrous sodium carbonate required to completely neutralise the spilled acid?. The balanced equation of the reaction is Na 2 CO 3 + 2HCl 2NaCl + H CO What is the mass of moles of sodium carbonate? XRMM = moleS X 106 = g g anhydrous sodium carbonate are required to completely neutralise the spilled acid

. Bottle of contents 2.5L of concentrated hydrochloric acid were spilled. It was neutralised with sodium carbonate. The spilled acid was 36%(w/v) (i) What volume of carbon dioxide in L ( at STP) would be produced in this neutralisation reaction? The balanced equation of the reaction is Na 2 CO 3 + 2HCl 2NaCl + H CO What is the volume of moles of carbon dioxide? moles X 24 = Gas volume X 24L = L of carbon dioxide gas( at STP) would be produced in this neutralisation reaction

The following reaction may be used to reduce emissions of sulfur dioxide in waste gases 2CaCO 3 + 2SO 2 + O 2 2CaSO 4 +2CO 2 What volume of sulfur dioxide ( measured at S.T.P) could be removed from the waste gases by this reaction for every kilogram of calcuim carbonate used? Q294 (i)Find how many moles of calcuim carbonate are reacted ( grams – moles) (ii)Find how many moles of sulfur dioxide would react with this calcuim carbonate ( use balanced equation) (iii)Find what the volume of sulfur dioxide gas would be reacted (moles – volume) 1000g/ RMM = moles of calcuim carbonate 100/ 100 = moles of calcuim carbonate would be used

The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: 2CaCO 3 + 2SO 2 + O 2 2CaSO 4 +2CO 2 I f 126g of glucose are consumed.. Q293 Answer: 2CaCO 3 + 2SO 2 + O 2 2CaSO 4 +2CO 2 Moles in B.E: therefore… moles of sulfur dioxide would react with this much caluimcarbonate The following reaction may be used to reduce emissions of sulfur dioxide in waste gases 2CaCO 3 + 2SO 2 + O 2 2CaSO 4 +2CO 2 What volume of sulfur dioxide ( measured at S.T.P) could be removed from the waste gases by this reaction for every kilogram of calcuim carbonate used? Find how many moles of sulfur dioxide would react with this calcuim carbonate ( use balanced equation) (i)Find what the volume of sulfur dioxide gas would be reacted (moles – volume) Q294

10 moles of SO 2 x 22.4 = Volume of gas at stp (10)(22.4) = 224L 224L of sulfur dioxide would react, so thats how much would be removed by doing this reaction The following reaction may be used to reduce emissions of sulfur dioxide in waste gases 2CaCO 3 + 2SO 2 + O 2 2CaSO 4 +2CO 2 What volume of sulfur dioxide ( measured at S.T.P) could be removed from the waste gases by this reaction for every kilogram of calcuim carbonate used? Find what the volume of sulfur dioxide gas would be reacted (moles – volume) Q294

295. (ii) A solution of sodium hypochlorite NaOCl is labelled as having a concentration of 5% (w/v). Express the concentration in grams per litre. 5%w/v means 5g in 100cm 3. How much in a litre? (5/ 100) x 1000 = 50g Answer: There are 50g of NaOCl in a litre

100cm3 of this 5%w/v solution were reacted with excess chloride ion according to the equation OCl - + Cl - + 2H + Cl 2 + H 2 0 (iii) How many molecules of chlorine gas were liberated? 5%w/v sodium hypochlorite solution means 5g in 100cm 3. How many moles? 5g / RMM = Moles NaOCl - 5g/ 74.5 g = There are moles of NaOCl that reacted There must be moles of Cl 2 that are made in the reaction

(iii) How many molecules of chlorine gas were liberated? moles of Cl 2 x 6 x (0.0671)(6 x ) = 1x x = x x molecules of carbon dioxide would be formed

296. Sulfur dioxide was prepared by heating excess dilute hydrochloric acid with 6.3g of sodium sulfite according to the equation: Na 2 SO 3 + 2HCl NaCl + SO 2 + H 2 0 (i) How many moles of sodium sulfite were used? 6.3g of sodium sulfite / RMM = Moles 6.3/ 126 = moles of sodium sulfite would be used

.05 moles of sulfur dioxide would be obtained 296. Sulfur dioxide was prepared by heating excess dilute hydrochloric acid with 6.3g of sodium sulfite according to the equation: Na 2 SO 3 + 2HCl NaCl + SO 2 + H 2 0 (i) What volume of sulfur dioxide would be obtained? Na 2 SO 3 + 2HCl NaCl + SO 2 + H 2 0 Moles in B.E: therefore… moles of SO 2 x 22.4L = Volume at stp (.05)(22.4) = L of sulfur dioxide would be obtained

g of carbon dioxide for every kilometre travelled Car is used for 8km every day (i) Each day, what mass of CO 2 are released? ( 143) x 8 = 1,144g of carbon dioxide used per day 1144g of carbon dioxide = x moles 44g of carbon dioxide = 1 mole (1)(1144) = 44x 26 = x 26 moles of carbon dioxide are used every day ( ii) Each day, how many moles of CO 2 are released?

26 moles of CO 2 = xL 1 Mole of CO 2 = 22.4L (26)(22.4) = 1x 582.4L = x 582L of carbon dioxide would be produced per day g of carbon dioxide for every kilometre travelled Car is used for 8km every day ( iii) Each day, what volume of CO 2 is released?

300. Large SUV emits 264g of carbon dioxide per kilometre, how many more litres of CO2 would be released into the environment (i) Each day, what mass of CO 2 are released? ( 264) x 8 = 2,112g of carbon dioxide used per day 2112g of carbon dioxide = x moles 44g of carbon dioxide = 1 mole (1)(2112) = 44x 26 = x 48 moles of carbon dioxide are used every day ( ii) Each day, how many moles of CO 2 are released?

48 moles of CO 2 = xL 1 Mole of CO 2 = 22.4L (48)(22.4) = 1x L = x L of carbon dioxide would be produced per day g of carbon dioxide for every kilometre travelled Car is used for 8km every day ( iii) Each day, what volume of CO 2 is released? 300. Large SUV emits 264g of carbon dioxide per kilometre, how many more litres of CO2 would be released into the environment

( iii) Each day, how much more Co2 is released in litres when the SUV is used? L – 582L = 493.2L more carbon dioxide was released when driving the SUV

Q298. An indigestion table contains a mass of 0.30g of magnesium hydroxide. Balanced equation for reaction: Mg(OH) 2 + 2HCl MgCl 2 + 2H 2 0 (i) Calculate the volume of 1.0M HCl neutralised by two of these digestion tablets. Give your answer to the nearest cm 3 Mass of Mg(OH) 2 in two digestion tablets = 0.30 x 2 = 0.60g Moles of Mg(OH) 2 in 0.60g of tablets 0.6g / RMM = mole 0.6/58 = moles The balanced equation : Mg(OH) 2 + 2HCl MgCl 2 + 2H moles of HCl would be needed

1.0mole / moles = cm 3 of HCl would be needed What volume of 1.0M HCl would contain moles?

ii) What mass of salt is formed in this neutralisation? What is the mass of moles of the salt? moles /RMM = mass / 95 g = g X = g of the salt is formed The balanced equation : Mg(OH) 2 + 2HCl MgCl 2 + 2H

iii) How many magnesium ions are present in this amount of salt? How many ions in moles of the salt? moles x 6 x = x For every molecule of MgCl 2 there will be one Mg ion.

Another remedy of Mg(OH) 2 in water is marked 6%(w/v). What volume of this second remedy would have the same effect as the tablets from before? We need to find what volume of the second remedy would contain mole of Mg(OH) 2 Remedy has 6g of Mg(OH) 2 in 100cm 3. How many moles are in 6g? This remedy has 6g in 100cm3 6g/ 58 = moles There are moles of Mg(OH) 2 in 100cm 3 So 10cm3 OF THE SOLUTION WOULD BE NEEDED

Q299. A mass of 13g of granulated zinc was reacted with 100cm3 of a 2M solution of nitric acid. The equation for the reaction is 3Zn + 8NO 3 3Zn(NO 3 ) 2 + 2NO + 4H 2 0 (i) Show clearly that the zinc is in excess in the reaction 13g of zinc / RMM = moles of Zinc present

Q299. A mass of 13g of granulated zinc was reacted with 100cm3 of a 2M solution of nitric acid. The equation for the reaction is 3Zn + 8NO 3 3Zn(NO 3 ) 2 + 2NO + 4H 2 0 (i)Show clearly that the zinc is in excess in the reaction 2moles/ 1000 X100 = 0.2 moles X = 0.2 moles of NO 3 present in the reaction

3Zn + 8NO 3 3Zn(NO 3 ) 2 + 2NO + 4H So Zinc is in excess in the reaction

What mass of zinc nitrate was formed?

Extra material

Converting moles to mass The mass of one mole of carbon dioxide = 44g How many moles of carbon dioxide in 22g? Number of moles = Total mass____ Mass of one mole Number of moles = 22g 44g Number of moles = 0.5 Answer: 0.5 moles

Calculations involving balanced equations Magnesium reacts with oxygen to produce magnesium oxide according to the equation: 2Mg + O 2 2MgO In an experiment a student burns 9g of Mg in oxygen. Question 1 – how many moles of Mg were reacted? 9g of Mg = ? moles of Mg Number of Moles = Given mass Mass of one mole Number of moles = 9g 24g Number of moles = moles

Question 2 – how many moles of Mgo would be formed? 2Mg + O 2 2MgO Moles in B.E: therefore… Ans: moles of MgO would be formed in this reaction Question 3 - How many grams of MgO is formed in this reaction? I mole of MgO = 40g moles of MgO = ( x 40g) = 15g Answer: 15g of MgO will be formed

Calcium oxide is manufactured in a lime kiln by heating calcium carbonate to bring about the following reaction: CaCO 3 CaO +CO 2 In this process 5,000g of calcium carbonate(CaCO 3 ) were reacted. Question 1 – how many moles of CaCO3 were reacted? 5000g of CaCO 3 = ? moles of CaCO 3 Number of Moles = Given mass Mass of one mole Number of moles = 5000g 100g Number of moles = 50

Q2: CaCO 3 CaO +CO 2 Moles in B.E: therefore… Moles of CaO would be formed Q3:How many grams of CaO will be formed? 50 moles of CaO formed Mass of one mole= 56g Mass of 50 moles = 50 x 56g = 2800g Answer: 2,800 g of CaO will be formed How many moles of CaO will be formed?

Calculations involving one reactant in excess Reagent in excess = More of this reactant is present that is needed in the reaction Limiting reagent = reactant that is not in excess ( will be totally used up) – this will determine how much of each product is generated

Ammonia gas can be prepared from ammonium chloride and sodium hydroxide: NH 4 Cl + NaOH NH 3 + NaCl + H 2 0 If 20.7g of ammonium chloride and 10g of sodium hydroxide are used (i) which reagent is in excess? (ii) which reagent is the limiting reagent? (ii) Calculate the mass of sodium chloride and the volume of ammonia formed at s.t.p

Another example.. Worksheet 24.5 A,B Worksheet 24.7 a

Percentage yield Percentage Yield = Actual Yield x 100 Theoretical Yield

Example 1 – In an experiment to prepare ethane 10.2g of ethanol was heated with aluminium oxide and 1.7g of ethene was formed. Calculate the percentage yield of ethene C 2 H 5 OH C 2 H 4 + H 2 0

Try now 2012 Q2 (D), (E) Worksheet 24.2 Worksheet 24.3

Question 340e)assume all other features allow maximum yield of ethanol, what mass of ethanol would be made from 8.94g of sodium dichrommate, and a 75% yield was obtained? 3C 2 H 5 OH + Cr 2 O H + 3CH 3 CHO + 2Cr H 2 0 Na 2 Cr 2 O g X = 8.94/ 298 X = 0.03 moles 0.03 moles

3C 2 H 5 OH + Cr 2 O H + 3CH 3 CHO + 2Cr H moles 0.03 moles x 3 = moles theoretical yield 75% Yield from this reaction 0.09 moles x0.75 = moles of ethanol would be produced moles x 44g X = 2.97g 2.97g of ethanol would be produced

Question 341a) Which is the limiting factor? 3C 2 H 5 OH + Cr 2 O H + 3CH 3 CHO + 2Cr H 2 0 Na 2 Cr 2 O g = ? moles X = 8.84/ 298 X = moles a) So which is the limiting factor? Na 2 Cr 2 O 7 is the limiting reactant as it will be totally used up in the reaction C 2 H 5 OH is in excess as some of this will be left over in the reaction ( only x 3 = moles used up ) molesC 2 H 5 OH 5.52g = ? moles X = 5.52/ 46 X = 0.12

Question 341b) what is the % yield? 3C 2 H 5 OH + Cr 2 O H + 3CH 3 CHO + 2Cr H moles moles x 3 = moles theoretical yield Actual yield 1.6 of ethanal was actually formed 1.6/ 44 = moles of ethanal actual yield % Yield = Actual yieldx 100 Theoretical yield x = = 40.86%

Na 2 Cr 2 O g X = 17.88/ 298 X = 0.06 Question 344d) Which is the limiting factor? 3C 2 H 5 OH + Cr 2 O H + 3CH 3 CHO + 2Cr H molesC 2 H 5 OH 11.04g = x moles X = 11.04/ 46 X = moles a) So which is the limiting factor? Na 2 Cr 2 O 7 is the limiting reactant as it will be totally used up in the reaction C 2 H 5 OH is in excess as some of this will be left over in the reaction ( only 0.06x 3 = 0.18 moles used up )

Question 344e) what is the % yield? 3C 2 H 5 OH + Cr 2 O H + 3CH 3 CHO + 2Cr H moles0.06 moles x 3 = 0.18 moles theoretical yield Actual yield 2.97 of ethanal was actually formed 2.97g/ 44g = x moles of ethanal actual yield % Yield = Actual yieldx 100 Theoretical yield x = 37.5 = 37.5%

C 2 H 5 OH 0.8g = ? cm x 0.8 = 6.4g of ethanol 6.4 g = ? moles 6.4/ 46 X = moles Na 2 Cr 2 O 7. 2H g = ? moles X = 29.8/ 298 X = 0.1 moles 331 Show clearly that the ethanol was the limiting reagent when 8.0cm 3 of ethanol (density 0.80g cm -3 ) was added to 29.8g of sodium dichromate, Na 2 Cr 2 O 7.2H 2 0. There was excess sulfuric acid present. 3C 2 H 5 OH + Cr 2 O H + 3CH 3 CHO + 2Cr H moles moles If there was 0.1 moles of sodium dichromate reacting it would need (3 x 0.1 moles) of ethanol = 0.3 moles There is not enough ethanol for this!!! Ethanol is limiting reactant, Sodium dichromate is in excess

C 2 H 5 OH 0.8g = ? cm x 2.3 = 1.84g of ethanol 1.84 g = ? moles X = 1.84/ 46 X = 0.04moles g of sodium dichromate, Na 2 Cr 2 O 7.2H 2 0, 2.3cm 3 of ethanol (density 0.80g cm -3 ) were reacted.. There was excess sulfuric acid present. 1.7g of ethanoic acid was formed. (i) Show clearly that the ethanol was the limiting reagent 3C 2 H 5 OH + Cr 2 O H + 3CH 3 CHO + 2Cr H moles 0.04moles If there was moles of sodium dichromate reacting it would need There is not enough ethanol for this!!! Ethanol is limiting reactant, Sodium dichromate is in excess Na 2 Cr 2 O 7. 2H g = ? moles X = 8.84/ 298 X = moles

g of sodium dichromate, Na 2 Cr 2 O 7.2H 2 0, 2.3cm 3 of ethanol (density 0.80g cm -3 ) were reacted.. There was excess sulfuric acid present. 1.7g of ethanoic acid was formed. (ii) Calculate the percentage yield of ethanoic acid 3C 2 H 5 OH + Cr 2 O H + 3CH 3 COOH + 2Cr H moles 0.04moles = Theoretical yield Actual yield of ethanoic acid: 1.7g = ? moles X = 1.7/60 X=.0283 moles This is the actual yield % Yield = Actual yieldx 100 Theoretical yield x = = 70.75%

C 2 H 5 OH 1.29 g = ? moles X = 1.84/ 46 X = moles Na 2 Cr 2 O 7. 2H g = ? moles X = 6.62/ 298 X = moles g of sodium dichromate, Na 2 Cr 2 O 7.2H 2 0, 1.29g of ethanol were reacted.1.2g of ethanoic acid was formed. (i) Show clearly that the sodium dichromate was in excess 3C 2 H 5 OH + 2Cr 2 O H + 3CH 3 COOH + 4Cr H moles moles If there was moles of sodium dichromate reacting it would need ( moles) of ethanol = There is not enough ethanol for this!!! Ethanol is limiting reactant, Sodium dichromate is in excess

% Yield = Actual yieldx 100 Theoretical yield x = C 2 H 5 OH + 2Cr 2 O H + 3CH 3 COOH + 4Cr H = Theoretical yield Actual yield of ethanoic acid: 1.2g = ? moles X = 1.2/60 X=.02 moles This is the actual yield moles