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Book 2

- Amount of Substance

...................The MOLE!!!

ValencyThis is a number assigned to each element that indicates the number of electrons that it has lost/gained in order to become

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stable. Valencies are used to write chemical formulae and to determine the ions formed

Sodium tends to lose electron where possible. We know this because sodium is in group so has electron in its outer shell.We would say that sodium has a VALENCY of 1

Valencies of elements in groups

Group Number of outer shell electrons

Valency Ion formed

Name of ion

1 1 1 +1 Sodium2 2 2 +2 Magnesium3 3 3 +3 Aluminium4 4 4 -5 5 3 -3 Nitride6 6 2 -2 Sulphide7 7 1 -1 chloride

Group 8 elements are called the noble gases. They do not react with other elements as they have a full octet

Valencies of elements not in groups (transition metals)

Metal Valency Ion formed/ nameCopper 1 or 2 Cu+, copper (I), Cu2+, copper

(II)Zinc 2 Zn2+, ZincIron 2 or 3 Fe2+, Iron (II), Fe3+, Iron (III)

Manganese 2, 4 or 7 Mn2+, Manganese (II)Chromium 3 or 6 Cr3+, Chromium (III)

Silver 1 Ag+, SilverLearn these!!

Complex ions contain more than one element, you will probably be familiar with some of these. You must learn the formula, charge & name of the ion.

Name formula Valency

Carbonate CO32- 2

Hydrogencarbonate HCO3- 1

Sulphate SO42- 2

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Sulphite SO32- 2

Hydrogensulphate HSO4- 1

Nitrate NO3- 1

Hydroxide OH- 1Dichromate Cr2O7

2- 2Manganate MnO4

- 1Ammonium NH4

+ 1Phosphate PO4

3- 3

complete the following table

Name Formula of ion

Valency Name Formula of ion

Valency

iodide I- 1 magnesium

Mg2+ 2

Suphate SO42- 2 ammonium

NH4+ 1

sulphide S2- 2 nitride N3- 3aluminium Al3+ 3 hydroxi

deOH- 1

dichromate Cr2O72- 2 Iron (II) Fe3+ 3

When we know what ion an element forms we can write equations to show the process. These are called half ionic equationse.g. magnesium

Mg Mg2+ + 2e-

write ½ eqns to show the following forming ions

a) Lithium Li Li+ + e-

b) Chlorine Cl2 + 2e- 2Cl-

c) Sulphur S + 2e- S2-

Reading and writing chemical formulae

Naming compounds Rules1. metal name goes first and doesn’t change2. non-metal elements become ‘ides’ to show that they are

negative ions e.g. chloride3. ate means oxygen is present e.g. sulphate4. ite also means oxygen present but LESS than ate5. when non-metals form positive ions they are called ‘ium’ e.g.

ammonium6. if a metal has more than one valency it has a Roman numeral

in the name of the compound to show which ion is present.

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e.g. Cu2O = copper (I) oxide

Reading

Formula Name of compound Composition K2O Potassium oxide 2K and 1 OCaSO4 Calcium sulphate 1 Ca, 1 S and 4 OMg(NO3)2 Magnesium

nitrate1 Mg, 2N and 6 O

CuSO4 .5H2O Copper (II) sulphate

1 Cu, 1S 9O and 10 H

Na2CO3.10H2O Sodium carbonate 2 Na, 1C, 13 O and 20 H

Writing

There are two methods, one involves using valencies, the other - ions.

Example: magnesium chloride

Valency method ion method

Mg Cl Mg2+ Cl-

2 1 to be neutral need 2 Cl-

swap = MgCl2 therefore = MgCl2

Complete the table

Name formulaStrontium fluoride Sr F

2 1 = SrF2Sodium sulphate Na SO4

1 2 = Na2SO4Silicon chloride Si Cl

4 1 = SiCl4Aluminium carbonate Al CO3

3 2 = Al2(CO3)3Ammonium phosphate NH4 PO4

1 3 = (NH4)3PO4Copper (II) oxide Cu O

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2 2 = CuO

Formula nameCsF Caesium fluorideCaCO3 Calcium carbonateFeI3 Iron (III) iodideCuSO4 Copper (II) sulphateNaHCO3 Sodium hydrogen carbonateFeO Iron (II) oxide

WRITING EQUATIONSChemical equations must be BALANCED and often include state symbols

(g) Gas e.g. H2, N2, O2, CO2, Cl2, F2 and group 8(l) Liquid H2O, Hg, Br2 (s) Solid all metals, C , S, P, I2, Si (aq) Aqueous (dissolved in water), all acids and alkalis, most

salts,

Write and balance the following equations and add state symbols:-Magnesium + hydrochloric acid magnesium chloride + hydrogenMg(s) + 2HCl (aq) MgCl2(aq) + H2(g)nitrogen + hydrogen ammonia (NH3)N2(g) + H2(g) NH3(g)sodium hydroxide+ sulphuric acid sodium sulphate + waterNaOH(aq) + H2SO4(aq) Na2SO4(aq) H2O(l)

THE MOLE!

An amount of substance is the quantity whose unit is the mole

A mole is the amount of any substance containing as many particles as there are carbon atoms in exactly 12g of carbon-12 Symbol = n

A mole is an actual amount, it is 6.02 x 1023. (Compare: a dozen = 12, a score = 20, a pair = 2) This is called the Avogadro constant.

The Avagadro constant, NA, is the number of atoms per mole of the carbon-12 isotope.

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So one mole of magnesium contains 6.02 x10 23 magnesium atoms.Two moles of magnesium would contain 1.2. x 1024 Mg atoms

BEWARE! Because oxygen exists as O2 if we have one mole of O2 we would have 2 x 6.02 x 1023 = 1.2 x 1024 atomsQuestions:How many atoms are there in

a) 3 moles of Na3 x 6.02 x 1023 = 1.81 x 1024

b) 0.25 moles of Al0.25 x 6.02 x 1023 = 1.51 x 1023

c) 2 moles of H22 x 2 x 6.02 x 1023 = 2.41 x 1024

Relative molecular mass, Mr is the average mass of a molecule compared to carbon-12This is calculated using relative atomic masses. It is used for simple molecular (covalent) moleculese.g. CO2

Relative formula mass is also calculated using ArThis is used for giant molecular (covalent) molecules. e.g. diamondCalculating moles: The relationship between moles, mass and Mr is used continuously in chemical calculations

in grams

Moles = mass (n) Mr

Calculated from the periodic tableCalculate the Mr ofa) H2SO4 b) Mg(OH)2(2 x 1) + 32.1 + (4 x 16) = 98.1 24.3 + {(16 + 1) x 2] = 58.3

Calculate the mass of a) 6 moles of rubidium

6 x 85.5 = 513g mass =b) 0.25 moles of CO2 moles x

molar mass0.25 x 44 = 11g

c) 0.01 moles of H2SO4

No. of atoms =

moles of x 6.02 x 1023

atoms

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0.01 x 98.1 = 0.981g

how many moles are present in a) 1.2g of Mg

n = 1.2/24.3 = 0.0494 b) 30g of C

n = 30/12 = 2.5c) 6.7g of PCl5Mr PCl5 = 208.5 n = 6.7/208.5 = 0.0321

Molar massMolar mass is the mass of one mole of a substance, it is the same as Mr except that Mr does NOT have units, molar mass has units of g mol-1 (grams per mole)

Molar mass is the mass per mole of a substance Units are g mol-1 (g/mol)To calculate the molar mass we use the relative atomic masses of the atoms from the periodic table.Using moles in calculationsSteps:

1. calculate moles of the reactant2. use the balanced eqn to deduce moles of product that

should be made (theory)3. calculate the mass of product that should be made

(theory)

Zinc reacts with hydrochloric acid producing zinc chloride and hydrogen gas. Write a balanced symbol equation, taking care with the formula of zinc chloride.

Zn + 2HCl ZnCl2 + H2

The equation tells us that:

one mole of zinc reacts with 2 moles of HCl

mole ratio is 1 : 2 : 1 : 1so 1 mole of zinc makes 1 mole of ZnCl2

If 6.5g of zinc are used, how many moles of HCl are needed to react with it? n of Zn = 6.5 = 0.0994 ( to 3 s.f.)

65.4As ratio is 1 : 2

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Zn : HCl n of HCl = 2 x 0.0994 = 0.199

How many moles of zinc chloride should be made?As ratio Zn : ZnCl2

1 : 1 n of ZnCl2 = 0.0994What mass of zinc chloride should we make (in theory) in this reaction?Mass = n x Mr = 0.0994 x [65.4 + (2 x 35.5)] = 13.6g

If only 7.3g of zinc chloride is ACTUALLY made, what is the percentage yield?

% yield = 7.3 x 100 = 53.7% 13.6

Percentage yield calculations

given in question

% Yield = actual mass x 100theoretical mass

the one you calculate

Questions

1. when 20g of calcium carbonate are thermally decomposed, only 4g of calcium oxide are actually made. Calculate % yield.CaCO3 CaO + CO2Step 1 n of CaCO3 = 20 = 0.200 (3s.f.)

100.1

step 2 ratio is 1 : 1 so moles of CaO = 0.200

step 3 theoretical mass CaO = 0.2 x 56.1 = 11.2g

step 4 % yield = 4 x 100 = 35.7% 11.2

2. when 30g of chlorine react directly with excess iron, 25g of the product iron (III) chloride is actually made. Write a balanced symbol eqn and calc % yield.

2Fe + 3Cl2 2FeCl3n of Cl2 = 30 = 0.423 mol

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70n of FeCl3 = 0.423 x 2/3 = 0.282 mol

theoretical mass = 0.282 x 162.3 = 45.8g

% yield = 25 x 100 = 54.9% 45.8

MOLES AND GASES

The ideal gas equation Volume Moles

Temp 1000

Pressure PV = nRT cm3 dm3

R = Universal gas constant 8.31jk-1mol-1

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When using the ideal gas equation we must use certain units of volume, pressure and temperature we call these SI units

Temperature: Kelvin

oC (+273) k

Pressure: Pa (Pascals) 100kpa = 1000,00pa

Kpa x103 pa

Volume: m3

cm3 x10-3 dm3 x10-3 m3

1. Rearrange the ideal gas equation to find Volume

V = nRT P 2. Rearrange the ideal gas equation to find Pressure

P = nRT V

3. Rearrange the ideal gas equation to find Moles

n = PV RT

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4. Rearrange the ideal gas equation to find Temperature

T = PV nR

Question: Potassium Nitrate decomposes on strong heating forming Oxygen and Potassium Nitrite as the only products

a)A 1.00g sample of KNO3 (Mr 101.1) was heated strongly until fully decomposed. Calculate the number of moles of KNO3 in the sample

b)At 25OC and 100kpa the oxygen gas produced in the decomposition occupied a volume of 12.6cm3. State the ideal gas equation and use it to calculate the number of moles of O2 produced.(R = 8.31J k-1mol-1)

1. n KNO3 = 1.00 / 101.1 = 9.89X10-3

2. Convert units (k, Pa and m3)

25OC + 273 = 298k

12.6cm3 x10-6 = 1.26x10-5m3

100kpa x103 = 100,000pa

4.Apply the ideal gas equation (rearranged) n = PV RT 100,000pa X 1.26x10 -5 m 3 = 5.088x10-4

8.31JK-1mol-1 X 298k C) Calculate the % yield of Oxygen in the reactionKNO3KNO2 + ½ O2

% yield = Actual moles / Theoretical moles x 100

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5.088x10-4 / 4.945x10-3 (9.89x10-3 / 2 due to 1:0.5 ratio) x 100 = 10.3% MOLES AND CONCENTRATIONThe concentration of a solution is the amount of solute, in mol, dissolved in 1 dm3 (1000cm3) of solution.

The units are mol dm-3 ( mol/dm3)

We can calculate the concentration of any solution if we know number of moles (n) volume in dm3

concentration molesC = n

v volume in dm 3 [ n = c x v ]

Acids are said to be concentrated or dilute.

A concentrated acid is one in which a large amount of acid is in each dm3 of the solution.

A dilute acid is one in which a small amount of acid is in each dm3 of the solution.

Calculate the concentration of the following:a) 9.4g of NaCl dissolved into 500cm3 of water. 9.4 = 0.161 mol c = n = 0.161 = 0.322 mol58.5 v 0.500

b) 10g of KOH dissolved into 2dm3 watern = 10 = 0.178 c = 0.178 = 0.0890 mol dm-3

56.1 2

What mass of solid must be added to make the following?

a) 200cm3 of a 0.1 mol dm-3 solution of NaOHn = 0.1 x 200/1000 = 0.0200 molmass = 0.0200 x 40 = 0.8g

b) 750cm3 of a 2 mol dm-3 solution of Na2CO3.10H2O

n = 2 x 750 / 1000 = 1.5 mol

[Mr for Na2 CO3.10H2O = (23 x 2) + 12 + (16 x 3) + (10 x 18) = 286 ]

calc moles n = mass

Mr

Calculate conc.C = n/v

Calc molesn = c x v/1000

Calc massMass = n x Mr

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mass = 1.5 x 286 = 429g

TITRATIONSThere are several types of titration; we will only focus on ACID-BASE titration’s.Titration’s are used to find the concentration of an acid or alkali (base)An indicator is used to show the ‘end-point’ which is when the acid has reacted with all of the alkali

Acid name FormulaHydrochloric acid HCl

Sulphuric acid H2SO4Nitric acid HNO3

Note: all acids and alkalis are given the state symbol (aq). The salts made when these react are also usually (aq).

Example25cm3 of 0.05mol dm-3 NaOH was added to a conical flask. This required exactly 17.5cm3 of HCl to react with it. Calculate the concentration of the HCl used.Steps

1. write a balanced symbol eqn2. calculate moles of the KNOWN (i.e. the thing we know

conc. and vol for)3. use the stoichiometry of the eqn to DEDUCE the number

of moles of UNKNOWN used4. calc the conc. of the unknown

calculation:1. NaOH + HCl NaCl + H2O

2. n of NaOH = 0.05 x 25/1000 = 1.25 x 10-3 ( 0.00125)

3. as ratio 1 : 1 n of HCl = 1.25 x 10-3

4. conc of HCl = 1.25 x 10-3/ 0.0175 = 0.0714 mol dm-3

Questions: 1. 25cm3 of 0.2moldm-3 NaOH reacts with 11.2cm3 of H2SO4. Calculate the conc. of the sulphuric acid. (work out the formula of the salt carefully)

2. A student has 50cm3of 0.01 moldm-3 KOH, she wishes to exactly neutralise this with 0.1 moldm-3 HNO3. What volume of the acid will she need?

Answer on the back of a page.

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1. 2NaOH + H2SO4 Na2SO4 + 2H2O25 cm3 11.2 cm3

0.2 mol dm-3 ?

n of NaOH = 0.2 x 25 = 5.0 x 10-3 ( 0.00500)

1000

n of H2SO4 ratio 2 : 1 = 0.00500 = 0.00250 2

conc H2SO4 = 0.00250 = 0.223 mol dm-3

0.0112

2. KOH + HNO3 KNO3 + H2O 50 cm3 ?0.01 mol dm-3 0.100 mol dm-3

n of KOH = 0.01 x 50 = 5.0 x 10 –4

(0.000500) 1000

n of HNO3 = 5.0 x 10-4 due to 1.1 ratio

vol = 0.00050 = 5 x 10-3 dm-3 (0.00500) 0.1 or 5cm3

Calculating formulae:

Copper has a valency of 1 or 2. Use the following data to calculate the correct formula for an oxide made when copper is heated in air.4.3g of copper was heated in air and 4.85g of solid oxide was formed.

Steps:1. subtract the mass of the metal from the mass of the

solid oxide formed to find mass of oxygen.2. calculate the number of moles of each element.3. divide the number of moles of each element by the

smallest value.

Mass of oxygen 4.85 – 4.3 = 0.55g

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Cu : O4.3 : 0.55 ( Ar)63.5 : 16

0.0677 : 0.0344 ( smallest) 1.97 : 1 formula = Cu2O

Iron has two possible valencies. If 7.21g of iron was heated in air the solid made weighed 10.29g. Calculate the formula of the iron oxide.

Mass of oxygen = 10.29 – 7.21 = 3.08g

Fe : O 7.21 3.08 55.8 16

0.129 : 0.193 1 : 1.5

2 : 3 formula = Fe2O3

Calculating Stoichiometric relationships.Sodium hydroxide, NaOH, reacts with an acid to form a salt and water. It is known that the acid is either sulphuric acid or hydrochloric acid. Using the information below, calculate the stoichiometric relationship between the acid and NaOH and therefore identify the acid used.

NaOH AcidConcentration/mol dm-3 0.01 0.005Volume/cm3 25 25

Moles of NaOH = 0.01 x 25 / 1000 = 2.50 x 10-

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Moles of acid = 0.005 x 25 / 1000 = 1.25 x 10-4

Ratio of NaOH : acid is 2.5 x 10-4 : 1.25 x 10-4

2 : 1so reacting ration is 2 NaOH :1 acid

acid must be H2SO4

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2NaOH + H2SO4 Na2SO4 + H2O

Potassium hydroxide reacts with either nitric acid or phosphoric acid, H3PO4. Use the data in the table below, calculate the stoichiometric relationship between the acid and KOH and therefore identify the acid used.

KOH AcidConcentration/mol dm-3 0.2 0.16Volume/cm3 20 25

Moles of KOH = 0.2 x 20 / 1000 = 4.0 x 10-3

Moles of acid = 0.16 x 25 / 1000 = 4.0 x 10-3

Ratio of KOH : acid is 4.0 x 10-3 : 4.0 x 10-3

1 : 1so reacting ration is 1 KOH :1 acid must be HNO3KOH + HNO3 KNO3 + H2OEMPIRICAL AND MOLECULAR FORMULAE

Empirical formula is the simplest whole number ratio of atoms of each element in a compound.

Molecular formula is the actual number of atoms of each element in a compound.

Determining the Empirical and Molecular formula.

A compound contains 6.75g of Al and 26.63g of Cl. What is its empirical formula? The molar mass of the compound is 267gmol-1 , calculate the molecular formula of this compound.

Steps1. find moles of each element (from mass or from % of element)2. divide by the smallest number3. use the whole number ratio to deduce Empirical formula4. use molar mass/Mr to calc molecular formula

Al : Cl6.75 : 26.63 27 : 35.5

0.25 : 0.750 smallest0.25 : 0.25

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1 : 3 Empirical formula = AlCl3 mass = 133.5

Mr = 267 so Ef x 2 = Mf = Al2Cl6

Questions (answer over the page)1. an organic molecule is found to contain 90.6% C and 9.4% H, calculate the empirical formulae. The Mr of the molecule is 106, use this to deduce the molecular formula.2. A compound contains only carbon, hydrogen and oxygen. Its composition by mass is 40%C and 6.7% H. calculate the Empirical formula. The molar mass is 180 gmol-1. Use this data to calculate Molecular formula.Answers to questions.ANSWERS

1 C : H90.6 : 9.412 : 1

7.55 : 9.41 : 1.25(can’t round up, x 4 to

get whole number)4 : 5

empirical formula = C4H5Er mass = 53 Mf = emp. Formula x 2Mr 106 106 / 53 = 2 =C8H10

2. C : H : O40 : 6.7 : 53.312 : 1 : 16

3.33 : 6.7 : 3.331 : 2 : 1

empirical formula = CH2O Er mass = 30 Mf = emp. Formula x 6Mr = 180 = C6H12O6

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Compounds containing water

Some compounds can exist in both the anhydrous and hydrated forms.

For example copper (II) sulphate is a white powder when anhydrous and forms blue crystals when hydrated.

The difference between the two forms is due to the presence of water of crystallisation.

CuSO4 + 5H2O CuSO4. 5 H2O

e.g. CuSO4 is anhydrousCuSO4. 5H2O is hydrated as it contains 5 waters of crystallisation.

Water of crystallisation: water molecules which make up the part of the crystal structure of a compound, that can be removed by heating.

Anhydrous: the compound which remains when the water is removed from a hydrated salt.Means: being without water.

Hydrated: a complex molecule which has water molecules incorporated

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Questions involving water of crystallisation

1. Calculate the percentage mass of water of crystallisation in copper (II) sulphate-5-water [CuSO4. 5 H2O]

Mass of CuSO4 . 5H2O = 63.5 + 32.1 + (4 x 16) + 10 + ( 5 x 16)

= 249.5mass of water in molecule = 5 x 18 = 90

% mass of H2O = 90 x 100 = 36% 249.5

2. Weighed samples of the following crystals were heated to drive off the water of crystallisation. When they reached constant mass, the following masses were recorded. Deduce the empirical formulae of the hydrates

a) 0.942g of MgSO4. a H2O gave 0.461g of residuemass of water removed= 0.942 – 0.461 = 0.481g

MgSO4 : H2O 0.461 : 0.481 120.4 180.00383 0.0267 1 7 formula = MgSO4 . 7 H2O

b) 1.124g of CaSO4. b H2O gave 0.889g of residuemass of water removed = 1.124 – 0.889 = 0.235g

CaSO4 : H2O 0.889 : 0.235 136.1 180.00653 0.0130 1 2 formula = CaSO4 . 2H2O

c)1.203g of Hg(NO3)2. c H2O gave 1.172g of residuemass of water removed = 1.203 – 1.172 = 0.031gHg(NO3)2 : H2O 1.172 : 0.031 325 180.00360 0.00172 2 1 1 ½ formula = Hg(NO3)2 . ½ H2O

ATOM ECONOMY

Make sure equation is balanced or you will get the wrong answer!!!!

atom economy = mass of desired product___ x 100 sum of masses of all products (or reactants)

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Mass of products = Mass of reactants as long as the equation balances.

e.g. ethene reacts with steam, the desired product is ethanol.

A single product is made in an addition reaction, giving 100% atom economy.

e.g. 2-iodobutane undergoes a hydrolysis reaction with aqueous sodium hydroxide; the desired product is butan-2-ol. Write a balanced equation for this reaction and calculate the atom economy.

Step 1: calculate the Mass of the desired productStep 2: divide this by the sum of the Mass of all the products

CH3CH2CHICH3 + NaOH CH3CH2CHOHCH3 + NaI

Mr = 74 Mr = 149.9

Atom economy = 74 x 100 = 33.1% (74 + 149.9)

The majority of the starting materials have turned in to waste.

Even if the reaction proceeded with a 100% yield, a large proportion of the mass of atoms used would be wasted.

A high atom economy benefits society as it reduces the amount of waste produced, and it reduces the cost to companies. Reactions are more sustainable.

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“Waste products” are the by-products formed during a reaction. Their disposal can be costly. However, some by-products can be sold on or used elsewhere in the chemical plant.

NB. Percentage yield and atom economy are not the same thing!!

Atom economy tells you the proportion of reactants that is converted into useful products.

Percentage yield tells you the efficiency of this i.e of converting reactants into products

Do not confuse them!!

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