1.1 Mole Concept and Avagadro S Constant (SL/HL)

Topic 1 Quantitative

1.1 Mole Concept and Avagadro’s Constant (SL/HL)

23 Avagadro’s Constant (NA or L) = 6.02 x 10

 Atoms of elements have too small a mass to weigh.

 It is more convenient to weigh a certain number of atoms.

 6 x 1023 atoms of carbon 12 atoms weigh 12g

 This number of particles is called a mole.

 Moles can be applied to atoms, molecules, electrons or (formula units) ions.

Number of Moles = Number of particles / Avagado’s Constant

Or Number of particles = Number of moles x Avagadro’s Constant

E.g. 2 moles of hydrogen atoms will contain 2 x 6.02 x 1023 = 12.04 x 1023 atoms.

E.g. If a sample of water contains 3 x 10 23 molecules, the water therefore contains

3 x 1023 / 6 x 1023 = 0.5 moles of water molecules E.g. 0.5 moles of NaCl will contain 3 x 1023 ions of Na+ and 3 x 10 23 ions of Cl-.

E.g. How many electrons are needed to produce one mole of Al during electrolysis?

Al3+ + 3e = Al

Therefore, 3 moles of electrons are need. i.e. 18 x 1023 electrons

Tok: Think for a while about what the Avogadro Number. (How long would it take you to count to that figure?) The magnitude of this figure is beyond the scale of our everyday experience. If this is the case, how do we know that the figure is accurate? What are the characteristics of the Scientific Method that enables scientists to make calculations like Avogadro performed, accurate and precise? 1.2 Formulas

Atomic mass (A) : Mass of an atom of the element on a scale where C12 has a mass of 12.

Molecular mass (M) : Add together all the atomic masses in a covalent molecule.

Relative atomic mass (Ar) : Most elements are made up of atoms with slightly different masses (called isotopes). The actual mass of the element is therefore an average of the mass of all the isotopes. (Cl35.5)

Relative molecular mass (Mr) : Add together all the relative atomic masses.

Formula Mass : Add together the relative atomic masses of all the particles in an ionic species (can include water of crystallisation)

E.g. CuSO4. 5H2O

63.55 + 32.06 + (4 x 16.0) + (10 x 1.01) + (5 x 16.0) = 249.71

NOTE: Since all of these terms are a comparision to the mass of C12 then none of them have any units.

TOK: think for a while about the effect of assigning numbers to the masses of chemical elements and how this allowed Chemistry to move forward in its’ development into a Physical Science by being able to use Mathematics to express relationships between reactants and products. It seems so simple to us now, but at that time it was a huge step forward! Molar Mass

 This is a general term that can be used for any of the above.  It is the mass of 1 mole of the element, molecule or compound.  Since it is an actual mass rather than a comparison to a mass, molar mass has units of grams per mole (g/mol, g mol-1).

Moles = Mass / Molar mass

n = m / Mr

Try some examples of how to use this equation. Empirical Formula and Molecular Formula

 The empirical formula is the simplest formula of a substance.  It is the simplest ratio of elements present in the substance.

 The molecular formula is therefore the actual formula of the substance.  It is the actual number of atoms of each element present.  It is a multiple of the empirical formula.

E.g. A compound has a molar mass of 88. It is found to contain 54.5% C, 9.1% H and 36.4% O.

C : H : O

54.5 9.1 36.4 12 1 16 4.54 9.1 2.275

4.54 9.1 2.275 2.275 2.275 2.275

1.99 4 1

Empirical formula = C2 H4 O

Molar mass = 44

Actual molar mass = 88

Molecular formula = C4 H8 02

E.g. 2 A compound only contains C, H and O. It has a molar mass of 172. Calculate the empirical and molecular formula if it contains 69.76% C, and 11.63% H.

E. g.3 A hydrocarbon has an empirical formula mass of 13. The molar mass is 78. What is its molecular formula? 1.3 Chemical Equations

 Reactants and products.  Single arrows go to completion, double arrows show reversible reactions.

 Coefficients are the numbers that go in front of the formula of the compound or molecule to balance the equation.(Sometimes questions ask for the SUM of coefficients. This means that all the coefficients in the equation need to be added up.)  Subscripts indicate the number atoms within a given molecule. E.g. O2 means that a molecule of Oxygen contains 2 atoms of Oxygen within it.  State symbols show the physical state of the reactants and products where s = solid, l = liquid, g = gas and aq = aqueous solution in water.

ToK: think about when these symbols are necessary in aiding understanding and when are they redundant? Is there a difference between:

NH3(g) and HClg) = NH4Cl(g) NH3(aq) and HCl(aq) = NH4Cl(aq)

What do you think the following equation represents?

NH3 and HCl = NH4Cl

Discuss this with your partner Is it always essential to use state symbols? What about this equation? Are they needed?

Fe + H2SO4 = FeSO4 + H2

 Ionic equations show what happens to aqueous ions during reaction

- - Cl2(g) + 2Br (aq) 2Cl (aq) + Br2(l)

 - What is the mole ratio of Cl2 to Br in the above equation? 1.4 Mass and Gaseous volume relationship

CaCO3(s) CaO(s) + CO2(g) When 3.52 g of calcium carbonate are decomposed, calculate: a) the mass of calcium oxide produced b) the volume of carbon dioxide released.

a) Moles of CaCO3 = 3.52 / 100

Moles of CaO produced = 3.52 / 100

Mass of CaO = 3.52 / 100 x 56 = 1.97 g

b) Moles of CO2 produced = 3.52 / 100

Reactions involving gases:

Whenever gases are involved you need to remember that they both have mass and volume and that the molar volume of an ideal gas under standard conditions is 2.24 x 10 -2 m3 mol-1 (22.4 dm3 mol-1)

3 This means that 2g of H2 has a volume of 22.4 dm 3 and 32g of O2 has a volume of 22400 cm

Volume of CO2 = 3.52 / 100 x 22.4= 0. 79 litres

Note that the above calculation assumes a 100% yield. What quantities would be produced if there was only a 75% yield?

E.g. 2 Calculate a) the volume of H2 produced b) the mass of salt produced When 1g of calcium metal is neutralised by an excess of sulphuric acid under standard conditions

Balanced Eq:

Moles of Ca;

Moles of H2

Volume of H2

Moles of CaSO4 salt:

Mass of CaSO4 salt:

Limiting Reactant  In a chemical reaction it is unusual for there to be exactly the same amounts of each of the reactants.  Usually one of the reactants will run out first. This is called the limiting reactant.  This is the most important reactant because as soon as it runs out then the reaction will stop.  Some of the other reactant will be left over at the end. It is said to be in excess (more than enough).

Fe + S FeS

If 5 g of Fe and 5 g of S are reacted together, calculate;

a) Which is the limiting reactant? b) How much FeS will therefore be formed? c) By how much is the other reactant in excess?

a) Fe : S 5/56 5/32 0.089 0.156

Therefore Fe is the limiting reactant as it will run out first, S will be in excess as there is more than enough.

b) Moles of limiting reactant used = 0.0893 Moles of FeS formed = 0.0893 Mass of FeS = 0.0893 x 88 = 7.85 g

c) Moles of S left over = 0.156 – 0.0893 = 0.0667 Mass of S in excess = 0.0667 x 32 = 2.15g

For Reactions involving only gases: Avogadro’s hypothesis (why is this not a Law?) tells us that at the same temperature and pressure, equal volumes of all gases contain the same number of particles. The Avogadro Constant is the number of molecules of any gas present in a volume of 22.41 L and is the same for the lightest gas (hydrogen) as for a heavy gas such as carbon dioxide or bromine. This actually makes calculations very easy!

E.g. What volume of sulfur trioxide would be produced by the complete reaction of 100cm3 sulphur dioxide with oxygen?

2SO + O2 2SO3

Ratios 2 1 2 i.e. 1 ½ 1

100cm3 50cm3 100cm3

What would be the composition of the final product in the above question if 100cm3 of oxygen was used instead of 50cm3

In the following question the volume of one of the gases produced is given. Calculate the volumes of the gases which react and are produced. All gas volumes are measured at the same temperature and pressure.

2N2O 2N2 + O2 3 15cm

Behaviour of Gases:  Ideal gases should have no forces between the particles, the particles volume is zero in comparison to the volume of the gas, and they have completely elastic collisions between the particles.

 When gases behave in this way the 1. volume halves if the pressure is doubled, 2. volume doubles if their temperature is doubled, 3. pressure doubles if their temperature doubles.

 This is represented by the following equation

PV/T(start) = PV/T(end)

 Where temperature is measured in Kelvin(K).

Eg Calculate the molar volume at STP for carbon dioxide given that 2.50g occupies 0.450 dm3 at 3.00atm and 160c ( Show all your workings and check the answer = 22.4 dm3 mol-1)

Ideal Gas Equation  To relate this to the number of moles of gas(n) we can use the ideal gas equation. PV = nRT

 Where temperature is in Kelvin(K)  Pressure is in Pascals(Pa) or newtons / metre2 (N / m2)  Volume in m3 (1 cm3 = 1 * 10-6 m3)  N is the moles of gas (mass/molar mass)  R is called the gas constant = 8.314 J K-1 mol-1

Since n = m / Mr this equation is often rewritten including the molar mass.

PV = mRT / Mr

E.g. Calculate the molar mass and therefore identity of a gas when 1.45g of this gas occupies a volume of 500 cm3 at 250C and 1.01 x 105 N/m2 pressure.

Analysing graphs related to the ideal gas equation: In the space below sketch the graphs you would expect when plotting the volume of a gas (dm3) against pressure (atm).  Which Law does this represent?

In the space below sketch the graphs you would expect when plotting the volume of a gas (dm3) against temperature (atm).  Which Law do this represent?  At which temperature does the volume of gas fall to zero?  What is the name given to this value? Has that temperature ever been reached in a Science lab?

TOK Consider the two scales of Kelvin and Celsius. The standard measure of temperature (SI units) in the is the Kelvin (K) scale, on which the only point established by arbitrary definition is the unique temperature at which the liquid, solid, and vapour forms of water can be maintained simultaneously. The interval between this temperature and absolute zero is defined as 273.16 kelvins, and the temperature of this “triple point” is designated 273.16 K (since 1967, no longer written °K). In essence, the Kelvin scale is the Celsius shifted by 273.15 degrees (because the triple point of water is actually 0.01 °C), with the same size unit of temperature. A temperature scale must be based on one or two standard temperatures, called fixed points. The Celsius scale uses the freezing point and boiling point of water.

 Why do you think these were originally adopted?  Is the freezing point REALLY zero degrees Celsius or is this an arbitrary scale on which we can compare other values.  Is the Celsius scale a natural scale or an artificial scale?>  Is this the same for the Kevin scale? Discuss these bullet points with your partner

1.5 Reactions in Solutions  When a solid (solute) is dissolved in a liquid (solvent) a solution is formed.  The concentration of this solution is defined as either the mass of or the number of moles dissolved in 1 decimetre cubed (dm3) which is the same as 1 litre.  Units of concentration are ( g dm-3 ) or ( mol dm-3)

Concentration(in mol dm-3) = Moles / Volume(in dm3)

C = n / V (in dm3)

Solution Calculations

1. Calculating concentration:

-3 Calculate the concentration in mol dm when 9.8 g of H2SO4 is dissolved in 0.4 dm3 of water.

3 Moles of H2SO4 in 400 cm = 9.8 / 98 = 0.1 moles

Concentration = n / V

= 0.1 / 0.4

= 0.25 mol dm-3

2. Calculating moles of solute: Calculate the amount of moles of NaCl in 215cm3 of 0.3 mol dm-3 solution.

Moles = C x V (in dm3)

= 0.3 x 0.215

= 0.0645 moles

3. Calculating Mass of Solute:

3 -3 What mass of NH4Cl is required to prepare 21.5 cm of 0.10 mol dm solution?

Moles = C x V

= 0.1 x 0.0215

= 0.00215 moles

Mass = n x Mr

= 0.00215 x 53.5

= 0.115 g (sig figs?)

4. Calculating Volumes:

Calculate the volume of 2 mol dm-3 HCl required to make 0.5 dm3 of 0.25 mol dm-3 diluted HCl.

Since moles of solute stays constant n = C1V1 = C2 V2

-3 3 -3 0.25 mol dm x 0.5 dm = 2 mol dm x V2

3 3 V2 = 0.0625 dm or 62.5 cm Titrations  This is an experimental procedure to accurately determine the concentration or volume of an unknown solution using a standard solution of prepared solution of known concentration.

1. Balance the equation between the reactants in solution.

2. Find the concentration and volume of one of the reactants. This is your starting point.

3. Find the number of moles of this first reactant used.

4. Use the balanced equation to work out the number of moles of the second reactant used.

5. Calculate either the concentration or volume of this second reactant.

E.g. 1 If 25 cm3of 0.1 mol dm-3 HCl requires 21.50 cm3of NaOH to reach neutralisation, calculate the concentration of NaOH.

HCl + NaOH NaCl + H2O

HCl : conc = 0.10 mol dm-3, vol. = 25.0 cm3

Moles of HCl = C x V = 0.10 x 0.025 = 0.0025 moles

From equation: 1 mole HCl reacts with 1 mole NaOH

Moles of NaOH in 21.5 cm3 = 0.0025 moles

Concentration of NaOH = n / V = (0.0025 / 0.0215)

= 0.116 mol dm-3

3 E.g. 2 Determine the concentration of H2SO4 if 16.35 cm is required to neutralise 25 cm3 of 0.11 mol dm-3 KOH.

H2SO4 + 2 KOH K2SO4 + 2 H2O KOH: conc. = 0.11 mol dm-3 vol. = 25 cm3

Moles of KOH = C x V = 0.11 x 0.025 = 0.00275 moles

From equation 2 moles of KOH react with 1 mole of H2SO4

3 Moles of H2SO4 in 16.35 cm = 0.00275 x 0.5 = 0.001375

Concentration of H2SO4 = n / V = 0.001375 / 0.01635

= 0.084 mol dm-3

E.g. 3 Determine the volume of 0.1 mol dm-3NaOH required to 3 -3 neutralise 20.0 cm of 0.055 mol dm H2SO4.

2 NaOH + H2SO4 Na2SO4 + 2 H2O

Moles of H2SO4 = C x V = 0.055 x 0.020 = 0.0011

From eq. 1 mole of H2SO4 reacts with 2 moles of NaOH

Moles of NaOH used = 0.0011 x 2 = 0.0022

Volume of NaOH required = n / C = 0.0022 / 0.1 = 0.022 dm3

Back Titrations (AHL ONLY)

20 g of marble is dissolved in 0.25 dm3 of 2.0 mol dm-3 nitric acid. A 25.0 cm3 portion of the remaining acid required 17.00 cm3 of 1.0 mol dm-3 NaOH for complete neutralisation.  Calculate the moles of NaOH used.

Moles of NaOH = C x V = 1.0 x 0.017 = 0.017 moles

 Calculate the moles of nitric acid neutralised by the NaOH.

NaOH + HNO3 NaNO3 + H2O

From eq. 1 mole of NaOH reacts with 1 mole of HNO3

3 moles of HNO3 in 25 cm = 0.017 moles

 Calculate the moles of nitric acid which have reacted with the marble.

3 Moles of HNO3 left in 250 cm = 0.017 x 10 = 0.17 moles

3 Moles of HNO3 in original 250cm = C x V = 2.0 x 0.25 = 0.5 moles

Moles of HNO3 reacted with marble = Difference between these

0.5 – 0.17 = 0.33 moles

 Calculate the % by mass of CaCO3 in the marble

2 HNO3 + CaCO3 Ca(NO3)2 + H2O + CO2

From eq. 2 moles of HNO3 react with 1 mole of CaCO3 …….. ….1 mole……………………..0.5 moles………..

Moles of CaCO3 in 20 g of marble = 0.5 x 0.33 = 0.165 moles

Mass of CaCO3 in 20g of marble = n x Mr = 0.165 x 100 = 16.5 g

% by mass = (16.5g / 20g) x 100% = 83 %