Chap 1 and 2 - Mole and Redox

Define relative atomic mass and relative isotopic mass

1. Relative atomic mass (Def.) : Defined as the ratio of the average mass of one atom of the element to 1/12 of the mass of an atom of 12C isotope

• Average includes relative abundance of ALL possible isotopes of that element 

2. Relative isotopic mass (Def.) : Of a particular isotope is defined as the ratio of the mass of one atom of the isotope to 1/12 of the mass of an atom of 12C isotope

• Deals specifically with atomic mass of 1 of the possible isotopes of an element

Define relative molecular mass and relative formula mass

1. Relative molecular mass (Def.) : Of a molecule is defined as the ratio of the average mass of one molecule of the substance to 1/12 of the mass of an atom of 12C isotope

2. Relative formula mass (Def.) : Of an ionic compound is defined as the ratio of the average mass of one formula unit of the compound to 1/12 of the mass of an atom of 12C isotope

State the unit for avogadros constant, molar mass, relative molecular mass

Avogadro's constant : /mol , molar mass : g , relative molecular mass : no unit

Define Avogadros constant, molar mass and molar gas volume

1. Avogradro’s constant  : (Def.) : number of particles in 1 mole of a substance (6.02 x 10^23)

2. Molar mass : (Def.) : mass of one mole of a substance and has a unit of g/mol 

3. Molar gas volume  : (Def.) : volume that one mole of a gas occupies at a particular set of temperature and pressure

What is ppm and ppb

• Concentrations also given as parts per million (ppm) or parts per billion (ppb) : used as a measure of small level (concentrations) of pollutants in air, water, soil

• When a solution is diluted (by adding more solvent), the concentration of the solution decreases but the number of moles of the solute in the diluted solution remains unchanged

State the ideal gas equation

pV = nRT (Ideal Gas Equation)

This equation is the Ideal Gas Law: 

• p = pressure (in Pascals, Pa) (Eg. p=1atm=101325 Pa (1 atm = 101325 Pa)

• V = volume (in m³)

• n = number of moles of gas

• R = universal gas constant (8.31 J mol⁻¹ K⁻¹)

• T = temperature (in Kelvin, K)

Define empirical and molecular formula

• Empirical formula (Definition) : of a compound is the simplest formula which shows the ratio of the atoms of the different elements in the compound

• Molecular formula (Definition) : of a compound is the formula which shows the actual number of atoms of each element in one molecule of the compound

State the standard combustion equation of a hydrocarbon

CxHy + (x + y/4) O2 -> xCO2 + y/2 H2O

Complete combustion of a hydrocarbon (a compound that contains hydrogen and carbon only) gives 0.352 g of carbon dioxide and 0.072 g of water. Calculate its empirical formula and its molecular formula, given that it has a molar mass of 104 g mol−1

Answer C8H8

A mixture of CuSO4•5H2O and MgSO4•7H2O is heated until a mixture of the anhydrous salt is obtained. If 5.00 g of the mixture gives 3.00 g of the anhydrous salts, what is the percentage by mass of CuSO4•5H2O in the mixture?

73.8%

Define aliquot, standard solution, equivalence point

• Aliquot : measured with a pipette and placed in a conical flask 

• Standard solutions : solution whose concentration is accurately known 

• Equivalence point : occurs when the reactants in the two solutions react according to the stoichiometry of the reaction

Define titre volume, end point, working range

• Titre volume : the volume of solution added from the burette 

• End point : where indicator changes colour 

• Working range : pH where indicator changes colour

Describe dilution of a solution

• Involves the addition of more solvent to a given solution

• The number of moles of solute present does not change -> the solute is dispersed in a larger final volume -> lower concentration

Describe sampling a solution

• Collection of a portion from a given solution

• The concentration of the sample is the same as the original solution

• The number of moles of solute present changes

State when back titration is used

• Used when : direct titration not possible 

  • Eg. One of the reactants is an insoluble solid (carbonates), when direct titration will be too slow, or when one of the reagents is volatile (Eg. ammonia)

Explain how back titration is used to find concentration of a solution

• A known excess of one reagent A is allowed to react with an unknown amount of B

• The amount of unreacted A is then determined by titration with a reagent C of known concentration

• From the titration results, the amount of unreacted A and the amount of B can be found by simple stoichiometric calculations

Define redox reaction, reduction, oxidation

1. Redox : A reaction that involves reduction and oxidation simultaneously

2. Reduction : A process whereby a substance gains electrons, resulting in a decrease in oxidation number

3. Oxidation : A process whereby a substance loses electrons, resulting in an increase in oxidation number

Define reducing agent, oxidising agent, disproportionation, oxidation state

1. Reducing agent : A substance that undergoes oxidation and is an electron doner 

2. Oxidising agent : A substance that undergoes reduction and it is an electron acceptor 

3. Disproportionation : A redox reaction in which the same substance is both oxidised and reduced

4. Oxidation state : The total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom

Explain how to find the oxidation number

1. Oxidation number 

• This oxidation number is average oxidation number of each element (NOT oxidation number is individual atoms) (assuming that all the atoms of the element has the same electron environment)

• Oxidation number have no physical or structural significance 
  

1. For compound with no net charge , all oxidation states must add up to 0

2. Ion with net charge , OS will add up to that net charge 

3. Ox no of elements in ionic compound is equal to the charge of its ion 

4. For a compound, the more electronegative element is given the negative oxidation number

• Very electropositive and electronegative elements always retain the same ON in all their compounds 

• In PT , electronegativity (ability of atom in a molecule to attract shared electrons in a bond) increases from left to right and down to up 

• Eg. CIF : F = -1 , CI = +1 

Find the oxidation state of oxygen in F2O

O : +2, F : -1

Explain how to find the oxidation number on individual atoms

• Pick a bond between 2 atoms.

• When the bond is broken , the more electronegative atom will gain one electron from the other atom and have a −1 ON 

• The less electronegative atom will lose one electron and have a +1 ON 

• If the bond is a double bond, the oxidation number will be –2 and +2

• If 2 elements are the same and bonded together -> there is no difference in electronegativity between the bonded atoms , both atoms are assigned a “0” oxidation number.

• Repeat steps 1 to 4 for all the bonds around the atom, and sum up all the oxidation number around an atom to get its overall oxidation number

State some of the common oxidising agents

State some of the common reducing agents

Describe the KMnO4 titration

• KMnO4 usually placed in burette and substance to be oxidised + H2SO4 is added to conical flask 

• Nitric acid and hydrochloric acid are both not suitable as nitric acid is itself an oxidising agent while hydrochloric acid can be oxidised by manganate(VII) ion to give chlorine

• MnO4 − + 8H+ + 5e− → Mn2+ + 4H2O

• Hydrogen peroxide oxidised to oxygen 

• When all the reductant is used up, solution turns pink

Explain why volume of H2SO4 used in KMnO4 titration need not be accurate

• Dilute H₂SO₄ is in excess and hence need not be measured accurately.

• This is to ensure the solution is sufficiently acidic for the complete reduction of KMnO₄

• An insufficient acidic medium will result in the formation of a brown precipitate of MnO₂ due to incomplete reduction of MnO₄– ions leading to a lower titre volume and unreliable results

  • More acid can be added with vigorous mixing to force the reduction of MnO₂ to Mn2+

Describe iodometric titration

• 2S2O3 2− + I2 → 2I − + S4O6 2− 

• Only sodium thiosulfate is stable enough to be used as RA in titration as the other RA are easily oxidised by O2 in the air 

• Only the weak OA, iodine, I2, will oxidise thiosulfate ion, S2O32−, quantitatively to tetrathionate ion, S4O62−

• Other stronger oxidising agents would oxidise thiosulfate to a variety of sulfur containing products

1. The thiosulfate solution is usually placed in the burette and the brown iodine solution in the conical flask

2. The solution gradually fades to pale yellow

3. Starch indicator is added into the flask -> Solution turns blue-black 

4. Titration then continues till all iodine is used up (End point : colourless)

Explain why the iodine solution should be titrated as soon as possible once it is prepared

• Iodine is volatile and will vaporise easily at room temperature hence titration must be carried out as soon as possible. If this is not possible, cover all flasks containing iodine

Why is the starch indicator added only towards the end of titration when the iodine solution is pale yellow, instead of right at the start?

• Starch forms a blue-black water-soluble complex with iodine in which the iodine is trapped within the starch molecules

• Hence, starch should not be added at the beginning of the titration when there is a high concentration of iodine since some iodine may remain trapped in the spiral starch molecules under high I2 conc even at the equivalence point -> inaccurate results

After titration is complete, a slow return of the blue colour is observed. Why is this so? 

This is due to atmospheric oxidation of I − back to I2. Ignore any slow return of blue colour after the titration is done