3 x 16 = 48
= 160g
8g of Fe2O3 and 2.7g of Al
= 0.05 mol and 0.1 mol
Aluminium is oxidised – lost electrons
2Al → 2Al3+ + 3e-
Iron is reduced - gained electrons
2Fe3+ + 6e- → 2Fe
Ionic equations + Balanced equations
CaCl2(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2NaCl(aq)
Ions
Cu2+ 2Cl- + 2Na+ 2OH- → Cu(OH)2 + 2Na+ + 2Cl-
Take out spectator ions
Cu2+ + 2OH- → Cu(OH)2
Moles, Mass
Avogadro’s constant = same number of particles as there are carbon12 atoms in 12 grams of carbon12
Some terms
Solvent = the liquid in which a solid is dissolved
Solute = the solid which dissolves in a liquid
Solution = the liquid containing a solute and a solvent
Saturated solution = solution which will not let any more solute dissolve in it
Salt Preparation
Salt = an acid in which the hydrogen ions have been replaced by metal cations
All salts are soluble due to Na, K ammonium cations
All salts are soluble due to nitrate anion
How to obtain salts
-
Dry crystals
-
Evaporate some of the water
-
Allow to cool and crystallise
-
Filter off and dry with filter paper
Displacement reaction
Metal replaces less reactive metals from one of its salts
Anhydrous salt = no water of crystallisation
Acid = substance produces H ions in aq solution
Alkali = a soluble base
Endothermic reaction
= takes in energy from surroundings therefore appears to cool E.g. ice pack
Exothermic reaction
= requires energy but gives out energy, therefore heat given off E.g. combustion
Topic 2
Alkanes
Structural formula Displayed Formula
Methane CH4
Ethane CH3CH3
Propane CH3CH2CH3
Etc
Alkenes
Ethene
Etc
Alcohols
- Primary = 1 carbon joined to the carbon that is attached to the OH
E.g. Propan-1-ol
- Secondary
= Two carbons joined to the carbon attached to the OH
E.g. Propan-2-ol
- Tertiary
= Three carbons joined to the carbon which is attached to the OH
E.g. 2-methylpropan-2-ol
Aldehyde
E.g. Methanal
Ketone
E.g. Propanone
Structure
Molecular formula = number of atoms in a single molecule E.g. CH4
Structural formula = how the atoms are grouped in a molecule E.g. CH3CH2CH3
Displayed formula =
Structural Isomers
= where the molecular formula is the same but the structural formula differs
E.g. C2H6O
Ethanol methoxy methane
These have different functional groups, but, the functional group can be attached to the main chain at different points and be a structural isomer.
E.g. Propan-1-ol or propan-2-ol
Functional group
C – C alkane eg methane
C=C alkene eg ethane
C – OH alcohol eg methanol
ketone eg propanone
aldehyde eg methanal
carboxylic acid eg
Homologous series
= a series of organic compounds with the same functional group
- generally have a formula (CnHn+2 = alkane)
E.g. Methane, ethane, propane, butane
Oxidation
- Primary alcohols
a) Partial oxidation
Produces an Aldehyde
Eg CH3CH2 → CH3
Ethanol → ethanal (aldehyde)
b) Full oxidation
Produces carboxylic acid
E.g. CH3CH2OH → CH3
Ethanol → ethanoic acid
Aldehydes are oxidated to carboxylic acids.
- Secondary alcohols
Oxidised to produce ketone
E.g. CH3COH2CH3 → CH3C CH3
propan-2-ol → propanone
Tests
Ketones fully oxidised therefore +ve with Benedicts*
Aldehydes not fully oxidised so +ve with Benedicts*
Carboxylic acid fully oxidised so –ve with Benedicts*
*of Fehlings solution
For a chart off all reactions relating to these reactions see “Smithers’ Supreme”
Dehydration
-Alcohol dehydrated → alkene
uses Al2O3 ore H3PO4
→ C=C-C + H-O-H
Propan-1-ol → propene + water
This is an elimination reaction (condensation when water is removed) as a small molecule is removed.
Behaviour with water
Alcohols – miscible with water
The longer the chain the less soluble.
Percentage Yield
- Amount produced = yield
percentage yield = X 100
= 1.8g/2.8g x 100
% yield = 64.3%
E.g. C4H9OH C4H8 + H2O
But-1-ene
Theoretical yield = x molar mass
C4H9OH → C4H8 + H2O
Ratio
mass 3.7 : ? : ?
no. of moles = 0.05 x RMM
= 0.05 x 56
mass = 2.8g
percentage yield = X 100
Topic 3
The periodic table
Emission spectra
- Elements get “excited” at different energy levels and increase in energy –when they return to “ground state emit photons
- Each element has its own emission spectra
- In line emission the lines average as frequency increases
- Electrons can only exist at certain E levels
- If enough energy is supplied ionisation occurs (electrons don’t fall back to ground state)
- This is called Em or ionisation
(Successive Em 's can occur in one atom, hence, Em1, Em2 etc.)
Em is the amount of energy required to remove 1 mole of electrons from 1 mole ions.
- This shows that the Em gradually increases in shells but jumps between shells
Quantum Shells
= a shell is: 2.8.1 (Na)
a shell
Em1 increases across period – more protons and neutrons to attract and the electrons are in the same shell so are no further from the nucleus.
Em1 decreases down a group – outer electrons are further from the nucleus so easier to pull off due to a weaker attraction from the nucleus.
Sub shells and orbital
Made from orbitals Px, y, z
Filling the sub shells with electrons
When an atom fills up with electrons the lowest energy levels fill up first.
1s 2s 2p 3s 3p 4s 3d 4p 5s
Increase in energy
Showing electron configuration
Another way of representing it is
1s2 2s2 2p6 3s2 3p4 = sulphur
Ionic Bonding
When an atom is ionised energy is required. The electrons are removed by another atom, e.g. chlorine. Its electron affinity Cl + e- → Cl- releases energy. The electron is transferred from a metal to a non-metal e.g. Na – Cl. This forms ions, Na + Cl → Na+ + Cl- The solid contains a giant lattice of Na+ + Cl- ions.
Metallic Bonding
Bonding in metals = giant structure
Atoms loose outer shell electrons to form cations. Therefore electrons become free and form a sea of electrons. Structure is held together by attraction between cations and electrons.
About the same number of electrons and cations
Mobile electrons -
Fixed cations
Topic 4
Acids and Bases
An acid is a proton c=donor
A base is a proton acceptor
E.g. HCl is an acid
It reacts with water by donating a proton to the water molecule.
HCl + H2O → Cl- + H3O+
H3O+ is the oxonium ion, often written as H+
NH3 is a base, it accepts protons from water
NH3 + H2O → NH4+ + OH-
Water is amphoteric = reacts with acids or bases
Amphiprotic = accepts or donates protons
If an acid reacts completely with water and nearly all of the H+ are donated, the acid is called a strong acid (weak base also)
E.g. HCl, H2SO4 HNO3. They dissociate into ions fully in solution.
Acids which do not dissociate into there ions fully in solution are called weak acids (they have strong bases) Often only 1% is dissociated.
E.g. Ethanoic acid, carbonic acid, phosphoric acid.
The same idea applies to weak bases which do not dissociate fully.
Acid – base reactions are a competition for protons
OH- - this is the strongest base
OH- ions strongly attract H+ ions to make water, NEUTRALISATION
Concentration
This is a measure of amount of substance dissolved in a volume of solution.
E.g. 1 mole of NaCl in 1dm3 solution has a concentration of 1 mol dm-3 or its molarity is 1 M.
E.g. If 117g of NaCl is used to make 500 cm3 of solution find its concentration.
RMM of NaCl = 58.5
No. of moles of NaCl = 117/58.5 = 2
Therefore 2 moles in 500 cm3 so 4 moles in 1000 cm3 = 4 M
Topic 5
Energy change and reactions
Energy exchanged = specific heat capacity x mass of solution, x temp change,
Between Reactants of the solution, c (J g-1K-1) m (g) ΔT (K)
and surroundings,
Q (J)
Q = cmΔT
With assumptions this becomes,
Q (J) = 4.18 J g-1K-1 x volume of solution (cm3) x temp change(K)
In this diagram, the system is the reaction mixture
The surroundings are the area outside of the system.
The enthalpy change of the reaction is the energy exchanged with the surroundings at constant pressure.
Hess’s Law
Energy cannot be created or destroyed
Therefore the total enthalpy change accompanying a chemical reaction is independent of the route taken
This is Hess’s Law.
So, ΔHө is the difference in energy of products and reactants, or,
ΔHө = Hөproducts - Hөreactants
E.g.
ΔHө = Hөproducts - Hөreactants
(3 x 157.3) – (-146)
Calculating Energy Changes Using Enthalpies of formation
The standard enthalpy change of formation of a compound (ΔHfө ) is the enthalpy change that takes place when one mole of the compound is formed from its elements under the standard conditions.
(Standard conditions being, 298K, 1 atm, 1 mol-3 solutions)
ΔHf(NH3) = -46 KJ mol-1
ΔHf(HCl) = -92 KJ mol-1
ΔHf(NH4Cl) = -314 KJ mol-1
ΔHfө = ΣΔHfө(products) - ΣΔHfө(reactants)
ΔHfө = -313 –[(-46) + (-92)]
= -176 KJ mol-1
