# AS & A2 OCR Chemistry Notes

## 1TODO AS Notes

### 1.1TODO Inorganic

#### 1.1.1 Atomic structures and isotopes

Particle Abbreviation Relative charge Relative mass
proton p+ 1+ 1
neutron n 0 1
electron e- 1- $$\frac{1}{1836}$$
• Atoms => same number of neutrons and protons
• Larger nucleus $$\therefore$$ more neutrons needed
• Every atom of the same element contains the same number of protons
1. Isotopes
• Isotopes are atoms of the same element with different numbers of neutrons and different masses
• $$^{16}_{8}O$$
• The upper left number is the mass number
• The bottom left number is the atomic number
• The letter is the chemical symbol
Isotope Protons,p+ Neutrons, n Electrons, e-
$$^{16}_{8}O$$ 8 8 8
• Isotopes can also be referred to by using simply the mass number or even just as element-mass.
2. Ions
• An ion is a charged atom (positive or negative)
• Positive ions are known as cations and have fewer electrons than protons $$\therefore$$ they have a positive charge
• Negative ions are known as anions and have more electrons than protons $$\therefore$$ they have a negative charge

#### 1.1.2 Relative mass

1. Carbon-12
• Carbon-12 is the isotope around which relative masses of ions, atoms, and isotopes are defined $$\therefore$$ the mass of a carbon-12 isotope is defined as exactly 12 atomic mass units

Relative isotopic mass

• Relative isotopic mass is the mass of an isotope relative to $$\frac{1}{12}$$ th of the mass of an atom of carbon-12.

Relative atomic mass

• Relative atomic mass Ar is the weighted mean mass of an atom of an element relative to $$\frac{1}{12}$$ th the mass of an atom of carbon-12.
• Relative atomic mass can be found with an element in the periodic table along with its atomic number

Determination of relative atomic mass

• A mass spectromoter is used to find the percentage abundances of the isotopes in a sample of an element
• You can then use the percentage abundances of the isotopes to calculate the relative atomic mass of an element with the following formula

Relative atomic mass = $$\frac{(PercentageAbundance_{1} * IsotopicMass{1}) + (PercentageAbundance_{2} * IsotopicMass{2}) + ...}{100}$$

#### 1.1.3 Polyatomic ions

1+ 1- 2- 3-
Ammonium (NH4+) Hydroxide (OH-) Carbonate (CO32-) Phosphate (PO43-)
Nitrate (NO3-) Sulfate (SO42-)
Nitrite (NO2-) Sulfite (SO32-)
Hydrogencarbonate (HCO3-) Dichromate(VI) (Cr2O72-)
Manganate(VII) (permanganate) MnO4-

#### 1.1.4 Amount of substance and the mole

• Avogadro constant NA = $$6.02 * 10^{23}$$
• One mole of an element has a mass in grams equal to the relative atomic mass

#### 1.1.5 Chemical formulae

Molar mass

number of moles n = $$\frac{mass m}{molar mass Mr}$$

Empirical formulae

• The empirical formula is the simplest whole number ratio of atoms of each element in a compound
1. More relative masses
• Some compounds exist as simple molecules whereas some exist as giant crystalline structures $$\therefore$$ two terms are needed for relative mass, one for simple molecules and another for giant structures

Relative molecular mass

• The relative molecular mass Mr compares the mass of a molecule with the mass of an atom of carbon-12
• Mr is calculated by summing the relative atomic masses of the elements which make up the molecule

Relative formula mass

• The relative formula mass compares the mass of a formula unit with the mass of an atom of carbon-12
• Relative formula mass is calculated by summing the relative atomic masses of the elements in an empirical formula
2. Finding empirical formula from mass example

In an experiment, 1.203g of calcium combines with 2.13g of chlorine to form a compound [Ar : Ca, 40.1; Cl, 35.5].

• Step 1: Convert mass into moles using n = $$\frac{m}{M_{r}}$$
 n(Ca) = $$\frac{1.203}{40.1}$$ = 0.030mol n(Cl) = $$\frac{2.13}{35.5}$$ = 0.060mol
• Step 2: To find the smallest whole-number ratio, divide by the smallest whole number
 n(Ca):n(Cl) = $$\frac{0.03}{0.030}$$ : $$\frac{0.060}{0.030}$$ = 1 : 2
• Step 3: Write the empirical formula: CaCl2

#### 1.1.6 Moles and volumes

Volume measurements

• Cubic centimetre (cm3): 1cm3 = 1ml
• Cubic decimetre (dm3): 1dm3 = 1l

Converting between moles and solution volumes

• n = number of moles
• V = volume in dm3
• c = concentration in moldm-3
• n = $$c * v$$
• If the volume is given in cm3, you must convert to dm3 by dividing by 1000
• n = $$\frac{c * v(cm^3)}{1000}$$

Standard solutions

• A standard solution is a solution of known concentration
1. Worked example

Calculate the mass of Na2CO3 required to prepare 100cm3 of a 0.250moldm-3 standard solution

• Step 1: Work out the amount in moles required
 n(Na2CO3) = $$\frac{c * V(cm^3)}{1000}$$ = $$\frac{0.250 * 100}{1000}$$ = 0.0250mol
• Step 2: Work out the molar mass of Na2CO3
 n(Na2CO3) = $$23.0 * 2 + 12.0 + 16.0 * 3$$ = 106.0gmol-1
• Step 3: Rearrange n = $$\frac{n}{M_{r}}$$ to calculate the mass of Na2CO3 required
 m = $$n * M$$ = $$0.0250 * 106.0$$ = 2.65g

Converting between amount in moles and gas volumes

n (mol) = $$\frac{volume (V)}{molar gas volume (V_{m})}$$ $$\therefore$$ at RTP (room temperature), n = $$\frac{V}{24}$$

Ideal gas equation

• p = pressure (Pa)
• V = volume (m3)
• n = amount of gas molecules (mol)
• R = ideal gas constant = 8.31Jmol-1K-1
• T = temperature (K)
• pV = nRT

#### 1.1.7 Reacting quantities

Stochiometry

• In a balanced equation, the balancing numbers give the ratio of the amount, in moles, of each substance
• The ratio is called the stochiometry of the reaction

Percentage yield

Percentage yield = $$\frac{actual yield}{theoretical yield} * 100$$

Limiting reagent

In a reaction with excess of a reactant used, the other reactant is known as the limiting reagant because it will be completely used up in the reaction

Atom economy

Atom economy = $$\frac{\Sigma Desired}{\Sigma All} * 100$$

• Reactions with high atom economies produce a large portion of desired products and are important for sustainability as they make best use of natural resources

#### 1.1.8 Acids, bases, and neutralisation

Acids

• An acid is a proton donator
• When in water, acids release H+ ions

Strong and weak acids

• Strong acids fully dissociate in aqueous solution and release all of its hydrogen atoms in solution as H+ ions
• e.g. HCl(aq) $$\rightarrow$$ H+(aq) + Cl-(aq)
• Weak acids only release a small proportion of their available ydrogen atoms into solution as H+ ions
• Weak acids partially dissociate in aqueous solution
• e.g. CH3COOH(aq) $$\rightleftharpoons$$ H+(aq) + CH3COO-(aq)

Bases and alkalis

• A base is a proton accepter
• An alkali is a base that dissolves in water releasing hydroxide ions (OH-) into the solution

Neutralisation

• In neutralisation of an acid, the protons in the acid are replaced by metal ions from the base
• When an acid is neutralised by a metal oxide or metal hydroxide, only a salt and water are formed
• When an acid is neutralised by a carbonate, carbon dioxide gas is also formed

#### 1.1.9 Acid-base titrations

Titrations

• A titration is a technique used to measure the volume of one solution that reacts exactly with another solution
• Titrations are used for:
 Finding the concentration of a solution Identification of unknown chemicals Finding the purity of a substance
1. Titration calculations
• From the results of a titration, you will know both the concentration and reacting volume of one of the solutions
• You will also know only the reacting volume of the other solution
• Step 1: Work out the amount, in mol, of the solute in the solution for which you know both the concentration and the volume
• Step 2: Use the equation to work out the amount, in mol, of the solute in the other solution
• Step 3: Work out the unknown information about the solute in the other solution

#### 1.1.10 Redox

Oxidation number

1. Rules for elements
• The oxidation number is always 0 for elements
2. Rules for compounds and ions
• Each atom in a compound has an oxidation number
• An oxidation number has a sign, which is placed before the number
Combined element Oxidation number Examples
O -2 H2O, CaO
H +1 NH3, H2S
F -1 HF
Na+, K+ +1 NaCl, K2O
Mg2+, Ca2+ +2 MgCl2, CaO
Cl-, Br- I- -1 HCl, KBr, CaI2
Special cases:
H in metal hydrides -1 NaH, CaH2
O in peroxides -1 H2O2
O bonded to F +2 F2O
• Σ (Oxidation numbers) = total charge

Roman numerals

• Roman numerals are used in the names of compounds of elements that form ions with different charges
• The Roman numeral shows the oxidation state (oxidation number) of the element, without a sign
• e.g. iron(II) = Fe2+ with an oxidation number +2

Redox reactions

3. Reduction and oxidation
 Reduction Gain of electrons Decrease in oxidation number Oxidation Loss of electrons Increase in oxidation number

#### 1.1.11 Electron structure

Electrons and shells

Shell number $$n$$ Number of electrons
1 2
2 8
3 18
4 32
• Shells are regarded as energy levels
• The energy increases as the shell number increases
• The shell number of energy level number is called the principal quantum number $$n$$

Atomic orbitals

• Shells are made up of atomic orbitals
• An orbital is a region around the nucleus that can hold up to two electrons with opposite spins
• "An orbital is a region in space which can hold up to two electrons"
1. s-orbitals
• Sphere shape
• Each shell contains one s-orbital
• The greater the shell number $$n$$, the greater the radius of its s-orbital
2. p-orbitals
• Dumb-bnell shape
• Three in each shell from $$n$$ = 2 onwards at right-angles to one another
• Referred to as px, py, pz
3. d-orbitals and f-orbitals
• Each shell from n = 3 contains five d-orbitals
• Each shell from n = 4 contains seven f-orbitals

Sub-shells

• Within a shell, orbitals of the same type are grouped together as sub-shells
 Shell s p d f Sub-shells present Number of electrons in sub-shells Number of electrons in shell 1 1 1s 2 2 2 1 3 2s + 2p 2 + 6 8 3 1 3 5 3s + 3p + 3d 2 + 6 + 10 18 4 1 3 5 7 4s + 4p + 4d + df 2 + 6 + 10 + 14 32

Filling of orbitals

4. Orbitals fill in order of increasing energy
• The sub-shells that make up shells have slightly different energy levels
• Within each shell, the new type of sub-shell added has a higher energy
• In the $$n$$ = 2 shell, the order of filling is 2s, 2p
• In the $$n$$ = 3 shell, the order of filling is 3s, 3p, 3d
• In the $$n$$ = 4 shell, the order of filling is 4s, 4p, 4d, 4f
• The 3d sub-shell is at a higher energy level than the 4s sub-shell $$\therefore$$ the 4s sub-shell fills before the 3d sub-shell $$\therefore$$ the order of filling is 3p, 4s, 3d

Electrons pair with opposite spins

• Electrons are negatively charged and repel one another
• Electrons have a property called spin - either up or down
• An arrow, $$\uparrow$$ for up and $$\downarrow$$ for down is used to indicate its spin
• The two electrons in an orbital must have opposite spin to help counteract the repulsion between the charges of the two electrons

Orbitals of the same energy are occupied singly first

Electron configuration

• Electron configuration can be written as 1sa2sb2pc… etc.
• e.g. For krypton (Z = 36) the electron configuration would be 1s22s22p63s23p64s23d104p6
• Electron configuration can also be expression in terms of the previous noble gas in the periodic table
• e.g. For Li the electron configuration could be written as 1s22s{1} or more simply [He]2s1
• Cations are formed when atoms lose electrons
• Anions are formed when atoms gain electrons

## 2TODO A2 Notes

### 2.2TODO Inorganic

#### 2.2.1 How Fast

1. Arrhenius equation
• k = Ae-Ea/RT

#### 2.2.2 Acids and Bases

 pH = -log[H+] [H+] = 10-pH Kw = [H+][OH-] For H2O: Kw = [H+]2 KA = $$\frac{[H^{+}][A^{-}]}{[HA]}$$ For weak acid: KA = $$\frac{[H^{+}]^2}{[HA]}$$ pKa = -log[KA] KA = 10-pKa
1. Finding pH of a strong base
1. Calculate [OH-]
2. Use Kw to find [H+]
3. Calculate pH
2. Buffer solutions and weak acids
1. pH of a weak acid
• e.g CH3COOH $$\rightleftharpoons$$ H+ + CH3COO-
• HA $$\rightleftharpoons$$ H+ + A-
2. pH of a buffer solution
• HA + NaOH $$\rightarrow$$ NaA + H2O
HA NaOH NaA
Start 0.5 0.2 0
Change -0.2 -0.2 +0.2
End 0.3 0 0.2
3. Artificial construction of a buffer solution
1. HA $$\rightarrow$$ vol 25cm3 and [0.1] = $$\frac{25 * 0.1}{1000}$$
2. A- $$\rightarrow$$ vol 10cm3 and [0.1] = $$\frac{10 * 0.1}{1000}$$
3. New conc. [HA] = $$\frac{2.5 * 10^{-3} * 1000}{35}$$ = 0.0714
4. [A-] = $$\frac{1*10^{-3} * 1000}{35}$$ = 0.0236
5. KA = $$1*10^{-4}$$
6. $$\frac{K_{A} * [HA]}{[A^{-}]}$$ = 3.6
3. Water
• H2O $$\rightleftharpoons$$ H+ + OH- ; Δ H = +ve
• While the pH of water changes with temperature, water cannot become more acidic or alkaline. Instead, the position of neutral changes.

#### 2.2.3 Entropy

• Δ G = Δ H - TΔ S

Created: 2019-03-09 Sat 18:10

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