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Title: AS Chemistry Class notes - Physical chemistry - Atomic structure
Description: This is class notes for the first year/AS chemistry. It covers the first chapter of physical chemistry - atomic structure including: Fundamental physics Mass number, atomic number and isotopes The arrangement of electrons The mass spectrometer Ionisation energy

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Physical chemistry 1
Amount of substance
Relative atomic and molecular masses, the Avogadro constant and the mole
Relative atomic mass Ar


The mass in grams of any atom or molecule is too small to find by weighing, so the masses of
atoms are compared, and relative masses are used
...
Now 1/12 or Carbon has an Ar of exactly 1
...




Ar = (average mass of 1 atom of an element *12) / (mass of 1 atom of carbon-12)
...




You find the Mr by adding the Ar of every atom present in the molecule
...
g
...


Relative formula mass


This also has the symbol Mr but is used for ionic compounds as they don’t exist as molecules
...




To count atoms, we weigh large numbers of them
...




If you weigh out the Mr of a molecule, it will have the same number of molecules
...




Entities is a general word for a particle
...




6
...


The mole

Physical chemistry 1
Amount of substance


The amount of substance that contains 6
...


Mass

Mr * moles


The Ar of any element (in grams) contains 1 mole of atoms
...




Number of moles = Mass * Mr
...




To get a solution with a concentration of 1 mol dm-3, add solvent to 1 mole of solvent
...




It’s measured in mol dm-3
...
of mole (n) / volume (V) (dm3)

Number of moles in a given solution
...
It changes with pressure and temperature
...


Boyle’s law


The product of pressure and volume is constant as long as temperature remains constant
...


Physical chemistry 1
Amount of substance

Charles’ law


Volume is proportional to the temperature as long as pressure remains constant
...




Volume (V) / temperature (T) = constant
...




Pressure (p) α temperature (T)
...




Combining these relationships gives:
o

Pressure (p) * volume (V) / temperature (T) = constant for fixed mass of gas
...




For n moles of gas:
o

Pressure (p) * volume (V) = no
...


o

PV = nRT
...
31 Jk-1mol-1
...
No gases obey it exactly, but at room temperature and pressure it
holds well for many gases
...


Units


P must be in Pa (Nm-2)
...




T must be in K (˚C + 273)
...


Using the ideal gas equation

Physical chemistry 1
Amount of substance


None of the terms in the equation refer to a particular gas, so the volume will be the same for
any gas
...




If you know the number of moles in a given mass of gas, you can find the mass of 1 mole of gas,
telling us the Mr
...


o

Stp is standard temperature and pressure which is 35 ˚C and 100 kPa
...


1
...

2
...

3
...



E
...
4
...
20g C, 4
...

o

Ca = 4
...
1 = 0
...
20/12
...
10 mol

o

O = 4
...
0 = 0
...
10:0
...
30 = 1:1:3

o

CaCO3



If the number of moles in each element (step 2) is very close to 0
...
2, 0
...
you can times
by 2, 5, 3 etc
...


Finding the molecular formula


It gives the actual number of each element in 1 molecule of the compound (this only applies to
substances that exist as a molecule)
...




To find the number of units of the empirical formula in the molecular formula: relative
molecular mass / relative mass of the empirical formula
...
g
...
17% C, 13
...
79% O, Mr=46
...
Find the empirical formula
o

C = 52
...
0 = 4
...
04/1 = 13
...
79/16 = 2
...
348:13
...
174 = 2:6:1

o

C2H6O

2
...

o

(2*12) + (6*1) + (1*16) = 46

3
...

o

46/46=1

o

There’s 1 unit of the empirical formula
...




It involves burning the unknown compound in excess oxygen and measuring the amount of
water, CO2, and other oxides
...




The basic method measures carbon, hydrogen, sulfur and nitrogen, assuming that the oxygen
makes up the difference after the other 4 elements have been weighed
...




The sample is burnt completely in oxygen
...




The unreacted nitrogen is measured by thermal conductivity
...




The molecular formula can be found if the relative molecular mass has been found using a mass
spectrometer
...
g
...
58 X compound with 1
...
54g H2O gave 1
...
54g H2O
...
The relative molecular mass is 58
...

1
...


Physical chemistry 1
Amount of substance
o

o

o

C = 1
...
32/44 = 0
...
03 moles of C
...
54g of H2O


Mr = 18



0
...
03 mol of H2O



Each mole has H2O has 2 moles of H so there’s 0
...




0
...
Mass = 0
...
36g



0
...
Mass = 0
...
06g



0
...
06 = 0
...
C and H
...
58 – 0
...
16) is O
...
0g O (Ar = 16)
...
16/16 = 0
...


O

o

Sample contains: 0
...
06 moles H and 0
...


o

C:H:O = 0
...
06:0
...


o

C3H6O

2
...


o

C3H6O

Balanced equations and related calculations


Equations represent what happens in a chemical reaction
...




Reactions → products
...




Atoms react together in ratios, called the stoichiometry
...


Physical chemistry 1
Amount of substance
Balanced symbol equations


Use the formulae of reactants and products
...




They tell us about the amounts of substances that react together and are produced
...

o

(s) means solid
...


o

(g) means gas
...


Writing balanced equations
1
...

2
...

3
...

4
...

5
...



E
...
Aluminium + oxide → aluminium oxide
1
...
1

2

2 3

3
...
(4*27) + (3*2*16) → (2*27*2) + (2*3*16)
108 + 96 → 108 + 96
204 → 204
This is balanced
...




Sometimes ions don’t take part in the reaction they’re called spectator ions
...
g
...


The total of charges on each side must be equal
...




E
...
0
...
Mg(s) + HCl(aq) → MgCl2(aq) + H2(g)
2
...
1 mole

2 mole

1 mole

1 mole

4
...
3
5
...
12g of Mg = 0
...
3 = 0
...

o

1 mole of Mg gives 1 mole MgCl2 therefore there’s 0
...


o

Mr of MgCl2 = 24
...
5) = 95
...


o

Mass of MgCl2 = 95
...
0049 = 0
...


Finding concentrations using titrations


Need to know concentration of reactant 1 and the equation for the reaction
...


1
...

2
...

3
...
Find the number of moles pf reactant 2 (V*conc
...
Deduce the moles of reactant 1 (from equation)
...
Find the concentration of reactant 1 (moles/V)
...




Watch out for dilution, done if the reactants are dangerous or if titration needs repeating
...
g
...
4mol dm-3 of NaOH in 25cm3, xmol dm-3 of HNO3 in 18
...

o

NaOH = 25cm3 = 0
...


o

V*conc
...
025*0
...
01 mole
...
01 moles

o

Mole/v = 0
...
0187 = 0
...




Most chemical reactions produce 2 or more products, but usually only 1 is required
...


Atom economy


Found directly from the balanced equation
...




Percentage atom economy = (mass of desired product / total mass of reactants)*100
...
g
...


o

From the equation calculate mass of each reactant and product involved
...


o

2*(23 + 16 + 1) = 80g
...


o

2*35
...
0g
...


o

23+16+35
...
5g
...
0g
...
5/151 * 100 = 49
...


o

Only 49
...


Some reactants, in theory, have no wasted atoms
...

o

The atom economy tells us, in theory, how many atoms must be wasted in a reaction
...
The practical process
...
As a result of reactants that don’t go to completion
...




In practice you never get the amount of product the equation predicts
...
of moles of a specified product / theoretical maximum number
of moles of the products *100
...


Physical chemistry 1
Amount of substance


Another problem is when reactions are reversible and don’t go to completion, which is common
...



Title: AS Chemistry Class notes - Physical chemistry - Atomic structure
Description: This is class notes for the first year/AS chemistry. It covers the first chapter of physical chemistry - atomic structure including: Fundamental physics Mass number, atomic number and isotopes The arrangement of electrons The mass spectrometer Ionisation energy