Recap – Chemical Equations
Formula equation – gives overview, balance for
numbers and types of atoms
eg Mg + 2HCl  MgCl2 + H2
Molecular equation – include states of reactants
and products
eg C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
Net ionic equation – focus on just what is
involved and balance charges
eg Ca2+(aq) + CO32-(aq)  CaCO3(s)
1
Masses of Atoms
1H
atom
4He
7Li
atom
atom
238U
atom
1p
1 nucleon
2p + 2n
4 nucleons
4 times the
mass of H
3p + 4n
7 nucleons
7 times the
mass of H
92p + 146n 238 nucleons
238 times the
mass of H
2
Isotopes and Average Mass
1H
atom
1p
99.99%
2H
atom
1p + 1n
0.01%
6Li
atom
3p + 3n
7.5%
7Li
atom
3p + 4n
92.5%
12C
atom
6p + 6n
98.9%
13C
atom
6p + 7n
1.1%
35Cl
atom
17p + 18n
75.8%
37Cl
atom
17p + 20n
24.2%
1.008
(6 x 7.5%) + (7 x 92.5%) = 6.9
(12 x 98.9%) + (13 x 1.1%) = 12.01
(35 x 75.8%) + (37 x 24.2%)= 35.5
3
Avogadro Number, NA
• Actual mass of H is 1.67 x 10-24 g
• Use a scaling factor: the Avogadro number, NA
NA = 6.022  1023
Definition:
The Avogadro number is the number of
atoms present in exactly 12 g of 12C.
• One ‘mole’ of anything contains the Avogadro
number, NA, of items
12C
4
Avogadro Number, NA
Atomic mass
mass of NA atoms
‘Molar mass’
Hydrogen
1.008 amu
1.008 g
Lithium
6.94 amu
6.94 g
12C
12.00 amu
12.00 g
Carbon
12.01 amu
12.01 g
Chlorine
35.45 amu
35.45 g
5
Molar Masses
Relative masses of molecules is simply the sum of the
atomic masses of the component atoms.
eg H2
H2O
C6H12O6
2 x 1.01 = 2.02 g mol-1
(2 x 1.01) + 16.0 = 18.02 g mol-1
(6 x 12.01) + (12 x 1.01) + (6 x 16.0)
= 180.18 g mol-1
For ionic solids, no molecules, use the empirical formula.
eg NaCl
23.0 + 35.5
= 58.5 g mol-1
6
The Mole
So now we can relate the number of grams of a
substance we weigh out in the lab to the
number of moles, and thus the number of
particles of the substance that we have:
no. of moles (n) = mass (m) / molar mass (M)
or
no. of moles (n) = no. of items / 6.022 x 1023
7
The Mole
n = mass / molar mass
n = no. of items / NA
Q: How many atoms present in 1.0 g of silver?
Moles = mass / molar mass
= 1.0 / 107.9
= 0.0093 mol
No. atoms = moles  NA
= 0.0093  6.022  1023
= 5.6  1021 atoms
8
The Mole
n = mass / molar mass
n = no. of items / NA
Q: Calculate the no of moles of water in 1000 g
water. Molar mass = 18.02.
Moles = mass / molar mass
= 1000 / 18.02
= 55.49 moles
Q: How many water molecules would be
present?
No. molecules = moles  NA
= 55.49  6.022  1023
= 3.342  1025 molecules
9
Learning Outcomes:
• By the end of this lecture, you should:
−
−
−
−
−
−
−
recognise the mass reported is a weighted average of the
mass of the individual isotopes of an element
be able to calculate an average mass from isotope data
understand that NA is a scaling factor
understand the difference between atomic mass and molar
mass
be able to calculate the molar mass of a compound
be able to convert between mass, moles and numbers of
atoms/ions/molecules
be able to complete the worksheet (if you haven’t already
done so…)
10
Questions to complete for next lecture:
1. Give the definition of a mole of substance.
2. How many atoms are there in 0.25 mole of silver?
3. Silver contains two isotopes: 107Ag (51.8%) and
109Ag(48.2%). What is the average atomic mass of
silver?
4. Table sugar (sucrose) has the formula C12H22O11.
a) What is the molar mass of sucrose?
b) How many moles of sucrose are present in a sugar lump with
mass 5.0 g?
c) How many molecules of sucrose are present in the sugar lump?
11