1. Recap – Chemical Equations
Chem 1001 Lecture 4
Recap – Chemical Equations
Formula equation – gives overview, balance for numbers and types of atoms
eg Mg HCl  MgCl 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)

2. Masses of Atoms 1H atom 1p 1 nucleon 4He atom 2p + 2n 4 nucleons
Chem 1001 Lecture 11
Masses of Atoms
1H atom 1p
1 nucleon
4He atom
2p + 2n
4 nucleons
4 times the mass of H
7Li atom
3p + 4n
7 nucleons
7 times the mass of H
238U atom
92p + 146n
238 nucleons
238 times the mass of H
Dr Chiara Neto

3. Isotopes and Average Mass
Chem 1001 Lecture 11
Isotopes and Average Mass
1H atom
2H atom 1p
1p + 1n
99.99%
0.01%
1.008
6Li atom
7Li atom
3p + 3n
3p + 4n
7.5%
92.5%
(6 x 7.5%) + (7 x 92.5%) = 6.9
12C atom
13C atom
6p + 6n
6p + 7n
98.9%
1.1%
(12 x 98.9%) + (13 x 1.1%) = 12.01
35Cl atom
37Cl atom
17p + 18n
17p + 20n
75.8%
24.2%
(35 x 75.8%) + (37 x 24.2%)= 35.5
Dr Chiara Neto

4. Avogadro Number, NA NA = 6.022  1023
Chem 1001 Lecture 11
Avogadro Number, NA
Actual mass of H is 1.67 x g
Use a scaling factor: the Avogadro number, NA
NA =  1023
Definition: The Avogadro number is the number of 12C atoms present in exactly 12 g of 12C.
One ‘mole’ of anything contains the Avogadro number, NA, of items
Dr Chiara Neto

5. Avogadro Number, NA Atomic mass mass of NA atoms ‘Molar mass’ Hydrogen
Chem 1001 Lecture 11
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
Dr Chiara Neto

6. Chem 1001 Lecture 12
Molar Masses
Relative masses of molecules is simply the sum of the atomic masses of the component atoms.
eg H2 2 x 1.01 = 2.02 g mol-1
H2O (2 x 1.01) = g mol-1
C6H12O6 (6 x 12.01) + (12 x 1.01) + (6 x 16.0) = g mol-1
For ionic solids, no molecules, use the empirical formula.
eg NaCl = 58.5 g mol-1
Dr Toby Hudson

7. The Mole no. of moles (n) = mass (m) / molar mass (M) or
Chem 1001 Lecture 12
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 / x 1023
Dr Toby Hudson

8. The Mole Q: How many atoms present in 1.0 g of silver?
Chem 1001 Lecture 12
n = mass / molar mass
The Mole
n = no. of items / NA
Q: How many atoms present in 1.0 g of silver?
Moles = mass / molar mass
= 1.0 / 107.9
= mol
No. atoms = moles  NA
=   1023
= 5.6  1021 atoms
Dr Toby Hudson

9. Chem 1001 Lecture 12
n = mass / molar mass
The Mole
n = no. of items / NA
Q: Calculate the no of moles of water in 1000 g water. Molar mass =
Moles = mass / molar mass
= 1000 / 18.02
= moles
Q: How many water molecules would be present?
No. molecules = moles  NA
=   1023
=  1025 molecules
Dr Toby Hudson

10. 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…)

11. Questions to complete for next lecture:
Chem 1001 Lecture 4
Questions to complete for next lecture:
Give the definition of a mole of substance.
How many atoms are there in 0.25 mole of silver?
Silver contains two isotopes: 107Ag (51.8%) and 109Ag(48.2%). What is the average atomic mass of silver?
Table sugar (sucrose) has the formula C12H22O11.
What is the molar mass of sucrose?
How many moles of sucrose are present in a sugar lump with mass 5.0 g?
How many molecules of sucrose are present in the sugar lump?