1. Mole Calculations Some Mathematical Functions What is a mole?
Avogadro’s Number
Converting between moles and mass
Calculating mass % from a chemical formula
Determining empirical and molecular formulae from mass

2. Carrying out Calculations
In chemistry, must deal with several mathematical functions.
Scientific Notation
Makes it easier to deal with large numbers, especially concentrations
Written as A ×10b, where A is a decimal number and b is a whole number
Example: Avogadro’s number
It is very inconvenient to write this. Instead, use scientific notation:
6.022 × 1023
Calculators:
Sharp & Casio Type in 6.022
Press the exponential function [EXP]
Key in 23

3. (a) 784000000 Questions How would you write the following: 7.84 × 108
2.3 × 10-4
9.22 × 106
1.5 × 10-8
(a)
(b)
(c)
(d)
Calculate the following:
(a) (1.38 × 104) × (8.21 × 106)
1.13 × 1011
(b) (8.56 × 10-8) × (2.39 × 104)
2.05 × 10-3

4. Common Decimal Prefixes
Symbol
Number
Word
Exponential Notation
tera T
1,000,000,000,000
trillion
1012
giga G
1,000,000,000
billion
109
Mega M
1,000,000
million
106
kilo k
1,000
thousand
103
hecto h
100
hundred
102
deca da 10
ten
101
deci d
0.1
tenth
10-1
centi c
0.01
hundredth
10-2
milli m
0.001
thousandth
10-3
micro 
millionth
10-6
nano n
billionth
10-9
pico p
trillionth
10-12
femto f
quadrillionth
10-15

5. Calculate the common logarithms of the following:
Calculators
Sharp: Press the [LOG] function
Type the number
Hit answer
Casio: Key in the number
Press the [LOG] function
Questions
Calculate the common logarithms of the following: 10
1,000,000
0.001
853
log 10
log
log 0.001
log 853 1 6 -3
2.931

6. Calculations: The Mole
Stoichiometry is the study of quantitative aspects of chemical
formulas and reactions
Mole: SI unit of the amount of a substance
Definition: A mole is the number of atoms in exactly 12g of the
carbon-12 isotope
This number is called Avogadro’s number and is given by ×1023
The mole is NOT just a counting unit, like the dozen, which specifies only the number of objects. The definition of a mole specifies the number of objects in a fixed mass of substance.
Mass spectrometry tells us that the mass of a carbon-12 atom is
1.9926×10-23g.
No. of carbon-12 atoms = atomic mass (g)
mass of one atom (g)
= g _
1.9926×10-23g
= ×1023 atoms

7. Other definitions of the Mole
One mole contains Avogadro’s Number (6.022 x 1023)
A mole is the amount of a substance of a system which contains as many elementary entities as there are atoms in 0.012kg (or12g) of Carbon-12
A mole is that quantity of a substance whose mass in grams is the same as its formula weight
E.g. Fe55.85
Iron has an atomic mass or 55.85g mol-1, so one mole of iron has a mass or 55.85g

8. Calculating the number of particles
One mole of any object always means × 1023 units of those objects. For example, 1 mol of H2O contains × 1023 molecules 1 mol of NaCl contains × 1023 formula units
Calculating the number of particles
Avogadro’s number is used to convert between the number of
moles and the number of atoms, ions or molecules.
Example
0.450mol of iron contains how many atoms?
Number of atoms = number of moles × Avogadro’s number (NA)
Therefore No. of atoms = (0.450mol) × (6.022 × 1023)
= 2.7 × 1023 atoms

9. Example
How many molecules are there in 4 moles of hydrogen peroxide (H2O2)?
No. of molecules = no. of moles × Avogadro’s number (NA)
= 4mol × (6.022 × 1023 mol-1)
= 24 ×1023 molecules
= 2.4 × 1024 molecules
Questions
How many atoms are there in 7.2 moles of gold (Au)?
Answer: 4.3 × 1024 atoms
The visible universe is estimated to contain 1022 stars. How many moles of stars are there?
Answer: 1022 stars = = 0.17 mol.
6.022×1023

10. Calculating the mass of one molecule
Example: What is the mass of one molecule of water?
Step 1: Calculate the molar mass of water
Molar mass of water = (2 × atomic mass H) + (1 × atomic mass O)
Molar mass H2O = (2 × 1.008g mol-1) + (1 ×16.000g mol-1)
= g mol-1
Step 2: Employ Avogadro’s number
Mass of one molecule = Molar mass
Avogadro’s no.
= g mol ×1023mol-1
= 2.992×10-23g
Note: Always check the units you have in your answer to ensure you are correct

11. Example
Calculate the mass of one molecule of ammonium carbonate [(NH4)2CO3]
Step 1: Calculate the molar mass
2 Nitrogen atoms
8 Hydrogen atoms
1 Carbon atom
3 Oxygen atoms
2 × 14.01gmol-1
8 × gmol-1
1 ×12.01gmol-1
3 × gmol-1
= gmol-1
= gmol-1
= gmol-1
= gmol-1
Total = gmol-1
Step 2: Employ Avogadro’s Number, NA
Mass of one molecule =
96.09 gmol-1 .
6.022×1023mol-1
= 1.59 × 10-22g
Questions
Calculate the mass of one molecule of:
Ethanoic acid (CH3COOH)
Methane (CH4)
Potassium dichromate (K2Cr2O7)
9.96 × g
2.66 × g
4.89 × g

12. Converting between mass and moles
In the lab, we measure the mass of our reactants in grams using a
balance. However, when these react they do so in a ratio of moles.
Therefore, we need to convert between the mass we measure and the
number of moles we require.
The expression relating mass and number of moles is:
Mass of sample (g) = no. of moles (mol) × molar mass (gmol-1)
Example
Calculate the mass in grams in 0.75mol of sodium hydroxide, NaOH
Step 1: Find the molar mass of the compound
Na: gmol-1
Mr: gmol-1
O: gmol-1
H: gmol-1
Step 2: Substitute into the above expression
Mass of sample =
0.75mol × gmol-1
= 30g

13. Questions Calculate the mass in grams present in:
(a) 0.57mol of potassium permanganate (KMnO4)
Answer: Molar mass KMnO4 = gmol-1
Mass in grams = 0.57mol × gmol-1
= g
(b) 1.16mol of oxalic acid (H2C2O4)
Answer: Molar mass H2C2O4 = gmol-1
Mass in grams = 1.16mol × gmol-1
= g
(c) 2.36mol of calcium hydroxide (Ca(OH)2)
Answer: Molar mass Ca(OH)2 = 74.1 gmol-1
Mass in grams = 2.36mol × 74.1 gmol-1
= g

14. Converting between moles and mass
Number of moles = mass of sample (g)
molar mass (gmol-1)
Example
Convert 25.0g of KMnO4 to moles
Step 1: Calculate the molar mass K Mn O
1 × gmol-1
1 × gmol-1
4 × gmol-1
39.10 gmol-1
54.93 gmol-1
64.00 gmol-1
Mr = gmol-1
Step 2: Substitute into above expression
25.0g .
158.03gmol-1
No. of moles =
= mol

15. Questions Calculate the number of moles in: (a) 1.00g of water (H2O)
Answer: Molar mass water = gmol-1
1.00g H2O = 0.055mol
(b) 3.0g of carbon dioxide (CO2)
Answer: Molar mass carbon dioxide = 44 gmol-1
3.0g CO2 = 0.068mol
(c) 500g of sucrose (C12H22O11)
Answer: Molar mass sucrose = gmol-1
500g C12H22O11 = 1.46mol
(d) 2.00g of silver chloride (AgCl)
Answer: Molar mass silver chloride = gmol-1
2.00g AgCl = 0.014mol

16. Important formulae so far….
Defining the mole:
No. of carbon-12 atoms = atomic mass (g)
mass of one atom (g)
Calculating the number of atoms or molecules, given the number of moles:
No. of atoms = No. of moles × Avogadro’s number (NA)
No. of molecules = No. of moles × Avogadro’s number (NA)
Calculating the mass of an individual molecule:
Mass of one molecule = Molar mass
Avogadro’s no.
Most important equation:
Mass of sample (g) = no. of moles (mol) × molar mass (gmol-1)
Number of moles = mass of sample (g)
molar mass (gmol-1)

17. Calculating mass percentage from a chemical formula
Many of the elements in the periodic table of the elements occur in
combination with other elements to form compounds.
A chemical formula of a compound tells you the composition of that compound in terms of the number of atoms of each element present.
The mass percentage composition allows you to determine the fraction of the total mass each element contributes to the compound.
Example
Ammonium nitrate (NH4NO3) is an important compound in the fertiliser industry. What is the mass % composition of ammonium nitrate?
Step 1: Calculate the molar mass of ammonium nitrate
Two N atoms: gmol-1
Four H atoms: gmol-1
Three O atoms: gmol-1
Molar mass NH4NO3 = gmol-1

18. Step 2: Determine the mass % composition for each element
Nitrogen: g N in one mol of ammonium nitrate
Mass fraction of N = g
80.05g
Mass % composition of N = g × 100%
80.05g
= 34.99% ≈ 35%
Hydrogen: 4.032g H in one mol of ammonium nitrate
Mass fraction of N = g
80.05g
Mass % composition of H = g × 100%
80.05g
= 5.04% ≈ 5%
Oxygen: 48.00g O in one mol of ammonium nitrate
As above, the mass % composition of O is found to be 60%

19. Therefore, the mass % composition of ammonium nitrate (NH4NO3) is:
% Nitrogen: 35%
% Hydrogen: 5%
% Oxygen: 60%
To check your answer, make sure it adds up to 100%
Question
What is the mass % composition of C12H22O11?
Answer: % Carbon: 42.1%
% Hydrogen: 6.5%
% Oxygen: 51.4%

20. Determining empirical formula from mass
The empirical formula of a compound tells you the relative number of
atoms of each element present in that compound. It gives you the
simplest ratio of the elements in the compound.
For example, the empirical formula of glucose (C6H12O6) is CH2O, giving the C:H:O ratio of 1:2:1
If you know the mass % composition and the molar mass of elements present in a compound, you can work out the empirical formula
Example
What is the empirical formula of a compound which has a mass % composition of 50.05% S and 49.95% O?
Step 1: Find the atomic masses of the elements present
Sulfur (S) : gmol-1
Oxygen (O) : gmol-1

21. Step 2: Determine the number of moles of each element present
Since we are dealing with percentages, we can express the mass % as
grams if we assume we have 100g of the compound.
Therefore, 100g of our compound contains 50.05g of sulfur and 49.95g of oxygen.
Convert number of grams to number of moles
Number of mol Sulfur = mass of sulfur in sample (g)
atomic mass of sulfur (gmol-1)
= g .
gmol-1
= 1.56 mol
Similarly, the no. of mol of Oxygen is found to be 3.12mol
Step 3: Determining the ratios of elements
Sulfur: 1.56mol
Oxygen: 3.12mol
Ratio : 3.12
Ratio must be in whole numbers. Here we must divide across by 1.56
Therefore, we have a ratio of 1:2 giving an empirical formula of SO2

22. Question
Determine the empirical formula of a compound that contains 27.3
mass% Carbon and 72.7 mass% Oxygen.
Answer: No. of mol Carbon = 2.27mol
No. of mol Oxygen = 4.54mol
Ratio 1:2 Empirical formula CO2
Monosodium glutamate (MSG) has the following mass percentage composition: 35.51% C, 4.77 % H, 37.85% O, 8.29% N, and 13.60% Na. What is its molecular formula if its molar mass is 169 gmol-1? 
Answer: C5H8O4NNa

23. Important calculations
Calculating mass percentage from a chemical formula
Step 1: Calculate the molar mass
Step 2: Determine the mass % composition for each element
Determining empirical formula from mass
Step 1: Find the atomic masses of the elements present
Step 2: Determine the number of moles of each element present
Step 3: Determining the ratios of elements

24. Molarity
Some chemical reactions involve aqueous solutions of reactants
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution
This concentration may be expressed in terms of molarity (M) or molar concentration:
M = Molarity = no. of moles
volume in Litres
Molarity is the number of moles of solute in 1 Litre (L) of solution

25. Example
What is molarity of an 85.0mL ethanol (C2H5OH) solution containing 1.77g of ethanol?
Step 1: Determine the number of moles of ethanol
Molar mass of ethanol, C2H5OH:
2 × carbon atoms
2 × gmol-1
24.02 gmol-1
1 × oxygen atom
1 × gmol-1
16.00 gmol-1
6 × hydrogen atoms
6 × gmol-1
6.048 gmol-1
46.07 gmol-1
No. of moles = mass in g
molar mass
No. of moles ethanol = g .
46.07 gmol-1
= mol

26. Step 2: Convert to molarity
Have 85.0mL ethanol
1 L = 1000mL
 Have L of ethanol
Molarity = no. of moles
volume in L
= mol
0.085 L
= 0.45 molL-1
≡ 0.45 M
Questions
Calculate the molarities of each of the following solutions:
(a) 2.357g of sodium chloride (NaCl) in 75mL solution
Answer: M
(b) 1.567mol of silver nitrate (AgNO3) in 250mL solution
Answer: M
(c) 10.4g of calcium chloride (CaCl2) in 2.20 × 102 mL of solution
Answer: M

27. Example
An antacid tablet is not pure CaCO3; it contains starch, flavouring, etc.
If it takes 41.3mL of M HCl to react with all the CaCO3 in one
tablet, how many grams of CaCO3 are in the tablet. You are given the
following balanced equation:
2HCl(aq) + CaCO3(s)  CaCl2(aq) + H2O(l) + CO2(g)
Step 1: Determine the no. of moles of HCl that react
Have M HCl solution  have mol in one litre
Have 41.3 mL of HCl solution  have L of HCl solution
Molarity = no. of moles
volume in L
 no. of moles = Molarity × volume in L
= molL-1 × L
= mol
≡ 8.5 × 10-3 mol HCl

28. Step 2: Determine no. of moles of CaCO3 used in the reaction
2HCl(aq) + CaCO3(s)  CaCl2(aq) + H2O(l) + CO2(g)
From the balanced equation, we can see that 2 moles of HCl are required to react with one mole of CaCO3
Therefore, if 8.5 × 10-3 mol of HCl are present in the reaction, we must have 4.25 × 10-3 mol of CaCO3 present.
Molar mass of CaCO3:
1 × calcium atom
1 × gmol-1
40.08 gmol-1
1 × carbon atom
1 × gmol-1
12.01 gmol-1
3 × oxygen atoms
3 × gmol-1
48.00 gmol-1
100 gmol-1
No. of mols = mass in g
molar mass
 Mass in g = no. of mols × molar mass
= (4.25 × 10-3 mol) × (100 gmol-1)
= g CaCO3 present in tablet

29. Questions
(a) How many moles of NaCl are present in 25.00mL of 1.85M NaCl(aq)?
Answer: 4.62 × 10-2 mol NaCl
(b) What volume of a 1.25 × 10-3 M solution of C6H12O6(aq) contains
1.44 × 10-6 mol of glucose?
Answer: 1.15 mL
(c) If stomach acid, given as 0.1 M HCl, reacts completely with an antacid tablet containing 500mg of CaCO3, what volume of acid in millilitres will be consumed? The balanced equation is:
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
Answer: 100mL acid

30. Important formulae… Calculating the number of moles:
No. of moles = mass in g
molar mass
Calculating the molarity or concentration:
Molarity = no. of moles
volume in L