The mole is the amount of a substance that contains the same number of particles as there are atoms in 12.00g of carbon-12. This number of atoms is 6.02 x 1023 and is called the Avogadro constant.
One mole of carbon atoms.Return to AS and A2 GCE Chemistry
1.2a Empirical formulae from reacting masses
An empirical formula is the simplest formula which shows the ratio
of each type of atom present in a compound.
A molecular formula is a simple multiple of the empirical formula showing
how many of each type of atom are present in a compound.
For 80g of carbon (molar mass = 12gmol-1) reacting with 20g of hydrogen
(molar mass 1gmol-1)
amount = mass/molar mass
| symbols | C | H |
| mass/g | 80 | 20 |
| amount/mol | 80/12 = 6.67 | 20/1=20 |
| ratio of atoms | 6.67/6.67=1 | 20/6.67=3 |
| empirical formula | CH3 | |
If the molar mass of the compound is 30gmol-1 the molecular formula
can be calculated:
Molecular formula = n*(empirical formula)
where n = molar mass of molecular formula/molar mass of empirical formula
n = 30/15 = 2
molecular formula = 2*CH3 =C2H6
Task 1.2a
1.2b Full and ionic equations
When iron is added to copper (II) sulphate, copper and iron (II) sulphate
is formed. The full equation is
Fe(s) + CuSO4(aq) ---> Cu(s) + FeSO4(aq)
In terms of ions this is:
Fe(s) + Cu2+(aq) + SO42-(aq) ----> Cu(s) +Fe2+(aq) + SO42-(aq)
Sulphate ions are spectator ions as they are unaffected by the
reaction. Therefore they are left out. The ionic equation is:
Fe(s) + Cu2+ (aq) ----> Fe2+ (aq) + Cu(s)
Both the number of atoms and charges balance.
Task 1.2b
1.2c Using reacting masses and chemical equations
What mass of sodium carbonate (molar mass =
106g/mol) can be made by heating 16.8g of sodium hydrogencarbonate (molar mass =
84g/mol)?
Method 1 (recommended)
2NaHCO3(s) ----> Na2CO3(s) + CO2(g)
+ H2O(g)
mass NaHCO3 =16.8g
amount NaHCO3 = mass NaHCO3/molar mass NaHCO3
=16.8g/84gmol-1 = 0.2mol
from equation: amount Na2CO3/amount NaHCO3 =1/2
amount Na2CO3 =1/2*0.2mol = 0.1 mol
mass Na2CO3
=
amount Na2CO3*molar mass Na2CO3
=0.1mol*106gmol =10.6g
Method 2
formula mass 2NaHCO3 = 168 formula mass Na2CO3
= 106
so 168g of sodium hydrogen carbonate forms 106g of sodium carbonate
so 1g of sodium hydrogen carbonate forms 106/168g of
sodium carbonate
so 16.8g of sodium hydrogen carbonate forms 16.8*106/168g of sodium carbonate
so 16.8g of sodium hydrogen carbonate forms 10.6g of
sodium carbonate
Task 1.2c
1.2d Avogadro constant and the amount of substance
The mole is the amount of a substance that contains the same
number of particles as there are atoms in 12.00g of carbon-12. This number
of atoms is 6.02 x 1023 and is called the Avogadro constant.
One mole of carbon atoms.
| amount of substance = number of particles/Avogadro constant |
Chemist cannot find numbers of particles easily so mass is
used. The mass of a substance which contains the Avogadro number of
particles is called the molar mass.
Molar mass can be found. The number of g in one mole is the same number as the
relative formula mass of a substance E.g. 23g of Na atoms and
28g of N2 molecules all contain the same number of particles,
they are all one mole of atoms or molecules.
| amount of substance = mass of substance/molar mass of substance |
Task 1.2d
1.2e Solution concentration data
The concentration of a solution can be stated as the mass of solute
per cubic decimetre of solution (g/dm3) or the amount in moles
of a solute present in 1dm3 of solution (mol/dm3).
Concentration = mass of solute/volume
| Concentration (Molarity) = amount of solute /volume of solution |
1dm3 = 1000cm3
Task 1.2e.1
A solution of known concentration is called a standard solution.
This
is used in volumetric analysis (e.g. acid /base titration) to determine the
amount of solute in a solution or it's molarity (concentration in mol/dm3).
A standard solution is made from a primary standard which should be:
easy to obtain and purify
stable
readily soluble
completely react
have a high molar mass
A good example is the the acid potassium hydrogenphthalate C8H5O4K
Task 1.2e.2
1.2f Volumetric calculations
A typical problem is to calculate the concentration of one solution given
its volume and the volume and concentration of a solution that it reacts
with.
For example what is the concentration of a solution of sodium hydroxide if
25.0cm3 of it in a flask react with 10.25cm3 of a
0.10moldm-3 solution of sulphuric acid.
amount of H2SO4 reacting = volume H2SO4
* Concentration of H2SO4
= 10.25 /1000dm3 * 0.10moldm-3 =
1.025*10-3 mol
H2SO4 + 2NaOH ---> Na2SO4
+ 2H2O
From equation
amount of NaOH/amount of H2SO4 = 2/1
amount of NaOH = 2/1* 1.025*10-3 mol = 2.05 *10-3 mol
concentration of NaOH=amount/volume = 2.05 *10-3 mol/25.0/1000dm3=
0.082mol dm-3
Task
1.2f
1.2g Molar volume of a gas
Avogadro's Law states that equal volumes of all gases under the same
conditions (temperature and pressure) contain the same number of particles.
One mole of any ideal gas occupies 22.4dm3 at stp (Standard
temperature and pressure) and 24dm3 at rtp (Room temperature
and pressure). This is the molar volume of a gas.
| amount of gas = volume of gas/molar volume |
H2(g) + Cl2(g) ----> 2HCl(g) (At stp)
22.4dm3 of hydrogen reacts with 22.4dm3 of chlorine
forms 44.8dm3 of hydrogen chloride
1 volume of hydrogen reacts with 1 volume of chlorine to form 2 volumes
of hydrogen chloride
1 mole of hydrogen molecules reacts with 1 mole of chlorine molecules
to form 2 moles of hydrogen chloride molecules.
task 1.2g
1.2h Reacting gas volumes
Equations can be used to deduce reacting gas volumes:
What volume of nitrogen dioxide can be formed when 16.4g of Calcium
nitrate (molar mass 164g/mol) decomposes on heating?
2Ca(NO3)2(s)----> 2CaO(s) + 4NO2(g) + O2(g)
amount of calcium nitrate = mass/molar mass = 16.4/164 = 0.100mol
from equation amount of nitrogen dioxide/amount of calcium nitrate
= 4/2
so amount of nitrogen dioxide = 0.200mol
volume of nitrogen dioxide = amount *molar volume
volume of nitrogen dioxide = 0.200*22.4dm3= 4.48dm3
.
Task 1.2h