Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. Masses of the reactants/products can be measured in grams or moles. A mole of a substance is the amount that contains the same number of units as the number of carbon atoms in 12 grams of carbon-12.
The Avogadro constant
12 g of carbon-12 contains 602 000 000 000 000 000 000 000 carbon atoms! This number is called the Avogadro constant and it is usually written in a short way as 6.02 x 1023. The mole can also be described as the amount of substance that contain 6.02 x 1023 particles. The particles can either be atoms, ions, or molecules depending on the nature of the substance.
So this means 1 mole of:
-carbon contains 6.02 x 1023 atoms.
-hydrogen gas cotains 6.02 x 1023 molecules.
-sodium chloride chloride contains 6.02 x 1023 ions of sodium and chlorine.
Each molecule of hydrogen gas contains 2 atoms of hydrogen making the chemical formulae of hydrogen gas H2. This means that 1 mole of H2 gas contains 2 x 6.02 x 1023 = 12.04 x 1023 atoms. Hydrogen is therefore called a diatomic gas.
Molar mass, Relative atomic mass and Relative mloecular mass
The relative atomic mass(Ar) of an element is the mass of the element is compared with the mass of 1 atom of the carbon-12 atom. The Ar of a substance is roughly equal to the nucleon number (the total number of neutrons and protons in an atom).
| Element | Symbol | Ar | Element | Symbol | Ar |
|---|---|---|---|---|---|
| hydrogen | H | 1 | chlorine | Cl | 35.5 |
| carbon | C | 12 | potassium | K | 39 |
| nitrogen | N | 14 | calcium | Ca | 40 |
The relative molecular mass of a compound (Mr) is the mass of the compound compared to the mass of the mass of the carbon-12 atom. It is called the relative formula mass if the compound is ionic. It is equal to the sum of the relative atomic masses of the atoms that make up the compound. However elements such as Hydrogen, Nitrogen, Chlorine naturally exist as molecules of 2 atoms each which means
they have Mr which is twice the Ar. They are said to be diatomic.
| Element | Symbol | Mr |
|---|---|---|
| hydrogen | H2 | 2 |
| chlorine | Cl2 | 71 |
| nitrogen | N2 | 28 |
| Substance | Formula | Mr of the compund |
|---|---|---|
| Sodium Chloride | NaCl | 23 + 35.5 = 58.5 |
| Sulphuric Acid | H2SO4 | 2(1) + 32 + 4(16) = 98 |
| Methane | CH4 | 12 + 4(1) = 16 |
The mass of 1 mole of a compound is called its molar mass and it is equal to the relative atomic mass(Ar) or relative molecular mass (Mr) of a substance. It is numerically equal to the Mr or Ar expressed in grams. The relative molecular mass of water is 18 and so its molar mass is 18 g.
Calculations involving moles
Mass of a given substance = Number of moles x mass of 1 mole
Example 1
Calculate the mass of
a) 3 moles
b) 0.2 moles of
carbon dioxide gas, CO2. (Ar: C = 12; O = 16)
a) One mole of CO2 contains 1 mole of carbon
atoms and 2 moles of oxygen atoms. Therefore:
mass of 1 mole of CO2 = (1 × 12) + (2 × 16)
= 44g
mass of 3 moles of CO2
= number of moles × mass of 1 mole of CO2
= 3 × 44
= 132g
b) mass of 0.2 mole of CO2
= number of moles × mass of 1 mole of CO2
= 0.2 × 44
= 8.8g
Example 2
Calculate the
a) mass of 0.5 moles of bromine atoms.
b) mass of 0.5 moles of bromine molecules.
c) number of moles of oxygen molecules are in 64 g of oxygen gas.
a) The Ar of bromine is 80, so 0.5 moles of bromine atoms
has a mass of 0.5 x 80 g = 40 g.
b) A bromine molecule contains 2 atoms (it’s diatomic), so its Mr is 160.
Therefore 0.5 moles of bromine molecules has a mass of 0.5 x 160 g = 80 g.
c) The Mr of oxygen is 32.
Number of mole = mass/ molar mass.
= 64/32
= 2 moles of oxygen molecules.
Empirical formula
The empirical formula is a formula which shows the simplest ratio in which atoms are combined. Different compounds with the same atomic ratio e.g C2H4, C3H6, C4H8 and C5H10 have the same empirical formula ( CH2 in this case). All of these formulae show the same ratio of carbon atoms to hydrogen atoms.
Empirical formula is calculated as follows:
- Write down the masses that
combine (in grams). - Convert the masses to moles of the atoms.
- Calculate the molar ratio in its simplest form.
- Write the empirical formula using molar ratio and the symbols of the atoms
Example 1
64 grams of sulfur combine with 64 grams of oxygen to form an oxide of sulfur. What is its empirical formula?
Answer
| Elements | Sulphur | Oxygen |
|---|---|---|
| Masses | 64g | 64g |
| Relative atomic masses (Ar) | 32 | 16 |
| Number of moles | 64/32 = 2 | 64/16 = 4 |
| Molar ratio (devide by the smallest number of moles) | 2/2 = 1 | 4/2 = 2 |
Therefore the ratio of Sulphur to Oxygen is 1:2 and so the empirical formula of the oxide that forms is SO2.
Example 2
It is given that compound A contains 80% carbon and 20% hydrogen. What is its empirical formula?
Anwser
Y contains 80% carbon and 20% hydrogen. So 100 g of A contains 80 g of carbon and 20 g of hydrogen
| Elements | Carbon | Hydrogen |
|---|---|---|
| Masses | 80g | 20g |
| Relative atomic masses (Ar) | 12 | 1 |
| Number of moles | 80/12 = 6.67 | 20/1 = 20 |
| Molar ratio (devide by the smallest number of moles) | 6.67/6.67 = 1 | 20/6.67 = 3 |
Therefore the empirical formula of A is CH3.
Molecula formula
The molecular formula shows the actual numbers of atoms that
combine to form a molecule. For some compounds the empirical formula is the same as molecular formula.
| Compound | Molecular formula | Empirical formula |
|---|---|---|
| methane | CH4 | CH4 |
| ethane | C2H6 | CH3 |
Molecular formula is calculated as follows:
- Calculate the empirical mass of the compound by adding the masses of the atom shown by the empirical formula.
- Calculate the number of formula units as follows
n = Mr of the compound/empirical mass for the compound. - Multiply the numbers in the empirical formula by n.
Example
A molecular compound has the empirical formula HO.
Its relative molecular mass is 34. What is its molecular formula?
(Ar : H = 1, O = 16.)
Answer
For the empirical formula HO, the empirical mass = 17. But Mr = 34.
So Mr / empirical mass =
34/17 = 2
So the molecular formula is 2 x HO = H2O2.
So the compound is hydrogen peroxide.
Concentration
The concentration of a solution is the amount of solute, in grams or moles, that is dissolved in a unit volume of solution. When 1 mole of sodium chloride is dissolved in 1 dm3 (1000cm3) of water the concentration of the solution is 1 mol/dm3.
Concentration is calculated as follows:
- concentration = mass/volume
the units can be g/cm3 or g/dm3 depending on the units of volume used. - concentration = number of moles/ volume.
the units are usually mol/dm3 in this case.
Example 1
How many moles of solute are in 500 cm3 of solution, of concentration 2 mol/dm3?
Answer
- First convert the volume in cm3 to dm3
volume = 500/1000
= 0.5 dm3
then use the formula:
number of moles = concentration x volume
= 2 x 0.5
= 1 mole.