#### Relative Atomic Mass

• ###### Calculate relative formula mass given relative atomic masses.
• The symbol for the relative formula mass is Mr and it refers to the total mass of the molecule.
• To calculate the Mr of a substance, you have to add up the relative atomic masses of all the atoms present in the formula.
• The relative atomic masses of every element is available on The Periodic Table, it is the larger of the two numbers.

Examples: #### Formulae of Simple Compounds

• ###### Calculate the formulae of simple compounds from reacting masses and understand that these are empirical formulae.
• The empirical formula gives the simplest whole number ratio of atoms of each element in the compound.
• It is calculated from knowledge of the ratio of masses of each element present.
• Use the equation for calculating the number of moles from mass:
• Moles = mass ÷ Mr
• The mole is a unit of measurement used in chemistry.
• The relative atomic mass of an element is exactly equal to the mass of one mole of that element.
• The same applies for a compound.
• For example the relative molecular mass of carbon dioxide is: 12 + (16 x 2) = 44.
• So one mole of CO2 has a mass of 44g.

Example 1:

Show that a compound which contains 10g of hydrogen and 80g of oxygen has an empirical formula of H2O.

Amount of hydrogen atoms = mass in grams ÷ Ar of hydrogen = (10 ÷ 1) = 10 moles

Amount of oxygen atoms = mass in grams ÷ Ar of oxygen = (80 ÷ 16) = 5 moles

The ratio of moles of hydrogen atoms to moles of oxygen atoms:

Hydrogen            Oxygen

Moles                 10             :            5

Ratio                   2              :            1

Since equal numbers of moles of atoms contain the same number of atoms, the ratio of hydrogen atoms to oxygen atoms is 2:1

Hence the empirical formula is H2O

Example 2:

What is the empirical formula of a compound that contains 7.83g of iron and 3.37g of oxygen?

Amount of iron atoms = mass in grams ÷ Ar of iron = (7.83 ÷ 55.85) = 0.14 moles

Amount of oxygen atoms = mass in grams ÷ Ar of oxygen = (3.37 ÷ 16) = 0.21 moles

The ratio of moles of iron atoms to moles of oxygen atoms:

Iron                  Oxygen

Moles                 0.14             :            0.21

Divide across by the smallest number (0.14):

Ratio                   1                 :            1.5

Simplify to the nearest whole number (multiply across by 2):

Ratio                   2                 :            3

Hence the empirical formula is Fe2O3

#### Empirical and Molecular Formulae

• ###### Deduce: a) the empirical formula of a compound from the formula of its Molecule b) the molecular formula of a compound from its empirical formula and its relative molecular mass
• The empirical formula tells you the simplest whole number ratio of atoms in a compound.
• The molecular formula tells you the actual number of atoms of each element in one molecule of the compound or element e.g. H2 has 2 hydrogen atoms, HCl has 1 hydrogen atom and 1 chlorine atom.

Deducing empirical formula

• The empirical formula can be deduced from the molecular formula.
• Do this by simplifying the number of each atom by dividing across by the largest whole number that will divide evenly into each number in the molecular formula.

Examples: Deducing molecular formula

• Divide the relative formula mass of the molecular formula by the relative formula mass of the empirical formula.
• Then multiply the number of each element present in the empirical formula by this number to find the molecular formula.

Example:

The Empirical Formula of X is C4H10S1 and the Relative Formula Mass of X is 180. What is the Molecular Formula of X?

Relative Formula Mass:       Carbon : 12      Hydrogen : 1      Sulfur : 32

Step 1 Calculate Relative Formula Mass of Empirical Formula

(C x 4) + (H x 10) + (S x 1)    =   (12 x 4) + (1 x 10) + (32 x 1)   =   90

Step 2 Divide Relative Formula Mass of X by Relative Formula Mass of Empirical

Formula

180 / 90 = 2

Step 3 Multiply Each Number of Elements by 2

(C4 x 2) + (H10 x 2) + (S1 x 2)     =    (C8) + (H20) + (S2)

Molecular Formula of X = C8H20S2

#### Plan an Investigation to Determine Empirical Formula

• ###### Describe an experiment to determine the empirical formula of a simple compound such as magnesium oxide.
• There are two methods to carry out this investigation:
• Method A: Combustion of magnesium oxide
• Method B: Reduction of magnesium oxide

Method A: Combustion of Magnesium Oxide Diagram of the apparatus to find the formulae of a metal oxide by combustion

Method:

• Measure and record the mass of crucible and lid
• Add sample of metal into crucible and measure mass with lid and record
• Strongly heat the crucible over a Bunsen burner for several minutes
• Frequently lift the lid to allow sufficient air into the crucible for the metal to fully oxidise but without letting any of the gaseous metal oxide to escape
• Continue heating until the mass of crucible remains constant
• Measure the mass of crucible and contents and record

Calculation of empirical formula:

• Mass of metal = (mass of crucible + lid + metal) – (mass of crucible + lid)
• Mass of metal oxide = (mass of crucible + lid + oxide) – (mass of crucible + lid)
• Mass of oxygen = mass of metal oxide – mass of metal
• Step 1 Divide each of the two masses by their Relative Atomic Masses
• Step 2 Simplify the ratio

Metal            Oxygen

Mass               x                       y

Mole              x / Mr               y / Mr

= a                   = b

Ratio                a          :           b

• Step 3 Write out the formula using the ratio e.g. for magnesium: MgaOb

Method B: Reduction of Magnesium Oxide Diagram of the apparatus to find the formulae of a metal oxide by reduction

Method:

• Measure and record the mass of the metal oxide
• Use a clamp to hold boiling tube horizontally, and place the metal oxide at the end of the tube
• Heat using a Bunsen burner until all  the oxide has completely changed colour, indicating that all the oxygen has been reduced
• Measure and record the mass of the remaining powder

Calculation of empirical formula:

• Mass of metal = mass of remaining metal powder
• Mass of oxygen = (mass of metal oxide) – (mass of metal powder)
• Step 1 Divide each of the two masses by their Relative Atomic Masses
• Step 2 Simplify the ratio

Metal            Oxygen

Mass               x                       y

Mole              x / Mr               y / Mr

= a                   = b

Ratio                a          :           b

• Step 3 Write out the formula using the ratio e.g. for copper: CuaOb

## Want to aim for a Level 9?

See if you’ve got what it takes. Test yourself with our topic questions. ### Author: Morgan

Morgan’s passion for the Periodic Table begun on his 10th birthday when he received his first Chemistry set. After studying the subject at university he went on to become a fully fledged Chemistry teacher, and now works in an international school in Madrid! In his spare time he helps create our fantastic resources to help you ace your exams.