Stoichiometry and the Mole: notes

Formulae and Equations

• Chemical formula: shows the atoms in a substance. Water is H2O: two hydrogen atoms and one oxygen atom.
• Balancing equations: each element must have the same number of atoms on both sides. Hydrogen plus oxygen gives water, so 2H2 + O2 gives 2H2O.
• Relative atomic mass (Ar): the mass of an atom compared with carbon. Oxygen has Ar 16.
• Relative formula mass (Mr): add the Ar values in the formula. H2O has Mr = (2 x 1) + 16 = 18.

The Mole

• Mole: the amount of substance containing 6.02 x 10^23 particles (Avogadro's constant).
• Moles from mass: moles = mass / Mr. Example: 36 g of water is 36 / 18 = 2 moles.
• Molar volume: one mole of any gas occupies 24 dm3 at room temperature and pressure. Example: 2 moles of oxygen occupy 48 dm3.
• Concentration: mol/dm3 = moles / volume in dm3. Example: 0.5 moles in 0.5 dm3 gives 1 mol/dm3.
• Percentage composition: (mass of element / Mr) x 100. Oxygen in H2O is 16 / 18 x 100 = 88.9 percent.

Calculations

• Empirical formula: the simplest whole number ratio of atoms. Divide each mass by Ar, then divide by the smallest answer.
• Molecular formula: the true number of atoms. An empirical formula of CH2 with Mr 42 gives C3H6.
• Reacting masses: 2Mg + O2 gives 2MgO, so 48 g of magnesium gives 80 g of magnesium oxide.
• Percentage yield: (actual yield / theoretical yield) x 100. Example: 6 g made from an expected 8 g gives 75 percent.