key concepts 3 - mass calculations

Question 1

Describe an experiment to determine the empirical formula of a
simple compound such as magnesium oxide

Answer 1

• get crucible and heat (makes it clean)
• leave to cool and weigh it and its lid
• add magnesium ribbon to crucible and reweigh
• heat the crucible and put lid on to stop solid bits escaping
• but leave small gap to allow oxygen in
• heat until magnesium ribbon is white
• let cool and reweigh

Question 2

how to work out the empirical formula of magnesium oxide from this experiment

Answer 2

1. extra mass in magnesium oxide came from ribbon so work out mass of oxygen
2. work out number of moles of magnesium, and oxygen atoms involved in reaction
3. work out lowest whole number ratio between Mg and O by dividing moles of both by the smallest number

Question 3

explain the law of conservation of mass applied to a closed system including a precipitation reaction in a closed flask

Answer 3

• no atoms are destroyed or created
• this is shown by closed system having the total mass staying the same before and after reaction

Question 4

explain the law of conservation of mass applied to a non-enclosed system including the reaction in an open flask that takes or gives out a gas

Answer 4

• if mass increases = at least one of the reactants is a gas that is found in air and the products are l, aq or s
  a) gas is there before in air but not contained in vessel so you couldn’t measure its mass
• if mass decreases = some or all of reactants are s, l or aq and at least one product is a gas
  a) as gas could escape

Question 5

what is one mole of particles of a substance defined as

Answer 5

a) avo’s constant number of particles (6.02 *10 to the power of 23) of that substance
b) a mass of “relative particle mass” g

Question 6

Explain why, in a reaction, the mass of product formed is
controlled by the mass of the reactant which is not in
excess

Answer 6

• the amount of product formed is directly proportional to the amount of limiting reactant used
• because if you add more of the limiting reactant there will be more reactant particles to take part in the reaction
  = which means more product particles are made (as long as the other reactants are in excess)