Chemical Measurements, Conservation of Mass and the Quantitative Interpretation of Chemical Equations

1. State the total number of atoms present in the formula of ammonium sulfate, (NH4)2SO4 .

Answer: 15 atoms


2. Which of the following symbol equations is correctly balanced?

• C3H8(g) + 4O2(g)  →  3CO2(g) + 4H2O(l)
• 2Al(s) + 6HCl(aq)  →  2AlCl3(aq) + 3H2(g)
• CuCO3(s)  →  CuO(s) + 2CO2(g)
• Mg(s) + AgNO3(aq)  →  Mg(NO3)2(aq) + Ag(s)

Answer: 2Al(s) + 6HCl(aq)  →  2AlCl3(aq) + 3H2(g)


3. Use the periodic table to calculate the relative formula mass (Mr) of magnesium nitrate, Mg(NO3)2 .

Answer: 24 + (2 × 14) + (6 × 16) = 148


4. Give one example of when the law of conservation of mass may appear not to be obeyed during a chemical reaction.

Answer: When metals react with oxygen gas (or chlorine gas), or during thermal decomposition reactions.


5. Calculate the mean value and its uncertainty for the following mass measurements: 2.95 g, 2.89 g, 2.65 g, 2.99 g

Answer: Anomalous value (2.65 g) discarded.
Average = (2.95 + 2.89 + 2.99) ÷ 3 = 2.94 g
The average value is no more than 0.05 g from the actual values, so the uncertainty = ± 0.05 g


6. Sodium hydrogencarbonate (NaHCO3 ) thermally decomposes to form sodium carbonate (Na2CO3 ), carbon dioxide gas and water. The equation for the reaction is: 2NaHCO3(s)  →  Na2CO3(s) + CO2(g) + H2O(l)

The table below shows the relative formula masses (Mr) of the products of this reaction.


a) Explain why the symbol equation above is balanced.

Answer: There is the same number of atoms of each element on both sides of the equation.


b) Use the periodic table to calculate the relative formula mass of sodium hydrogencarbonate.

Answer: Mr = 23 + 1 + 12 + (3 × 16) = 84


c) Show that mass is conserved during this reaction.

Answer: Relative mass of the reactants = 2 × 84 = 168; Relative mass of the products = 106 + 44 + 18 = 168. The sum of the relative formula masses of the reactants is the same as for the products, so mass is conserved.


d) A student investigated the thermal decomposition of sodium hydrogencarbonate. They followed this method:

•  Place 2.40 g of the solid into a crucible with a mass of 25.00 g
•  Heat the crucible strongly for 5 minutes and then allow it to cool.
•  Reweigh the crucible and the solid it contains.


Calculate the mass of the products in the crucible, if the total mass when reweighed is 26.51 g

Answer: 26.51 g – 25.00 g = 1.51 g


e) Suggest a reason why the mass of the products was less than the total mass of reactants used.

Answer: The water and carbon dioxide will have escaped from the crucible, so their masses will not have been measured/only the mass of the solid (sodium carbonate) has been measured.


7. Calcium metal and oxygen gas react to form calcium oxide (CaO).

a) Give the balanced symbol equation for the reaction between calcium and oxygen.

Answer: 2Ca + O2  →  2CaO


b) The table below shows the results from a series of experiments where 2.00 g of calcium was heated in a crucible with a mass of 28.00 g. Calculate the mean mass of calcium oxide and its uncertainty produced in these four experiments.


Answer: Masses of calcium oxide: 2.81 g , 2.79 g , 2.41 g , 2.84 g
Identification and removal of anomalous value (2.41 g)
Correct mean value calculated: 2.81 g
Correct uncertainty stated: ± 0.03 g


c) Explain why these results do not appear to follow the law of conservation of mass

Answer: The calcium gains oxygen as calcium oxide forms, but this mass of oxygen is not measured/only the mass of calcium is measured as a reactant by the mass balance.


8. When lead nitrate [Pb(NO3)2 ] is heated it breaks down to form a mixture of products. The equation for the reaction is: 2Pb(NO3)2(s)  →  2PbO(s) + 4NO2(g) + O2(g)

The table below shows the relative formula masses of some of the substances in the reaction.


a) When carried out as an experiment, the mass of the lead nitrate always seems to be greater than the mass of the three products. Explain why.

Answer: When the mass of the products is measured it will not include the mass of NO2 and O2 as these are both gases. This makes the mass of reactants appear to be greater than the mass of products.


b) Show that the relative formula mass of nitrogen dioxide (NO2 ) is equal to 46. Explain your answer.

Answer: Mass is conserved during a reaction, so the relative masses of the reactants = the relative mass of all the products. 2(331) = 2(223) + 32 + 4(NO2)  →  662 = 478 + 4(NO2)  →  NO2 = 46