Chem!stry Class: _______ Date: ____ / ____ / ____ Masses of Reacting Solids Two essential things to remember: Converting a mass in grams into a number of moles: Converting a number of moles into a mass in grams:
mass of substance (g) = amount of substance (in moles) × relative molecular mass of substance
The diagram shown below is a visual representation of the reaction between hydrogen and oxygen to
Mass of chemical on the left-hand-side: Mass of chemicals on the right-hand-side:
= (2 × (1.0 + 1.0)) + (1 × (16.0 + 16.0))
So, 2 mol of H2 (weighing 4.0 g) react with 1 mol of O2 (weighing 32.0 g) to produce 2 mol of H2O
This illustrates the law of conservation of mass. Mass is not lost or gained during a chemical reaction,
i.e. the total mass of the reactants is exactly equal to the total mass of the reaction products. This
allows chemists to perform useful mole mass and mass mole calculations employing balanced
For example, a chemist might calculate the mass of product that should be produced by a chemical
reaction if a certain mass of reactant is used.
Consider the balanced chemical equation for the reaction between sodium and chlorine:
This means that 2 mol of sodium will react with 1 mol of chlorine to produce 2 mol of sodium chloride:
2 mol of sodium weighing 2 × 23.0 = 46.0 g
1 mol of chlorine weighing 1 × (35.5 + 35.5) = 71.0 g
2 mol of sodium chloride weighing 2 × (23.0 + 35.5) = 117 g
But what if, while performing this reaction in the laboratory, a chemist only weighed out and used
1 mol of sodium instead of 2 mol? According to the balanced chemical equation, if only 1 mol of sodium
is used, it will react with half-as-much chlorine (i.e. 0.5 mol) to produce 1 mol of sodium chloride:
1 mol of sodium weighing 1 × 23.0 = 23.0 g
0.5 mol of chlorine weighing 0.5 × (35.5 + 35.5) = 35.5 g
1 mol of sodium chloride weighing 1 × (23.0 + 35.5) = 58.5 g
Example Calculation:
When an aqueous solution of sodium carbonate is added to an aqueous solution of silver nitrate, a
precipitate of silver carbonate is formed:
sodium carbonate + silver nitrate silver carbonate + sodium nitrate
Na2CO3(aq) + 2AgNO3(aq) Ag2CO3(s) + 2NaNO3(aq)
Calculate the mass of silver carbonate that will be produced from 850 g of silver nitrate:
Information has been provided about the mass of silver nitrate that was used. This information
should be used to calculate the amount (moles) of silver nitrate that was used.
Use the balanced chemical equation to deduce the relationship between moles of silver nitrate
and moles of silver carbonate (i.e. the mole ratio).
Finally, convert the amount (moles) of silver carbonate into a mass in grams.
amount of substance (mol) = mass of substance (g) relative molecular mass of substance
amount of AgNO3 (mol) = mass of AgNO3 (g) relative molecular mass of AgNO3
amount of AgNO3 (mol) = 850 (108 + 14.0 + (3 × 16.0))
from the balanced chemical equation, 2 mol of AgNO3 produce 1 mol of Ag2CO3
therefore, 5 mol of AgNO3 will produce 1/2 × 5 mol of Ag2CO3
therefore 5 mol of AgNO3 will produce 2.50 mol of Ag2CO3
mass of substance (g) = amount of substance (mol) × relative molecular mass of substance
mass of Ag2CO3 (g) = 2.5 × ((2 × 108) + 12.0 + (3 × 16.0))
Question 1 – Pharmaceutical Chemistry:
The anticancer drug cisplatin (formula: Pt(NH3)2Cl2) was discovered in the 1960s and has since been
used to successfully treat a wide range of tumours. It is synthesised in the laboratory according to the
K2PtCl4(aq) + 2NH3(aq) Pt(NH3)2Cl2(aq) + 2KCl(aq)
A patient suffering from lung cancer requires 6.00 g of cisplatin for chemotherapy. Calculate the mass
of K2PtCl4 that a pharmaceutical chemist must use to synthesise the drug:
Question 2 – Pharmaceutical Chemistry:
The pain killer aspirin can be synthesised in the laboratory according to the following word equation:
2-hydroxybenzoic acid + ethanoic anhydride aspirin + ethanoic acid
With reference to the graphical formulae shown above, write the balanced chemical equation for
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Aspirin tablets typically contain 0.50g of the drug. Calculate the mass of 2-hydroxybenzoic acid
that a pharmaceutical chemist must use to synthesis 1080 aspirin tablets:
Question 3 – Forensic Science:
In 1991 an American named George Trepal was convicted and sentenced to death for the fatal
poisoning of his neighbour using the toxic compound thallium(I) nitrate. Trepal added the thallium(I)
nitrate to bottles of Coca-ColaTM which he then left for his neighbour to drink.
The amount of thallium(I) nitrate present in the bottle of Coca-ColaTM was deduced by adding
potassium chromate to the soda and then measuring the mass of thallium(I) chromate that precipitated
thallium(I) nitrate + potassium chromate thallium(I) chromate + potassium nitrate
2TlNO3(aq) + K2CrO4(aq) Tl2CrO4(s) + 2KNO3(aq)
If 2.62 g of thallium(I) chromate precipitated from the solution, calculate the mass of thallium(I) nitrate
that was added to one bottle of Coca-ColaTM:
Question 4 – Environmental Science:
The fuel propane (formula: C3H8) burns completely in air to produce carbon dioxide and water.
Write the balanced chemical equation to describe the complete combustion of propane:
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The carbon dioxide gas that is produced may be harmful to the environment if it is released into
the atmosphere in large quantities. However, the carbon dioxide gas can be removed by
combining it with lithium hydroxide to form lithium carbonate:
lithium hydroxide + carbon dioxide lithium carbonate + water
2LiOH(s) + CO2(g) Li2CO3(s) + H2O(l)
Calculate the mass of lithium carbonate formed when 88.0 g of propane burn completely in air:
Question 5 – Analytical Chemistry:
A Group 1 metal carbonate has the general formula of M2CO3 where M represents the symbol of any
Group 1 metal. The Group 1 metal carbonate reacts with hydrochloric acid according to the following
metal carbonate + hydrochloric acid metal chloride + water + carbon dioxide
M2CO3(s) + 2HCl(aq) 2MCl(aq) + H2O(l) + CO2(g)
If 69.1 g of M2CO3 produce 22.0 g of carbon dioxide gas, deduce the identity of the Group 1 metal, M:
Converting a mass in grams into a number of moles: Converting a number of moles into a mass in grams:
mass of substance (g) = amount of substance (in moles) × relative molecular mass of substance
Question 1.
amount of substance (mol) = mass of substance (g) relative molecular mass of substance
amount of Pt(NH3)2Cl2 (mol) = mass of Pt(NH3)2Cl2 (g) relative molecular mass of Pt(NH3)2Cl2
amount of Pt(NH3)2Cl2 (mol) = 6.00 (195 + (2 × 14.0) + (6 × 1.0) + (2 × 35.5))
amount of Pt(NH3)2Cl2 (mol) = 6.00 300
from the balanced chemical equation, 1 mol of Pt(NH3)2Cl2 is produced from 1 mol of K2PtCl4
therefore, 0.02 mol of Pt(NH3)2Cl2 will be produced from 0.02 × 1 mol of K2PtCl4
therefore, 0.02 mol of Pt(NH3)2Cl2 will be produced from 0.0200 mol of K2PtCl4
mass of substance (g) = amount of substance (mol) × relative molecular mass of substance
mass of K2PtCl4 (g) = amount of K2PtCl4 (mol) × relative molecular mass of K2PtCl4
mass of K2PtCl4 (g) = 0.02 × ((2 × 39.1) + 195 + (4 × 35.5))
Question 2.
2-hydroxybenzoic acid + ethanoic anhydride aspirin + ethanoic acid
the total mass of aspirin in 1080 tablets, each containing 0.50 g of the drug = 1080 × 0.50 = 540 g
amount of substance (mol) = mass of substance (g) relative molecular mass of substance
amount of aspirin (mol) = mass of aspirin (g) relative molecular mass of aspirin
amount of aspirin (mol) = 540 ((9 × 12.0) + (8 × 1.0) + (4 × 16.0))
from the balanced chemical equation, 1 mol of aspirin is produced from 1 mol of 2-hydroxybenzoic acid
therefore, 3 mol of aspirin will be produced from 3 × 1 mol of 2-hydroxybenzoic acid
therefore 3 mol of aspirin will be produced from 3.00 mol of 2-hydroxybenzoic acid
mass of substance (g) = amount of substance (mol) × relative molecular mass of substance
mass of 2-hydroxybenzoic acid (g) = amount of 2-hydroxybenzoic acid (mol) × relative molecular mass of 2-hydroxybenzoic
mass of 2-hydroxybenzoic acid (g) = 3 × ((7 × 12.0) + (6 × 1.0) + (3 × 16.0))
mass of 2-hydroxybenzoic acid (g) = 3 × 138
Question 3.
amount of substance (mol) = mass of substance (g) relative molecular mass of substance
amount of Tl2CrO4 (mol) = mass of Tl2CrO4 (g) relative molecular mass of Tl2CrO4
amount of Tl2CrO4 (mol) = 2.63 ((2 × 204) + 52.0 + (4 × 16.0))
from the balanced chemical equation, 1 mol of Tl2CrO4 is produced from 2 mol of TlNO3
therefore, 0.005 mol of Tl2CrO4 will be produced from 0.005 × 2 mol of TlNO3
therefore 0.005 mol of Tl2CrO4 will be produced from 0.0100 mol of TlNO3
mass of substance (g) = amount of substance (mol) × relative molecular mass of substance
mass of TlNO3 (g) = amount of TlNO3 (mol) × relative molecular mass of substance TlNO3
mass of TlNO3 (g) = 0.01 × (204 + 14.0 + (3 × 16.0))
Question 4.
amount of substance (mol) = mass of substance (g) relative molecular mass of substance
amount of C3H8 (mol) = mass of C3H8 (g) relative molecular mass of C3H8
amount of C3H8 (mol) = 88.0 ((3 × 12.0) + (8 × 1.0))
from the first balanced chemical equation, 1 mol of C3H8 produces 3 mol of CO2
therefore, 2 mol of C3H8 will produce 2 × 3 mol of CO2
therefore 2 mol of C3H8 will produce 6.00 mol of CO2
from the second balanced chemical equation, 1 mol of CO2 produces 1 mol of Li2CO3
therefore, 6 mol of CO2 will produce 6 × 1 mol of Li2CO3
therefore 6 mol of CO2 will produce 6.00 mol of Li2CO3
mass of substance (g) = amount of substance (mol) × relative molecular mass of substance
mass of Li2CO3 (g) = amount of Li2CO3 (mol) × relative molecular mass of Li2CO3
mass of Li2CO3 (g) = 6 × ((2 × 6.9) + 12.0 + (3 × 16.0))
Question 5.
amount of substance (mol) = mass of substance (g) relative molecular mass of substance
amount of CO2 (mol) = mass of CO2 (g) relative molecular mass of CO2
amount of CO2 (mol) = 22.0 (12.0 + (2 × 16.0))
from the balanced chemical equation, 1 mol of CO2 is produced from 1 mol of M2CO3
therefore, 0.5 mol of CO2 will be produced from 0.5 × 1 mol of M2CO3
therefore 0.5 mol of CO2 will be produced from 0.500 mol of M2CO3
relative molecular mass of substance = mass of substance (g) amount of substance (mol)
relative molecular mass of M2CO3 = mass of M2CO3 (g) amount of M2CO3 (mol)
relative molecular mass of M2CO3 = 69.1 0.5
relative molecular mass of M2CO3 = 138.2
subtract the mass of the carbonate group (CO3) to determine the relative atomic mass of 2 M
relative atomic mass of 2 M = 138.2 – (12.0 + (3 16.0))
relative atomic mass of 2 M = 138.2 – 60.0
therefore the relative atomic mass of M = 78.2 2
hence the Group 1 element , M, is potassium and the formula of the carbonate is K2CO3