Chemical Calculations

Relative Atomic Mass The relative atomic mass of an element is the average mass of one atom of the element when compared with 1/12th of the mass of an atom of carbon-12. Ar = average mass of one atom of an element mass of 1/12th of an atom of carbon-12 Ar = symbol for relative atomic mass Ar has no units

Relative Molecular Mass The relative molecular mass or Mr of a molecule is the average mass of one molecule of a substance when compared with 1/12th of the mass of one atom of carbon-12. Mr is derived for any molecule by determining the sum total of the relative atomic masses of all the individual atoms present in the molecule. Mr has no units.

Questions Calculate the Mr values for the following substances : Cu(NO3)2 CuSO4. 5H2O

Mr of Cu(NO3)2 = 64 + 2(14 + (16 × 3) ) = 188 Mr of CuSO4. 5H2O = 64 + 32 + (16 × 4) + 5((2 × 1) + 16) = 250

Calculating Percentage Composition of A Compound The percentage by mass of an element/ ion can be determined by calculating the sum total of the relative masses of the element / ion present in the compound, dividing this value by the Mr of the compound and expressing the value as a percentage.

Questions Calculate the percentages of the elements present in sodium carbonate. Mr of Na2CO3 = (23 × 2) + 12 + (16 × 3) = 106 % of sodium in sodium carbonate = 46 / 106 × 100 = 43.4 %

% of carbon in sodium carbonate = 12 / 106 × 100 = 11.3 % % of oxygen in sodium carbonate = 48 / 106 × 100 = 45.3 % OR = 100 – 43.4 – 11.3 = 45.3 %

Calculating the Mass of an Element in a Compound Calculate the mass of nitrogen present in 200 g of ammonium nitrate. Mr of NH4NO3 = 14 + 4 + 14 + (16 × 3) = 80 Mass of nitrogen in 200 g of ammonium nitrate = 28 / 80 × 200 = 70.0 g

Question A sample of copper (II) sulphate has the formula CuSO4. xH2O. The composition of water in the compound is 36.0 %. What is the value of x ? Mr of CuSO4 = 64 + 32 + (16  4) = 160 % of CuSO4 in CuSO4.xH2O = 100 – 36.0 = 64.0 %

Relative mass of water in the salt = 36.0  160 = 90 64.0  x = 90 = 90 = 5 Mr of H2O 18  Formula of salt = CuSO4. 5H2O

The Mole One mole of any matter (atoms, ions or molecules) contains 6 × 10 23 particles. There are no units for the mole. Symbol = mol The mass of one mole of any substance is called the molar mass. The value of molar mass is equal to the Ar or Mr of the substance (either element or compound) in grams.

The volume of one mole of any gas at room temperature and pressure is 24 dm3 or 24 000 cm3. This is called the molar volume at r.t.p. d) Avogadro’s Law states that equal volumes of gases contain the same number of molecules under the same conditions of temperature and pressure. e) The volume of a gas is directly proportional to the number of moles of the gas.

Molar mass / mass of one mole of 23 Na = 23 g Mass of one mole of CO2 = 12 + (16 × 2) = 44 g Mass of one mole of H2 = 2 g Mass of one mole of O2 = 32 g

In 23 g of Na, there are 6 × 10 23 atoms of In 44 g of CO2, there are 6 × 10 23 molecules of CO2. In 2 g of H2, there are 6 × 10 23 molecules of H2. In 32 g of O2, there are 6 × 10 23 molecules of O2.

44 g of CO2 occupy 24 dm3 at r.t.p. 2 g of H2 occupy 24 dm3 at r.t.p. 32 g of O2 occupy 24 dm3 at r.t.p.  10 cm3 of O2 have the same no. of molecules as 10 cm3 of CO2. 10 cm3 of O2 and 10 cm3 of CO2 consist of the same number of moles.

Calculating number of moles of a substance Amount of an element in mol = Mass of sample (g) Ar b) Amount of a compound in mol Mr

c) Amount of an element / compound in mol = Number of particles 6 × 10 23 Amount of a gas in mol = Volume of gas in dm3 (or cm3) 24 (or 24 000)

Mass of sample (g) = Amount in mol  Ar (or Mr) Mr of molecule = Mass of sample in grams Amount in mol

Questions Determine the mass of 3 moles of ammonia. No. of moles of ammonia = mass of sample (g) Mr of NH3  Mass of ammonia = 3  17 = 51.0 g

0.25 moles of ethene has a mass of 7 g. What is the relative molecular mass of ethene ? Mr of ethene = Mass of sample No. of moles = 7 / 0.25 = 28

Determine the number of moles present in a 100 g sample of calcium oxide. Mr of CaO = 40 + 16 = 56 Amount of CaO in mol = Mass of sample Mr = 100 / 56 = 1.79 mol

Calculate the number of moles of nitrogen gas present in 1500 cm3 sample of the gas. Amount of nitrogen gas in mol = 1500 / 24 000 = 0.0625 mol.

Determine the volume occupied by 0.2 moles of sulphur dioxide gas at r.t.p. Volume of SO2 gas at r.t.p. = Amount of SO2 in mol  Molar Volume = 0.2  24 = 4.80 dm3

What is the number of moles of hydrogen atoms in 54 g of water ? Amount of water in sample in mol = 54 / 18 = 3 mol Ratio of moles of H2O : H : O = 1 : 2 : 1  Amount of hydrogen atoms in mol = 3  2 = 6.00 mol.

Find the mass of 60 dm3 of oxygen gas at r.t.p. Amount of oxygen gas in mol = Volume (dm3) = 60 = 2.5 mol 24 24 Mass of oxygen sample = 2.5  (16  2) = 80.0 g

How many atoms are there in 36 g of carbon ? Amount of carbon atoms in mol = Mass in g = 36 = 3 mol Ar 12 No. of oxygen atoms in sample = 3  6  10 23 = 1.8  10 24 atoms

What is the mass in grams of 5.4  10 24 molecules of chlorine gas ? Amount of chlorine gas in mol = No. of particles = 5.4  10 24 Avogadro’s No. 6  10 23 = 9 mol Mass of chlorine gas = No. of moles  Mr = 9  (35.5  2) = 639 g

What is the volume at r.t.p of 66 g of carbon dioxide ? Amount of carbon dioxide in mol = Mass in g = 66 = 1.5 mol Mr 44 Volume of CO2 gas at r.t.p. = Amount of CO2 in mol  Molar Volume = 1.5  24 = 36.0 dm3

Empirical Formulae It is the simplest formula of the compound. It shows the ratio of atoms of each element present in each molecule. It does not indicate the number of atoms present per molecule. e.g. Ethene has a molecular formula of C2H4. Empirical formula = CH2

Calculating Empirical Formulae To calculate the empirical (simple) formula of a substance, you must know the masses or percentages of the individual elements / ions present in the substance. Two steps are involved : The mass / % of the element / ion divided by the Ar (or Mr). (Amount in mol is determined) Each value obtained is divided by the smallest of the values obtained. (Ratio of amounts in mol is determined)

A hydrocarbon P of mass 28 g consists of 24 g of carbon. Determine the empirical formula of P. The empirical formula is CH2. Element Carbon Hydrogen Mass 24 4 Mass  Ar 24 / 12 = 2 4 / 1 = 4 Divide by smallest no. 2 / 2 = 1 4 / 2

Find the empirical formula of a compound with composition 48.6 % carbon, 43.2 % oxygen and 8.1 % hydrogen by mass. Empirical formula = C3H6O2 Element Carbon Hydrogen Oxygen % 48.6 8.1 43.2 %  Ar 48.6 / 12 = 4.05 8.1 /1 = 8.1 43.2 /16 = 2.7 Divide by smallest no. 4.05 / 2.7 = 1.5 8.1 / 2.7 = 3 2.7 / 2.7 = 1

Molecular Formula It is the exact formula of the compound. It shows the exact number of atoms present per molecule.

Determining Molecular Formulae To determine the molecular formula : the empirical formula must be determined first the relative molecular mass / molar mass must be known Molecular formula = n  Empirical formula  n = relative molecular mass (Mr) relative mass of empirical formula

Question Butene has an empirical formula of CH2. The relative molecular mass of butene is 56. Determine the molecular formula of butene. Relative mass of CH2 = 12 + 2 = 14  n = relative molecular mass (Mr) relative mass of empirical formula = 56 / 14 = 4  Molecular formula = 4  CH2 = C4H8

Calculations from Equations Equations tell us : a) what the reactants and products of a reaction are. the ratio of amount in mol of the reactants to the products. the ratio of volumes of reactants to products in reactions involving gases.

Questions Copper (II) oxide reacts with ammonia gas as follows : 3CuO + 2NH3 3Cu + 3H2O + N2 If 4.0 g of CuO was reacted completely, calculate : i) the amount of CuO in mol that reacted

Amount of CuO in mol = Mass in g Mr = 4.0 = 0.05 mol 64 + 16 The amount of NH3 in mol that reacted with the CuO

Ratio of mol of CuO : NH3 = 3 : 2 Amount of NH3 in mol = 0.05  2 3 = 0.03333 mol The volume of NH3 that reacted = Amount in mol  Molar volume = 0.03333  24 = 0.800 dm3 (3sf)

the volume of nitrogen gas produced. Ratio of mol of CuO : N2 = 3 : 1 Amount of N2 in mol = 0.05  1 3 = 0.01666 moles Volume of N2 = No. of moles  Molar Vol. = 0.01666  24 = 0.400 dm3

the mass of copper produced Ratio of mol of CuO : Cu = 1 : 1 Amount of Cu in mol = 0.05 mol Mass of Cu = Amount in mol  Ar = 0.05  64 = 3.20 g

Alternative method : 3CuO + 2NH3 3Cu + 3H2O + N2 3(64 + 16)g CuO 3(64) g Cu 4.0 g CuO ? Mass of copper formed = 4.0  3(64) 3(80) = 3.20 g

Nitrogen combines with hydrogen to form ammonia as shown below : N2 + 3H2 2NH3 30 cm3 of nitrogen gas is reacted with 60 cm3 of hydrogen gas at r.t.p. What is the volume of ammonia gas formed ? What is the total volume of gaseous products formed ?

Ratio of mol of N2 : H2 = 1 : 3 Therefore, 20 cm3 of nitrogen reacts with 60 cm3 of hydrogen gas. The hydrogen gas is the limiting reagent & the nitrogen gas is in excess. Ratio of mol of H2 : NH3 = 3 : 2 Volume of NH3 gas formed = 2  60 3 = 40.0 cm3

Volume of excess nitrogen = 30 – 20 = 10.0 cm3 Total volume of gaseous products = 10 + 40 = 50.0 cm3

Calculating Percentage Yield in a Reaction During experiments, the amount of chemicals produced is always less than the maximum amount you would expect. This is expressed as the percentage yield. Percentage yield of substance = Experimental mass of substance  100 % Maximum possible mass of product

Questions Zinc reacts with sulphur according to the equation : Zn + S ZnS 6.5 g of zinc was reacted with sulphur to make zinc sulphide. 9.0 g of zinc sulphide was obtained. Calculate the % yield.

Amount of Zn in mol = Mass = 6.5 Ar 65 = 0.1 mol Amount of ZnS in mol = 0.1 mol Expected mass of ZnS = Amount in mol  Mr = 0.1  (65 + 32) = 9.70 g

% yield of ZnS = 9.0  100 % = 92.8 % 9.7

Calculating Percentage Purity Very often, the reactants used in the reaction may not be pure. As such, the actual mass of the reactant is less than that reported. As a result, the yield of products will be less than the calculated value. Thus, to determine the percentage purity of the reactants, we work out the actual mass of the reactant based on the mass of the product formed.

Thus, % purity is calculated as follows : % purity of substance Mass of pure substance present  100% = Reported mass of sample

Questions An impure sample of calcium carbonate contains calcium sulphate as an impurity. When excess hydrochloric acid was added to 6 g of the sample, 1200 cm3 of gas were produced at r.t.p. Calculate the % purity of the calcium carbonate sample. Mass of impure calcium carbonate = 6 g Volume of CO2 = 1200 cm3

Amount of CO2 in mol = 1200 = 0.05 mol 24 000 CaCO3 + 2HCl CaCl2 + H2O + CO2 Ratio of mol of CaCO3 : CO2 = 1 : 1  Amount of CaCO3 in mol actually present = 0.0500 mol

Mass of CaCO3 actually present in sample = Amount in mol  Mr = 0.05  100 = 5.00 g % purity of CaCO3 = 5.0  100 % 6.0 = 83.3 %

Concentration of Solutions The concentration of a solution refers to the amount of solute (either in grams or in moles) dissolved in a fixed volume of the solvent. Concentration can be expressed in : grams per cm3 / grams per dm3 moles per dm3

Relevant Formulae : a) Concentration = Mass (Units : g/dm3) Volume b) Concentration = Amount in mol (Units : mol/dm3) Concentration = Concentration  Mr (in g/dm3) (in mol/dm3)

Concentration = Concentration (in g/dm3) (in mol/dm3) Ar or Mr Amount in mol = Concentration  Volume (in mol/dm3) (in dm3) f) Mass of sample = Concentration  Volume in a fixed (in g/dm3) (in dm3) volume, V

Questions A student dissolved 5.3 g of sodium nitrate in 50 cm3 of water. Determine : the concentration of the solution in g/cm3 the concentration of the solution in g/dm3 the concentration of the solution in mol/dm3 Concentration of solution = Mass (g) Volume (cm3)

= 5.3 = 0.106 g/cm3 50 ii) Concentration of solution = Mass (g) Volume (dm3) = 5.3 = 106 g/dm3 (50  1000)

iii) Concentration of solution = Amount in mol Volume (dm3) = (5.3 ÷ 85) = 1.25 mol/dm3 (50 ÷ 1000) OR Concentration of solution (in mol/dm3) = Concentration (in g/dm3) Mr = 106 ÷ 85 = 1.25 mol/dm3

You are provided with a solution of sodium chloride of concentration 0.8 mol/dm3. Calculate the concentration in g/dm3. Concentration of solution (in g/dm3) = Concentration (in mol/dm3)  Mr = 0.8  (23 + 35.5) = 46.8 g/dm3

Calculate the number of moles of calcium chloride present in 50 cm3 of 0.5 mol/dm3 solution of calcium chloride. Amount of CaCl2 in mol = Conc. (mol/dm3)  Vol. (dm3) = 0.5  (50 ÷ 1000) = 0.0250 moles. 1000 cm3 0.5 mol 50 cm3 ?

What mass of sodium hydroxide must be dissolved in 200 cm3 of water to prepare a 5.0 mol/dm3 solution? Mr of NaOH = 23 + 16 + 1 = 40 Amount of NaOH in mol that is reqd = Conc. (mol/dm3)  Vol. (dm3) = 5.0  (200 ÷ 1000) = 1.0 mol

 Mass of NaOH reqd = Amount in mol  Mr = 1.0  40 = 40 g OR 1000 cm3 5.0  40 g 200 cm3 ? Mass of NaOH reqd = 200  5.0  40 1000 = 40.0 g

Express the concentration of a 0.064 mol/dm3 solution of sodium hydroxide in g/dm3. Mr of NaOH = 23 + 16 + 1 = 40  Concentration of NaOH in g/dm3 = Conc. (mol/dm3) × Mr = 0.064 × 40 = 2.56 g/dm3.

Express the concentration of a 20 g/dm3 solution of sodium hydroxide in mol/dm3. Concentration of NaOH in mol/dm3 = Conc. (g/dm3) ÷ Mr = 20 ÷ 40 = 0.500 mol/dm3

Volumetric Analysis A method of determining the quantity of a substance present in a solid or in a solution using a technique called titration.

Terminology Standard solution : A solution of known concentration. Molar solution : A solution that contains one mole of a substance in one cubic decimeter of solution.

Formulae 1. Concentration = Concentration  Mr (in g/dm3) (in mol/dm3) 2. Amount in mol = Concentration  Volume (in mol/dm3) (in dm3) When diluting a solution, the no. of moles of the substance is the same before & after dilution. M1V1 = M2V2

In titration, M1V1 = a M2V2 b where M represents the concentration of the solutions in mol/dm3 & V represents the volumes of solution in cm3 & a and b represent the mole ratio of substances 1 and 2

Questions 200 cm3 of water was added to 300 cm3 of 0.1 mol/dm3 hydrochloric acid. Calculate the new concentration of the solution in mol/dm3. M1V1 = M2V2 0.1 × 300 = M2 × 500 M2 = 0.1 × 300 = 0.0600 mol/dm3 500

20 cm3 of 0.100 mol/dm3 hydrochloric acid was used to neutralise 15.6 cm3 of calcium hydroxide solution. Determine the concentration of the alkali in mol/dm3. 2HCl + Ca(OH)2 CaCl2 + 2H2O M1 = 0.1 M2 = ? V1 = 20 V2 = 15.6 a = 2 b = 1 0.1 × 20 = 2 M2 × 15.6

M2 = 0.1 × 20 = 0.0640 mol/dm3 15.6 × 2

Choice of indicators for VA Reaction Indicator Strong acid – strong Litmus alkali Methyl orange Phenolphthalein 2. Weak acid – strong Phenolphthalein alkali Litmus Strong acid – weak Methyl orange Strong acid – soluble Methyl orange carbonate