- Relative Formula Mass (Mr) is the sum of the Relative Atomic Mass (Ar) in a chemical formula
Calculation Examples
E.g. 1: Find the Mr of Mg(NO3)2 E.g. 2: FInd the Mr of MgCl2
Mg = 24 Mg = 24
N x 2 = 14 x 2 =28 Cl x 2 = 35.5 x 2 = 71
O x 6 = 16 x 6 = 96 MgCl2 = 24 + 71 = 95
Mg(NO3)2 = 24 + 28 + 96 = 148 Mr = 95
Mr = 148
Tip: Do not forget to multiply out the brackets when calculating.
1.16 Understand the use of the term mole to represent the amount of substance
1.17 UNderstand the term mole as the Avogradro number of particles (atoms, molecules, formulae, ions or electrons)in a substance
Mole: A quantity or amount of particles (6 x 10^23) or Avogradro number of particles which is constant for any substance (atoms, molecules, formulae, ions or electrons).
1.18 Carry Out mole calculations using relative atomic mass and relative formula mass
A very important formula that you must remember and tattooed on to your soul…
Moles = Mass / Molar Mass (Relative formula mass) or in symbol form: n = M / Mr
E.g. Calculate the mass of Na2CO3 produced from 4.2g of NaHCO3.
2NaHCO3 --> Na2CO3 + H2O + CO2
n of NaHCO3 = m / Mr = 4.2 / (23+1+12+48) = 4.2 / 84 = 0.05 moles
2NaHCO3 --> 1Na2CO3
therefore… nNa2CO3 = 0.05 x0.5 = 0.025
Mass of Na2CO3 = n x Mr = 0.025 x (46+12+48) = 0.025 x 106 = 2.65g
Moles = Mass / Molar Mass (Relative formula mass) or in symbol form: n = M / Mr
E.g. Calculate the mass of Na2CO3 produced from 4.2g of NaHCO3.
2NaHCO3 --> Na2CO3 + H2O + CO2
n of NaHCO3 = m / Mr = 4.2 / (23+1+12+48) = 4.2 / 84 = 0.05 moles
2NaHCO3 --> 1Na2CO3
therefore… nNa2CO3 = 0.05 x0.5 = 0.025
Mass of Na2CO3 = n x Mr = 0.025 x (46+12+48) = 0.025 x 106 = 2.65g
4.12 Calculate Molar Enthalpy Change From Heat Energy Change
Another formula to remember… but not as important as the previous one…
Molar Enthalpy Change (KJ/mol) = Enthalpy Change (KJ) / Moles (mol)
E.g. A chemical reaction releases 25.2 KJ of heat energy when 0.25 moles of reagent react. Calculate the molar enthalpy change.
Enthalpy change: -25.2KJ (negative because heat energy is released)
Moles: 0.25
therefore… -25.2 / 0.25 = -100.8 KJ/mol
Molar Enthalpy Change (KJ/mol) = Enthalpy Change (KJ) / Moles (mol)
E.g. A chemical reaction releases 25.2 KJ of heat energy when 0.25 moles of reagent react. Calculate the molar enthalpy change.
Enthalpy change: -25.2KJ (negative because heat energy is released)
Moles: 0.25
therefore… -25.2 / 0.25 = -100.8 KJ/mol
1.23 Calculate Empirical and Molecular Formulae from experimental Data
E.g. a) A hydrocarbon contains 7.7% hydrogen and 92.3% carbon. Find its empirical formula.
Hydrogen: Carbon
7.7/1 : 9.23/12 (Divide percentage/mass by relative atomic mass)
7.7 : 7.7
1 : 1 (Divide both sides of the ratio by the smallest number, 7.7 in this case.)
therefore... empirical formula of this compound is CH.
b) The formula mass of the hydrocarbon is 78. Find its molecular formula.
12 + 1 = 13 (Find the relative molecular formula of the empirical compound)
78/13 = 6 (Divide the molecular formula with the number obtained above)
hence….answer is: C6H6
Hydrogen: Carbon
7.7/1 : 9.23/12 (Divide percentage/mass by relative atomic mass)
7.7 : 7.7
1 : 1 (Divide both sides of the ratio by the smallest number, 7.7 in this case.)
therefore... empirical formula of this compound is CH.
b) The formula mass of the hydrocarbon is 78. Find its molecular formula.
12 + 1 = 13 (Find the relative molecular formula of the empirical compound)
78/13 = 6 (Divide the molecular formula with the number obtained above)
hence….answer is: C6H6
1.25 Calculate Percentage Yield
A third formula to remember…..
% yield = Actual mass obtained / Theoretical mass obtained x 100%
E.g. When 1.24g of CuCO3 is fully decomposed it produces only 0.72g of CuO. Calculate the % yield of this experiment. CuCO3 --> CuO + CO2
0.72g = actual mass obtained
n of CuCO3 = m / Mr = 1.24 / (64+12+48) = 1.24/124 = 0.01 moles
1CuCO3 --> 1CuO
Mass of CuO = n x Mr = 0.01 x (64+16) = 0.8g (theoretical mass obtained)
% yield: 0.72/0.8 x 100% = 90%
% yield = Actual mass obtained / Theoretical mass obtained x 100%
E.g. When 1.24g of CuCO3 is fully decomposed it produces only 0.72g of CuO. Calculate the % yield of this experiment. CuCO3 --> CuO + CO2
0.72g = actual mass obtained
n of CuCO3 = m / Mr = 1.24 / (64+12+48) = 1.24/124 = 0.01 moles
1CuCO3 --> 1CuO
Mass of CuO = n x Mr = 0.01 x (64+16) = 0.8g (theoretical mass obtained)
% yield: 0.72/0.8 x 100% = 90%
1.19 Understand the term molar volume of a gas and use its values at room temperature and pressure in calculations
The volume of any gas at room temperature is 24dm^3. Remember the following formula.
Mole = Volume (dm^3) / 24 or n = v / 24
E.g. What volume of oxygen is required for the complete combustion of 11.5g of sodium?
4Na(s) + O2(g) --> 2Na2O(s)
n of Na = m / Mr = 11.5 / 23 = 0.5 mole
4Na --> 1O2
0.5/4 = 0.125 mole
0.125 = v / 24
V = 24 x 0.125
V = 3dm^3
Mole = Volume (dm^3) / 24 or n = v / 24
E.g. What volume of oxygen is required for the complete combustion of 11.5g of sodium?
4Na(s) + O2(g) --> 2Na2O(s)
n of Na = m / Mr = 11.5 / 23 = 0.5 mole
4Na --> 1O2
0.5/4 = 0.125 mole
0.125 = v / 24
V = 24 x 0.125
V = 3dm^3
1.26 Carry out mole calculations using volumes and molar concentrations
4.9 describe experiment to carry out acid-alkali titration
The last formula to remember for this topic…
Mole (mol) = Concentration (mol/dm^3) x Volume (dm^3) or n = cv
E.g. If 25cm^3 of NaOH can be neutralized by 17.5cm^3 of 0.5mol/dm^3 HCl, what is the concentration of NaOH? HCl + NaOH --> NaCl + H2O
n of HCl = cv = 0.0175 x 0.5 = 0.00875 mole
1NaOH --> 1HCl
0.00875 = c 0.025
c = 0.35 mol/dm^3
Remember that the units of volume in all calculations must be in dm^3 not cm^3.
Conversion: 1000cm^3 = 1dm^3
Titration
If the average volume was 25.1cm^3 then…
Inaccuracies
Mole (mol) = Concentration (mol/dm^3) x Volume (dm^3) or n = cv
E.g. If 25cm^3 of NaOH can be neutralized by 17.5cm^3 of 0.5mol/dm^3 HCl, what is the concentration of NaOH? HCl + NaOH --> NaCl + H2O
n of HCl = cv = 0.0175 x 0.5 = 0.00875 mole
1NaOH --> 1HCl
0.00875 = c 0.025
c = 0.35 mol/dm^3
Remember that the units of volume in all calculations must be in dm^3 not cm^3.
Conversion: 1000cm^3 = 1dm^3
Titration
- Carefully use pipette and filler to measure 25cm^3 of NaOH of unknown concentration into a conical flask
- Add 2 or 3 drops of phenolphthalein indicator and swirl
- Prepare the 0.5 mol/dm^3 HCl in the curette
- Take initial reading on the burette
- Put the conical flash under the burette on a white tile
- Carefully open the tap on the burette and mix the acid into the alkali whilst swirling the solution
- When the indicator changes color, stop adding the acid and note down the final burette reading.
- Calculate the amount of HCl required to neutralize 25cm^3 of NaOH.
- Repeat the experiment for more reliable results and find an average.
If the average volume was 25.1cm^3 then…
- NaOH + HCl --> NaCl + H2O
- n of HCl = cv = 0.5 x 0.0251 = 0.01255
- 1HCl --> 1NaOH
- c of NaOH = n / v = 0.01255 / 0.025 = 0.502 = 0.50 mol/dm^3
Inaccuracies
- Colour change is subject
- Too much phenolphthalein was used
- Concentration of acid only given to 1 d.p.