Chapter 9: Chemical Calculations

download report

Transcript Chapter 9: Chemical Calculations

Chemical Calculations
What does an equation tell us?
A balanced chemical equation shows important
facts about a reaction:
Known as mole ratio
a) The reactants
b) The products
c) The ratio of the amounts (in moles) of the
reactants and the products
d) The state of each reactants/products if indicated
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Stoichiometry
It is the relationship between the amounts
(measured in moles) of reactants and
products involved in a chemical reaction.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Stoichiometry
Mg(s) +
2HCl(aq)
1 mol of
magnesium
2 mol of
hydrochloric
acid

MgCl2(aq) + H2(g)
1 mol of
magnesium
chloride
1 mol of
hydrogen
1 mole of a substance is equal to its Mr in grams. Hence,
24 g of
magnesium
73 g of
hydrochloric
acid
95 g of
magnesium
chloride
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
2 g of
hydrogen
Example 1 Calculate the mass of the solid
obtained when 16.8 g of sodium hydrogencarbonate
is heated strongly until there is no further change.
2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(l)
Mr of NaHCO3 = 23 + 1 + 12 + (3 × 16) = 84
Mr of Na2CO3 = (2 × 23) + 12 + (3 × 16) = 106
168
Number of moles of NaHCO3 used = 84 = 0.2 mol
2 mol of NaHCO3 produce 1 mol of Na2CO3.
1
Number of moles of Na2CO3 obtained = 0.2 ×
= 0.1 mol
2
 Mass of solid obtained
= number of moles of Na2CO3 × Mr of Na2CO3
= 0.1 × 106 = 10.6 g
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 63.0 dm3 of carbon monoxide,
measured at r.t.p., was used to react with
iron(II) oxide. What mass of iron was produced
at the end of the reaction? [Ar of Fe = 56]
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
Number of moles of carbon monoxide used
= 63/24 = 2.625 mol
3 mol of carbon monoxide produce 2 mol of iron.
Number of moles of iron produced in the reaction
= 2.625  2/3 = 1.75 mol
 Mass of iron produced
= number of moles of Fe × Ar of Fe = 1.75 × 56 = 98.0 g
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we do calculations involving
gases?
In chapter 8, we learnt that 1 mole of any gas
occupies 24 dm3 at room temperature and
pressure.
Hence, the volume of gas is proportional
to the number of moles of the gas, and
vice versa.
Thus, we can change mole of gas to
volume of gas.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we do calculations involving
gases?
N2(g) + 3H2(g)
OR

2NH3(g)
24 dm3 of N2(g) + 72 dm3 H2(g)  48 dm3 NH3(g)
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1
N2(g) + 3H2(g)  2NH3(g)
a) Calculate the volume of ammonia gas produced when 20 cm3
of nitrogen gas was reacted with an excess of hydrogen gas.
b) What is the volume of hydrogen required for this reaction?
a) 1 mol of nitrogen produces 2 mol of ammonia,
i.e. 1 volume of nitrogen produces 2 volumes of ammonia.
 Volume of ammonia produced = 2 × 20 = 40 cm3
b) Also, 1 volume of nitrogen reacts with 3 volumes of
hydrogen.
 Volume of hydrogen required
= 3 × 20 = 60 cm3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2
At room temperature and pressure, 25 cm3 of butane (C4H10)
exploded after it was mixed with an excess of oxygen.
Calculate (a) the volume of carbon dioxide produced, and (b) the
volume of oxygen required for the reaction.
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(l)
a) According to the equation,
2 volumes of butane produce 8 volumes of carbon dioxide.
 Volume of carbon dioxide produced = 8 × 25 = 100 cm3
2
b) 2 volumes of butane react with 13 volumes of oxygen.
13
 Volume of oxygen required =
× 25 = 162.5 cm3
2
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Limiting Reactants
Ideally, reactions should be carried out using exact
quantities of reactants to reduce wastage.
However, many reactions are carried out using an excess
amount of one reactant. Why?
This ensures that the more expensive reactant is
completely used up.
To do so, we make use of the idea of limiting reactants.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Limiting Reactants
H2(g) + Cl2(g)  2HCl(g)
1 mole of hydrogen
reacts with
to produce
2 moles of hydrogen chloride
1 mole of chlorine
The reactants are in stoichiometric proportion.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
All reactants are
used up as they
reacted in
stoichiometric
proportion.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
reactants
product
The reactants that are not
used up are called the
excess reactants.
The reactant that is completely used up in a reaction is known
as the limiting reactant. In this case, it is chlorine.
It is called the limiting reactant because it determines or
limits the amount of products formed.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
reactants
product
Here, hydrogen is the limiting reactant. Chlorine is
the excess reactant.
The amount of products formed in a reaction is always
determined by the amount of the limiting reactant.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1 Zinc reacts with hydrochloric acid according to the
equation Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g).
If 0.05 mol of zinc was added to 0.075 mol of hydrochloric acid,
a) identify the limiting reactant.
b) calculate the amount (in moles) of the reactant that remained.
a) 2 mol of HCl will react with 1 mol of Zn, hence
0.10 mol of HCl will react with 0.05 mol of Zn.
0.075 mol of HCl. Hence, amount of zinc used
= (0.075/2) = 0.0375 mol of Zn.
This is
simple
proportion
Since 0.05 mol of Zn were used but only 0.0375 mol
used, the zinc must be in excess and HCl is the limiting
reactant.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
(continued)
Example 1 Zinc reacts with hydrochloric acid according to the
equation Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g).
If 0.05 mol of zinc was added to 0.075 mol of hydrochloric acid,
a) identify the limiting reactant. (done)
b) calculate the amount (in moles) of the reactant that remained.
b) Amount of zinc which remained unreacted
= 0.05 – 0.0375
= 0.0125 mol
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 Ethene, C2H4, burns in oxygen as shown in the equation.
In an experiment, 10 cm3 of ethene was burnt in 50 cm3 of oxygen.
a) Which gas was supplied in excess? Calculate the volume of the excess
gas remaining at the end of the reaction.
b) Calculate the volume of carbon dioxide produced.
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
a) 1 volume of ethene reacts with 3 volumes of oxygen.
 Volume of oxygen used = 3 × 10 = 30 cm3
However, 50 cm3 of oxygen was used.
Oxygen gas was in excess.
Hence, volume of O2 remaining
= initial volume – volume used = 50 – 30 = 20 cm3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 (continued)
Ethene, C2H4, burns in oxygen as shown in the equation. In an
experiment, 10 cm3 of ethene was burnt in 50 cm3 of oxygen.
a) Which gas was supplied in excess? Calculate the volume of the
excess gas remaining at the end of the reaction. (done)
b) Calculate the volume of carbon dioxide produced.
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
b) 1 volume of ethene produces 2 volumes of
carbon dioxide.
 Volume of carbon dioxide produced
= 2 × 10 = 20 cm3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Why is it important to identify the
limiting reactant?
Chemists choose the most expensive reactant to be
the limiting reactant.
This ensures all the expensive reactant is used up
and not wasted.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
The Concentration of a Solution
Chemical reactions often involve solutions of
substances in water. It is therefore important to
know the concentration of a solution.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we determine the
concentration of a solution?
Concentration (g/dm3) =
Mass of solute in grams
Volume of solvent in dm3
1 dm3 is equivalent to 1000 cm3.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1
A solution of glucose contains 0.45 g of glucose in
75 cm3 of solution. What is the concentration of the
glucose solution in g/dm3?
Volume of glucose solution
75
=
dm3 = 0.075 dm3
1000
Concentration of glucose solution
mass
=
volume
= 0.45 / 0.075
= 6.00 g/dm3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Molar Concentration
The concentration of a solution can also be expressed in
mol/dm3.
When concentration is expressed in mol/dm3, it is called
molar concentration.
Concentration (mol/dm3)
=
Number of moles of solute
Volume of solvent in dm3
or
Concentration
(mol/dm3)
=
Concentration (g/dm3)
Mr
1 mol/dm3 means
1 mole of solute dissolved in 1 dm3 of solution.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1 A solution of sodium hydroxide was prepared by
dissolving 3.5 g of sodium hydroxide in distilled water and making
the volume up to 100 cm³. What is the concentration of the sodium
hydroxide solution in mol/dm³?
Mr of sodium hydroxide (NaOH) = 23 + 16 + 1 = 40
Number of moles of NaOH used
= mass / Mr = 3.5 / 40 = 0.0875 mol
Volume of NaOH = 100 cm3 = 100 / 1000 = 0.10 dm3
Concentration of NaOH solution in mol/dm3
= number of moles of NaOH / volume of solution (dm3)
= 0.0875 / 0.10
= 0.875 mol/dm3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 The concentration of a bottle of hydrochloric acid
is 73 g/dm3. How many moles of HCl are there in 250 cm3 of
this acid?
Mr of HCl = 1 + 35.5 = 36.5
Moles of HCl in 73 g = 73 / 36.5 = 2 mol
 1000 cm3 of the HCl solution contains
2 mol of HCl.
How many moles
250 cm3 will contain
(250 / 1000)  2 = 0.5 mol of HCl
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
are there in 250 cm3
since there are 2 mol
in 1000 cm3?
Example 3 The concentration of a solution of HCl is
30 g/dm3. What is its concentration in mol/dm3?
Concentration of HCl in mol/dm3
= (concentration in g/dm3) / Mr of HCl
=
30
1  35.5
= 0.822 mol/dm3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Volumetric Analysis
Many products that we use contain chemicals
dissolved in water. To check the concentration of
these substances, a chemist performs volumetric
analysis.
To do volumetric analysis, we use a method
called titration.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Titration - General Steps
burette
conical flask
In titration, we determine the volume of a
solution required to completely react
with a known volume of another
solution.
By calculation, we can then determine
the concentration of a solution.
solution of
unknown
concentration
Titration experiments that involve the use of an acid, e.g. HCl, and a
base, e.g. NaOH, are called an acid-base titration.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we perform titration?
The titrant is placed in the burette.
The solution of unknown
concentration is introduced into a
conical flask using a pipette.
One or two drops of indicator are
then added to the conical flask.
The titrant is allowed to react
completely with a known volume
(usually 25 cm3) of the solution of
unknown concentration.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we perform titration?
Titration is stopped at the end-point.
The end-point is reached when the indicator permanently
changes colour.
The volume of titrant used is noted.
The concentration of the unknown solution can then be calculated.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1 25 cm3 of the ammonia solution required 21.9 cm3 of
0.11 mol/dm3 sulphuric acid to achieve the end-point of titration.
Calculate the concentration of ammonia, in g/dm3.
H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
Number of moles of sulphuric acid used in 25 cm3
= 0.11 × 21.9 / 1000 = 2.409 × 10–3 mol
1 mol of sulphuric acid reacts with 2 mol of ammonia solution.
 Number of moles of ammonia in 25 cm3
= 2 × 2.409 × 10–3 = 4.818 × 10–3 mol
The ratio is 1:2. Hence the number of moles of
ammonia is always twice in this reaction.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1 (continued)
25 cm3 of the ammonia solution required 21.9 cm3 of 0.11 mol/dm3
sulphuric acid to achieve the end-point of titration. Calculate the
concentration of ammonia, in g/dm3.
There are 4.818  10-3 mol in 25 cm3.
There are 0.1927 mol in
1000 cm3.
Conc = 0.1927 mol/dm3
 Number of moles of ammonia in 1000 cm3
= 4.818 × 10–3 × 1000/25 = 0.1927 mol
Mr of NH3 = 14 + (3 × 1) = 17
 Concentration of ammonia in g/dm3
= 0.1927 × 17
= 3.28 g/dm3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Conc (in mol/dm3)
= conc (in g/dm3) / Mr
Example 2
16.6 g of a metal carbonate, M2CO3, was made up to 1000 cm3
of aqueous solution. 25 cm3 of this solution required 30 cm3 of
0.20 mol/dm3 HCl for complete reaction.
a) Calculate the number of moles of HCl used in this reaction.
b) Write the equation for the reaction between M2CO3 and HCl.
c) Calculate the number of moles of M2CO3 present in
i) 25 cm3 of solution.
ii) 1 dm3 of solution.
d) Calculate
i) the relative atomic mass of M.
ii) the relative formula mass (Mr) of M2CO3.
e) Identify the metal M.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 (continued)
16.6 g of a metal carbonate, M2CO3, was made up to 1000 cm3
of aqueous solution. 25 cm3 of this solution required 30 cm3 of
0.20 mol/dm3 HCl for complete reaction.
a) Calculate the number of moles of HCl used in this reaction.
b) Write the equation for the reaction between M2CO3 and HCl.
a) Number of moles of HCl used
= 0.2 × 30 = 0.006 mol
1000
b) The equation for the reaction is
M2CO3(aq) + 2HCl(aq)  2MCl(aq) + H2O(l) + CO2(g)
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 (continued)
16.6 g of a metal carbonate, M2CO3, was made up to 1000 cm3
of aqueous solution. 25 cm3 of this solution required 30 cm3 of
0.20 mol/dm3 HCl for complete reaction.
c) Calculate the number of moles of M2CO3 present in
i) 25 cm3 of solution.
ii) 1 dm3 of solution.
M2CO3(aq) + 2HCl(aq)  2MCl(aq) + H2O(l) + CO2(g)
c) i)1 mol of M2CO3 reacts with 2 mol of HCl.
Since the number of moles of HCl used was 0.006,
the number of moles of M2CO3 in 25 cm3 = 0.003 mol
1000
ii)  There is 0.003 × 25 in 1000 cm3 (1 dm3)
= 0.12 mol of M2CO3 in 1 dm3 of the solution.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 (continued)
d) Calculate
i) the relative atomic mass of M.
ii) the relative formula mass (Mr) of M2CO3.
e) Identify the metal M.
d) i) Let Ar of M be Q.
The relative formula mass of M2CO3
= (2 × Q) + (1 × 12) + (3 × 16) = 2Q + 60
Number of moles of M2CO3 in 16.6 g = 16.62Q + 60
 From (c)(ii),
0.12 = 16.62Q + 60
0.24Q + 7.2 = 16.6
Q = 39.2
ii) Relative formula mass = 2 × 39.2 + 60 = 138.4
e) The Ar of M is 39.2. M is potassium
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.