Transcript Document
West Midlands Chemistry
Teachers Centre
Tuesday 23rd November 2010
Mole Calculations
Presenter: Dr Janice Perkins
Reacting ratio in equations
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
1 ‘formula’ 3 molecules
10 ‘formulae’ 30 molecules
1106
‘formulae’
1dozen
‘formulae’
3106
molecules
3 dozen
molecules
2 atoms 3 molecules
20 atoms 30 molecules
2106
atoms
3106
molecules
2 dozen
atoms
3 dozen
molecules
Funny numbers
Dozen = 12
Gross = 12 12 = 144
Score = 20
Mole = 6.023 1023
That’s
602300000000000000000000
The Avogadro Constant (L)
602300000000000000000000
Or 6.02 1023
It is just a number – no more special
than a ton, a score or a dozen – its
just a bit bigger!
Reacting Ratio
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
1 formula
3 molecules
10 ‘formulae’ 30 molecules
2 atoms 3 molecules
20 atoms 30 molecules
1106
‘formulae’
3106
molecules
2106
atoms
3106
molecules
6.021023
‘formulae’
18.06 1023
molecules
12.04 1023
atoms
18.06 1023
molecules
3 moles
2 moles
3 moles
1 mole
Mole Ratio (Reacting Ratio)
1:2
3:2
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
1:3
3:3
Fe2O3 : CO = 1:3
CO : CO2 = 1:1
Fe2O3 : Fe = 1:2
CO : Fe
= 3:2
Sufficient data to calculate moles?
Calculation
based on
Mass
Solution
Gases
Data needed
Mass and Mr or Formula or Name
Volume
and
Concentration
P and T and V and R
Mass
Mole Calculations
n= m
Mr
Volume of
solution
n=vc
Volume of
gas
Moles of
known
substance
Mole
ratio
from
equation
Mass
Mr
Moles of
unknown
substance
n = pV
RT
Ideal Gas
Moles
Mass
Equation
volume
= P(in
Pa)conc
V(in m3)
pV = nRT
RMr T(in Kelvin)
Rearranging the formula
Moles = Mass
Mr
Mr = Mass
Moles
Mass = Moles Mr
Mass
Moles
Mr
Mass
Mole Calculations
n= m
Mr
Volume of
solution
n=vc
Volume of
gas
n = pV
RT
Moles of
known
substance
Mole
ratio
from
equation
Mass
Mr
Moles of
unknown
substance
m = n Mr
Mr = m
n
Vol
Conc
Rearranging the formula
Moles = Volume Concentration
Vol = Moles
Conc
Conc = Moles
Vol
Moles
Vol
Conc
Mass
Mole Calculations
n= m
Mr
Volume of
solution
n=vc
Volume of
gas
n = pV
RT
Moles of
known
substance
Mole
ratio
from
equation
Mass
Mr
Moles of
unknown
substance
m = n Mr
Mr = m
n
Vol
Conc
V
P
T
v= n
c
c = n
v
Rearranging the formula
Moles = pV
RT
Volume = nRT
p
Pressure = nRT
V
Temperature = pV
nR
Units are vital:
‘V’ always in m3
‘P’ always in Pa
‘T’ always in Kelvin
Mass
Mole Calculations
n= m
Mr
Volume of
solution
n=vc
Volume of
gas
n = pV
RT
Moles of
known
substance
Mole
ratio
from
equation
Mass
Mr
Moles of
unknown
substance
m = n Mr
Mr = m
n
v= n
Vol
c
Conc c = nv
V V = nRT
p
P
T
p = nRT
V
T = pV
nR
Example 1:
Calculating moles from masses
Calculate the number of moles in 300g of CaCO3
Use the equation Mass = Mr x moles
Mr = 100
Rearrange equation
Moles = mass/Mr
= 300
100
= 3 moles
Example 2:
Calculating concentration of solution
Calculate the concentration, in mol dm-3, of the solution
formed when 19.6 g of hydrogen chloride, HCl, are
dissolved in water and the volume made up to 250 cm3.
First we need to calculate the number of moles of HCl.
Moles HCl
Moles = mass/Mr
= 19.6
= 0.537 mol
36.5
Conc of HCl(aq) = moles
= 0.537
volume (in dm3) 250 10-3
= 2.15 mol dm-3
Example 3: Calculating moles of solution and solid,
then Mr and Ar (Jan 09 chem1)
A metal carbonate MCO3 reacts with HCl as in the following
equation.
MCO3 + 2HCl MCl2 + H2O + CO2
A 0.548 g sample of MCO3 reacted completely with
30.7 cm3 of 0.424 mol dm-3 HCl.
(a) Calculate the amount, in moles, of HCl which reacted
with the 0.548 g of MCO3
Moles = vol x conc (in dm3)
= 30.7/1000 x 0.424
= 0.0130 mol
(b) Calculate the amount, in moles, of MCO3 in 0.548 g
Look at equation again
MCO3 + 2HCl MCl2 + H2O + CO2
There is a 2:1 ratio of reactants.
We have calculated that there are 0.0130 mol of HCl
so there must be half that amount of MCO3
= 0.0130/2
= 0.0065 mol = 6.50 x 10-3 mol
(c) Calculate the Mr of MCO3
A mass of 0.548g of MCO3 contains 0.0065 mol.
Use the equation Mr = mass/moles
Mr = 0.548/0.0065 = 84.3
(d) Use your answer to deduce the Ar of M
We have just calculated the Mr of MCO3 to be 84.3
Since there is 1xC and 3xO this makes 12 + 48 = 60.
So this means M must have an Ar of 84.3 – 60 = 24.3
This is Magnesium.
Example 4: Identity unknown – less structured calculation
The
carbonate of
of metal
metal M
M has
has the
the
formula
M
CO
The
The carbonate
the formula
formula M
M222CO
CO333... The
The
equation
for the
the reaction
reaction of
of this
this carbonate
carbonate
with
hydrochloric
equation for
carbonate with
with hydrochloric
hydrochloric
acid
acid is
is given
given below.
below.
M
M2CO
CO3 +
+ 2HCl
2HCl
2MCl
2MCl +
+ CO
CO2 +
+ H
H2O
O
2
3
2
2
A
sample of
of M
M2CO
, of mass 0.394 g, required the addition
A sample
2CO3
3, of mass 0.394 g, required the addition
33 of 0.263 mol dm-3-3 solution of hydrochloric acid
of
21.7
cm
of 21.7 cm of 0.263 mol dm solution of hydrochloric acid
for complete reaction.
Calculate the Ar of metal M and deduce its identity.
(a) Find moles of known substance - in this case HCl
Moles HCl(aq) = volume concentration
= 21.7 10-3 0.263
= 5.71 10-3 mol
M2CO3 + 2HCl 2MCl + CO2 + H2O
(b) Calculate the moles of the other substance- in this case M2CO3
The mole ratio is 2:1
We have calculated the moles HCl = 5.71 10-3 mol
So the moles M2CO3 = 5.71 10-3 /2 = 2.85 x 10-3 mol
(c) Now find Mr of M2CO3
Mr of M2CO3 =
=
mass
moles
138
=
0.394
2.85 10-3
(d) find Ar of metal M and hence deduce its identity
2 Ar(M)
Ar(M)
M
=
=
=
=
Mr(M2CO3) - Mr(‘CO3’)
138 – 60 = 78
78/2 =
39
Potassium
Example 5:
Gases – calculating the volume
A sample of ethanol vapour, C2H5OH (Mrr == 46), was
maintained at a pressure of 100 kPa
kPa and at a temperature of
K. Use
Use the
the ideal
ideal gas
gas equation
equation to
to calculate the volume, in
366 K.
cm33, that 1.36 g of ethanol vapour would occupy under these
R == 8.31
8.31 JJ K
K-1-1 mol
mol-1-1))
conditions. (The gas constant R
Moles ethanol = 1.36
46
= n = 0.0296 mol
Use pV
pV == nRT
nRT
Use
V == nRT
nRT ==
0.0296 8.31
8.31 366
366
V
0.0296
100000
pp
100000
8.992 10
10-4-4 m
m33
== 8.992
== 8.992 10-4 106 = 899 cm3
Example 6:
Empirical formula
Jun 09
An oxide of nitrogen contains 30.4% by mass of nitrogen,
Calculate the empirical formula.
First calculate the % of O
100 – 30.4 = 69.6%
Now put in columns
N
O
mass
30.4
69.6
Ar
14
16
= moles
2.17
4.35
Ratio
2.17/2.17 = 1
4.35/2.17= 2
Empirical formula = NO2
Example 7:
Water of crystallisation
The Mr of a hydrated sodium carbonate was found to be
250. Use this value to calculate the number of molecules
of water of crystallisation, x, in this hydrated sodium
carbonate, Na2CO3 • xH2O.
Mr(Na2CO3 • xH2O) = 250
x Mr(H2O)
x 18
x
=
=
=
=
Mr(Na2CO3) = 106
Mrr((Na
Na22CO
CO33• •xH
xH2O)
(Na22CO
CO33))
2O)- Mrr(Na
250 – 106 = 144
144
144/18
=
8
Example 8:
Water of crystallisation – less structured
Sodium carbonate forms a number of hydrates of general
formula Na22CO33 • xH22O. A 3.01 g sample of one of these
hydrates was dissolved in water and the solution made up to
250 cm33.
In titration, a 25 cm33 portion of this solution required 24.3
cm
cm33 of
of 0.200 mol
mol dm
dm-3-3 hydrochloric
hydrochloric acid
acid for
for complete
reaction
Na22CO33 + 2HCl 2NaCl + H22O + CO22
Calculate the Mr of Na22CO33 • xH22O
(a) Calculate moles HCl since you can do this
Moles HCl =volume concentration
= 24.3 10-3 0.200
= 4.86 10-3 mol
(b) Now calculate the moles of Na2CO3 in the 25 cm3 sample of
solution.
Na2CO3 + 2HCl 2NaCl + H2O + CO2
Moles Na2CO3 =Mole Ratio Moles HCl
=
1/2
½
4.86 10-3
= 2.43 10-3
(c) deduce the moles of Na2CO3 in 250 cm3 of solution
Moles
Moles Na
Na22CO
CO33 (250)
(250)
=
= Moles
Moles Na
Na22CO
CO33 (25)
(25)
250
25
= 2.43 10-2
(d) Calculate the Mr of hydrated Na2CO3 [Mass = 3.01 g]
Mr
Mr= Mass / Moles
= 3.01/2.43 10-2 = 124
Example 9:
Percentage yield
Jan 09
In an experiment 165 g TiCl4 were added to an excess of water.
The equation for the reaction is as follows
TiCl4 + 2H2O TiO2 + 4HCl
(a) Calculate the amount, in moles, of TiCl4 in 165 g
Moles = mass = 165
Mr
189.9
= 0.869
(b) Calculate the maximum amount, in moles, of TiO2 which
can be formed in this experiment.
Look at the equation.
1 mole of TiCl4 gives 1 mole of TiO2
So 0.869 moles of TiCl4 gives 0.869 moles of TiO2
(c) Calculate the maximum mass of TiO2 formed in this experiment.
mass = Mr x moles
= 79.9 x 0.869
= 69.4 g
(d) In this experiment only 63.0 g of TiO2 were produced. Calculate
the percentage yield.
Percentage yield = actual (experimental) mass of product
theoretical ( calculated) mass of product
= 63.0
69.4
= 90.8%
Example 10 :
% atom economy
You should learn this formula and be able to use it correctly.
% atom economy = mass of desired product x 100
total mass of reactants
Calculate the % atom economy for the formation
of CH2Cl2 in this reaction.
CH4 + 2Cl2 CH2Cl2 +2HCl
There are no numbers given
You only need the Mr values
Desired product = CH2Cl2 Mr =(12+ 2 +71) = 85
Reactants =CH4+2Cl2 Mr = (12+4)+(2 x71)= 158
%atom economy = 85 x 100 = 53.8%
158
Example 11: Moles HCl, moles & identity ‘Z’
The chloride of an element Z reacts with water according to
the following equation.
ZCl4(l) + 2H2O(l) ZO2(s) + 4HCl(aq)
A 1.304 g sample
sample of
of ZCl
ZCl44 was added to water. The solid ZO2
was removed by filtration and the resulting solution was
made up
made
uptoto250
250cm
cm3 3ininaavolumetric
volumetricflask.
flask.
25.0 cm3 of this solution was titrated
titrated against
against 0.112moldm
0.112moldm-3
NaOH(aq), and 21.7cm3 were needed to reach the end point.
Use this information
information to
to calculate
calculate the
the number
number of
of moles of
HCl produced
HCl
produced and
and
and hence
hence
hence the
the number
number
number of
of moles of ZCl
ZCl4
present in the sample.
Calculate the
Calculate
theMM
ZCl
. From
From
your
your
your
answer
answer
answer
deduce
deduce
deduce
thethe
the
Ar A
ofr
A
r of
r ofZCl
4. 44From
element
of
element
element
Z and
ZZ and
and
hence
hence
hence
its its
identity.
itsidentity.
identity.
identity.
• The only complete data we have is for NaOH(aq)
• From this we can calculate the moles of HCl in 25.0 cm3
• We scale this to give the number of moles of HCl, in
250 cm3, formed in the reaction
Moles NaOH(aq) =
=
=
volume concentration
21.7 10-3 0.112
2.43 10-3 mol
Moles HCl(25) = mole ratio moles of NaOH
= 1 2.43 10-3 = 2.43 10-3
Moles HCl(250) = 2.43 10-3 250/25 = 0.0243 mol
ZCl44(l)
ZCl
(l) ++ 2H
2H22O(l)
O(l)
ZO
ZO22(s)
(s) ++ 4HCl(aq)
4HCl(aq)
Find HCl : ZCl4 mole ratio and hence find moles ZCl4
Moles ZCl4
= Mole Ratio Moles HCl
= 1/4 or
1/44/1 0.0243
= 6.076 10-3
Find Mr of ZCl4 – then Ar of Z and hence identify Z
M
Mrr of
of ZCl
ZCl44 =
= mass
mass
moles
Ar of Z
Element Z
=
=
1.304
in q)
1.304 (given
6.076 10-3
=
= 214.6
214.6
= Mrr (ZCl
(ZCl44)) -– 44 AArr(Cl)
(Cl)= 214.6 - 142 = 72.6
= Ge
So, ZCl4 = GeCl4