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2013
Chemistry NCEA L2
2.1 Quantitative Analysis
GZ Science resources 2013
Version 1.1
The mole and molar mass
The number of particles in a substance is measured in moles
1 mole of particles = 6.02214 x 1023 particles for any
substance! called Avogadro’s number

Molar mass is calculated by adding up the
atomic mass (or mass number ) or each
individual atom.
E.g. H2O Molar mass
M = 1 + 1 + 16 = 18
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Converting number of particles to amount, n
If the actual number of particles is known (atoms, ions or molecules) then the
amount, n, is easily calculated since
n=
number of particles
6.02x10 23
For example 12 x 1024 sodium ions is equivalent to
24
12x10
23
6.02x10
mol = 20 mol of sodium ions.
How many moles of water molecules are there in 3 x 1022 molecules of H2O?
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Relative atomic and molecular mass Ar and M
Sometimes information is provided in terms of the relative atomic mass, Ar,
of an element rather than its molar mass. This has exactly the same
numerical value but has no units e.g. Ar of oxygen is 16 and the M of oxygen
is 16 g mol-1.
Note: The relative atomic mass is, in fact, the mass of an atom relative to the mass of
an atom of the isotope carbon -12 which has a mass of 12.
This means an atom of oxygen-16 is 16
times heavier than an atom of carbon-12.
12
The relative molecular mass, Mr, is the mass of a molecule relative to carbon-12.
e.g. Mr of CuSO4 is 159.5, the same numerical value as the molar mass but having
no units, (M = 159.5 g mol-1).
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Calculating Molar Mass of a compound
1. Identify how many
of each atom
2. Write a sum using
the molar mass for
each atom
3. Calculate the total
e.g. H2SO4
2 x M(H) + M(S) + 4 x M(O)
M(H2SO4)
= 98 gmol-1
2H
1S
4O
e.g. CuSO4.5H2O
1 Cu
1S
9O
(2 x 1) + 32
+ ( 4 x 16)
M(Cu) + M(S) + 9 x M(O) + M(CuSO4.5H2O)
10 x M(H)
=249.5 gmol-1
10 H
63.5 + 32 + (9 x 16)
+ (10 x 1)
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One step calculations
EXAMPLE
Calculate the amount (in moles) of methane, CH4, in 12.5 g of the gas?
M(CH4) = 16.0 g mol1
n=m/
M
n=c.v
a) Select one of the above equations according to what information is in the
question and rearrange if necessary. (i.e. mass and molar mass)
b) Write down the equation
n(CH4) = m/M
c) Write in the values (with units) underneath and calculate
n(CH4) =12.5g/16.0g mol-1
n(CH4) = 0.781mol
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One step calculations
1. Mass of
x moles? Given m or M
/M
M
n=m/
n=m
n = moles (mol)
m = mass (grams)
m=n.
M
M = molar mass (gmol-1)
m = 0.241mol x 142.1gmol-1
m = 34.2g
2.
How many moles? Given v or c
n = moles (mol)
c = concentration (molL-1)
nn == cc. .vv
n = 0.100molL-1 x 0.010L
n = 1 x 10-3 moles
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v= volume (L)
Concentration
Concentration is a quantitative measure of how much dissolved substance there is in a
given volume of solvent. The concentration can be expressed or measured in 3
different ways:
1) grams per litre, g L-1
If the mass is in grams and the volume in litres then
concentration =
mass
volume
2)
% w/v - or percentage weight per volume, is the mass of solute in 100mL of
solvent.
3)
moles per litre, mol L-1
This is the most common concentration unit used in chemistry. It is
calculated using
n
amount (in moles)
concentration =
=
volume (in litres)
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V
One step calculations
The mass (m) of a substance
Symbol = m unit = grams
The amount n is
related to mass m
using relationship
1. Identify what is asked
for, what is given and
calculate the molar mass
of the substance
2. Write an equation
e.g how much SO2 is
present in 1.8kg of SO2?
n = ? m = 1800g
M(SO2) = 64gmol
e.g. calculate mass of 0.5
moles of CuSO4.5H2O
e.g. calculate the amount
of Na+ ions in 25g of
Na2CO3
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n = m/M
mol =
mass
molar mass
3. Calculate the quantity
n = m/M
n = 1800/64
n = 28.1 moles
m = ? n =0.5g
M(CuSO4.5H2O) =
249.5gmol -1
m=nxM
m = 0.5 x 249.5
m = 125g
n(Na+) = ? m = 25g
First find n(Na2CO3)
M(Na2CO3)=106gmol
n = m/M
n=25/106
n=0.236
n(Na+)=0.472moles
-1
-1
Two step calculations
EXAMPLE
If 2.3 g of sucrose (C12H22O11) is dissolved in 12mL of distilled water, what will the
concentration be?
M(C12H22O11) = 342 g mol-1
n=m/
M
n=c.v
a) Select one of the above equations depending on what 2 values you
have (i.e you have mass – 2.3g, and M – 342gmol-1 but only volume –
12ml)
b) Write down the equation chosen and calculate
n(C12H22O11) = m/M
c) Use the calculated moles to insert in the other equation and calculate
c(C12H22O11) = n / v
information given in question (convert to litres)
Two step calculations
EXAMPLE
500 mL of 0.253 mol L1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L1 NaHCO3
solution.
What is the concentration of the final solution?
n=c.v
c=n/v
a) Calculate n of each solution (n = c.v) and add together to get n (total)
b) Add volumes together to get v (total)
c) Calculate concentration of final solution
concentration (final) = n (total) / v (total)
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Two step calculations
3. What is the concentration? ( M and m given)
c=n/v
a) n = 2.12g /142.1gmol-1
n = 0.015mol
b) c = 0.015mol / 0.250L
c = 0.059molL-1
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but
n=m/M
Two step calculations
4. How much mass of x to make up solution? Given v + c
n=c.v
conc = 0.200molL-1 vol = 250ml
n=c.V
x = CuSO4
M(CuSO4) = 249.5gmol-1
a) n = 0.200molL-1 x 0.250L
n = 5.00 x 10-3 mol
b) m = M . n
m = 249.5gmol-1 x 5.00 x 10-3 mol
m = 1.248g
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Percentage composition
EXAMPLE
Allicin is the compound responsible for the characteristic smell of garlic. Its formula is
C6H10S2O. What is the % of sulfur (S) in this compound?
m=n x
M
a) Place each element from the formula into a column and write the above
equation under each and calculate
C
H
m(C) = n x M
S
m(H) = n x M
C6 H10 S20
n (moles) is found from compound
C=12
m(S) = n x M
H=1 S= 32.1
M(molar mass) is found in periodic table
b) Calculate the mass of the compound by adding all the masses together
c) Calculate % of element in the compound
i.e. % S = mass (S) ÷ Mass (compound) x 100
d) All of the percentages should add up to 100%
Percentage composition
5a. What is the Percentage Composition of x in a compound?
Na2B4O7
x=B
Na
B
O
m=n.M
m=n.M
m=n.M
m = 2 x 23.0
m = 4 x 10.8
m = 7 x 16.0
m(Na) = 46g
m(B) = 43.2g
m(O) = 112.0g
m(Na) + m(B) + m(O) = 201.2g
%B = 43.2g / 201.2g x 100
%B = 21.5
Shortcut Note – you can calculate the Molar mass of the element in question (B) and
divide it by the Molar mass of the compound but you lose the opportunity to gain
partial marks if you make a mistake.
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Percentage composition
The mass of the different elements within a compound can be expressed as a percentage
of the total mass – this can be used to identify a compound
1. From the chemical
formula find the amount
of each element present
(in moles)
2. Use the molar masses
of each element to find
the mass of each element
present.
3. Express the amount of
each element as a
percentage
4. Check to see that
adding the percentages
gives 100%
e.g. 1 mole of C2H6
contains 2 mol of C and 6
mol of H
The mass of C present is 2
x 12.0 = 24.0g
The mass of 1 mole of
C2H6 is 30.0g, so the
percentage composition of
C is
80.0% + 20.0%=100%
The mass of H present is 6
x 1.0 = 6.0g
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C= 24.0/30.0 x 100=80.0%
H=6.0/30.0 x 100=20.0%
checked
Mass and Percentage composition
EXAMPLE
A precipitate of 1.34 g of Fe2O3 was obtained by treating a vitamin supplement dissolved in
water with NaOH and then heating. What is the mass of iron in the tablet?
m=n x
M
a) Place each element from the formula into a column and
write the above equation under each and calculate mass.
Fe
m(Fe) = n x M
O
m(O) = n x M
b) Calculate the mass of the compound by adding all the masses from the
elements together i.e. m (Fe2O3) = m(Fe) + m(O)
c) Calculate % of element in the compound
i.e. % Fe = mass (Fe) ÷ Mass (compound) x 100
d) Find the % Fe in the mass given (i.e 1.34g)
(mass given ÷100) x %
Make sure your calculated mass
is less than the given mass
(unless 100% then it will be the
same.
Use units, especially in your
final answer.
Mass and Percentage composition
5b. Find Mass of element in compound given mass of compound
% of O in BaSO4
a) Find Percentage Composition of each atom in a given compound
Ba
S
O
m(Ba) = n . M
m(S) = n.M
m(O) = n. M
m(Ba) = 1mol x 137.3gmol-1
m(S) = 1mol x 32.0gmol-1 m(O) = 4mol x 16.0
m(Ba) = 137.3g
m(S) = 32.0g
m(O) = 64g
b) Calculate m of compound - m(BaSO4) = m(Ba) + m(S) + m(O) = 233.5g
m(BaSO4) = 233.5g
c) Calculate Percentage % O = 64g/233.5g x 100
%O = 27.4%
d) Calculate percentage of O in given mass of compound mass = 0.192g
m(O) = 27.4% of 0.192g
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m(O) = 0.0526g
Empirical formula
EXAMPLE
Caffeine is a stimulant which has a mass composition of 49.5% carbon, 5.20% hydrogen,
28.9% nitrogen and 16.5% oxygen.
Calculate the empirical formula of caffeine.
n=m/
M
a) All % should add to 100%, change each into grams so they all total 100g (i.e.
28.9% = 28.9g)
b) Calculate the number of moles of each element (place each element in a
column and write formula underneath, then calculate n)
c) Divide each n value by the smallest value – You may have to multiply all of the
numbers up until you get the lowest common whole number of each
i.e n(C) = 4.13mol n(H) = 5.20mol n(N) = 2.06mol n(O) = 1.03mol -1.03mol
1.03mol
1.03mol
1.03mol
d) Write the empirical formula underneath with the element followed by the
number of moles i.e C4H5N2O
Empirical formula
6. Empirical Formula ( Find ratio of each atom in a compound)
a) Convert % of each atom to Mass (grams) - total g = 100g
C = 40%
H = 6.67%
O = 55.33%
b) Calculate n of each
C
H
O
n(C) = m/ M
n(H) = m /M
n(O) = m/ M
n(C) = 40g/12.0gmol-1
n(H) = 6.67g/1.0gmol-1
n(O) 55.33g/16.0gmol-1
n(C) = 3.33mol
n(H) = 6.67mol
n(O) = 3.33mol
c) Divide each n by the smallest n value to obtain ratio
3.33/3.33
1
d) Convert to formula
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:
:
6.67/3.33
2
C H 2O
:
:
3.33/3.33
1
Empirical formula
An empirical can be calculated from the percentage composition. It is the simplest whole
number ratio of the atoms in a compound.
1. Find the mass of
each element
present in 100g of
the compound
2. Change these
masses to amounts
using
n = m/M
3. Divide each result
by the smallest
number of moles
4. Make small
approximations
or multiply all
results by a constant
to get a simple
whole number ratio
5. This is the
empirical formula
e.g. Analysis of a
compound showed
that it contained
14.29% H and
85.71% C
in 100g of
compound there is
14.29g of H
85.71g of C
In 100g of
compound there is
Moles of H
14.29/7.14 = 2.001
Moles of H
2.001≈ 2
The empirical
formula of the
compound is CH2
Moles of C
7.14/7.14=1
Moles of C
1
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14.29/1=14.29mol H
since
M(H)=1gmol
-1
85.71/12 =7.14mol C
Since M(C)=12gmol
-1
Molecular formula
EXAMPLE
The porphryn molecule forms an important part of the haemoglobin structure. This
porphryn molecule is 78.5% carbon, 3.2% hydrogen and 18.3% nitrogen.
(i) Calculate the empirical formula of porphryn.
(ii) If the molar mass of porphryn is 306 g mol1, calculate the molecular formula.
a) Calculate the empirical formula so you have a whole number formula
i.e porphryn = C10H5N2
molar mass of carbon
b) Calculate the molar mass of the empirical formula i.e C = (10 x 12) + …
c) Divide the molar mass of the molecular formula (given in question) by
the molar mass of the empirical formula to calculate the multiplier
i.e (molecular)306gmol-1 / (empirical) 153gmol-1 = 2
d) If the multiplier value is 1 then the empirical formula is the same as the
molecular formula. Otherwise multiply each element by the value
C10H5N2
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x 2=
C20H10N4
Molecular formula
7. Molecular Formula ( Find moles of each atom in a compound)
Given the Molecular M = 180.0gmol-1
a) Find M of the empirical formula (add up all M x each atom)
b) Calculate multiplier = Molecular M
Empirical
M
=6
c) Multiply empirical formula by multiplier
CH2O x 6 =
GZ Science resources 2013
C6H12O6
=
180.0gmol-1
30gmol-1
Molecular formula
The Molecular Formula can be determined from the empirical formula and molar mass It
gives the actual number of atoms present
1.Find empirical formula
and
Calculate molar mass
2.Given molar mass of the
molecular formula
calculate
units present (x)
3.Multiply empirical
formula by unit (x)
4. Write molecular
formula
Example.
The empirical formula of a
compound was found to
be CH2O
molar mass is 60.0gmol
X=
=(CH2O)x
=(CH2O)2
The molecular formula is
C2H4O2
Molar mass of
empirical formula
X=
X=
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molar mass
60.0
30.0
2
Gravimetric Analysis
This is an experimental procedure involving the measurement of masses of substances
to determine the composition of unknown substances.
The technique is:
Highly accurate (since masses can be measured accurately)
Useful for substances involved in chemical reactions that go to completion.
Calculations should be carried out to three significant figures.
If the masses of the elements combined to make a compound are known accurately,
than the empirical formula can be calculated.
Masses of compounds can be determined from precipitation.
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Water of Crystalisation





The crystals of some salts contain molecules of water.
These molecules fit into the crystal lattice.
The ratio of water to salt is always a whole number
The water present in hydrated salts is known as water of crystallisation
When a hydrated salt is heated, the water is driven off and the salt is said to
become anhydrous
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Water of Crystalisation
EXAMPLE
Hydrated magnesium sulfate is heated in a crucible. The following data is collected.
mass of crucible and lid = 26.49 g
mass of hydrated magnesium sulfate = 2.12 g
mass crucible, lid and magnesium sulfate after first heating = 27.55 g
mass of crucible, lid and magnesium sulfate after second heating = 27.52 g
Use these results to determine the formula of hydrated magnesium sulfate.
M(MgSO4) = 120 g mol1 M(H2O) = 18.0 g mol1
a) Write down two columns : Anhydrous salt and Water
b) Calculate the mass of each from the data given.
> Anhydrous salt = (mass of crucible, lid and MgSO4 after second heating) –
(mass of crucible and lid)
> Water = (mass of hydrated salt) – (mass of salt calculated)
c) Calculate n of each (n=m/M)
d) Divide each by smallest value and write down formula
MgSO4 . (number of moles) H2O
-
Water of Crystalisation
Anhydrous salt
Water
mass = m(heated) – m(crucible)
mass = m(hydrated) – m(heated)
m(CaCO3) = 1.56g
m(H2O) = 1.52g
n= m/M
n = m/M
n=1.56g/90.08gmol-1
n= 1.52g/18.0gmol-1
n=0.017mol
n = 0.080mol
Data from
experiment
or given in
question
n(CaCO3) : n(H2O)
0.017mol : 0.080mol
0.017mol : 0.017mol
.
The dot ( )
means + water
1
:
.
5
CaCO3 5H2O
GZ Science resources 2013
Divide each by the
smallest value –this
will be n (anhydrous
salt)
Equations and mole ratios n = m/M
EXAMPLE
Wine makers convert sugars, such as glucose, to ethanol and carbon dioxide.
C6H12O6  2C2H5OH + 2CO2
Starting with 540g of glucose what is the maximum amount of ethanol, in moles and also
in grams that could be produced.
M (C2H5OH) = 46 g mol-1
M (C6H12O6) = 180g mol-1
C6H12O6
M = 180gmol-1
m = 540g
-
Write down the
equation with
the known
information
underneath

n = m/M
1
n = 540g / 180gmol-1
n = 3.0mol
2
2C2H5OH + 2CO2
M = 46gmol-1
m=?
m= nxM
1. Calculate moles
of known. 2. convert
mole ratios. 3.
calculate mass of
unknown.
m = 6.0 mol x 46gmol-1
m = 276g
1 mole of C6H12O6 forms 2 moles of C2H5OH
So
n (C2H5OH) = n(C6H12O6) x U/K = 3.0mol x 2/1 = 6.0mol
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3
Equations and mole ratios n = m/M
9. Calculate mass of x given equation and mass of a compound y
y = Cl2
m(Cl2) = 12.8g
Cl2 + 2NaI
2NaCl
x = NaCl
+ I2
a) Calculate moles of compound y
n
= m/M
n(Cl2) = 12.8g/71.0gmol-1
n(Cl2)=0.180moles
b) Convert moles y to x
1mol Cl2 forms 2mol NaCl so n(NaCl) = 2 x n(Cl2) = 0.361moles
c) Calculate mass (or concentration) of NaCl
m
=n.M
c
= n/v
m(NaCl) = 0.361mol x 58.5gmol-1
c(NaCl) = 0.361mol x 0.100L
m(NaCl) = 21.12g
c(NaCl) = 3.61molL-1
Equations and mole ratios n = m/M
9. Calculate Mass of x given Formula and equation and mass of y
m(CH4) = 6800g
m(CH3OH) = ?
3CH4
+
CO2
+
2H2O
4CH3OH
a) Calculate using mass ratios – write under formulas
3CH4
+
Mass=
6800g
M
48gmol-1
=
CO2
m(CH3OH) = m(CH4)
+
2H2O
=x
=128.0gmol-1
x
M(CH3OH)
M(CH4)
m(CH3OH) = 6800g
x
128.0gmol-1
48.0gmol-1
m(CH3OH) = 18133.3g
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4CH3OH
or
18.133kg
Equations and mole ratios n = m/M
EXAMPLE
20.0 mL of an aqueous solution of 0.105 mol L1 sodium carbonate is titrated with a 0.250
mol L1 hydrochloric acid solution.
What volume of hydrochloric acid is required to reach the equivalence point of the
titration.
2 HCl (aq) + Na2CO3(aq) →
2 NaCl(aq) + CO2(g) + H2O(l)
Na2CO3
c = 0.105molL-1
v = 0.020L
+
n=cxv
1
n = 0.105 x 0.020
n = 2.10 x 10-2mol
2
2HCl
→ 2NaCl + CO2 + H2O
c = 0.250molL-1
Write down the equation with the
v=?
known information underneath
1. Calculate moles
of known. 2. convert
mole ratios. 3.
calculate volume of
unknown.
v= n/c
v = 4.20 x 10-2 mol / 0.250
v = 0.0168L or 16.8mL
1 mole of Na2CO3 forms 2 moles of HCl
So
n (HCl) = 2 x n(Na2CO3) = 2 x (2.10 x 10-2)mol = 4.20 x 10-2mol
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3
Equations and mole ratios n = m/M
13. Calculate concentration of x given conc. of y and formula
Na2CO3 + 2HCl
2NaCl
+ H 2O
+ CO2
10.0ml of 0.106molL-1 sodium carbonate reacted with 21.56ml HCl
a) Calculate mols of y (Na2CO3)
n(Na2CO3) = c x v
= 0.106molL-1 x 0.0100L
= 0.00106mol
b) Convert mols y to x
1mol Na2CO3 = 2molHCl
n(HCl) = 2 x n(Na2CO3)
=0.00212mol
c) Calculate concentration
c = n/v
c(HCl) = 0.00212mol / 0.02156L
= 0.0983molL-1
Empirical formula and mole ratios n = m/M
EXAMPLE
A chemist found that 4.69 g of sulfur combined with fluorine to produce 15.81 g of gas.
Determine the empirical formula of the compound.
?S
+

?F
m = 4.69g
m = m(SF?) – m(S) = 11.12g
M = 32.1gmol-1
M = 19.0gmol-1
1
n(S) = m/M
n(F) = m/M
n(S) = 4.69/32.1
n(F) = 11.12/19.0
n(S) = 0.146mol
n(F) = 0.585mol
n(S) = 0.146
n(F) = 0.585mol
0.146
= 1
0.146mol
:
S F4
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2
3
m = 15.81g
Write down the equation with
the known information
underneath
Calculate number of moles of each element
making up compound
Divide each number of moles by the
smallest value to get the smallest common
whole number
4
4
SF?
Use those numbers
to write formula of
compound
Empirical formula and mole ratios n = m/M
10. Calculate compound Formula given mass and equation
m(product) = 43.5g
m(Fe) =15.0g
Fe
+
Cl2
Fe Clx
a) Calculate moles of reactant (for which given mass)
n(Fe) = m/M
n(Fe) = 15.0g/55.9gmol-1
n(Fe)=0.268moles
b) Convert moles reactant to product
1mol Fe forms 1mol FeClx
c) Calculate mass of each possible molecular formula of FeClx
m
=n.M
m
=n.
M
m(FeCl2) =0.268mol x 126.9gmol-1
m(FeCl3)= 0.268mol x 162.3gmol-1
M(FeCl2) = 34.1g
m(FeCl3) = 43.5g
d) Match m(product) with calculated mass = FeCl3
Dilution calculations
EXAMPLE
20.0 mL of a 0.00500 mol L–1 FeCl3.6H2O. solution was pipetted into a volumetric flask and
diluted to 100 mL. What is the concentration of the diluted solution?
concentration (new) = concentration (original) x
volume (original)
volume (new)
c(new) = 0.00500molL-1 x (0.020L / 0.100L)
c(new) = 0.00500molL-1 x 0.2
c(new) =
0.00100molL-1
Because this is a dilution the new
concentration must be less than the
original
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Make sure you convert all of
your volumes into litres first
Dilution calculations
11. Calculate new concentration given old conc. and volumes
e.g. 40ml of 2.0molL-1 diluted by adding 60ml water
c(new) = c(old)
x
old volume
new volume
c(new) = 2.0molL-1
x
40ml
(40ml + 60ml)
C(new) =0.80molL-1
Also can calculate new concentration through evaporation
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Mixed solution calculations
EXAMPLE
500 mL of 0.253 mol L1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L1 NaHCO3
solution. What is the concentration of the final solution?
n(solution 1)= c1 x v1
+
n(solution 2)= c2 x v2 =
total moles
Final concentration = total mols / total volume (v1 + v2)
n(solution 1 ) = 0.253molL-1 x 0.50L =
c(new) = 0.00500molL-1 x 0.2
c(new) =
0.00100molL-1
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Make sure you convert all of
your volumes into litres first
Volumetric Analysis
Volumetric analysis involves a titration.
>In a titration, a solution, of unknown concentration, is typically
titrated (added) from a burette into a flask.
>The flask typically contains a known amount of solution. A pipette
is used to measure a precise set volume called an aliquot.
>In an acid-base titration the end point of the titration is reached
when there is a neutralisation. Indicator (usually phenolphthalein)
is added into the flask and a colour change (lasting more than 10
seconds) will indicate a neutralisation.
>Using c=n/v and a balanced equation, the concentration of the
solutions can be calculated.
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Titration equipment
Pipette
>calibrated to provide an exact
volume.
>clean and rinsed with solution to
be pipetted.
>reading from the bottom of the
meniscus.
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Titration equipment
Burette
>burette must be cleaned and rinsed
with small amount of solution that will
go into it.
>The volume of the solution in the
burette is read from the bottom of the
meniscus. Volume is recorded before
and after the titration.
>the volume of solution delivered by
the burette is called a titre.
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Titration equipment
Conical Flask
>rinsed with distilled water before
use.
>aliquot of solution (unknown or
known concentration) placed into the
flask from the pipette.
>a few drops of indicator are placed
into the flask
>the flask is swirled during the
titration.
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Reading the burette
Reading is to be taken
from the bottom of the
meniscus curve.
a) Read the ml
29ml
b) Read the 1/10 of ml
29.9ml
c) Divide the 1/10ml up
into 0.00, 0.02,
0.04…0.08 and read to
closest
29.98ml
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Titration equipment setup
Make sure you label your beaker
with the solutions they contain
Rinse all glassware appropriately
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Indicator selection
Before starting the titration a suitable pH indicator must be chosen. The endpoint of the
reaction, when all the products have reacted, will have a pH dependent on the relative
strengths of the acids and bases. The pH of the endpoint can be roughly determined using
the following rules:
A strong acid reacts with a strong base to form a neutral (pH=7) solution.
A strong acid reacts with a weak base to form an acidic (pH<7) solution.
A weak acid reacts with a strong base to form a basic (pH>7) solution.
When a weak acid reacts with a weak base, the endpoint solution will be basic if the base is
stronger and acidic if the acid is stronger. If both are of equal strength, then the endpoint
pH will be neutral.
A suitable indicator should be chosen, that will experience a change in colour close to the
end point of the reaction.
Indicator
Acid
Base
Methyl Orange
Red
Yellow
Phenolphthalein
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Colourless
Pink
pH range
3.1 - 4.4
8.3 - 10.0
Standard solutions
A standard solution is a solution whose concentration is known accurately.
Its concentration is usually given in mol L–1. When making up a standard solution it is
important that the correct mass of substance is accurately measured and all of this is
successfully transferred to the volumetric flask used to make up the solution.
Making a 1molL-1 of NaCl solution
a) Calculate the mass of 1mole of NaCl
m (NaCL) = n x M
m (NaCl) = 58.5g
b) Determine the volume of the volumetric flask being used and mass of
substance added.
1L flask = 1mol
100mL flask = 1/10mol
c) Weigh and add substance to the volumetric flask with wash bottle
d) Continue filling flask with distilled water to line – bottom of meniscus
must touch line
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Titration method
1. First, the burette should be rinsed with the standard solution, the pipette with the
unknown solution, and the conical flask with distilled water.
2. Secondly, a known volume of the unknown concentration solution should be taken
with the pipette and placed into the conical flask, along with a small amount of the
indicator chosen. The burette should be filled to the top of its scale with the known
solution with a funnel. Remove the funnel after the burette is filled. Record the
starting volume.
3. The known solution should then be allowed out of the burette, into the conical flask. At
this stage we want a rough estimate of the amount of this solution it took to
neutralize the unknown solution. Let the solution out of the burette until the indicator
changes colour and then record the value on the burette. This is the first titre and
should be excluded from any calculations.
4. Perform at least three more titrations, this time more accurately, taking into account
roughly of where the end point will occur. Record of each of the readings on the
burette at the end point. Endpoint is reached when the indicator just changes colour
for more than 10 seconds. The titres must be concordant to 0.2ml to gain Excellence (
0.4ml Merit and Achieved)
Recording Titrations
1. Read your instructions carefully. Understand clearly which solution goes into the
conical flask and which goes into the burette. Make sure you convert all volumes into
litres i.e. 10ml aliquots from a pipette is 0.010L
2. Record the given concentration on your sheet
Concentration of standard sodium hydroxide solution = _____________ mol L1
3. Rule up a table to record your titrations. Titre = final reading – start reading
initial
titration
Final
reading
Start
reading
titre
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1st
titration
2nd
titration
3rd
titration
4th
titration
Concordant titres
1. At least 3 titrations must be within 0.2ml of each other to be concordant at Excellence
level
2. Any titres not falling within this range must be discarded and not used in the average.
3. If more there are more than 3 concordant titres within this range they will be included
in the average calculation
EXAMPLE
successive titrations – 24.50 mL, 23.25 mL, 23.35 mL and 23.28 mL.
23.35mL – 23.25mL = 0.10ml
This difference is less than 0.2ml allowed, and 23.28 falls within this range so all
three are used. 24.50mL falls outside this range so is discarded.
v = (23.35mL + 23.28mL + 23.25mL) ÷ 3 = 23.29mL = 0.0233L
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Titration calculations c = n/v
EXAMPLE
A standard solution of 0.180 mol L1 hydrochloric acid was titrated against 25.0 mL samples of a solution of
sodium carbonate. The following volumes of hydrochloric acid solution were used in successive titrations –
24.50 mL, 23.25 mL, 23.35 mL and 23.28 mL.
The equation for the reaction is
Na2CO3 + 2 HCl
2 NaCl + CO2 + H2O
Use this information to determine the concentration of the sodium carbonate solution.
Give your answer to three significant figures.
Na2CO3
v = 0.0250L
c= ?
+
2HCl
→ 2NaCl + CO2 + H2O
v = 0.0233L (concordant titres)
c = 0.180molL-1
Write down the equation with the
known information underneath
n=cxv
1
n = 0.180molL-1x 0.0233L
n = 4.19 x 10-3mol
2
1. Calculate moles
of known. 2. convert
mole ratios. 3.
calculate
concentration of
unknown.
c= n/v
c =2.10 x 10-3 mol /0.0250L
c = 8.39 x 10-2molL-1
2 moles of HCl forms 1 mole of Na2CO3
So
n (Na2CO3) = n(HCl) ÷ 2 = (4.19 x 10-3) ÷ 2 = 2.10 x 10-3mol
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Titration calculations c = n/v
EXAMPLE
20ml ethanoic acid solution (unknown) is titrated with a solution of 0.105 mol L1 sodium
hydroxide (known) What is the concentration of hydrochloric acid.
CH3COOH (aq) + NaOH(aq) →
NaCH3COO(aq) + H2O(l)
NaOH
+
c = 0.105molL-1
v = titration (eg 0.018L)
n=cxv
1
n = 0.105 x 0.018
n = 1.89 x 10-3mol
CH3COOH
c=?
v = 0.020L
→
1. Calculate moles
of known. 2. convert
mole ratios. 3.
calculate volume of
unknown.
NaCH3COO + H2O
Write down the equation with the
known information underneath
c= n/v
c = 1.89 x 10-3 mol / 0.020
c = 0.0945molL-1
2
So
n (CH3COOH) = n(NaOH) x U/K (1/1) = 1.89 x 10-3mol
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