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Titrations

Titrations

Titrations are used to determine the concentration of an acid or alkali solution.


Method: Titration

Set up a clamp stand with a burette fixed parallel to it. Place a conical flask underneath the burette
Rinse the conical flask and the burette with distilled water
Fill the burette with the acid, ensure that the bottom of the meniscus sits on the maximum measuring line of the burette. The concentration of this acid must be known.
Fill the conical flask with a set volume of the acid solution using a pipette.
Add an appropriate pH indicator to the alkali
First, there should be a rough titration to form a ‘general’ idea to where the end point is (or the exact point of neutralisation). To perform this, add the acid in the burette to the alkali solution, swirling the mixture until the solution turns clear.
To do an accurate titration, take the initial reading for the amount of acid in the burette that remains. Allow the acid to run through the burette until 2cm3 of the end point to which now add the acid dropwise. Stop adding the acid when the solution turns changes colour.
Work out the total amount of acid used in the accurate titration to neutralise the alkali.
Total amount of acid used = Final reading – Initial Reading

Repeat the titration again multiple times, until there are at least 3 concordant results
Use these values to find the mean volume of acid used (remove any anomalous results if any)

Method: Standard Solution

Work out the amount in moles of the alkali
Moles = Concentration x Volume

Work out the amount in grams of the alkali
Mass = Moles x Mr

Using a weighing boat and digital balance, measure the required mass and pour this into a beaker
Add distilled water to the beaker until all of the solution has been dissolved
Using a funnel, pour the solution in the beaker into a 250cm3 volumetric flask
Rinse the beaker and funnel with distilled water and add this to the flask too (ensures all of the solute is in the flask)
Fill the rest of the flask upto the 250cm3 with distilled water (ensure the meniscus sits on the measurement line)
Place a bung on the flask and shake the solution
Calculate the concentration of the solution*

*e.g. 40cm3 of 0.250 mol dm-3 KOH was used to neutralise 22.0 cm3 of HNO3 solution. Calculate the concentration of the nitric acid in mol dm-3

KOH + HNO3 -> KNO3 + H2O

For the KOH, there is a known volume and concentration but an unknown amount of moles.
Moles = Concentration x volume

First convert the 40cm3 into dm-3 (which is the same as concetration) to do this multiply the volume in cm3 by x10-3
40cm3 x 10-3 = 0.04

Multiply the volume (in dm-3) by the concentration
0.04 dm-3 x 0.250 mol dm-3 = 0.01 moles

*notice how the dm-3 can cancel, therefore leaving just the moles

Now the number of moles for the KOH is known, as it is a 1:1 ratio (as there is one mole of KOH and HNO3) the moles can be copied to the HNO3. Therefore there is a known volume and a known amount of moles, but an unknown concentration
Concentration = moles / volume

Convert the volume of HNO3 from cm3 to dm-3 by multiply it by x10-3
22cm3 x 10-3 = 0.022 dm-3

0.01 moles / 0.022dm-3 = 0.455 moles

Indicators

Indicators must be used that changes colour quickly over a small pH range. These must be used to determine the exact point of neutralisation compared to other indicators such as Universal Indicator that changes gradually
Methyl orange – red in acid, yellow in alkali
Phenolphthalein – colourless in acid, pink in alkali

Empirical Formula

Empirical Formula

A formula that gives the simplest whole number ration of atoms in a compound found from an experiment. Empirical Formula means ‘from experiment’.

Molecular formula is the actual ration for a molecule

For example for Benzene:

– Actual Formula: C6H6

– Empirical Formula: CH

Examples

  • Determine the Empirical Formula of methane given that 6.0g of Methane can be decomposed into 4.5g of carbon and 1.5g of hydrogen. Work out the empirical formula.

Carbon

Hydrogen

Mass

4.5g

1.5g

Ar

12

1

Moles

4.5g / 12 = 0.375

1.5g /1 = 11.5

Divide by smallest

0.375 / 0.375

1.5 / 0.375

Ratio

1

4

= CH4

Empirical Formula from percentages

Empirical formula can be used using percentages, just using the percentage instead of it’s mass.

Examples

  • The composition of a compound is 40% sulphur and 60% oxygen by weight. What is its empirical formula?

Sulphur

Oxygen

Mass

40%

60%

Ar

32.1

16.0

Moles

40 / 32.1 = 1.25

60% / 16.0 = 3.75

Divide by smallest

1.25 / 1.25

3.25 / 1.25

Ratio

1

3

= SO3

  • Calculate the percentage of iron present in Fe2O3

2Fe 2 x 55.8 x100 = 69.92%

Fe2O2 x 55.8 + 48

  • Percentage composition of 81.8% carbon, 6.1% hydrogen and 12.1% oxygen.

Carbon

Oxygen

Hydrogen

Mass

81.8%

12.1%

6.1%

Ar

12.0

16.0

1

Moles

81.8 / 12.0 = 6.16

12.1% / 16.0 = 0.75625

6.1 / 1 = 6.1

Divide by smallest

6.16 / 0.75625 

 0.75625/ 0.75625

6.1/0.75625

Ratio

9

1

8

= C9H8O

  • Blue Sulphate crystals yield the following percentage composition, Cu – 25%, S – 12.8%, O – 25.6% and water crystallisation – 36.1%. Find the empirical formula of the crystals

Copper

Sulphur

Oxygen

H2O

Mass

25.0%

12.1%

25.6%

36.1%

Ar

63.5

32.1

16.0

18

Moles

25.0 / 63.5 = 0.39

12.1% / 32.1 = 0.37

25.6 / 16.0 = 1.60

36.1 / 18 =2.01

Divide by smallest

0.39 / 0.37 

0.37 / 0.37

1.60 / 0.37

2.01/ 0.37

Ratio

1

1

4

5

= CuSO4 • 5H2O

  • A sugar solution contains 40% carbon, 6.7% hydrogen and 53.3% oxygen.

Carbon

Oxygen

Hydrogen

Mass

40.0%

53.3%

6.7%

Ar

12.0

16.0

1

Moles

40.0 / 12.0 = 3.3

53.3% / 16.0 = 3.3

6.7 / 1 = 6.7

Divide by smallest

3.3 / 3.3 

3.3 / 3.3

3.3 / 3.3

Ratio

1

1

2

= CH2O

  • 24% Carbon and 76% fluorine

Carbon

Fluorine

Mass

24%

76%

Ar

12

19

Moles

24 / 12 = 2

76% / 19 =4

Divide by smallest

2 / 2 

1.5 / 0.375

Ratio

1

2

Calculating molecular formula from empirical formula

This can be deduced from the empirical formula if the Mr is known (as the Mr can come from mass spectrometry)

In mass spectrometry the molecular ion is the last peak on the graph. The Mr is this value.

21% in this example graph is the Molecular Ion.

A compound has an empirical formula of ClCH2 and a molecular weight of 98.96 g/mol. What is its molecular weight formula

Mass of empirical unit?

Mass of Chlorine + Mass of Carbon + Mass of Hydrogen2

35.5 + 12 + 2 = 49.50 g/mol

Empirical Formula = Mass of Molecular unit = 98.96

Mass of Empirical Formula 49.50

= 2.00 empirical units per molecular unit

It takes two empirical units to make a molecular unit, so the molecular formula is Cl2C2H4

 

Atom Economy

Atom Economy

The atom economy of a chemical reaction is a measure of the amount of starting materials that become useful products. Inefficient, wasteful processes have low atom economies. Efficient processes have high atom economies, and are important for sustainable development, as they use fewer natural resources and create less waste.

Companies want to maximise as much product as possible to allow for maximum profits. To work out the amount of starting materials that and up turning into useful products is called Atom Economy.

% of atom economy = Mr of useful products x100

Total Mr of all products

e.g. C2H5OH -> C2H4 + H2O

Mr: C2H5OH = (12×2) +(1×6) + 16 = 46 

C2H4 = (12×2) + (1×4) = 28

H2O = (1×2) + 16 = 18

Atom Economy = 28 x100 = 61%

(28+18)

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Percentage Composition

It is useful to know how much an element is makes up a compound. This process is called Percentage Composition by Mass.

  • Find the molar mass of the compound by adding up the masses of each atom in the compound.
  • Calculate the mass due to the component in the compound you are for which you are solving by adding up the mass of these atoms.
  • Divide the mass due to the component by the total molar mass of the compound
  • Multiply by 100.

% Z = (number of atoms of Z) x (atomic mass of Z)

Formula Mass of the compound

e.g.

  1. Calculate the percent composition of carbon in each of the following:

CO2

Carbon = 12.00

Oxygen = (16.0 x 2) = 32

12 + 32 = 44g/mol

12.00 g/mol x100 = 27%

44 g/mol

  1. Calculate the percent composition of carbon in each of the following:

C6H12O6

C6 = (12 x 6) = 72

H12 = (12 x 1) = 12

O6 = (16×6) = 96

72 g/mol x100 = 40%

180 g/mol

Percentage Composition can be used to find unknown substances

  1. e.g. C21H22N2O2 was found on the side of a road. If it contains 83% Carbon, it is Strychnine. Work out if the chemical is Strychnine.

    C = 12 x 21 = 252

    H = 1 x 22 = 22

    N = 2 x 14 = 28

    O = 16 x 2 = 32

    +334

    252 x 100 = 75.449%

    334

    Therefore it is not Strychnine as it contains 75.449% where Strychnine contains 83%.

    1. Calcium Chloride Hexahydrate has a chemical Formula of

CaCl2 . 6H2O

What is the theoretical Percentage of water?

Ca = 40.1 amu

Cl2 = 35.5 x 2 = 71.0 amu

H = 2 x 6 = 12.0 amu

O = 16.0 x 6 = 96.0 amu

Formula weight of CaCl2 = 219.1 amu

Formula weight of H2O = 108.0 amu

108 ÷ 219.1 x 100 = 49.29%

Moles

Moles

A mole (or mol for short- such an abbreviation…) is a unit of measurement
For example 1 mole of HCl has the same number of molecules than C for example.

We know that atoms are small and an awful lot of them can fit into one place.

To work out amount of atoms in a substance we look at their atomic mass:

For example Carbon-12, the base that is used for comparison against.

12g of Carbon 12 has 1 mole of atoms.

So what is 1 mole of atoms?

1 mole = 6.02×1023 atoms.

The above figure is known as Avogadro’s number. (but don’t worry, you’re given this value in the (aqa) exam)