Specifications Core Practicals

# Quantitative Analysis

## Concentration (Higher Only)

The concentration of a solution can be calculated using:

• the moles of dissolved solute (in mol)

• the volume of solvent (in dm³)

concentration = moles of solute  ÷ volume of solvent

The units for concentration are mol/dm³, but they may also be written as moldm-³ - don't worry as these mean the same thing.

Converting from grams to moles

You need to be able to calculate concentration in gdm-³ and be able to convert these units into mol/dm³.

To do this, think back to our moles equations, where:

moles = mass ÷ Mr

For example, a solution of sodium hydroxide has the concentration of 5 g/dm³.

To convert it into mol/dm³, we divide 5 ("the mass") by 40 (the Mr of NaOH).

This gives a concentration of 0.125 mol/dm³. Volume unit conversions

• divide by 1000 to convert from cm³ to dm³

• multiply by 1000 to convert from dm³ to cm³

## Core Practical 5: Acid-Alkali Titration

A titration is carried out to determine the volume of hydrochloric acid required to neutralise a solution of unknown concentration of sodium hydroxide.

An indicator must be used to determine the end point. Standard procedure for a titration must be carried out, such as the use of a white tile and swirling the conical flask to obtain an accurate end point. The data must then be used to determine the concentration of the unknown solution. Method

1. use a pipette and pipette filler to add 25 ml of sodium hydroxide to a clean conical flask

2. add a few drops of phenolphthalein indicator and place the conical flask on a white tile

3. fill the burette with hydrochloric acid and record the starting volume

4. slowly open the tap of the burette, and add the acid to the conical flask, swirling to mix

5. stop adding the acid when the end-point is reached (when the colour permanently changes from pink to colourless) and record your final volume

6. repeat steps 1-5 until you get concordant titres (results are within 0.10 ml of each other)

Calculating average titres

When completing a titration, you will have multiple readings which you need to calculate an average reading for.

It is important that the titres you include in your calculation are concordant (within 0.10 ml). If they are not, you will lose marks in an exam.

Rough
Run 1
Run 2
Run 3
final (ml)
15.05
14.70
14.75
14.70
inital (ml)
0.05
0.05
0.00
0.00
titre (ml)
15.00
14.65
14.75
14.70

Here are some results from a titration experiment. The average titre for these results would only include the values from run 1, 2 and 3 as these are the only concordant results.

## Titration Calculations (Higher Only)

Worked Example

NaOH + HCl → NaCl + H₂O

During a titration, 25 ml of sodium hydroxide was used to neutralise hydrochloric acid. This is the same as 0.025 dm³.
If the concentration of the alkali is 0.12 mol/dm³, then the amount of moles used is 0.025 x 0.12 = 0.003 moles.

NaOH
HCl
concentration
0.12
?
volume
0.025
?
moles
0.003
?

If hydrochloric acid (unknown concentration) completely neutralises the alkali then we know that 0.003 moles of acid were also present (because the balanced symbol equation shows a 1:1 ratio of acid reacting with alkali).
If 26.8 ml of acid were needed to neutralise 25 ml of alkali, we know 0.0268 dm³ acid were used.

NaOH
HCl
concentration
0.12
?
volume
0.025
0.0268
moles
0.003
0.003

The concentration of the acid then uses the above equation again (but rearranged):

concentration = moles ÷ volume

So the concentration of the acid used = 0.003 ÷ 0.0268 = 0.11 mol/dm³

## Percentage Yield

A perfect reaction will 100% convert reactant into product. When we work out how much product we’ve made, compared to what we expect, it isn’t always 100% … for a few reasons:

• incomplete reactions (left over reactants)

• practical losses during the experiment (product still left on equipment)

• competing, unwanted reactions (side reactions)

We can work out a % of what we have made, compared to what we expected:

% yield = (actual yield ÷ theoretical yield) x 100

If a reaction should produce 50 g of product, but only produces 42 g, then we can use the above equation to calculate that the percentage yield is 84% ((42 ÷ 50) x 100). ## Atom Economy

Atom economy is a measure of how many of the atoms from the starting material end up in the desired product.

Many reactions make more than one chemical, but often we only really want one of them. We can use this calculation to work out how ‘green’ our reaction is:

% atom economy = (RFM of product ÷ RFM of all reactants) x 100

If we have a high atom economy, then the majority of the atoms we put into the equation are what we get out!

A reaction that only produces one product, will have a 100% atom economy. ## Reaction Pathways (Higher Only)

Usually, there is more than one way to make a particular chemical.

reaction pathway describes the sequence of reactions needed to produce the desired product.

The pathway chosen for a product depends on factors such as:

• percentage yield

• atom economy

• rate of reaction

• equilibrium position

• usefulness of by-products ## Molar Volume (Higher Only)

Avogadro's law states that equal volumes of different gases contain an equal number of molecules (this only applies at room temperature and pressure).

The molar volume is the volume occupied by one mole of molecules of any gas at room temperature and pressure. (The molar volume will be provided as 24 dm³ or 24000 cm³ in calculations where it is required).

Avogadro's law can be used to calculate the volumes of gases involved in reactions:

hydrogen reacts with chlorine to form hydrogen chloride
H₂(g) + Cl₂(g) → 2HCl(g)

The molar ratio of hydrogen to chlorine is 1:1, and of reactants to products is 1:2. This means, for example:

• 1 cm³ of hydrogen reacts exactly with 1 cm³ of chlorine, and will make 2 cm³ of hydrogen chloride

• 250 cm³ of hydrogen reacts exactly with 250 cm³ of chlorine, and will make 500 cm³ of hydrogen chloride

• 1 dm³ of hydrogen reacts exactly with 1 dm³ of chlorine, and will make 2 dm³ of hydrogen chloride

## Calculations with Gases (Higher Only)

Worked example

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

5.0 g of sodium reacts fully with water. Calculate the volume of hydrogen produced. (Ar of Na = 23, molar volume = 24 dm³)

Na
H₂
mass or volume
5.0 g
?
RAM or RFM
23
not needed
moles
0.217
?

If sodium completely reacts with water, then we know that 0.217 moles of sodium will make 0.109 moles of hydrogen (because the balanced symbol equation shows a 2:1 ratio of sodium reacting to make hydrogen). Na
H₂2
mass or volume
5.0 g
?
RAM or RFM
23
not needed
moles
0.217
0.109

1 mole of gas takes up a volume of 24 dm³

So the volume of hydrogen produced = 0.109 x 24 = 2.61 dm³