EL1 - Atomic Structure, Mass Spectrometry, and Nuclear Fusion

Question 1

How can relative masses be measured?

Describe what happens.

Answer 1

Using a mass spectrometer:

Vaporisation:
The sample is turned into a gas (vaporised) using an electrical heater.

Ionisation:
The gas particles are bombarded with high-energy electrons to ionise them. Electrons are knocked off the particles, leaving positive ions.

Acceleration:
The positive ions are accelerated by an electric field.

Detection:
The time taken for the positive ions to reach the detector is measured. This depends on an ion’s mass and charge – light, highly charged ions will reach the detector first, while heavier ions with a smaller charge will take longer.

For each sample analysed, a mass spectrum (graph) is produced.

Question 2

What is on each axis of a mass spectrum?

Answer 2

The y-axis gives the abundance of ions, often as a percentage. For an element, the height of each peak gives the relative isotopic abundance, e.g. 75.5% are the 35Cl isotope.

The x-axis units are given as a ‘mass/charge’ ratio. Since the charge on the ions is mostly 1+ you can often assume the x-axis is simply the relative mass.

If the sample is an element, each line will represent a different isotope of the element.

Question 3

How can you work out Ar from a mass spectrum?

Answer 3

Step 1:
For each peak, read the % relative isotopic abundance from the y-axis and the relative isotopic mass from the x-axis.
Multiply them together to get the total mass for each isotope.
e.g. 79 x 24 = 1896; 10 x 25 = 250; 11 x 26 = 286

Step 2:
Add up these totals.
1896 + 250 + 286 = 2432

Step 3: Divide by 100 (since percentages were used)
Ar (Mg) = 2432 / 100 = 24.32 = ~ 24.3

Question 4

How do you calculate number of moles?

Answer 4

moles = number of particles / particles in a mole (avogadro’s constant)

moles = mass / molar mass or mr

moles = (conc in mol dm-3 x vol in cm3) / 1000

Question 5

What is reaction stoichiometry?

Answer 5

How many moles of one reactant react with how many moles of another reactant.

Question 6

Calculate the mass of iron oxide produced if 28.0g of iron is burnt in air:

2Fe + O2 –> Fe2O3

Mr Fe = 55.8 g mol-1
Mr O = 16

Answer 6

moles in 28g of Fe = mass/Mr = 28 / 55.8 = 0.502 moles

From the equation:
2 moles Fe produces 1 mole Fe2O3,
so 0.502 moles Fe produces 0.251 moles Fe2O3.

Mr Fe2O3 = (2 x 55.8) + (3 x 16) = 159.6 g mol-1

Mass Fe2O3 = no. moles x Mr = 0.251 x 159.6 = 40.1g

Question 7

Aluminium reacts with oxygen to form aluminium oxide, Al2O3. Calculate the number of grams of Al2O3 that could be produced if 2.50g of aluminium and 2.50g of oxygen were allowed to react.

Mr Al = 27 g mol-1
Mr O2 = 32 g mol-1

Answer 7

First write the equation for the reaction:
Al (s) + O2 (g) –> Al2O3 (s)

Then balance it:
4Al (s) + 3O2 (g) –> 2Al2O3 (s)

Then work out the moles of aluminium and oxygen:

Mr Al = 27 g mol-1, so moles in 2.50g Al
= mass / Mr = 2.5 / 27 = 0.0926 mol

Mr O2 = 32g mol-1, so moles in 2.5g O2
= mass / Mr = 2.5 / 32 = 0.0781 mol

From the equation:
4 moles Al produces 2 moles Al2O3, so 0.0926 moles Al produces 0.0463 moles Al2O3.
3 moles O2 produces 2 moles Al2O3, so 0.0781 moles O2 produces (0.0781 / 3) x 2 = 0.0521 moles Al2O3.

This means Al is the limiting reagent, as it produces fewer moles of Al2O3, so 2.50g of Al and 2.50g of O2 can only produce 0.0463 moles of Al2O3.

Mr Al2O3 = (2 x 27) + (3 x 16) = 102 g mol-1

So mass of Al2O3 produced = no. moles x Mr
= 0.0463 x 102 = 4.72g.