Revision notes for AQA GCSE Chemistry Sizes of particles and their properties. Open the guide for explanations and worked examples. Written against the AQA GCSE Chemistry (8462) specification, so the content matches what's examinable rather than general Chemistry background.
Sizes of particles and their properties
What you'll learn
- What nanoscience means, and the size range for nanoparticles.
- How nanoparticles compare with atoms, molecules, fine particles and dust.
- Why smaller particles have a larger surface area to volume ratio.
- How this can make nanoparticles behave differently from the same material in bulk.
Why particle size matters
Chemistry is not only about what a substance is made from. Sometimes, the size of the particles also matters.
A lump of a material, a fine powder of the same material, and nanoparticles of the same material can have different properties because different amounts of their surface are exposed.
Definition
Bulk material
A bulk material is a normal, larger-scale sample of a substance, such as a lump, sheet, grain or ordinary powder particle, rather than nanosized particles of it.
Nanometres and standard form
Very small particle sizes are measured in nanometres.
Definition
Nanometre
A nanometre is one billionth of a metre: 1 nm=1×10−9 m1 \text{ nm} = 1 \times 10^{-9} \text{ m}1 nm=1×10−9 m.
Very small numbers are often written in standard form. Standard form means writing a number as a value from 1 up to, but not including, 10 multiplied by a power of 10. For example, 2.5×10−6 m2.5 \times 10^{-6} \text{ m}2.5×10−6 m is in standard form.
Example
Converting nanometres to metres
A nanoparticle has a diameter of 50 nm. Convert this to metres.
- Use the conversion 1 nm=1×10−9 m1 \text{ nm} = 1 \times 10^{-9} \text{ m}1 nm=1×10−9 m, so multiply the number of nanometres by 1×10−91 \times 10^{-9}1×10−9.
- Substitute the value: 50 nm=50×1×10−9 m50 \text{ nm} = 50 \times 1 \times 10^{-9} \text{ m}50 nm=50×1×10−9 m.
- Write the answer in standard form: 50×10−9 m=5.0×10−8 m50 \times 10^{-9} \text{ m} = 5.0 \times 10^{-8} \text{ m}50×10−9 m=5.0×10−8 m.
Nanoscience and particle categories
Definition
Nanoscience and nanoparticles
Nanoscience is the study of structures that are between 1 nm and 100 nm in size. A nanoparticle is a particle in this size range.
A typical atom is around 0.1 nm in size as a GCSE scale estimate, and a small molecule may be about 1 nm. So nanoparticles are bigger than individual atoms and many simple molecules, but much smaller than dust particles.
A diameter is the distance across a particle through its centre. In this topic, you need to know these size ranges:
- Nanoparticles: 1 nm to 100 nm.
- Fine particles, often called PM2.5: 100 nm to 2500 nm, which is 1×10−7 m1 \times 10^{-7} \text{ m}1×10−7 m to 2.5×10−6 m2.5 \times 10^{-6} \text{ m}2.5×10−6 m.
- Coarse particles, often called PM10 or dust: 2500 nm to 10000 nm, which is 2.5×10−6 m2.5 \times 10^{-6} \text{ m}2.5×10−6 m to 1×10−5 m1 \times 10^{-5} \text{ m}1×10−5 m.
PM stands for particulate matter, meaning tiny solid or liquid particles in the air.
Tip
Reading powers of ten
In metres, a more negative power of 10 means a smaller length. For example, 1×10−9 m1 \times 10^{-9} \text{ m}1×10−9 m is smaller than 1×10−6 m1 \times 10^{-6} \text{ m}1×10−6 m.
Example
Estimating atoms across a nanoparticle
A nanoparticle is 20 nm across. Estimate how many atom-sized widths fit across it, using 0.1 nm as the size of an atom.
- Check the units first: both values are in nm, so no conversion is needed.
- Divide the nanoparticle size by the atom size: 20 nm0.1 nm=200\frac{20 \text{ nm}}{0.1 \text{ nm}} = 2000.1 nm20 nm=200.
- The nanoparticle is about 200 atom-sized widths across, so it is on the scale of hundreds of atoms.
Common Mistake
Nano is not just another word for tiny
Do not call every small particle a nanoparticle. For GCSE Chemistry, nanoparticles are specifically in the range 1 nm to 100 nm.
Surface area and volume
To understand why nanoparticles can behave differently, you need two measurements.
Surface area is the total exposed outside area of an object.
Volume is the amount of space an object takes up.
Definition
Surface area to volume ratio
The surface area to volume ratio compares the exposed outside area with the space occupied: surface area to volume ratio=surface areavolume\text{surface area to volume ratio} = \frac{\text{surface area}}{\text{volume}}surface area to volume ratio=volumesurface area.
For a cube with side length aaa:
surface area=6a2volume=a3surface area to volume ratio=6a2a3=6a\begin{aligned} \text{surface area} &= 6a^2 \\ \text{volume} &= a^3 \\ \text{surface area to volume ratio} &= \frac{6a^2}{a^3} = \frac{6}{a} \end{aligned}surface areavolumesurface area to volume ratio=6a2=a3=a36a2=a6This means that as the side length gets smaller, the surface area to volume ratio gets larger.
Key Idea
The factor of 10 rule
If the side length of a cube decreases by a factor of 10, its surface area to volume ratio increases by a factor of 10.
Example
Splitting a cube into smaller cubes
A cube of material has side length 10 cm. It is cut into smaller cubes with side length 1 cm. Show that the total surface area to volume ratio increases by a factor of 10.
- For the original cube, calculate surface area and volume: 6a2=6×(10 cm)2=600 cm26a^2 = 6 \times (10 \text{ cm})^2 = 600 \text{ cm}^26a2=6×(10 cm)2=600 cm2 and a3=(10 cm)3=1000 cm3a^3 = (10 \text{ cm})^3 = 1000 \text{ cm}^3a3=(10 cm)3=1000 cm3.
- Calculate the original surface area to volume ratio: 600 cm21000 cm3=0.6 cm−1\frac{600 \text{ cm}^2}{1000 \text{ cm}^3} = 0.6 \text{ cm}^{-1}1000 cm3600 cm2=0.6 cm−1.
- Work out the number of smaller cubes: 1000 cm31 cm3=1000\frac{1000 \text{ cm}^3}{1 \text{ cm}^3} = 10001 cm31000 cm3=1000. Each small cube has surface area 6×(1 cm)2=6 cm26 \times (1 \text{ cm})^2 = 6 \text{ cm}^26×(1 cm)2=6 cm2, so the total surface area is 1000×6 cm2=6000 cm21000 \times 6 \text{ cm}^2 = 6000 \text{ cm}^21000×6 cm2=6000 cm2.
- Calculate the new ratio: 6000 cm21000 cm3=6 cm−1\frac{6000 \text{ cm}^2}{1000 \text{ cm}^3} = 6 \text{ cm}^{-1}1000 cm36000 cm2=6 cm−1. Compare the ratios: 6÷0.6=106 \div 0.6 = 106÷0.6=10, so the surface area to volume ratio has increased by a factor of 10.
Common Mistake
One particle vs the same mass
A single tiny cube has less surface area than a large cube. The important comparison is the total surface area when the same volume or mass is split into many smaller particles.
Why nanoparticles can have different properties
A property is a characteristic of a material, such as its reactivity, colour, strength, melting point, or how well it works as a catalyst.
Nanoparticles may have properties that are different from the same material in bulk because they have a very high surface area to volume ratio.
This matters because many processes happen at the surface of particles. For example, a catalyst works when reactant particles contact its surface. If more surface is available, more contact can happen.
Key Idea
Why smaller amounts can be effective
For the same mass of material, nanoparticles provide much more total surface area than larger particles, so smaller quantities may be needed to have the same effect.
Example
Explaining why less catalyst may be needed
A catalyst is changed from large particles into nanoparticles of the same material. Explain why a smaller mass of catalyst might still be effective.
- The catalyst works at its surface, because reactant particles must contact the catalyst surface.
- Nanoparticles have a higher surface area to volume ratio, so a larger fraction of their atoms are exposed at the surface.
- Therefore, a smaller mass of nanoparticle catalyst can provide enough surface area for the reaction, so less material may be needed.
Common Mistake
Do not overstate it
The GCSE wording is that nanoparticles may have different properties from the same material in bulk. Do not write that they always have completely different properties.
Exam technique
In the exam
- Learn the key size ranges: nanoparticles are 1–100 nm, fine particles are 100–2500 nm, and coarse particles or dust are 2500–10000 nm.
- For cube questions, use 6a26a^26a2 for surface area and a3a^3a3 for volume, then calculate surface areavolume\frac{\text{surface area}}{\text{volume}}volumesurface area.
- When explaining nanoparticle properties, always link your answer to their high surface area to volume ratio and increased surface contact.
Self review
Check yourself
- What size range, in nm, defines a nanoparticle?
- If the side length of a cube decreases by a factor of 10, what happens to its surface area to volume ratio?
- Why might nanoparticles of a catalyst allow a smaller mass of catalyst to be used?
Bulk and surface properties of matter including nanoparticles (chemistry only)
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