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Module 2: Chapter 3 (Amount of substance)

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Total cards: 24•Content version: v4•Public deck•Updated: 8/1/2026

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#1
Front (Question)

How is mol calculated?

Back (Answer)

amount, n = mass, m / molar mass, M

#2
Front (Question)

How is the number of ‘particles’ calculated?

Back (Answer)

n x Na (6.02 x 10^23)

#3
Front (Question)

What is the molecular formula?

Back (Answer)

the number of atoms of each element in a molecule

#4
Front (Question)

What is the empirical formulae?

Back (Answer)

the simplest whole-number ration of atoms in each element in a compound

#5
Front (Question)

What is the water of crystallisation?

Back (Answer)

• Many crystalline solids are hydrated, water molecules are incorporated into their crystalline structure • this water is known as the water of crystallisation

#6
Front (Question)

What is the approach to calculating the formula of a hydrated solid?

Back (Answer)

• subtract mass of crucible & anhydrous salt from mass of crucible; hydrated salt and use answer to calculate the moles of water • subtract the mass of crucible from mass of crucible & anhydrous salt and use the answer to calculate the number of moles in the salt • divide the moles of water by the moles of the compound

#7
Front (Question)

What is the equation to calculate mol of a gas at RTP?

Back (Answer)

mol (n) = volume of gas (dm^3) / 24dm^3

#8
Front (Question)

What is the ideal gas equation?

Back (Answer)

pV = nRT p -> pressure (Pa) V -> volume (m^3) n -> moles of gas R -> ideal gas constant (8.314) T -> temperature (Kelvin)

#9
Front (Question)

How do you convert between Pa and KPa?

Back (Answer)

Pa -(÷1000)-> KPa KPa -(x1000)-> Pa

#10
Front (Question)

How do you convert between cm^3, dm^3 and m^3?

Back (Answer)

cm^3 -(÷1000)-> dm^3 -(÷1000)-> m^3 m^3 -(x1000)-> dm^3 -(x1000)0-> cm^3 cm^3 -(÷1 000 000)-> m^3 m^3 -(x1 000 000)-> cm^3

#11
Front (Question)

How do you convert between K and Celsius?

Back (Answer)

C -(+273)-> K K -(-273)-> C

#12
Front (Question)

What is the concentration of a solution?

Back (Answer)

The concentration of a solution tells us the amount of solute in a particular volume of solution

#13
Front (Question)

How is concentration calculated?

Back (Answer)

concentration (moldm^-3) = mol (n) / volume (dm^3)

#14
Front (Question)

How do you convert between conc in gdm^-3 and moldm^-3?

Back (Answer)

gdm^-3 -(÷Mr)-> moldm^-3 moldm^-3 -(xMr)-> gdm^-3

#15
Front (Question)

What is meant by the ‘amount of substance’?

Back (Answer)

• n - counts the number of particles in a substance measured in the unit mole, mol.

#16
Front (Question)

Explain what a mole is

Back (Answer)

One mole is the amount of a substance that contains 6.02x10^23 particles (Avogadro constant)

#17
Front (Question)

What is Avogadro’s constant?

Back (Answer)

Is 6.02x10^23 mol^-1, the number of particles in each mole of carbon-12 - C-12 is the standard measurement of relative atomic masses

#18
Front (Question)

What is Molar Mass, M?

Back (Answer)

• molar mass is the mass per mole of a substance - the units of molar mass are gmol^-1

#19
Front (Question)

What is molar gas volume?

Back (Answer)

• the molar gas volume Vm is the volume per mole of gas molecules at a stated temperature and pressure • volume of gas depends on the pressure and temperature, most experiments are at RTP

#20
Front (Question)

What is meant by RTP?

Back (Answer)

• RTP is about 20 degrees celcius and 101 kPa (1 atm) • at RTP 1 mole of gas molecules has a volume of approx 24.0 dm^3 • therefore at RTP the molar gas volume = 24.0 dm^3

#21
Front (Question)

What is meant by the term hydrated?

Back (Answer)

a crystalline compound containing carbon and hydrogen only

#22
Front (Question)

What does anhydrous mean?

Back (Answer)

containing no water molecules

#23
Front (Question)

How do you calculate percentage yield

Back (Answer)

actual yield / theoretical yield x100

#24
Front (Question)

Why is theoretical yield hard to achieve?

Back (Answer)

• reaction may not have gone to completion • other reactions may have taken place alongside the reaction (side reactions) • purification of the product may result in the loss of some product