A particularly helpful and entertaining source of everything IB chemistry related is Richard Thornley, who has his own YouTube Channel which can be found here. His material is completely up to date, and even includes material for the new syllabus starting in 2016. His playlist for this topic is very detailed, and I find it useful to note and rewatch his videos at least twice.
Another source of good notes and a copy of the syllabus for this topic is ibchem.com. An easy method to study is to remember all the definitions that you are required to know off by heart. The best way to do this is rote learning: simply write, remember and repeat, until you’re 100% sure you can recall it in an instant.
This will help you a lot in topic tests, in which there are at least 3 easy marks asking you to Define, explain or how does certain things, and if you have these things pinned down, they’re essentially free marks. Each textbook also has it’s own summary questions, or something similar. I recommend working on these whenever you’re about to have a test, and to only look at the answers after you’ve completed the questions.
This topic is relatively straightforward, just be sure to memorise the definitions included in the topic; these are just easy marks in tests and exams. When it comes to equations, you don’t need to remember these, only need to know how to apply them. There aren’t that many equations in this topic, save for n = m/M, and the gas equations. When writing up chemical equations in this topic, make sure you include the equations of state, (g) for gas, (l) for liquid, and (s) for solids, as teachers can get quite picky about these things.
Like I said before, the best way to study this chapter is to simply go over the questions included in the textbook, and do any questions that your teacher provides you. Makes sure you raise questions about problems that you’re having, this is what teachers are for!
Chapter Notes:
1.1.1 – Apply the mole concept to substances
- a mole is the name given to a quantity – similar to a dozen etc.
- represents 6.02 x 10^23 particles
- this value is also known as the Avagadro’s constant
Application
- if we have 0.250 mol. of Carbon the number of atoms Carbon present are: 0.250 x 6.02 x 10^23
- if there is 1 mol of a substance then there are 6.02 x 10^23 particles of that substance
1.1.2 – Determine the number of particles and the amount of substance (in moles)
𝒏 = 𝑵/6.02 x 10^23
where:
n = number of moles
N = number of particles in moles
1.2.1 – Define the terms relative atomic mass (Ar) and relative molecular mass (Mr)
Ar: the weighted mean mass of all the naturally occurring isotopes of an atom
Mr: the sum of the relative atomic masses of constitute elements, i.e. in a substance, compound, etc
1.2.2 – Calculate the mass of one mole of a species form its formula & 1.2.3 – Solve problems involving the relationship between the amount of substance in moles, mass and molar mass
- the mass of one mol of a species is equal to the sum of all the Mr of its constitute atoms
- this is also known as the Molar Mass
n=m/M
where:
n = number of moles
m = mass in grams
M = Molar mass
1.2.4 – Distinguish between the terms empirical formula and molecular formula
Empirical Formula – The simplest whole number ratio of atoms of different elements in the compound
Molecular Formula – The actual number of atoms of different elements covalently bonded in a molecule
1.2.5 – Determine the empirical formula from the percentage composition or from other experimental data
1.2.6 – Determine the molecular formula when given both the empirical formula and experimental data
1.3.1 – Deduce chemical equations when all the reactants and products are given
1.3.2 – Identify the mole ratio of any two species in a chemical equation
the mole ration in a chemical equation is simple the ratio of moles of the molecules in the equation. This can be observed through the number preceding the chemical formula of substances in a chemical equation.
1.3.3 – Apply the state symbols (s), (l), (g) and (aq)
The state symbols show what state the substances in a chemical reaction are in when they react. this is done be writing the state symbol in the lower right corner of the formula for a substance.
s = solid
l = liquid
g = gas
aq = aqueous
1.4.1 – Calculate theoretical yields from chemical equations
it is possible to calculate the theoretical yields of a chemical reaction through a chemical equation by looking at the mol. ratio of the substances in the equation.
The theoretical yields can be calculated by determining the mol. of one substance in the reaction, the moles of the other substances (used or produced) are in compliance with mol. ratio of the chemical equation. This is not applicable if there are limiting reagents of the reaction is not a complete reaction.
1.4.2 – Determine the limiting reactant in excess when quantities of reacting substances are given
A limiting reactant(reagent) is the reactant that is fully consumed in the reaction. This can be calculated by determining the number of moles of each reactant available for the reaction then by using the mol. ratio for the chemical equation it is possible to observe how many mol. of one substance must be present due to there being x mol. of the other reactant. The substance that limits the amount of moles of the other reactants that react is the limiting reagent.
1.4.3 – Solve problems involving theoretical, experimental and percentage yield
1.4.4 – Apply Avogadro’s law to calculate reacting volumes of gases
1.4.5 – Apply the concept of molar volume at standard temperature and pressure in calculations
1.4.6 – Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas
1.4.8 – Analyse graphs relating to the ideal gas equation
1.5.1 – Distinguish between the terms solute, solvent, solution and concentration (g dm-3 and mol dm-3)
Fresh Questions & Good Talks 