Calculations in IB Chemistry SL and HL

Below is a list of the calculations are expected to be able to do in IB chemistry. This needs committing to memory,
and each type of calculation will need practice to master. Good luck!

Calculations shaded in grey are only required at HL.

Quantitative Chemistry
Purpose The Calculation Notes
1. Calculating a quantity # ! When calculating numbers of atoms
!"
in moles from a number $ within molecules, multiply the number
of particles of particles (N) by the number of
n = the quantity in moles atoms in the formula
N = the number of particles ! To find out numbers of particles,
L = Avogadro’, con,tant, 6.02x1023 rearrange to: N = n.L
2. Determine relative %& " ∑(!)*+,- /0 12/*34 12/*56 *1337 ! Mr has no unit as it is a relative value
molecular or formula ! To calculate molar mass, Mm, just stick
mass, Mr a ‘g’ for gram, on the end
3. Calculate a quantity in * ! To determine the mass of a given
!"
moles, from a mass of a %8 number of moles of a substance use:
substance * " %8 4 !
n = the quantity in moles !To determine the molar mass of a
m = the mass of substance you are given given mass of a substance:
in grams *
%8 "
Mm = the molar mass of the substance !
4. Determine empirical 1. Divide each % by the atomic mass
formula from % 2. Divide each answer to Step 1 by the smallest answer to Step 1
composition by mass 3. Multiply all answers to Step 1 to remove any obvious fractions
a. If there i, a ‘.5’ multiply everything by 2
b. If there i, a ‘.3’ multiply everything by 3 etc
5. Determine molecular 98 "
%&
49 ! This and the previous calculation are
formula from empirical *(9: 7 : often combined together in exam
formula questions
Fm = molecular formula
Mr = relative molecular mass
Fe = empirical formula
m(Fe) = empirical formula mass
6. Use mole ratios to
!(;7 " !(<7 4
!)*+,- /0 ; 5! ,=)125/! ! The second term in this equation is the
determine the number of !)*+,- /0 < 5! ,=)125/! mole ratio
moles of B that can be ! You must use a fully balanced equation
made from A ! This is the central step in many
stoichiometry calculations
7. Calculate theoretical 1. Use Calculation 6 to determine the !
yield expected quantity of product in
moles
2. Use a rearranged Calculation 3 to
determine the expected mass.
 Purpose The Calculation Notes
8. Determine limiting and Divide moles of each reactant by their ! You would often then need to combine
excess reactants coefficient in the balanced equation this with Calculation 6 to determine a
a. Smallest value ! limiting quantity of product (in moles).
b. Largest value ! excess
9. Calculating percentage > ?5,@A "
162)1@ B5,@A
4 DEE ! Actual and theoretical yield must have
yield. 2C,/-,2561@ B5,@A the same units.
! You might sometimes be required to
rearrange this equation, or use it to
work backwards from this to find the
amount of reactant you started with.
10. Apply Avogadro’, Law FG FH ! Only applies when temperature and
"
to calculate reacting !G !H pressure remain constant.
volumes of gases ! Units of V do not matter. But must be
V1 = the initial volume of gas the same.
n1 = the initial quantity of gas in moles ! This is really a special case of the Ideal
V2 = the final volume of gas Gas Law where the pressure,
n2 = the final volume of gas in moles temperature and gas constant terms
cancel each other out.
11. Calculate molar F ! Only applies at standard conditions:
!"
quantities of gases at II4J o 273 K (0oC)
standard temperature o 101,000 Pa (1.00 atm)
and pressure n = quantity in moles ! If volume of gas is given in m3, use
V = the volume of gas in dm3 2.24x10-5 as your molar volume.
22.4 is the molar volume of an ideal gas ! Molar volumes are given in the data
at 273K and 101,000 Pa booklet and do not need memorising.
12. The ideal gas KF " !LM ! In practice, you can often use:
equation o V in units of dm3
P = pressure in Pa o Pa in units of kPa
V = volume of gas in m3 ! You will need to be comfortable
n = quantity of gas in moles rearranging this equation to change
R = the gas constant, 8.31 the subject.
T = temperature in Kelvin (OC + 273) ! This takes time to use, so only use it in
non-standard conditions, or when the
laws in Calculation 13 would not be
quicker.
13. Relationship between Charles’ Law, at fixed pressure: ! These only work where the quantity in
temperature, pressure FG FH moles remains fixed.
"
and volume MG MH ! All of these are just special cases of the
ideal gas law, where the remaining
Gay-Lussac’s Law, at fixed volume: terms just cancel each other out.
KG KH
"
MG MH

Boyle’s Law, at fixed temperature:

KG FG " KH FH

V1 and V2 are initial and final volume

P1 and P2 are initial and final pressure
T1 and T2 are initial and final temperature
 Purpose The Calculation Notes
14. Molar Concentration ! ! Unit, are pronounced ‘mole, per
6"
F decimetre cubed’
! You need to be able to use any
c = concentration in mol dm-3
rearrangement of this equation
n = the quantity in moles
V = the volume in dm3 (litres)
15. Concentration by * ! Unit, are pronounced ‘gram, per
6"
mass F decimetre cubed’
! You need to be able to use any
c = concentration in g dm-3 rearrangement of this equation
m = the quantity in grams
! Generally, if you have a concentration
V = the volume in dm3 (litres)
like this, you should convert it into a
molar concentration before
proceeding.

Atomic Structure
Purpose The Calculation Notes
16. Determine relative > 1+)!A1!6, ! % abundance may be given in the
<& " ∑(12/*56 *1334 7
atomic mass from % DEE question, or you may need to read it
abundance data from a mass spectrum
! If you convert the percentages to
Ar = relative atomic mass decimals (i.e. 0.8 for 80%, 0.25 for
25%), there is no need to divide by
100.
17. Determine % If there are two isotopes, label one of ! Ar, Ia and Ib will be provided in the
abundance from relative them ‘a’ and one ‘b’. question, so you can plug the numbers
atomic mass in, and then rearrange to find x.
Now: ! To find y, simply do y=100-x
! If you have three isotopes, you must
NOP Q BOR know the abundance of at least one in
<& "
NQB order to find the other two. You would
also need to subtract the abundance
Ar = relative atomic mass of this one from the 100, before doing
Ia = the mass of isotope a the rest of the sum.
Ib = the abundance of isotope b
x and y is the abundance of each isotope

However, since x+y = 100%, y = 100-x so:

NOP Q (DEE S N7OR

<& "
DEE

Periodicity and Bonding

Purpose The Calculation Notes
No calculations. Huzzah!
Energetics
Purpose The Calculation Notes
18. Calculating the heat = " *6TM ! Be careful of the unit, of ma,,…you
change of a pure may need to convert kg into g
substance q = the heat change in Joules ! Be careful of the units for specific heat
m = the mass of substance in grams capacity, if it is J K-1 kg-1 you will need
c = specific heat capacity in J K-1 g-1 to convert your mass into kg.
∆T = temperature rise in K or OC
19. Calculating an TU " S*6TM ! The minus sign is needed to ensure
enthalpy change from that an exothermic reaction has a
experimental data ∆H = the enthalpy change in Joules negative enthalpy change.
m = the mass of solution in grams ! Units are J or kJ mol-1
c = specific heat capacity ! The mass of solution is assumed to be
of water: 4.18 J K-1 g-1 the same as its volume in cm3.
∆T = temperature rise in K or OC ! The specific heat capacity of the
reactants is ignored.
20. Calculating ∆Hr using Once you have produced you Hess cycle: ! See the ‘Energetic,’ PowerPoint for
a Hess cycle 1. Write the relevant ∆H onto each advice on constructing Hess cycles.
arrow
2. Multiply each ∆H in accordance
with the stoichiometry
3. To do your sum, add when you go
with an arrow, and subtract when
you go against one.
21. Calculating ∆Hr from Method 1: Make a Hess cycle, then do as ! It is more reliable to use Hess cycles
average bond enthalpies in Calculation 20. and you can easily forget whether it is
reactants – products or vice versa.
Method 2: ! Average bond enthalpies can be found
TU& " ∑(-,1621!2 +/!A37 S ∑(V-/A)62 +/!A37 in Table 10 of the data booklet.
! You only need to worry about the
bonds that broken and made. If a
bond, for example a C-H is present at
the start and finish, you can ignore
it….thi, can ,ave time in exam,.
22. Calculating ∆Hr from Method 1: Make a Hess cycle, then do as ! It is more reliable to use Hess cycles
enthalpies of formation in Calculation 20. and you can easily forget whether it is
products – reactants or vice versa.
Method 2: ! ∆Hof for elements in their standard
TU& " ∑ TUWX (V-/A)6237 S ∑ TUWX (-,1621!237 states is zero.
! ∆Hof values for many compounds can
∆Hr = enthalpy change of reaction be found in Table 11 of the data
∆Hof = enthalpy change of formation booklet.
! In some questions, you may also need
to take a state change into account, if
standard states are not used.
23. Calculating ∆Hr from Method 1: Make a Hess cycle, then do as ! It is more reliable to use Hess cycles
enthalpies of combustion in Calculation 20. and you can easily forget whether it is
reactants – products or vice versa.
Method 2: ! ∆Hoc for CO2 and H2O is zero.
TU& " ∑ TUYX (-,1621!237 S ∑ TUYX (V-/A)6237 ! ∆Hoc values for many compounds can
be found in Table 12 of the data
∆Hr = enthalpy change of reaction booklet.
∆Hoc = enthalpy change of combustion
 Purpose The Calculation Notes
24. Calculating lattice You need to build a Born-Haber cycle….,ee the Energetic, PowerPoint for help.
enthalpy
25. Calculating ∆S from TZ X " ∑ Z X (V-/A)6237 S ∑ Z X (V-/A)6237 ! Units are J K-1 mol-1
standard entropy values ! So values can be found in Table 11 of
∆So = standard entropy change of the data booklet
reaction ! You cannot assume that So of an
So = standard entropy of each substance element is zero. It is not.

26. Calculating ∆Gr Method 1: Make a Hess cycle, then do ! Units are J or kJ mol-1
,tandard Gibb’, Free similar to Calculation 20. ! It is more reliable to use Hess cycles
Energy of Formation and you can easily forget whether it is
values Method 2: products – reactants or vice versa.
T[& " ∑ T[WX (V-/A)6237 S ∑ T[WX (-,1621!237 ! ∆Gof for elements in their standard
states is zero.
∆Gr = Gibb’, free energy of reaction ! ∆Gof values for many compounds can
∆Gof = Gibb’, free energy of formation be found in Table 11 of the data
booklet.
27. Calculating ∆Gr from T[ " TU S MTZ ! If ∆H i, in kJ mol-1, you will need to
experimental data divide ∆S by 1000 to convert it to unit,
∆G = Gibb’, free energy of kJ K-1 mol-1
∆H = Enthalpy change ! You may fir,t need to calculate ∆H and
T = Temperature in Kelvin ∆S using Calculations 23 and 24.
∆S = Entropy change

Kinetics
Purpose The Calculation Notes
28. Calculate the rate of a ST\L] T\K] ! Units are mol dm-3 s-1
L12, " "
reaction T2 T2 ! The minu, ,ign in front of ∆[R] i,
because the concentration of
∆[R] = change in reactant concentration reactants decreases
∆[P] = change in product concentration
∆t = change in time
29. Calculate the gradient 6C1!^, 5! B ! Used for calculating rate from graph of
[-1A5,!2 "
of a slope 6C1!^, 5! N concentration (y-axis) over time (x-
axis)
30. Determine the order 1. Identify two experiments, where the ! Sometime,, you can’t find two ca,e,
of reaction with respect concentration of ‘x’ ha, changed, but where only [x] has changed, in which
to a reactant, x. all others have remained the same. case you may need to take into
2. Compare the change in [x] to the account the order of reaction with
change in rate: respect to other reactants.
a. If doubling [x] has no effect
on rate, then 0th order.
b. If doubling [x] doubles rate,
then 1st order.
c. If doubling [x] quadruples
rate, then 2nd order.
 Purpose The Calculation Notes
31. Deduce a rate `
L12, " _\<] \;] \b] a c
! Reactants with a reaction order of zero
expression can be omitted from the rate equation
k = rate constant (see below) ! Given suitable information, you may
[A/B/C] = concentration of each reactant need to calculate the value of the rate
x/y/z = order of reaction with respect to constant if given rates, concentrations
each reactant and reaction order, or the expected
rate given the other information.
32. Determining the units */@ A*ef 3 eG ! If you can’t under,tand thi,, try to
for the rate constant d!523 " memorise:
*/@ ` (A*ef 7`
o 0th order: mol dm-3 s-1
! Mol and dm-3 terms on the top and o 1st order: s-1
the bottom should be cancelled out o 2nd order: mol-1 dm3 s-1
! Remaining mol and dm-3 terms on the o 3rd order: mol-2 dm6 s-1
bottom should then be brought to
the top by inverting their indices.

x = the overall order of the reaction

33. Determining The Arrhenius equation: ! Units of Ea are kJ mol-1
egh
activation energy
_" <, ij ! From the graph, you might also be
asked to calculate A:
Where: o ‘ln A’, i, the y-intercept of the
o k = rate constant graph, ,o ,imply rai,e ‘e’ to the
o A = Arrhenius constant power of this intercept.
o e = exponential constant
o Ea = activation energy
o R = gas constant, 8.31
o T = temperature in Kelvin

This rearranges to:

SmP D
kl _ " 4 Q kl <
L M

Which is basically the equation for a

,traight line in the form ‘y=mx+c’ where:
o ‘y’ i, ln k
o ‘x’ i, 1/T
o ‘m’ i, –Ea/R
o ‘c’ i, ln A
So, if we do a reaction at a range of
temperatures and calculate ln k, then:
! Draw a graph with 1/T along the x-
axis, and ln k on the y-axis
! Draw a straight line of best fit.
! Determine the gradient of the line
! Then:
mP " S[-1A5,!2 n L
Equilibrium
Purpose The Calculation Notes
34. Determining the For the reaction: ! Only applies to reactants for which
equilibrium constant wA + zB ! yC + zD there is a concentration:
expression o Aqueous substances are
\b]a \p] c included
oY "
\<]q \;] c o Solids and pure liquids are
omitted as they do not have a
Kc = equilibrium constant concentration per se
! For reactions involving gases, use
partial pressures instead of
concentrations.
! At SL, you only need be able to
construct the expression.
! At HL, you may need to calculate Kc
from the expression and suitable data,
or to determine reactant equilibrium
concentrations of reactants, given Kc
and suitable information.

Acids and Bases

Purpose The Calculation Notes
35. Determining changes ! For each increa,e of ‘1’ on the pH ! At SL, you do not need to be able to
in [H+] given changes in scale, divide [H+] by 10. calculate pH, just to understand it in
pH ! For each decrea,e of ‘1’ on the pH relative terms.
scale, multiply [H+] by 10
36. Deduce [H+] and [OH-] oq " \U r ]4 \sU e ] ! At standard conditions, Kw = 1.00x10-14
given Kw ! Kw varies with temperature, so it is
So: important to know how to do this
oq calculation.
\U r ] "
\sU e ]

And:
oq
\sU e ] "
\U r ]

37. Calculating pH from VU " S@/^Gt \U r ] ! With strong acids, you can assume [H+]
[H+] and vice versa is the same as the concentration of the
\U r ] " DEeuv acid (adjusted for the stoichiometry)
! With weak acids, you will need to
calculate [H+] using Ka or pKa.
38. Calculating pH from VsU " S@/^Gt \sU e ] ! With strong bases, you can assume
[OH-] and vice versa [OH-] is the same as the concentration
\sU e ] " DEeuwv of the acid (adjusted for the
stoichiometry)
! With weak acids, you will need to
calculate [OH-] using Kb or pKb.
 Purpose The Calculation Notes
39. Determining Ka and Kb oq " oP 4 oR ! This is useful when trying of determine
of acids/bases and their the strength the conjugate base of a
conjugate bases/acids So: weak acid, and the conjugate acid of a
oq weak base.
oP "
oR

And:
oq
oR "
oP

40. Determining pKa and Voq " VoP Q VoR ! This is useful when trying of determine
pKb of acids/bases and the strength the conjugate base of a
their conjugate So: weak acid, and the conjugate acid of a
bases/acids VoP " Voq S VoR weak base.

And:
VoR " Voq S VoP

41. Determining pH from Voq " VoP Q VoR ! This is useful to quickly and easily
pOH and vice versa calculate one of pH/pOH from the
So: other (or [H+]/[OH-]).
VU " DJ S VsU

And:
VsU " DJ S VU

42. Calculating pH of a \U r ]\<e ] ! We assume that [HA] is equal to the

solution of a weak acid oP "
\U<] concentration stated in the question,
as only a very small amount has
Since [H+] = [A-]: dissociated.
\U r ]H ! You may need to work backwards from
oP " pH to work out Ka:
\U<]

So: \U r ] " DEeuv

\U r ] " √oP 4 \U<]
Then:
Then: \U r ]H
oP "
VU " S@/^Gt \U r ] \U<]

! You will not need to work out

problems for polyprotic weak acids
that would require quadratics.
 Purpose The Calculation Notes
43. Calculating pOH of a \;r ]\sU e ] ! We assume that [BOH] is equal to the
solution of a weak base oR "
\;sU] concentration stated in the question,
as only a very small amount has
Since [B+] = [OH-]: dissociated.
\sU e ]H ! You may need to work backwards from
oR " pOH to work out Kb:
\;sU]

So: \sU e ] " DEeuwv

\sU e ] " √oR 4 \;sU]
Then:
Then: \sU e ]H
oR "
VsU " S@/^Gt \sU e ] \;sU]

44. Determining pH of \U r ]\<e ] ! You may need to use your

acidic buffer solutions oP "
\U<] stoichiometry to calculate the [HA]
(and alkali buffers) and [A-] in the buffer solution first.
Now [H+] is not equal to [A-]: ! Since additional A- is added, we now
need to use the concentration of A-
So: from the buffer as our [A-].
oP 4 \<e ] ! [HA] should either be that stated in the
\U r ] "
\U<] question, or that calculated via
stoichiometry, depending on the
Then: context.
VU " S@/^Gt \U r ]
! For alkali buffers, do the same process
but with OH-, Kb etc.

Oxidation and Reduction

Purpose The Calculation Notes
X X (612C/A,7 X
!
o
45. Calculate E cell mY:yy "m S m (anode) The value should always be positive, so
if you get it the wrong way round, just
Eocell = cell potential take off the minus sign.
Eo = standard electrode potential ! Eo can be found in the data booklet.
! You do not need to take the
stoichiometry of the reaction into
account, just use the Eoas they are in
the data booklet.

Organic Chemistry
Purpose The Calculation Notes
No calculations. Huzzah!
Measurement and Processing
Purpose The Calculation Notes
46. Calculating relative ! Large measurements have lower
uncertainties (in %) L,@125z, )!6,-215!2B " relative uncertainties
PR{Xy|}: |~Y:&}P•~}a
.100
{•c: XW €Py|:

47. Calculating absolute ! This only works when adding and

uncertainty when M/21@ )!6,-215!2B subtracting values with the same units
addition/subtraction is " ∑(1+3/@)2, )!6,-215!25,37
involved

48. Calculating relative ! For use when you are

uncertainty when M/21@ )!6,-215!2B multiplying/dividing values with
multiplication/division is " ∑(-,@125z, )!6,-215!25,37 different units
involved

Human Biochemistry
Purpose The Calculation Notes
47. Calculating the energy Use: ! You may need to convert the energy
value of food from value into a value per mole by dividing
combustion data = " *6TM q by the number of moles of substance
burnt.
q = the heat change in Joules
m = the mass of substance in grams
c = specific heat capacity in J K-1 g-1
∆T = temperature rise in K or OC

48. Calculating iodine DEE ! All units should be grams

#(OH 7 " *(OH 74
numbers *(@5V5A7 ! You may need to convert from data
involving bromine to an ‘iodine
N(I2) = the iodine number equivalent’…ju,t u,e your
m(I2) = the mass of iodine reacting in g stoichiometry
m(lipid) = the mass of lipid involved in g
 Purpose The Calculation Notes
49. Calculate the number Calculate the quantity in moles of I2 ! This works because each mole of
of double bonds in a lipid corresponding to the iodine number: double bonds reacts with one mole of
using iodine number. #(OH 7 iodine.
!(OH 7 "
%& (OH 7

Calculate the quantity in moles of lipid in

100g:
DEE
!(@5V5A7 "
%& (@5V5A7

Then the number of double bonds is:

!(OH 7
p/)+@, +/!A3 "
!(@5V5A7

Where:
N(I2) = the iodine number
n = quantities in moles
Mm = molar masses

Medicines and Drugs

Purpose The Calculation Notes
No calculations. Huzzah!