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
1. Calculating a
quantity in moles
from a number of
particles

2. Determine
relative molecular or
formula mass, Mr

3. Calculate a
quantity in moles,
from a mass of a
substance

4. Determine
empirical formula
from % composition
by mass
5. Determine
molecular formula
from empirical
formula

6. Use mole ratios to

determine the
number of moles of
B that can be made
from A

The Calculation
n=

N
L

Notes

n = the quantity in moles

N = the number of particles
L = Avogadros constant,
6.02x1023

M r= (number of atoms . atomic mass)

n=

m
Mm

n = the quantity in moles

m = the mass of substance you
are given in grams
Mm = the molar mass of the
substance

When calculating numbers of

atoms within molecules,
multiply the number of
particles (N) by the number of
atoms in the formula
To find out numbers of
particles, rearrange to: N = n.L
Mr has no unit as it is a relative
value
To calculate molar mass, Mm,
just stick a g for grams on the
end
To determine the mass of a
given number of moles of a
substance use:

m=M m . n

To determine the molar mass

of a given mass of a substance:

M m=

m
n

1. Divide each % by the atomic mass

2. Divide each answer to Step 1 by the smallest answer to Step 1
3. Multiply all answers to Step 1 to remove any obvious fractions
a. If there is a .5 multiply everything by 2
b. If there is a .3 multiply everything by 3 etc
Mr
This and the previous
Fm =
. Fe
calculation are often combined
m(F e )
together in exam questions
Fm = molecular formula
Mr = relative molecular mass
Fe = empirical formula
m(Fe) = empirical formula mass

n ( B )=n ( A ) .

number of Bequation
number of A equation

The second term in this

equation is the mole ratio
You must use a fully balanced
equation
This is the central step in many
stoichiometry calculations

Purpose
7. Calculate
theoretical yield

8. Determine
limiting and excess
reactants

9. Calculating
percentage yield.

10. Apply
Avogadros Law to
calculate reacting
volumes of gases

The Calculation
1. Use Calculation 6 to
determine the expected
quantity of product in moles
2. Use a rearranged Calculation
3 to determine the expected
mass.
Divide moles of each reactant by
their coefficient in the balanced
equation
a. Smallest value
limiting
b. Largest value excess

Yield=

actual yield
.100
theoretical yield

V1 V2
=
n 1 n2
V1 = the initial volume of gas
n1 = the initial quantity of gas in
moles
V2 = the final volume of gas
n2 = the final volume of gas in
moles

11. Calculate molar

quantities of gases
at standard
temperature and
pressure

12. The ideal gas

equation

n=

V
22.4

n = quantity in moles
V = the volume of gas in dm3
22.4 is the molar volume of an
ideal gas at 273K and 101,000 Pa

PV =nRT
P = pressure in Pa
V = volume of gas in m3
n = quantity of gas in moles
R = the gas constant, 8.31
T = temperature in Kelvin (OC +
273)

Notes

You would often then need to

combine this with Calculation 6
to determine a quantity of
product (in moles).

Actual and theoretical yield

must have 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.
Only applies when temperature
and pressure remain constant.
Units of V do not matter. But
must be the same.
This is really a special case of
the Ideal Gas Law where the
pressure, temperature and gas
constant terms cancel each
other out.

Only applies at standard

conditions:
o 273 K (0oC)
o 101,000 Pa (1.00 atm)
If volume of gas is given in m3,
use 2.24x10-5 as your molar
volume.
Molar volumes are given in the
data booklet and do not need
memorising.
In practice, you can often use:
o V in units of dm3
o Pa in units of kPa
You will need to be comfortable
rearranging this equation to
change the subject.
This takes time to use, so only
use it in non-standard
conditions, or when the laws in
Calculation 13 would not be
quicker.

Purpose

The Calculation

13. Relationship
between
temperature,
pressure and volume

Charles Law, at fixed pressure:

V1 V2
=
T 1 T2

Notes

Gay-Lussacs Law, at fixed

volume:

P1 P2
=
T1 T 2

These only work where the

quantity in moles remains
fixed.
All of these are just special
cases of the ideal gas law,
where the remaining terms just
cancel each other out.

Boyles Law, at fixed

temperature:

P1 V 1=P 2 V 2

V1 and V2 are initial and final

volume
P1 and P2 are initial and final
pressure
T1 and T2 are initial and final
temperature
14. Molar
Concentration

c=

n
V

c = concentration in mol dm-3

n = the quantity in moles
V = the volume in dm3 (litres)
15. Concentration
by mass

c=

m
V

c = concentration in g dm-3
m = the quantity in grams
V = the volume in dm3 (litres)

Units are pronounced moles

per decimetre cubed
You need to be able to use any
rearrangement of this equation

Units are pronounced grams

per decimetre cubed
You need to be able to use any
rearrangement of this equation
Generally, if you have a
concentration like this, you
should convert it into a molar
concentration before
proceeding.

Atomic Structure
Purpose
16. Determine
relative atomic mass
from % abundance
data

The Calculation
abundance
atomic mass .
100
()
A r =

Ar = relative atomic mass

Notes

% abundance may be given in

the question, or you may need
to read it from a mass
spectrum
If you convert the percentages
to decimals (i.e. 0.8 for 80%,
0.25 for 25%), there is no need
to divide by 100.

Purpose
17. Determine %
abundance from
relative atomic mass

The Calculation
If there are two isotopes, label
one of them a and one b.

Notes

Now:

xI + yI b
Ar= a
x+ y

Ar = relative atomic mass

Ia = the mass of isotope a
Ib = the abundance of isotope b
x and y is the abundance of each
isotope

Ar, Ia and Ib will be provided in

the question, so you can plug
the numbers in, and then
rearrange to find x.
To find y, simply do y=100-x
If you have three isotopes, you
must know the abundance of
at least one in order to find the
other two. You would also need
to subtract the abundance of
this one from the 100, before
doing the rest of the sum.

However, since x+y = 100%, y =

100-x so:

Ar=

xI a +(100x )I b
100

Periodicity and Bonding

Purpose

The Calculation

Notes

No calculations. Huzzah!

Energetics
Purpose
18. Calculating the
heat change of a
pure substance

19. Calculating an
enthalpy change
from experimental
data

The Calculation
q=mc T
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

H=mc T
H = the enthalpy change in
Joules
m = the mass of solution in
grams
c = specific heat capacity
of water: 4.18 J K-1 g-1
T = temperature rise in K or OC

Notes

Be careful of the units of

massyou may need to
convert kg into g
Be careful of the units for
specific heat capacity, if it is J
K-1 kg-1 you will need to convert
your mass into kg.
The minus sign is needed to
ensure that an exothermic
reaction has a negative
enthalpy change.
Units are J or kJ mol-1
The mass of solution is
assumed to be the same as its
volume in cm3.
The specific heat capacity of
the reactants is ignored.

Purpose
20. Calculating Hr
using a Hess cycle

21. Calculating Hr
from average bond
enthalpies

22. Calculating Hr
from enthalpies of
formation

The Calculation
Once you have produced you
Hess cycle:
1. Write the relevant H onto
each 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.
Method 1: Make a Hess cycle,
then do as in Calculation 20.

Notes

See the Energetics PowerPoint

for advice on constructing Hess
cycles.

It is more reliable to use Hess

cycles and you can easily
forget whether it is reactants
Method 2:
products or vice versa.
Average bond enthalpies can
H r = ( reactant bonds ) ( product bonds)
be found 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.this
can save time in exams.
Method 1: Make a Hess cycle,
It is more reliable to use Hess
then do as in Calculation 20.
cycles and you can easily
forget whether it is products
Method 2:
reactants or vice versa.
products
Hof for elements in their
reactants
standard states is zero.
H of ()
Hof
values
for
many
o
compounds can be found in
H f ()
Table 11 of the data booklet.
H r=
In some questions, you may
also need to take a state
change
into
account,
if
Hr = enthalpy change of
standard states are not used.
reaction
o
H f = enthalpy change of
formation

Purpose
23. Calculating Hr
from enthalpies of
combustion

The Calculation
Method 1: Make a Hess cycle,
then do as in Calculation 20.

Notes

Method 2:

reactants
products
H oc ()
H oc ()
H r=

It is more reliable to use Hess

cycles and you can easily
forget whether it is reactants
products or vice versa.
Hoc for CO2 and H2O is zero.
Hoc
values
for
many
compounds can be found in
Table 12 of the data booklet.

Hr = enthalpy change of
reaction
Hoc = enthalpy change of
combustion
24. Calculating
lattice enthalpy
25. Calculating S
from standard
entropy values

You need to build a Born-Haber cycle.see the Energetics PowerPoint

for help.
products
Units are J K-1 mol-1
products
So values can be found in Table
S o ()
11 of the data booklet
o
You cannot assume that So of
S ()
o
an element is zero. It is not.

S =

So = standard entropy change of

reaction
So = standard entropy of each
substance
26. Calculating Gr
standard Gibbs Free
Energy of Formation
values

Method 1: Make a Hess cycle,

then do similar to Calculation 20.

Method 2:

products
reactants
Gof ()
Gof ()
G r=

Units are J or kJ mol-1

It is more reliable to use Hess
cycles and you can easily
forget whether it is products
reactants or vice versa.
Gof for elements in their
standard states is zero.
Gof
values
for
many
compounds can be found in
Table 11 of the data booklet.

Gr = Gibbs free energy of

reaction
Gof = Gibbs free energy of
formation
27. Calculating Gr
from experimental
data

G= HT S
G = Gibbs free energy
H = Enthalpy change
T = Temperature in Kelvin
S = Entropy change

If H is in kJ mol -1, you will

need to divide S by 1000 to
convert it to units of kJ K-1 mol-1
You may first need to calculate
H and S using Calculations
23 and 24.

Kinetics
Purpose
28. Calculate the
rate of a reaction

The Calculation
Rate=

[R ] [ P]
=
t
t

Notes

Units are mol dm-3 s-1

The minus sign in front of [R]
is because the concentration of
reactants decreases

Used for calculating rate from

graph of concentration (y-axis)
over time (x-axis)
Sometimes, you cant find two
cases where only [x] has
changed, in which case you
may need to take into account
the order of reaction with
respect to other reactants.

[R] = change in reactant

concentration
[P] = change in product
concentration
t = change in time
29. Calculate the
gradient of a slope

Gradient=

change y
change x

30. Determine the

order of reaction
with respect to a
reactant, x.

1. Identify two experiments,

where the concentration of x
has changed, but all others
have remained the same.
2. Compare the change in [x] to
the change in rate:
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.

31. Deduce a rate

expression

Rate=k [ A] x [B] y [C ] z

k = rate constant (see below)

[A/B/C] = concentration of each
reactant
x/y/z = order of reaction with
respect to each reactant

Reactants with a reaction order

of zero can be omitted from
the rate equation
Given suitable information, you
may need to calculate the
value of the rate constant if
given rates, concentrations and
reaction order, or the expected
rate given the other
information.

Purpose

The Calculation

32. Determining the

units for the rate
constant

3 1

Units=

33. Determining
activation energy

mol dm s
x
3 x
mol (dm )

Notes

If you cant understand this, try

to memorise:
o 0th order: mol dm-3 s-1
o 1st order: s-1
o 2nd order: mol-1 dm3 s-1
o 3rd order: mol-2 dm6 s-1

Units of Ea are kJ mol-1

From the graph, you might also
be asked to calculate A:
o ln A, is the y-intercept
of the graph, so simply
raise e to the power of
this intercept.

Mol and dm-3 terms on the top

and the bottom should be
cancelled out
Remaining mol and dm-3
terms on the bottom should
then be brought to the top by
inverting their indices.

x = the overall order of the

reaction
The Arrhenius equation:

k =A e
Where:
o
o
o
o
o
o

Ea
RT

k = rate constant
A = Arrhenius constant
e = exponential
constant
Ea = activation energy
R = gas constant, 8.31
T = temperature in
Kelvin

This rearranges to:

ln k =

Ea 1
. + ln A
R T

Which is basically the equation

for a straight line in the form
y=mx+c where:
o y is ln k
o x is 1/T
o m is Ea/R
o c is 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 yaxis
Draw a straight line of best fit.
Determine the gradient of the
line
Then:

Ea =Gradient R

Equilibrium
Purpose
34. Determining the
equilibrium constant
expression

The Calculation
For the reaction:
wA + zB yC + zD

z
[C] y
K c =
Kc = equilibrium constant

Notes

Only applies to reactants for

which there is a concentration:
o Aqueous substances are
included
o Solids and pure liquids
are omitted as they do
not have a 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
35. Determining
changes in [H+]
given changes in pH

The Calculation

For each increase of 1 on the

pH scale, divide [H+] by 10.
For each decrease of 1 on
the pH scale, multiply [H+] by
10

Notes

At SL, you do not need to be

able to calculate pH, just to
understand it in relative terms.

Purpose
+

36. Deduce [H ] and

[OH-] given Kw

The Calculation
+

OH
K w =

Notes

At standard conditions, Kw =
1.00x10-14
Kw varies with temperature, so
it is important to know how to
do this calculation.

So:

+
H

OH

And:

OH

+
H

37. Calculating pH
from [H+] and vice
versa

H
pH=log 10

+
H =10 pH

38. Calculating pH
from [OH-] and vice
versa

39. Determining Ka
and Kb of
acids/bases and
their conjugate
bases/acids

OH
pOH=log 10

OH =10pOH

K w =K a . K b

So:

K a=

Kw
Kb

And:

K b=

Kw
Ka

With strong acids, you can

assume [H+] is the same as the
concentration of the acid
(adjusted for the
stoichiometry)
With weak acids, you will need
to calculate [H+] using Ka or
pKa.
With strong bases, you can
assume [OH-] is the same as
the concentration of the acid
(adjusted for the
stoichiometry)
With weak acids, you will need
to calculate [OH-] using Kb or
pKb.
This is useful when trying of
determine the strength the
conjugate base of a weak acid,
and the conjugate acid of a
weak base.

Purpose
40. Determining pKa
and pKb of
acids/bases and
their conjugate
bases/acids

The Calculation
pK w = pK a + pK b

Notes

This is useful when trying of

determine the strength the
conjugate base of a weak acid,
and the conjugate acid of a
weak base.

This is useful to quickly and

easily calculate one of pH/pOH
from the other (or [H+]/[OH-]).

We assume that [HA] is equal

to the concentration stated in
the question, as only a very
small amount has dissociated.
You may need to work
backwards from pH to work out
Ka:

So:

pK a= pK w pK b
And:

pK b= pK w pK a

41. Determining pH
from pOH and vice
versa

pK w = pK a + pK b
So:

pH=14 pOH
And:

pOH=14 pH
42. Calculating pH
of a solution of a
weak acid

+
H

K a=

+
H =10 pH

Since [H+] = [A-]:

+
H

K a=

Then:

+
H

K a=

So:

Then:

+
H
pH=log 10

You will not need to work out

problems for polyprotic weak
acids that would require
quadratics.

Purpose
43. Calculating pOH
of a solution of a
weak base

The Calculation

+
B

OH

K b=

Since [B+] = [OH-]:

OH

K b=
So:

OH

Then:

OH
pOH=log 10

Notes

We assume that [BOH] is equal

to the concentration stated in
the question, as only a very
small amount has dissociated.
You may need to work
backwards from pOH to work
out Kb:

OH =10pOH

Then:

OH

K b=

Purpose
44. Determining pH
of acidic buffer
solutions (and alkali
buffers)

The Calculation

+
H

K a=

Notes

Now [H+] is not equal to [A-]:

So:

K a .

You may need to use your

stoichiometry to calculate the
[HA] and [A-] in the buffer
solution first.
Since additional A- is added, we
now need to use the
concentration of A- from the
buffer as our [A-].
[HA] should either be that
stated in the question, or that
calculated via stoichiometry,
depending on the context.
For alkali buffers, do the same
process but with OH-, Kb etc.

Then:

+
H
pH=log 10

Oxidation and Reduction

Purpose
45. Calculate E

o
cell

The Calculation
o
cell

E =E ( cathode )E

(anode)

Eocell = cell potential

Eo = standard electrode potential

Notes

Organic Chemistry

The value should always be

positive, so if you get it the
wrong way round, just take off
the minus sign.
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.

Purpose

The Calculation

Notes

No calculations. Huzzah!

Measurement and Processing

Purpose
46. Calculating
relative
uncertainties (in %)

The Calculation

Notes

Large measurements have

This only works when adding

absolute uncertainty lower relative uncertainties

Relative uncertainty=
value
.100

47. Calculating
absolute uncertainty
when
addition/subtraction
is involved
48. Calculating
relative uncertainty
when
multiplication/divisio
n is involved

and subtracting values with the

Total uncertainty= ( absolute uncertainties)
same units

For use when you are

multiplying/dividing values
Total uncertainty= ( relative uncertainties)
with different units

Human Biochemistry
Purpose
47. Calculating the
energy value of food
from combustion
data

The Calculation
Use:

Notes

You may need to convert the

energy value into a value per
mole by dividing q by the
number of moles of substance
burnt.

All units should be grams

You may need to convert from
data involving bromine to an
iodine equivalentjust use
your stoichiometry

q=mc T
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 numbers

N (I 2)=m ( I 2 ) .

100
m(lipid)

N(I2) = the iodine number

m(I2) = the mass of iodine
reacting in g
m(lipid) = the mass of lipid
involved in g

Purpose

The Calculation

49. Calculate the

number of double
bonds in a lipid
using iodine number.

Calculate the quantity in moles of

I2 corresponding to the iodine
number:

Notes

I
I
N ( 2)
M r ( I 2)
n( 2)=

This works because each mole

of double bonds reacts with
one mole of iodine.

Calculate the quantity in moles of

lipid in 100g:

n(lipid )=

100
M r (lipid )

Then the number of double

bonds is:

I
n( 2)
n(lipid )
Double bonds=
Where:
N(I2) = the iodine number
n = quantities in moles
Mm = molar masses

Medicines and Drugs

Purpose

The Calculation
No calculations. Huzzah!

Notes