# Calculations for IB Chemistry

```Mr Field
INTRODUCTION
ο§ The purpose of this presentation, is to collect all the calculations you need for IB
Chemistry in one place.
ο§ It will not help you to master them…the only shortcut to mastery is many hours of
dedicated practice.
ο§ Good luck!
PURPOSE:
CALCULATING A QUANTITY IN MOLES FROM A NUMBER OF PARTICLES
The Calculation
Notes
π
π=
πΏ
n = the quantity in moles
N = the number of particles
ο§ When calculating numbers of atoms
within molecules, multiply the number
of particles (N) by the number of atoms
in the formula
ο§ To find out numbers of particles,
rearrange to: N = n.L
STOICHIOMETRY
PURPOSE:
DETERMINE RELATIVE MOLECULAR OR FORMULA MASS, MR
The Calculation
Notes
ππ =
ο§ Mr has no unit as it is a relative value
(ππ’ππππ ππ ππ‘πππ . ππ‘ππππ πππ π )
ο§ To calculate molar mass, Mm, just stick a
‘g’ for grams on the end
STOICHIOMETRY
PURPOSE:
CALCULATE A QUANTITY IN MOLES, FROM A MASS OF A SUBSTANCE
The Calculation
π
π=
ππ
n = the quantity in moles
m = the mass of substance you are given
in grams
Mm = the molar mass of the substance
Notes
ο§ To determine the mass of a given
number of moles of a substance use:
ο§ π = ππ . π
ο§ To determine the molar mass of a given
mass of a substance:
ο§ ππ =
π
π
STOICHIOMETRY
PURPOSE:
DETERMINE EMPIRICAL FORMULA FROM % COMPOSITION BY MASS
The Calculation
1.
Divide each % by the atomic mass
2.
Divide each answer to Step 1 by the
3.
Multiply all answers to Step 1 to
remove any obvious fractions
Notes
ο§ You can follow the same method if you
have plain composition by mass rather
than % composition by mass.
a. If there is a ‘.5’ multiply everything by 2
b. If there is a ‘.33’ multiply everything by
3 etc
STOICHIOMETRY
PURPOSE:
DETERMINE MOLECULAR FORMULA FROM EMPIRICAL FORMULA
The Calculation
ππ
πΉπ =
.πΉ
π(πΉπ ) π
Notes
ο§ This and the previous calculation are
often combined together in exam
questions
Fm = molecular formula
Mr = relative molecular mass
Fe = empirical formula
m(Fe) = empirical formula mass
STOICHIOMETRY
PURPOSE:
USE MOLE RATIOS TO DETERMINE THE NUMBER OF MOLES OF ‘B’ THAT
The Calculation
ππ’ππππ ππ π΅ ππ πππ’ππ‘πππ
π π΅ =π π΄ .
ππ’ππππ ππ π΄ ππ πππ’ππ‘πππ
Notes
ο§ The second term in this equation is the
mole ratio
ο§ You must use a fully balanced equation
ο§ This is the central step in many
stoichiometry calculations
STOICHIOMETRY
PURPOSE:
CALCULATE THEORETICAL YIELD
The Calculation
Notes
ο§ Use mole ratios to determine the
ο§ n/a
expected quantity of product in moles
ο§ Use π = ππ . π to determine the
expected mass.
STOICHIOMETRY
PURPOSE:
DETERMINE LIMITING AND EXCESS REACTANTS
The Calculation
Notes
ο§ Divide moles of each reactant by their
ο§ You would often then need to use mole
coefficient in the balanced equation
ο§ Smallest value ο  limiting
ratios to determine a quantity of
product (in moles).
ο§ Largest value ο  excess
STOICHIOMETRY
PURPOSE:
CALCULATING PERCENTAGE YIELD.
The Calculation
πππ‘π’ππ π¦ππππ
% πππππ =
. 100
π‘βπππππ‘ππππ π¦ππππ
Notes
ο§ Actual and theoretical yield must have
the same units.
ο§ 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.
STOICHIOMETRY
PURPOSE:
APPLY AVOGADRO’S LAW TO CALCULATE REACTING VOLUMES OF GASES
The Calculation
π1
π2
=
π1
π2
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
Notes
ο§ Only applies when temperature and
pressure remain constant.
ο§ Units of V do not matter. But must be the
same.
ο§ This is really a special case of the Ideal
Gas Law where the pressure,
temperature and gas constant terms
cancel each other out.
STOICHIOMETRY
PURPOSE:
CALCULATE MOLAR QUANTITIES OF GASES AT STANDARD TEMPERATURE
AND PRESSURE
The Calculation
π
π=
22.4
Notes
ο§ Only applies at standard conditions:
ο§ 273 K (0oC)
ο§ 101,000 Pa (1.00 atm)
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
ο§ If volume of gas is given in m3, use
ο§ Molar volumes are given in the data
booklet and do not need memorising.
STOICHIOMETRY
PURPOSE:
THE IDEAL GAS EQUATION
The Calculation
ππ = πππ
Notes
ο§ In practice, you can often use:
ο§ V in units of dm3
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)
ο§ Pa in units of kPa
ο§ You will need to be comfortable
rearranging this equation to change the
subject.
ο§ This takes time to use, so only use it in
non-standard conditions, or when the
laws in Calculation 13 would not be
quicker.
STOICHIOMETRY
PURPOSE:
RELATIONSHIP BETWEEN TEMPERATURE, PRESSURE AND VOLUME
The Calculation
Charles’ Law, at fixed pressure:
Notes
π1
π1
=
Gay-Lussac’s Law, at fixed volume:
π2
π2
π1
π1
=
ο§ These only work where the quantity in
π2
π2
Boyle’s Law, at fixed temperature: π1 π1 = π2 π2
moles remains fixed.
ο§ All of these are just special cases of the
ideal gas law, where the remaining
terms just cancel each other out.
V1 and V2 are initial and final volume
P1 and P2 are initial and final pressure
T1 and T2 are initial and final temperature
STOICHIOMETRY
PURPOSE:
MOLAR CONCENTRATION
The Calculation
Notes
π
π=
π
c = concentration in mol
ο§ Units are pronounced ‘moles per
decimetre cubed’
dm-3
ο§ You need to be able to use any
rearrangement of this equation
n = the quantity in moles
V = the volume in dm3 (litres)
STOICHIOMETRY
PURPOSE:
CONCENTRATION BY MASS
The Calculation
Notes
π
π=
π
c = concentration in g
dm-3
m = the quantity in grams
V = the volume in dm3 (litres)
ο§ Units are pronounced ‘grams per
decimetre cubed’
ο§ You need to be able to use any
rearrangement of this equation
ο§ Generally, if you have a concentration
like this, you should convert it into a
molar concentration before
proceeding.
STOICHIOMETRY
PURPOSE:
DETERMINE RELATIVE ATOMIC MASS FROM % ABUNDANCE DATA
The Calculation
π΄π =
% πππ’ππππππ
(ππ‘ππππ πππ π .
)
100
Notes
ο§ % abundance may be given in the
question, or you may need to read it
from a mass spectrum
ο§ 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.
ATOMIC STRUCTURE
PURPOSE:
DETERMINE % ABUNDANCE FROM RELATIVE ATOMIC MASS
The Calculation
Notes
If there are two isotopes, label one of them ‘a’ and one ‘b’.
Now: π΄π =
ο§ Ar, Ia and Ib will be provided in the
question, so you can plug the numbers
in, and then rearrange to find x.
π₯πΌπ +π¦πΌπ
π₯+π¦
However, since x+y = 100%, y = 100-x so:
π΄π =
π₯πΌπ +(100−π₯)πΌπ
100
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
ο§ To find y, simply do y=100-x
ο§ 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.
ATOMIC STRUCTURE
PURPOSE:
CALCULATING THE HEAT CHANGE OF A PURE SUBSTANCE
The Calculation
π = ππβπ
Notes
ο§ Be careful of the units of mass…you may
need to convert kg into g
q = the heat change in Joules
m = the mass of substance in grams
ο§ Be careful of the units for specific heat
capacity, if it is J K-1 kg-1 you will need to
c = specific heat capacity in J K-1 g-1
βT = temperature rise in K or OC
ENERGETICS
PURPOSE:
CALCULATING AN ENTHALPY CHANGE FROM EXPERIMENTAL DATA
The Calculation
βπ» = −ππβπ
β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
βT = temperature rise in K or OC
Notes
ο§ The minus sign is needed to ensure that
an exothermic reaction has a negative
enthalpy change.
ο§ Units are J or kJ mol-1
ο§ The mass of solution is assumed to be
the same as its volume in cm3.
ο§ The specific heat capacity of the
reactants is ignored.
ENERGETICS
PURPOSE:
CALCULATING βHR USING A HESS CYCLE
The Calculation
Notes
Once you have produced you Hess
cycle:
ο§ See the ‘Energetics’ PowerPoint for
1.
Write the relevant βH onto each
arrow
2.
Multiply each βH in accordance with
the stoichiometry
3.
with an arrow, and subtract when you
go against one.
ENERGETICS
PURPOSE:
CALCULATING βHR FROM AVERAGE BOND ENTHALPIES
The Calculation
Notes
Method 1: Make a Hess cycle, then do as in
Calculation 20.
ο§ It is more reliable to use Hess cycles
Method 2:
βπ»π
ο§ Average bond enthalpies can be found
=
πππππ‘πππ‘ πππππ  −
and you can easily forget whether it is
reactants – products or vice versa.
in Table 10 of the data booklet.
(πππππ’ππ‘ πππππ ) ο§ 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.
ENERGETICS
PURPOSE:
CALCULATING βHR FROM ENTHALPIES OF FORMATION
The Calculation
Notes
Method 1: Make a Hess cycle, then do as in
Calculation 20.
ο§ It is more reliable to use Hess cycles
Method 2:
βπ»π
ο§ βHof for elements in their standard
=
βπ»ππ (πππππ’ππ‘π ) −
and you can easily forget whether it is
products – reactants or vice versa.
states is zero.
βπ»ππ (πππππ‘πππ‘π )
βHr = enthalpy change of reaction
βHof = enthalpy change of formation
ο§ βHof values for many compounds can
be found in Table 11 of the data
booklet.
ο§ In some questions, you may also need
to take a state change into account, if
standard states are not used.
ENERGETICS
PURPOSE:
CALCULATING βHR FROM ENTHALPIES OF COMBUSTION
The Calculation
Notes
Method 1: Make a Hess cycle, then do as in
Calculation 20.
ο§ It is more reliable to use Hess cycles
Method 2:
βπ»π
ο§ βHoc for CO2 and H2O is zero.
=
βπ»ππ (πππππ‘πππ‘π ) −
βπ»ππ (πππππ’ππ‘π )
and you can easily forget whether it is
reactants – products or vice versa.
ο§ βHoc values for many compounds can
be found in Table 12 of the data
booklet.
βHr = enthalpy change of reaction
βHoc = enthalpy change of combustion
ENERGETICS
PURPOSE:
CALCULATING LATTICE ENTHALPY
The Calculation
Notes
You need to build a Born-Haber
cycle….see the Energetics PowerPoint
for help.
ο§ n/a
ENERGETICS
PURPOSE:
CALCULATING βS FROM STANDARD ENTROPY VALUES
The Calculation
βπ π =
π π (πππππ’ππ‘π ) −
Notes
π π (πππππ’ππ‘π ) ο§ Units are J K-1 mol-1
ο§ So values can be found in Table 11 of
the data booklet
βSo = standard entropy change of
reaction
ο§ You cannot assume that So of an element
is zero. It is not.
So = standard entropy of each substance
ENERGETICS
PURPOSE:
CALCULATING βGR STANDARD GIBB’S FREE ENERGY OF FORMATION
VALUES
The Calculation
Notes
Method 1: Make a Hess cycle, then do
similar to Calculation 20.
ο§ Units are J or kJ mol-1
ο§ It is more reliable to use Hess cycles
and you can easily forget whether it is
products – reactants or vice versa.
Method 2:
βπΊπ
=
βπΊππ (πππππ’ππ‘π ) −
βπΊππ (πππππ‘πππ‘π )
βGr = Gibb’s free energy of reaction
βGof = Gibb’s free energy of formation
ο§ βGof for elements in their standard
states is zero.
ο§ βGof values for many compounds can
be found in Table 11 of the data
booklet.
ENERGETICS
PURPOSE:
CALCULATING βGR FROM EXPERIMENTAL DATA
The Calculation
βπΊ = βπ» − πβπ
βG = Gibb’s free energy
βH = Enthalpy change
Notes
ο§ If βH is in kJ mol-1, you will need to
divide βS by 1000 to convert it to units
of kJ K-1 mol-1
ο§ You may first need to calculate βH and
βS using Calculations 23 and 24.
T = Temperature in Kelvin
βS = Entropy change
ENERGETICS
PURPOSE:
CALCULATE THE RATE OF A REACTION
The Calculation
−β[π] β[π]
πππ‘π =
=
βπ‘
βπ‘
β[R] = change in reactant concentration
Notes
ο§ Units are mol dm-3 s-1
because the concentration of reactants
decreases
β[P] = change in product concentration
βt = change in time
KINETICS
PURPOSE:
CALCULATE THE GRADIENT OF A SLOPE
The Calculation
πβππππ ππ π¦
πΊπππππππ‘ =
πβππππ ππ π₯
Notes
ο§ Used for calculating rate from graph of
concentration (y-axis) over time (xaxis)
KINETICS
PURPOSE:
DETERMINE THE ORDER OF REACTION WITH RESPECT TO A REACTANT,
X.
The Calculation
1.
2.
Identify two experiments, where the
concentration of ‘x’ has changed, but
all others have remained the same.
Compare the change in [x] to the
change in rate:
a. If doubling [x] has no effect on rate,
Notes
ο§ Sometimes, you can’t 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.
then 0th order.
b. If doubling [x] doubles rate, then 1st
order.
3.
If doubling [x] quadruples rate, then
2nd order.
KINETICS
PURPOSE:
DEDUCE A RATE EXPRESSION
The Calculation
πππ‘π = π[π΄]π₯ [π΅]π¦ [πΆ] π§
Notes
ο§ Reactants with a reaction order of zero
can be omitted from the rate equation
k = rate constant (see below)
[A/B/C] = concentration of each reactant
x/y/z = order of reaction with respect to
each reactant
ο§ 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.
KINETICS
PURPOSE:
DETERMINING THE UNITS FOR THE RATE CONSTANT
The Calculation
πππ ππ−3 π  −1
ππππ‘π  =
πππ π₯ (ππ−3 )π₯
ο§ mol and dm-3 terms on the top and the
bottom should be cancelled out
ο§ Remaining mol and dm-3 terms on the
bottom should then be brought to the top
by inverting their indices.
Notes
ο§ If you can’t understand this, try to
memorise:
ο§ 0th order: mol dm-3 s-1
ο§ 1st order: s-1
ο§ 2nd order: mol-1 dm3 s-1
ο§ 3rd order: mol-2 dm6 s-1
x = the overall order of the reaction
KINETICS
PURPOSE:
DETERMINING ACTIVATION ENERGY
Notes
The Calculation
ο§ The Arrhenius equation: π = π΄π
ο§ This rearranges to: ln π =
−πΈπ 1
.
π π
−πΈπ
ππ
+ ln π΄
ο§ Which is basically the equation for a straight
line in the form ‘y=mx+c’ where: ‘y’ is ln k; ‘x’ is
1/T; ‘m’ is –Ea/R; ‘c’ is ln A
ο§ So, if we do a reaction at a range of
temperatures and calculate ln k, then:
ο§ Where:
ο§
ο§
ο§
ο§
ο§
ο§
k = rate constant
A = Arrhenius constant
e = exponential constant
Ea = activation energy
R = gas constant, 8.31
T = temperature in Kelvin
1.
Draw a graph with 1/T along the x-axis, and
ln k on the y-axis
ο§ Units of Ea are kJ mol-1
2.
Draw a straight line of best fit.
ο§ From the graph, you might also be asked
3.
Determine the gradient of the line
4.
Then: πΈπ =
−πΊπππππππ‘
π
to calculate A:
ο§ ‘ln A’, is the y-intercept of the graph, so
simply raise ‘e’ to the power of this
intercept.
KINETICS
PURPOSE:
DETERMINING THE EQUILIBRIUM CONSTANT EXPRESSION
The Calculation
For the reaction:
wA + zB ο  yC + zD
Notes
ο§ Only applies to reactants for which there is a
concentration:
ο§ Aqueous substances are included
ο§ Solids and pure liquids are omitted as they do not
have a concentration per se
[πΆ]π¦ [π·]π§
πΎπ =
[π΄]π€ [π΅]π§
ο§ For reactions involving gases, use partial
ο§ At SL, you only need be able to construct the
expression.
Kc = equilibrium constant
ο§ 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.
EQUILIBRIUM
PURPOSE:
DETERMINING CHANGES IN [H+] GIVEN CHANGES IN PH
The Calculation
Notes
ο§ For each increase of ‘1’ on the pH scale,
ο§ At SL, you do not need to be able to
divide [H+] by 10.
ο§ For each decrease of ‘1’ on the pH scale,
calculate pH, just to understand it in
relative terms.
multiply [H+] by 10
ACIDS AND BASES
PURPOSE:
DEDUCE [H+] AND [OH-] GIVEN KW
The Calculation
Notes
πΎπ€ = π» + . [ππ»− ]
ο§ At standard conditions, Kw = 1.00x10-14
ο§ Kw varies with temperature, so it is
So:
π»+
πΎπ€
=
[ππ»− ]
And:
ππ»−
important to know how to do this
calculation.
πΎπ€
=
[π» + ]
ACIDS AND BASES
PURPOSE:
CALCULATING PH FROM [H+] AND VICE VERSA
The Calculation
Notes
ππ» = −πππ10 [π» + ]
[π»+ ]
=
10−ππ»
ο§ With strong acids, you can assume [H+]
is the same as the concentration of the
ο§ With weak acids, you will need to
calculate [H+] using Ka or pKa.
ACIDS AND BASES
PURPOSE:
CALCULATING PH FROM [OH-] AND VICE VERSA
The Calculation
Notes
πππ» = −πππ10 [ππ»− ]
[ππ» − ]
=
10−πππ»
ο§ With strong bases, you can assume [OH-
] is the same as the concentration of the
ο§ With weak acids, you will need to
calculate [OH-] using Kb or pKb.
ACIDS AND BASES
PURPOSE:
DETERMINING KA AND KB OF ACIDS/BASES AND THEIR CONJUGATE
BASES/ACIDS
The Calculation
πΎπ€ = πΎπ . πΎπ
So:
And:
πΎπ€
πΎπ =
πΎπ
Notes
ο§ This is useful when trying of determine
the strength the conjugate base of a
weak acid, and the conjugate acid of a
weak base.
πΎπ€
πΎπ =
πΎπ
ACIDS AND BASES
PURPOSE:
DETERMINING PKA AND PKB OF ACIDS/BASES AND THEIR CONJUGATE
BASES/ACIDS
The Calculation
ππΎπ€ = ππΎπ + ππΎπ
So:
ππΎπ = ππΎπ€ − ππΎπ
Notes
ο§ This is useful when trying of determine
the strength the conjugate base of a
weak acid, and the conjugate acid of a
weak base.
And:
ππΎπ = ππΎπ€ − ππΎπ
ACIDS AND BASES
PURPOSE:
DETERMINING PH FROM POH AND VICE VERSA
The Calculation
ππΎπ€ = ππΎπ + ππΎπ
Notes
ο§ This is useful to quickly and easily
calculate one of pH/pOH from the other
(or [H+]/[OH-]).
So:
ππ» = 14 − πππ»
And:
πππ» = 14 − ππ»
ACIDS AND BASES
PURPOSE:
CALCULATING PH OF A SOLUTION OF A WEAK ACID
Notes
The Calculation
π»+ [π΄− ]
πΎπ =
[π»π΄]
ο§ We assume that [HA] is equal to the
concentration stated in the question, as only a
very small amount has dissociated.
ο§ You may need to work backwards from pH to
Since [H+] = [A-]:
So:
[π» + ] =
πΎπ =
π»+ 2
work out Ka:
[π» + ] = 10−ππ»
[π»π΄]
πΎπ . [π»π΄]
ο§ Then: πΎπ =
π»+ 2
[π»π΄]
ο§ You will not need to work out problems for
Then:
ππ» = −πππ10 [π» + ]
polyprotic weak acids that would require
ACIDS AND BASES
PURPOSE:
CALCULATING POH OF A SOLUTION OF A WEAK BASE
Notes
The Calculation
π΅ + [ππ» − ]
πΎπ =
[π΅ππ»]
Since [B+] = [OH-]:
ππ» − 2
πΎπ =
[π΅ππ»]
ο§ We assume that [BOH] is equal to the
concentration stated in the question, as
only a very small amount has dissociated.
ο§ You may need to work backwards from
pOH to work out Kb:
[ππ» − ] = 10−πππ»
So:
[ππ» − ] =
Then:
πΎπ . [π΅ππ»]
πππ» = −πππ10 [ππ» − ]
ο§ Then:
ππ» − 2
πΎπ =
[π΅ππ»]
ACIDS AND BASES
PURPOSE:
DETERMINING PH OF ACIDIC BUFFER SOLUTIONS (AND ALKALI BUFFERS)
Notes
The Calculation
π»+ [π΄− ]
πΎπ =
[π»π΄]
Now [H+] is not equal to [A-]:
So:
[π»+ ]
πΎπ . [π΄− ]
=
[π»π΄]
ο§ You may need to use your stoichiometry
to calculate the [HA] and [A-] in the buffer
solution first.
to use the concentration of A- from the
buffer as our [A-].
ο§ [HA] should either be that stated in the
question, or that calculated via
stoichiometry, depending on the context.
ο§ For alkali buffers, do the same process
Then:
ππ» = −πππ10
[π»+ ]
but with OH-, Kb etc.
ACIDS AND BASES
PURPOSE:
CALCULATE EOCELL
The Calculation
π
πΈππππ
= πΈ π πππ‘βπππ − πΈ π (anode)
Eocell = cell potential
Eo = standard electrode potential
Notes
ο§ The value should always be positive, so
if you get it the wrong way round, just
take off the minus sign.
ο§ Eo can be found in the data booklet.
ο§ You do not need to take the
stoichiometry of the reaction into
account, just use the Eoas they are in the
data booklet.
OXIDATION AND REDUCTION
PURPOSE:
CALCULATING RELATIVE UNCERTAINTIES (IN %)
The Calculation
πππππ‘ππ£π π’πππππ‘ππππ‘π¦ =
Notes
πππ πππ’π‘π π’πππππ‘ππππ‘π¦
.100
π ππ§π ππ π£πππ’π
ο§ Large measurements have lower
relative uncertainties
MEASUREMENT AND PROCESSING
PURPOSE:
IS INVOLVED
The Calculation
Notes
πππ‘ππ π’πππππ‘ππππ‘π¦
=
ο§ This only works when adding and
(πππ πππ’π‘π π’πππππ‘ππππ‘πππ )
subtracting values with the same units
MEASUREMENT AND PROCESSING
PURPOSE:
CALCULATING RELATIVE UNCERTAINTY WHEN
MULTIPLICATION/DIVISION IS INVOLVED
The Calculation
Notes
πππ‘ππ π’πππππ‘ππππ‘π¦
=
ο§ For use when you are
(πππππ‘ππ£π π’πππππ‘ππππ‘πππ )
multiplying/dividing values with
different units
MEASUREMENT AND PROCESSING
PURPOSE:
CALCULATING THE ENERGY VALUE OF FOOD FROM COMBUSTION DATA
The Calculation
Notes
Use:
ο§ You may need to convert the energy
π = ππβπ
value into a value per mole by dividing
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
βT = temperature rise in K or OC
HUMAN BIOCHEMISTRY
PURPOSE:
CALCULATING IODINE NUMBERS
The Calculation
Notes
100
ο§ π(πΌ2 ) = π πΌ2 .
π(πππππ)
ο§
ο§ N(I2) = the iodine number
ο§ m(I2) = the mass of iodine reacting in g
ο§ All units should be grams
ο§ You may need to convert from data
involving bromine to an ‘iodine
equivalent’…just use your
stoichiometry
ο§ m(lipid) = the mass of lipid involved in
g
HUMAN BIOCHEMISTRY
PURPOSE:
CALCULATE THE NUMBER OF DOUBLE BONDS IN A LIPID USING IODINE
NUMBER.
The Calculation
Notes
Calculate the quantity in moles of I2
corresponding to the iodine number:
π(πΌ2 )
π(πΌ2 ) =
ππ (πΌ2 )
N(I2) = the iodine number
Calculate the quantity in moles of lipid in 100g:
100
π(πππππ) =
ππ (πππππ)
Then the number of double bonds is:
π(πΌ2 )
π·ππ’πππ πππππ  =
π(πππππ)
n = quantities in moles
Mm = molar masses
ο§ This works because each mole of
double bonds reacts with one mole of
iodine.
HUMAN BIOCHEMISTRY
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