Method of calculating the %
percentage by mass of the elements in a compounds
The 'percent' % by mass composition of a compound in terms of its constituent elements is calculated
in three easy steps
Chemistry calculations 4.
How do you calculate the percent (%) by mass of
an element in a compound formula?
How do you calculate the percent (%) by mass of
water or an ion in a compound formula?
(i) Calculate the formula or molecular mass of the compound
see section 2.
Calculating relative formula/molecular mass (Mr) of a compound
(ii) Calculate the mass of the specified element
(for its %) in the compound, taking into account the number of atoms of the element in the compound formula
(iii) Calculate (ii) as a percentage of (i)
relative atomic mass of
element x number of atoms of element in formula
|% element in compound =
relative formula mass of the compound
% by mass of Z = 100 x Ar(Z) x
atoms of Z / Mr(compound)
It always seems complicated when stated
in this formal way, but the calculations are actually quite easy ..
as long as you can correctly read
- Calculation of % composition Example 4a.1
Calculation of % composition Example 4a.2
- Calculate the % of copper in copper sulphate,
- Relative atomic masses: Cu = 64, S = 32 and O = 16
- relative formula mass = 64 + 32 + (4x16) = 160
- only one copper atom of relative atomic mass 64
- % Cu = 100 x 64 / 160
40% copper by mass in the compound
- Note that similarly, you can calculate the % of the
other elements in the compound e.g.
- % sulfur = (32/160) x 100 =
- % oxygen = (64/160) x 100 =
- Also note that if you haven't made any errors, they
should add up to 100%,useful arithmetical !
Calculation of % composition Example 4a.3
- Calculate the % of oxygen in aluminium sulphate,
- Relative atomic masses: Al = 27, S = 32 and O = 16
- relative formula mass = 2x27 + 3x(32 + 4x16) = 342
- there are 4 x 3 = 12 oxygen atoms, each of relative atomic mass 16,
- giving a total mass of oxygen in the formula of 12 x 16 = 192
- % O = 100 x 192 / 342 =
56.1% oxygen by mass in aluminium sulphate
Calculation of % composition Example 4a.4
- The next two examples extend the idea of %
element composition to include % composition of part of a compound, in these
cases water in a hydrated salt and the sulfate ion in a potassium salt.
- Calculate the % of water in hydrated
magnesium sulphate MgSO4.7H2O
- Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1
- relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1
+ 16)] = 246
- 7 x 18 = 126 is the mass of water
- so % water = 100 x 126 / 246 =
51.2 % H2O
- Note: The determination
and calculation of the formula of a hydrated salt like MgSO4.7H2O
is covered in Calculations section
- Calculate the percentage by mass, of sulfate
ion in sodium sulfate
- formula of sodium sulfate Na2SO4,
atomic masses: Na = 23, S = 32, O = 16
- Formula mass Na2SO4 =
(2 x 23) + 32 + (4 x 16) = 142
- Formula mass of sulfate ion SO42-
(or just SO4 will do for the calculation) = 32 + (4 x 16) = 96
- Therefore % sulfate ion in sodium sulfate =
(96/142) x 100 = 67.6% SO4
type in answer
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percentage mass composition calculations including % of any component in a
compound or a mixture
Atomic masses used for 4b. questions: C = 12, Cl
= 35.5, Fe = 56, H = 1, Mg = 24, N = 14, Na =
23, O = 16, S = 32,
By now I assume you can do formula mass calculations and
read formula without any trouble, so ALL the detail of such calculations is NOT
shown, just the bare essentials!
Ammonium sulfate, (NH4)2SO4,
is an important ingredient in many artificial fertilisers supplying to
plants the essential mineral elements of nitrogen and sulfur.
(a) Calculate the percentage of nitrogen and the
percentage of sulfur in ammonium sulfate.
formula mass of ammonium sulfate = (2 x 18) + 32 +
64 = 132.
with two nitrogen atoms in the formula; % nitrogen
by mass = 100 x 28/132 = 21.2% N
with one sulfur atom in the formula, % sulfur by
mass = 100 x 32/132 = 24.2% S
(b) Calculate the percentage of sulfate ion in
To calculate the percentage of a 'part' of a
compound, you just use the formula mass of that 'part!
formula mass of sulfate, SO4, is 32
+ (4 x 16) = 96
therefore % sulfate by mass = 100 x 96/132 =
Note: If the question refers to the sulfate ion
itself, SO42-, its just the same % mass
What is the percentage of carbonate ion in sodium
formula mass of sodium carbonate = 46 + 12 + 48 = 106
formula mass of carbonate, CO3 = 12 + (3 x
16) = 60
therefore % carbonate ion by mass = 100 x 60/106 =
56.6% CO3 (for the CO32- ion)
Calculate the percentage water of crystallisation in
magnesium sulfate crystals, MgSO4.7H2O, known as Epsom
formula mass of Epsom salt = 24 + 32 + 64 + (7 x 18) =
formula mass of water = 18, mass of seven water
molecules is 7 x 18 = 126
therefore % of water of crystallisation in the
crystals = 100 x 126/246 = 51.2% H2O
Rock salt is mainly sodium chloride, NaCl
On analysis of an impure sample of rock salt, it was
found to contain by mass 57.5% of chlorine as chloride ion.
(a) Calculate the percentage purity of the salt.
the formula mass of sodium chloride is 58.5
the formula mass of chloride is 35.5
therefore you need to scale up from the % mass of
chloride ion to the % mass of sodium chloride.
the scale up factor must be 58.5/35.5 = 1.648
therefore percentage of sodium chloride in the
rock salt = 57.5 x 1.648 = 94.8% NaCl
(b) What assumption have you made in this calculation
to make this a valid calculation?
You have assumed that non of the impurities
contain the sodium or chloride ion.
There may be other sodium or chloride salts in the
rock salt mixture.
A mixture of sand and a compound based on iron(II)
sulfate (*), FeSO4. is used to treat grass e.g. lawns and bowling
greens to promote plant growth and kill moss.
What percentage by mass of iron(II) sulfate is
required in the mixture to give 15% by mass of iron(II) ions (Fe2+)?
You need to scale up from the mass of iron ions to
the mass of the compound FeSO4.
formula mass of FeSO4 = 56 + 32 + 64 =
atomic mass of iron Fe or iron(II) ion Fe2+
= 56 (note the atom and ion have the same mass!)
therefore the scaling up factor is 152/56 = 2.714
therefore % iron(II) sulfate required in the
mixture = 15 x 2.714 = 40.7% FeSO4
Note (*): The actual iron compound used in lawn
treatments is crystals of ammonium iron(II) sulfate,
old name ferrous ammonium sulphate, its a double salt, but I've just
based the calculation on the iron(II) sulfate part.
A baking powder mixture contains sodium
To get sufficient rising action from the carbon
dioxide gas (CO2) formed in baking, it should contain a minimum
of 50% carbonate ion (CO32-).
Calculate the minimum percentage of sodium hydrogen
carbonate that should be in the mixture.
You need to scale up from the formula masses of the
carbonate ion and that of sodium hydrogen carbonate.
formula mass of carbonate, CO3 = 12 +
48 = 60 (same for the carbonate ion)
formula mass sodium hydrogencarbonate = 23 + 1 +
60 = 84.
therefore the scale up factor is 84/60 = 1.4
so, minimum percentage sodium hydrogen carbonate
in the mixture should be 50 x 1.4 = 70% NaHCO3
In an experiment 6.0 g of metal M was burned in
a crucible, by heating in air, until there was no more gain in weight. Apart
from the mass of the crucible, the final mass of the residue was 10.0 g. The
oxide O formed was an essential ingredient in a ceramic pigment
mixture P for glazing pottery.
(a) What % of the oxide O is the metal M.
100 x 6.0/10.0 = 60% of M in the
(b) The mixture P must contain 25% by mass of
the metal M.
What mass of the oxide O is needed to make
12 g of the mixture P?
25% of 12 g is 12 x 25/100 = 3.0 g, so this is the
mass of metal M in 12 g of mixture P as the oxide O
You now have to scale up to the mass of the oxide
O needed which contains 60% of M.
Scaling up gives 3.0 x 100/60 = 5.0 g
of oxide M is needed.
Above is typical periodic table used in science-chemistry
courses for use in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
However, for calculations of percentage composition (and any other
quantitative chemistry calculations) note:
GCSE level, relative atomic masses are quoted as whole numbers (integers) e.g. C
= 12, Fe = 56, Ag = 108 etc.
from copper Cu = 63.5 and chlorine Cl = 35.5
advanced A level chemistry, (but still pre-university) relative atomic masses will
be quoted to one decimal place
e.g. H = 1.0, C = 12.0, Cl = 35.5, Fe = 55.8, Cu =
63.5, Ag 107.9 etc.
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OTHER CALCULATION PAGES
What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
formula/molecular mass of a compound or element molecule
Law of Conservation of Mass and simple reacting mass calculations
Composition by percentage mass of elements
in a compound
Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
Reacting mass ratio calculations of reactants and products
moles) and brief mention of actual percent % yield and theoretical yield,
and formula mass determination
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
Moles and the molar volume of a gas, Avogadro's Law
Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
Electrolysis products calculations (negative cathode and positive anode products)
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
Energy transfers in physical/chemical changes,
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
Radioactivity & half-life calculations including
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