Percentage % element by mass to calculate the
composition of a compound and other % composition calculations
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Brown's Chemistry  GCSE/IGCSE/GCE (basic A level)
O Level Online Chemical Calculations
4a.
Calculating the %
composition by mass of elements in compounds
Quantitative Chemistry
calculations online Help for problem solving
in doing percent of mass of elements in a given compound formula
calculations. Practice revision questions on % composition of an element
in compound, using experiment data, making predictions. This page describes,
and explains,
with fully worked out examples, how to calculate the composition of a
compound in terms of the % by mass of each element in a compound. It
doesn't matter what the nature of the compound is i.e. it is irrelevant
whether its an ionic compound or a covalent compound. Online practice exam chemistry CALCULATIONS and
solved problems for KS4 Science
GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB
courses. These revision notes and practice questions on how to do percentage by
mass of elements in a compound calculations and worked examples should
prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry
science courses.
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Selfassessment Quizzes
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questions)
type in answer
for
F and H tiers or
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4a.
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.
2.
Calculating relative formula/molecular mass (M_{r}) 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 = 

X 100 

relative formula mass of the compound 
% by mass of Z = 100 x A_{r}(Z) x
atoms of Z / M_{r}(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
a formula!
 Calculation of % composition Example 4a.1
 Calculate the % of copper in copper sulphate,
CuSO_{4}
 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 =
20% S
 % oxygen = (64/160) x 100 =
40% O
 Also note that if you haven't made any errors, they
should add up to 100%,useful arithmetical !
 Calculation of % composition Example 4a.2
 Calculate the % of oxygen in aluminium sulphate,
Al_{2}(SO_{4})_{3
}
 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.3
 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 MgSO_{4}.7H_{2}O
 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 % H_{2}O
 Note: The determination
and calculation of the formula of a hydrated salt like MgSO_{4}.7H_{2}O
is covered in Calculations section
14.4.

Calculation of % composition Example 4a.4
 Calculate the percentage by mass, of sulfate
ion in sodium sulfate
 formula of sodium sulfate Na_{2}SO_{4},
atomic masses: Na = 23, S = 32, O = 16
 Formula mass Na_{2}SO_{4} =
(2 x 23) + 32 + (4 x 16) = 142
 Formula mass of sulfate ion SO_{4}^{2}
(or just SO_{4} will do for the calculation) = 32 + (4 x 16) = 96
 Therefore % sulfate ion in sodium sulfate =
(96/142) x 100 = 67.6% SO_{4}
Selfassessment Quizzes:
type in answer
QUIZ or
multiple choice
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4b. Other
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!
Example 4b.1
Ammonium sulfate, (NH_{4})_{2}SO_{4},
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
ammonium sulfate.
To calculate the percentage of a 'part' of a
compound, you just use the formula mass of that 'part!
formula mass of sulfate, SO_{4}, is 32
+ (4 x 16) = 96
therefore % sulfate by mass = 100 x 96/132 =
72.7% SO_{4}
Note: If the question refers to the sulfate ion
itself, SO_{4}^{2}, its just the same % mass
calculation!
Example 4b.2
What is the percentage of carbonate ion in sodium
carbonate? (Na_{2}CO_{3})
formula mass of sodium carbonate = 46 + 12 + 48 = 106
formula mass of carbonate, CO_{3} = 12 + (3 x
16) = 60
therefore % carbonate ion by mass = 100 x 60/106 =
56.6% CO_{3} (for the CO_{3}^{2} ion)
Example 4b.3
Calculate the percentage water of crystallisation in
magnesium sulfate crystals, MgSO_{4}.7H_{2}O, known as Epsom
salt.
formula mass of Epsom salt = 24 + 32 + 64 + (7 x 18) =
246
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% H_{2}O
Example 4b.4
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.
Example 4b.5
A mixture of sand and a compound based on iron(II)
sulfate (*), FeSO_{4}. 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 (Fe^{2+})?
You need to scale up from the mass of iron ions to
the mass of the compound FeSO_{4}.
formula mass of FeSO_{4} = 56 + 32 + 64 =
152
atomic mass of iron Fe or iron(II) ion Fe^{2+}
= 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% FeSO_{4}
Note (*): The actual iron compound used in lawn
treatments is crystals of ammonium iron(II) sulfate,
(NH_{4})_{2}Fe(SO_{4})_{2}.6H_{2}O,
old name ferrous ammonium sulphate, its a double salt, but I've just
based the calculation on the iron(II) sulfate part.
Example 4b.6
A baking powder mixture contains sodium
hydrogencarbonate, NaHCO_{3}.
To get sufficient rising action from the carbon
dioxide gas (CO_{2}) formed in baking, it should contain a minimum
of 50% carbonate ion (CO_{3}^{2}).
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, CO_{3} = 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% NaHCO_{3}
Example 4b.7
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
oxide O
(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.
Example 4b.8
PLEASE NOTE:
Above is typical periodic table used in sciencechemistry
courses for use in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
However, for calculations of percentage composition (and any other
quantitative chemistry calculations) note:
(i) At
GCSE level, relative atomic masses are quoted as whole numbers (integers) e.g. C
= 12, Fe = 56, Ag = 108 etc.
apart
from copper Cu = 63.5 and chlorine Cl = 35.5
(ii) In
advanced A level chemistry, (but still preuniversity) 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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Composition by percentage mass of elements
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(relating reacting masses and formula
mass)

Using
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(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 GayLussac's Law (ratio of gaseous
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Molarity, volumes and solution
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Energy transfers in physical/chemical changes,
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