CHEM1011 Lecture Notes - Lecture 8: Ionic Compound, Iron Supplement, Fluorine
2 Jul 2018
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TOPIC 8.
CHEMICAL CALCULATIONS II:
% composition, empirical formulas.
Percentage composition of elements in compounds.
In Topic 1 it was stated that a given compound always has the same composition by
weight regardless of how it was produced. The reason for this is that each compound
has a fixed chemical formula which specifies the number of atoms of each of its
component elements. For example, the compound glucose of formula C6H12O6 always
has 6 C atoms, 12 H atoms and 6 O atoms in every one of its molecules. Using the
mole concept, if the formula of the compound is known, then from the atomic weights
of the component elements, the % by weight for each element in the compound can be
calculated.
Using glucose as an example,
Molecular formula: C6H12O6.
Each glucose molecule contains
6 carbon atoms 12 hydrogen atoms 6 oxygen atoms
an Avogadro number of glucose molecules (1 mole of glucose) contains
6 × Avogadro number (= 6 moles) of C atoms and
12 × Avogadro number (= 12 moles) of H atoms and
6 × Avogadro number (= 6 moles) of O atoms.
1 mole of glucose has a mass (its gram formula weight or molar mass) which is the
sum of the gram atomic weights of all of the constituent atoms.
i.e. mass of 1 mole of glucose, C6H12O6 = (6 × 12.01 + 12 × 1.01 + 6 × 16.00) g
= 180.18 g (using atomic weight data to 2 decimals)
1 mole of carbon atoms weighs 12.01 g and there are 6 moles of C atoms in 1 mole of
glucose, so the mass of carbon in 1 mole of glucose = 6 × 12.01 g = 72.06 g.
and % carbon in C6H12O6 = 72.06 × 100 % = 40.0 % by mass.
180.18
Similarly, the percentage of hydrogen and oxygen can be calculated as follows:
1 mole of H atoms weighs 1.01 g and there are 12 moles of H atoms in 1 mole of
glucose, so the mass of hydrogen in 1 mole of glucose = 12 × 1.01 g = 12.12 g.
% hydrogen = 12.12 × 100 % = 6.7 % by mass.
180.18
1 mole of O atoms weighs 16.00 g and there are 6 moles of O atoms in 1 mole of
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glucose, so the mass of oxygen in 1 mole of glucose = 6 × 16.00 g = 96.00 g.
% oxygen = 96.00 × 100 % = 53.3 % by mass.
180.18
[Note that the sum of the % of all the elements must add up to 100 %.]
Thus the % composition of glucose by mass is
carbon 40.0 % oxygen 53.3 % hydrogen 6.7 %
In this way, the % composition by mass of any compound can be calculated provided
that its formula is known.
Example: Calculate the % composition by mass of chloride ion in sodium chloride.
The molar mass of NaCl = 22.99 + 35.45 g = 58.44 g.
From the formula of sodium chloride,
1 mole of NaCl (58.44 g) contains 1 mole of Cl– (35.45 g).
Therefore % by mass of Cl– = 35.45 × 100 % = 60.67 %
58.44
Determination of empirical formulas of compounds.
Recall that the empirical formula of a compound is the simplest integer ratio of the
elements in that compound. The empirical formula of any pure compound can be
determined from analytical data giving the percentage composition by mass of the
elements present. The calculations involved are simply the reverse of those just done
above where, from the empirical or molecular formula, the % by weight of its
component elements was deduced. Now, given the % by weight of the component
elements, the empirical formula will be deduced. This is best shown by some
examples.
Example 1. Analysis of a compound returns the following data for % by mass:
iron (Fe): 63.5 % sulfur (S): 36.5 %
From the data, in 100.0 g of compound there would be 63.5 g of iron combined with
36.5 g of sulfur. The empirical formula expresses the simplest ratio of the relative
number of atoms of Fe and S present and as one mole of atoms of any element
contains NA atoms of that element, this will be the same as the relative number of
moles of Fe and S atoms present in the compound. To calculate the number of moles
of Fe and of S atoms present in the compound, divide the mass of each of these two
constituent elements by their atomic weights.
Moles of Fe = mass / atomic weight of Fe = 63.5/55.85 = 1.14 mole of Fe atoms
Moles of S = mass / atomic weight of S = 36.5/32.07 = 1.14 mole of S atoms
Thus the ratio of moles of Fe to moles of S in the compound is
1.14 moles of Fe atoms : 1.14 moles of S atoms.
i.e. 1.14 × NA Fe atoms: 1.14 × NA S atoms which on dividing by NA gives
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1.14 Fe atoms:1.14 S atoms.
However, the empirical formula must have integer quantities for all the numbers of
atoms in it. In this example, it is obvious that, within the usual allowable experimental
error in analytical data of about 0.3 %,
ratio of atoms of Fe : atoms of S = 1.00 : 1.00 and the empirical formula is FeS.
The above calculation is illustrated in the following flow diagram.
iron 63.5 % by mass sulfur 36.5 % by mass
= 63.5 g in 100 g of compound = 36.5 g in 100 g of compound
Calculate moles of Fe and S atoms:
63.5 / 55.85 moles of Fe atoms 36.5 / 32.07 moles of S atoms
= 1.14 moles Fe = 1.14 moles S
As ratio of moles =
ratio of atoms,
Fe1.14S1.14
Simplify: FeS
Example 2. Analysis returns the following data for an unknown compound:
nitrogen: 26.2 % chlorine: 66.4 % hydrogen: 7.5 % by mass.
Determine its empirical formula.
[Note that these percentages add up to 100 %, within the experimental error. If they do
not add up to 100 %, the difference is attributed to oxygen for which normally there is
no analytical data available.]
In 100.0 g of compound there are 26.2 g of nitrogen, 66.4 g of chlorine and 7.5 g of
hydrogen.
To convert to moles of N, Cl and H atoms, divide each mass by the atomic weight of
each element. [A common error is to divide by the molecular weight for species such
as N, Cl, O and H which occur in nature as diatomic molecules.]
Moles of N atoms = 26.2/14.01 = 1.87 mole of N atoms
Moles of Cl atoms = 66.4/35.45 = 1.87 mole of Cl atoms
Moles of H atoms = 7.5/1.008 = 7.4 mole of H atoms
ratio of the number of atoms of N : Cl : H in the compound is 1.87 : 1.87 : 7.4
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Document Summary
In topic 1 it was stated that a given compound always has the same composition by weight regardless of how it was produced. The reason for this is that each compound has a fixed chemical formula which specifies the number of atoms of each of its component elements. For example, the compound glucose of formula c6h12o6 always has 6 c atoms, 12 h atoms and 6 o atoms in every one of its molecules. Using the mole concept, if the formula of the compound is known, then from the atomic weights of the component elements, the % by weight for each element in the compound can be calculated. 6 carbon atoms 12 hydrogen atoms 6 oxygen atoms. An avogadro number of glucose molecules (1 mole of glucose) contains. 6 avogadro number (= 6 moles) of c atoms and. 12 avogadro number (= 12 moles) of h atoms and. 6 avogadro number (= 6 moles) of o atoms.
