Chapter 2: Analysis by Mass
The quantitative analysis by mass is called Gravimetric Analysis. It can also be used to determine the
chemical formula of a compound.
Water is a component in many products. It can be naturally occurring or added during the manufacturing
process. Cheaper or bigger products are not necessarily the best value because the mass of water in the
sample may disguise the actual amount of product being purchased.
The percentage of water in a sample can be determined
by heating the sample at 110oC until there is no change
in mass. This is referred to as “heating to constant
mass”.
Your Turn:
A 23.1g sample of soil is heated for 60 minutes at 110oC. It is then weighed and the process
repeated a number of times. A mass of 21.0g is eventually obtained, which does not change upon
further heating. Calculate the percentage of water in this sample of soil.
Initial mass of soil
Final mass of soil
Mass of water lost
% water =
mass of water
Initial mass of sample
x
100
Gravimetric analysis will often involve the following steps:
1) Weighing the sample to be analysed
2) Dissolving the sample in water
3) Adding a suitable chemical to form a precipitate
4) Filtering to collect a precipitate
5) Repeated drying and weighing until a constant mass of precipitate is formed.
***simple Stoichiometry can then be used to deduce the amount of the component in the original
mixture***
When designing a gravimetric procedure, knowledge of the solubilities of the precipitates likely to be
produced is important. Vacuum filtration is often used in gravimetric analysis in preference to gravity
filtration. This is because it is quicker and it assists in drying excess solvent off the precipitate.
Your Turn:
A technician takes a 5.0g sample of vegemite and dissolves it into water. Enough silver nitrate
solution is then added to precipitate all the chloride ions present. After filtering and drying to
constant mass, 1.05g silver chloride is obtained. Calculate the percentage sodium chloride in the
sample:
Sources of Error:
If certain things happen during the analysis it is important to be able to predict the effect this will have on
the final result.
Action
Insoluble materials not filtered
out before forming precipitate
Not enough of the precipitate
forming chemical added
Effect on result
Overestimated
Underestimated
Forming a precipitate that is too
soluble
Forming extra precipitate due to
presence of competing ions
Weighing precipitate before it is
dry
Not rinsing precipitate before
drying out
Underestimated
Adding too much of the
precipitate causing chemical
No effect
Using too much water for the
initial dissolving
No effect
Overestimation
Overestimated
Overestimated
Reason
Apparent mass of the precipitate
will increase
Not enough of the precipitate will
form as some of the ions being
analysed for will remain in
solution
Not all the ions being analysed
will be in the precipitate
Too much precipitate will form
The water present will increase
the apparent mass of precipitate
As the precipitate dries, other
soluble chemicals will begin to
crystallise out of the small
amount of solution still trapped
in the precipitate. These will add
to the mass.
This is a necessary part of the
method to make sure all required
ions are in the precipitate.
Chemical must be in excess.
This is a practical consideration –
the more water you have, the
longer the filtering step.
Empirical and Molecular Formula:
The empirical formula of a compound is determined by calculating the number of moles of each element
that is present. These figures are converted into the simplest whole number ratio. The molecular formula
is the actual number of atoms present in a molecule of a substance. If the molar mass is known, it is
possible to calculate the empirical formula.
Your Turn:
A compound contains 64.9g Carbon, 13.5g Hydrogen and 21.6g of Oxygen. If the molar mass of the
compound is found to be 148, find the empirical and molecular formula.