Gravimetric analysis
Four example problems
1. Determining salt in food
2. Determining sulfate and sulfur content in fertiliser
3. Determining the empirical formula of a hydrated
molecule
4. Determining the empirical formula of an alcohol
Question 1: Determining salt in food
A 6.67 g sample of potato chips was crushed, blended with water and
filtered. Excess silver nitrate was added to the filtrate. The dried
precipitate had a mass of 0.245 g. Calculate the amount of sodium
chloride that was in the potato chips (assume all chloride is in the form of
the sodium salt). Give your answer in mg of NaCl per 100 g of chips.
Where do we start?
• Write out the relevant chemical reaction:
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
• Identify what else we know:
The final mass of AgCl is 0.245 g (3 significant figures) taken from 6.67 g (3 significant
figures) of chips
• Think about where you want to get to:
Calculate how many mg of NaCl are present in 100 g of potato chips
Calculating the moles of AgCl
• NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
m(AgCl) = 0.245 g
M(AgCl) = 107.9 gmol-1 + 35.5 gmol-1
= 143.4 gmol-1
n(AgCl) = 0.245 g/143.4 gmol-1
= 0.001709 mol
Calculating the mass of NaCl
• NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
n(AgCl) = 0.001709 mol
From the chemical equation we know that the moles
of NaCl to the moles of AgCl is 1:1, therefore:
n(NaCl) = 0.001709 mol
M(NaCl) = 23.0 gmol-1 + 35.5 gmol-1 = 58.5 gmol-1
m(NaCl) = 0.001709 mol × 58.5 gmol-1 = 0.09998 g
Calculating the amount of NaCl in mg per 100 g
m(NaCl) 0.09998 g
−1
=
× 1000 mgg × 100
m(chips)
6.67 g
3
= 1.50 × 10
There are 1.50 ×
chips.
3
10
mg of NaCl per 100 g of
What does this really mean?
• If you look at the nutrition panel of any food, you will find the mass of
sodium reported in mg per 100 g.
• There are a few questions that you might ask:
• If there are 1.50 × 103 mg of NaCl per 100 g of chips, how mg of Na are there
per 100 g of chips?
• How would the presence of other chloride salts (e.g. KCl) impact calculations of
sodium content based on a silver chloride precipitation reaction?
• Other food additives also contain sodium (e.g. monosodium glutamate or
sodium bicarbonate). Would a AgCl precipitation identify other sources of
sodium?
Question 2: Determining sulfate and sulfur
content in fertiliser
A student took a 0.998 g sample of fertiliser and added it to 100 mL of
water. The solution was mixed well until all soluble fertiliser had dissolved.
The solution was then filtered. The filtrate was acidified with dilute
hydrochloric acid, then treated with excess barium chloride. The precipitate
was filtered, dried to a constant mass and weighed. The weight of the
precipitate was 0.387 g. Calculate the percentage of sulfate and the
percentage of sulfur in the sample of fertiliser.
Where do we start?
• Write out the relevant chemical reaction:
SO42-(aq) + Ba2+(aq)  BaSO4(s)
• Identify what is important and what is not for the calculation:
Is the volume of water added important? (more about this later)
Is the presence of HCl important? (more about this later)
Question 2: Determining sulfate and sulfur
content in fertiliser
A student took a 0.998 g sample of fertiliser and added it to 100 mL of
water. The solution was mixed well until all soluble fertiliser had dissolved.
The solution was then filtered. The filtrate was acidified with dilute
hydrochloric acid, then treated with excess barium chloride. The precipitate
was filtered, dried to a constant mass and weighed. The weight of the
precipitate was 0.3827 g. Calculate the percentage by mass of sulfate and
the percentage by mass of sulfur in the sample of fertiliser.
Where do we start (continued)?
• Identify what we know:
The final mass of BaSO4 is 0.3827 g (4 significant figures) taken from 0.998 g (3 significant figures)
of fertiliser
• Think about where you want to get to:
Calculate how many grams of sulfate are present in 100 grams of fertiliser. Do the same for sulfur.
Calculating the moles of BaSO4
• SO42-(aq) + Ba2+(aq)  BaSO4(s)
m(BaSO4) = 0.3827 g
M(BaSO4) = (137.3 + 32.1 + 4 × 16.0) gmol-1
= 233.4 gmol-1
n(BaSO4) = 0.3827 g/233.4 gmol-1
= 0.001640 mol
Calculating the mass of SO4
2-
• SO42-(aq) + Ba2+(aq)  BaSO4(s)
n(BaSO4) = 0. 001640 mol
From the chemical equation we know that the moles
of SO42- to the moles of BaSO4 is 1:1, therefore:
n(SO42-) = 0. 001640 mol
M(SO42-) = (32.1 + 4 × 16.0) gmol-1 = 96.1 gmol-1
m(SO42-) = 0. 001640 mol × 96.1 gmol-1 = 0.1576 g
Calculating the percentage by mass of SO42- in
the fertiliser
2−
m(SO4 )
0.1576 g
=
× 100
m(fertiliser)
0.998 g
= 15.8% (round to 3 sig figs)
The fertiliser contains 15.8% sulfate by mass.
Calculating the percentage by mass of sulfur in
the fertiliser
2The mole ratio of S to SO4 is 1:1.
n(SO4
2-)
= n(S) = 0.001640 mol
m(S) = 32.1 gmol-1 × 0.001640 mol
= 0.05263 g
m(S)
0.05263 g
=
× 100 = 5.27%
m(fertiliser)
0.998 g
Are there other considerations?
• Is knowing the exact volume of water used to dissolve the sulfate
containing compounds important?
• To answer this, consider the purpose of the water and whether it ever came
into play in any of the calculations.
• What is the purpose of the hydrochloric acid?
• To answer this, you need a bit more information.
• Fertiliser also contains carbonate containing compounds.
• Will carbonate precipitate in the presence of barium ions?
• What happens to carbonate in the presence of acid?
Question 3: Determining the empirical
formula of a hydrated molecule
Magnesium sulfate readily takes up water and is usually found in a hydrated form with a
set number of water molecules per molecule of magnesium sulfate. A student
conducted an experiment to calculate the empirical formula for hydrated magnesium
sulfate.
• Mass of empty crucible:
45.39 g
• Mass of crucible, cover and magnesium sulfate before heating: 50.14 g
• Mass of crucible, cover and magnesium sulfate after 1st heating: 47.70 g
• Mass of crucible, cover and magnesium sulfate after 2nd heating: 47.63 g
• Mass of crucible, cover and magnesium sulfate after 3rd heating: 47.62 g
Where do we start?
• Think about where you want to get to
 mole ratio of water to magnesium sulfate
Calculate the number of moles of water
• m(H2O) = starting mass – dried mass
= 50.14 g – 47.62 g
= 2.52 g
• M(H2O) = (2 × 1.0 + 16.0) gmol-1
= 18 gmol-1
• n(H2O) = 2.52 g/18 gmol-1
= 0.1400 mol
Calculate the number of moles of anhydrous
magnesium sulfate
• m(MgSO4) = dried mass – crucible
= 47.62 g – 45.18 g
= 2.44 g
• M(MgSO4) = (24.3 + 32.1 + 4 × 16.0) gmol-1
= 120.4 gmol-1
• n(MgSO4) = 2.44 g/120.4 gmol-1
= 0.0203 mol
Calculate the mole ratio and determine the
empirical formula
n(H2O)
0.1400 mol
=
n(MgSO4) 0.0203 mol
n(H2O)
6.9
=
n(MgSO4)
1
The empirical formula hydrated magnesium sulfate is MgSO47H2O
Question 4: Determining the empirical
formula of an alcohol
When 12 g of an unknown alcohol was fully combusted, it was found to
produce 26.4 g of carbon dioxide and 14.4 g of water. What is the empirical
formula for the alcohol?
Where do we start?
• Write out the relevant chemical reaction:
CxHyOz+ O2  CO2 + H2O
We know that alcohols contain carbon, hydrogen and oxygen, but we don’t know the numbers of
each, so we will just call them x, y and z.
• Think about what you know and how you can use it to get started:
Knowing the mass of the products will allow the calculation of the number of moles of the
products, which will give the number of moles of carbon and hydrogen. Oxygen will be a bit more
tricky.
Determining the number of moles of carbon
and hydrogen
CxHyOz + O2  CO2 + H2O
Mass (g):
26.4
14.4
Molar mass (gmol-1):
44.0
18.0
Moles (g/gmol-1 ):
0.60
0.80
Moles (C or H)
0.60
1.60
(note: the number of moles of carbon equals the
number of moles of CO2; the number of moles of
hydrogen is twice the number of moles of H2O)
Determining the number of moles of oxygen
CxHyOz + O2  CO2 + H2O
Knowing the number of moles of carbon and
hydrogen, we can work out the mass of carbon
and hydrogen in the alcohol and from that derive
the mass of oxygen in the alcohol.
Determining the number of moles of oxygen
CxHyOz + O2  CO2 + H2O
Moles of C and H(mol)
0.6
1.6
Molar mass (gmol-1)
12.0
1.0
Mass (mol × gmol-1)
7.2
1.6
Mass of oxygen = 12 g – 7.2 g – 1.6 g = 3.2 g
Moles of oxygen in the alcohol = 3.2 g/16.0 gmol-1
= 0.2 mol
Determining the empirical formula
CxHyOz
Mole ratio:
0.6:1.6:0.2
Divide through by lowest number in ratio:
0.6 1.6 0.2
: :
0.2 0.2 0.2
3:8:1
The empirical formula for the alcohol is C3H8O.
C3H8O – propyl alcohol (IUPAC name: propan-1-ol)