Year 12 Chemistry Chapter 2: Analysis by Mass
2.1 Measuring water content
Water is a component in many consumer products, including foodstuffs. In some cases water occurs naturally in the food and in others it is added as part of the manufacturing process.
Some brands are less concentrated than others and contain a higher water content.
Cheapest or biggest doesn’t always give best value for money because the mass of water in the may disguise the actual amount of product being purchased. For example Farmers need to check the water content of the grain they harvest.
The percentage of water in a sample can be determined by heating the sample 110  C until there is no change in mass. This is referred to as heating to constant mass.
Weigh a sample
Heat the sample in a chemical oven at
110  C
Cycle must be repeated until the mass is
constant
Allow the sample to cool in a dessicator
Reweigh
Example: an analytical chemist wanting to find the percentage of water in a can of soup obtains the following results. What is the percentage of water in the soup?
Initial mass of soup 233.1 g
Second mass, after heating
Third mass, after heating
33.6 g
24.3 g
Fourth mass, after heating 24.3 g
Mass of water = original weight – constant weight

Therefore,
Mass of water in soup =
% water in the soup =
Question: 5
Review:
THE MOLE
One of the most useful units that chemists use is the mole. If you remember the mole is a unit used for counting particles.
NOTE: 3 mol of H atoms is completely different from 3 mol of H
2
molecules.
N = amount of substance m = mass of substance
M = Molar mass of substance
Avogadro’s number (N
A
) = ___________ particles
X amount number of
Example: Calculate the amounts, in mol, of H
2
O molecules in 9.0 g of water.
Relative molecular mass of H2O =
Therefore, molar mass (M) = g/mol or _______
Since n = m/M n(H
2
O) =
So, 9.0 g of water contains _____ mol of H
2
O.
Amounts of Gases
Gas properties:
 Spread to fill the volume available
 Have low densities
 Are easily compressed
 Mix rapidly.

When the internal volume of a gas container is known, it is possible to calculate the amount of substance (n), measured mol, by using the general gas equation:
P =
V = n =
R =
T =
Pressure =
1.00 atm = 760mmHg = 101 x 10 3 Pa (101 kPa) = 1.01 bar
Volume =
1000 mL = L
1 mL = cm
1L = dm 3
1000 L = m
3
3
Molar Volume of a gas
The volume of one mole of a gas depends on the gas temperature and pressure.
Standard temperature and pressure (STP) = 0°C and 1 atmosphere
Molar volume of a gas = 22.4 L/mol
Standard Laboratory Conditions (SLC) = 25°C and 1 atmosphere
Molar volume of a gas = 24.5 L/mol n = n =
V =
V m
=
Example: A steel cylinder with a volume of 30.0 L is filled with nitrogen gas to a pressure of 2.00 atm at 25.0°C. What mass of nitrogen does the cylinder contain? pV = nRT
P =
V = n =
R =
T =

Example: Calculate the amount, in mol, of sulfur dioxide in 10.0 L of the gas, measured at STP.
QUESTIONS: 6, 7, 9.
2.2 Finding the Composition of a Compound
Empirical Formula
Chemic analysis enables us to determine the empirical formula of a compound. This shows the simplest whole-number ratio of the present in the compound.
Example:
 ethene
Molecular formula: C
2
H
Empirical formula: CH
2
4
 glucose
Molecular formula: C
6
H
12
O
6
Empirical formula:
Empirical formulas are determine experimental, usually by finding the mass of each element in a given mass of compound.
Example: chemical analysis of an oxide of sulfur present in the gaseous emissions from a factory shows it contains 40.0% sulfur. Find its empirical formula.
A 100 g sample of this compound will contain _____ g of sulfur and ____ g of oxygen.
The ratio of the amounts of sulfur and oxygen is:
m(S) m(O) n(S) : n(O) = M(S) : M(O)
40.0 g
= 32.0 g mol
60.0 g
-1 : 16.0 g mol -1
= 1.25 : 3.75
Simplify by dividing both values by the smaller number
= 1 : 3
So, the empirical formula of this oxide of sulfur is SO
3

Example: find the empirical formula of an alkane extracted from natural gas if, upon complete combustion, a sample of the alkane produces 7.75 L of carbon dioxide measured at SLC and 7.59 g of water.
Calculate the amount of carbon in mol
V n(CO
2
) = V m
=
There is only one carbon atom in each CO2 molecule, so the amount of carbon in the sample is also ____ mol.
Calculate the amount of hydrogen atoms in mol n(H
2
M(H
O) = M(H
2
2
O)
O) =
Since each water molecule has two hydrogen atoms then:
_____ x 2 = _______ mol
The ratio n(C) : n(H)
=
Divide both by the smallest number
=
This is still not a whole number so turn decimals into fractions:
= 1 : 2__
= 1 : ___
Multiple both number by the lowest common denominator: ____
=
The empirical formula is ________ name the alkane: ________
Example: After heating of 2.95 g of crystals of hydrated magnesium sulfate
(MgSO
4
.xH
2
O), the mass of the residue is found to be 1.44 g. Find the value of x in the formula of the compound.
The mass of water lost from the sample is 2.95 g – 1.44 g =
The residue is magnesium sulfate (MgSO
4
)
So the ratio n(MgSO
4
) : n(H
2
O) =
=
=
Divide each value by the smallest number
=
Since the ration n(MgSO
4
) : n (H
2
O) is equal to 1 : x, the formula of the hydrated salt is
MgSO
4
.__H
2
O.
Example: An organic compound formed by fermentation of sugar is known to contain only the elements carbon, hydrogen and oxygen. It is analysed by burning a 1.00 sample in air.

If 1.91 g of carbon dioxide and 1.17 g of water are produced, determine the empirical formula of the compound. n(CO
2
) = therefore, n(C) = n(H
2
O) = therefore n(H) = to find mol of oxygen you must first calculate the mass of carbon and hydrogen. Then subtract the mass of the two elements from the sample, giving you oxygen. m(O) =
m(O) n(O) = M(O) = = mol
The ratio n(C) : n(H) : n(O) =
Divide each value by the smallest number
= : :
The compound has the empirical formula =
Molecular formulas
A molecular formula gives the actual number of atoms of each element present in a molecule of a compound. The molecular formula is always a whole-number multiple of the empirical formula.
Empirical formula: CH
2
O
Molecular formula: C
6
H
12
O
6
, ____ times larger.
A molecular formula can be deduced from an empirical formula if the molar mass is known.
Example: A compound has the empirical formula CH and a molar mass of 78 g mol is the molecular formula?
-1 . What
The molecule must contain a whole number of CH units
M(CH) = ____ g mol -1 molar mass of compound
Number of CH units in a molecule = molar mass of one unit = =
Therefore the molecule formula is CH x ____ =
Questions: 10, 12, 13, 19, 20, 23, 24 & 34a.

2.3 Calculating masses of reactants and products
The mole is particularly useful for calculating the quantities of substances consumed or produced in chemical reactions.
Mole ratios can be used to calculate masses of reactants and products.
Define the term ‘precipitate’: ___________________________________________
__________________________________________________________________
Example: Calculate the mass of lead iodide that can be made from 30.0 g of potassium iodide.
Step 1: Write a balanced equation (nitrate is involved):
Step 2: Calculate the amount in mol of potassium iodide consumed (known)
M(KI) n(KI) =
Step 3: use the ratio of amounts of substances to calculate the number of moles of lead iodide produced.
unknown mol ratio = known
Step 4: Calculate the mass of lead iodide (unknown)
M(PbI
2
) = n(PbI
2
) =
Example: The active ingredient of an antacid powder is magnesium carbonate. When treated with excess hydrochloric acid, a 3.50 g sample of the antacid produced 714 mL of carbon dioxide, measured at 22.0C and 101.3 kPa pressure. Calculate the percentage of magnesium carbonate in the powder.
Step 1: Write a balanced equation
Step 2: Find the amount (in mol) of the known substance, in this case CO
2
Step 3: Mole ratio

Step 4: find the mass of the unknown substance
% is mass of magnesium carbonate over the sample mass x100.
=
Calculations involving excess reactants
Calculations become more complex if the reactants are not present in the stoichiometry ratio. In these cases you must determine which reactant is completely consumed in the reaction (limiting reactant) and which one is in excess. The amount of limiting reactant determines how much product is formed.
*Look at Figure 2.5
Example: Calculate the mass of silver bromide that can be formed if a solution containing 15.0 g of silver nitrate is allowed to react with a solution containing 10.0 g calcium bromide.
Step 1: Write a balanced reaction
Step 2: Calculate the amount in mol of both reactants
M(AgNO
3
) = n(AgNO
3
) = n(CaBr
2
) =
M(CaBr
2
) =
* Identify the limiting and excess reactant. n(CaBr
2
) = 1 n(AgNO
3
) 2 n(AgNO
3
) 2 n(CaBr
2
) = 1
So to react 0.0883 mol of AgNO
3
So to react 0.0500 mol of CaBr n(AgNO n(AgNO
3
3
) required
) available
2 n(CaBr
2
) required = ½ x 0.0883 mol
= 2 x n(CaBr
2
)
= 2 x 0.500 mol
= 0.04415 mol n(CaBr
2
) available = 0.0500 mol
= 0.1 mol
= 0.0883 mol -> limiting reactant, CaBr
2
in excess
Step 3: mole ratio using limiting reactant
Step 4: Calculate the mass of silver bromide.

Table of results: become familiar with using these tables
2 AgNO
3 (aq)
+ CaBr
2 (aq)
 2AgBr
(s)
+ Ca(NO
15.0 g 10.0 g
3
)
(aq)
Initially
Reacting (ratio from equation)
Finally
0.0883 mol
0.0883 mol
0.0500 mol
0.04415 mol  0.0883 mol
0.00585 mol 0.0883 x 187.80
= 16.6 g
0.04415 mol
0.04415 mol
Questions 15, 16, 26 & 28.
2.4 Finding the composition of a mixture
Most commercial products are mixtures. An analytical chemist may seek to find the percentage of one component in such a mixture.
Ions can be analysed by a technique called gravimetric analysis. This involves forming a suitable precipitate with the ion and using the amount of precipitate to calculate the amount of the ion in the sample:





Example: Infant cereals should not contain more the 0.3% sodium chloride. A 7.802 g sample of baby cereal was blended with water and filtered. The solution obtained was mixed with excess silver nitrate solution, causing silver chloride to precipitate. The precipitate was collected by filtration, dried and weighed. A mass of 0.112 g was obtained. What is the percentage of sodium chloride in the baby food, assuming that all the chloride is present as sodium chloride?
Step 1: Write a full balanced chemical equation
Write an ionic equation:
Step 2: Calculate the amount in mol of AgCl present in the precipitate:
M(AgCl) = n(AgCl) =
Step 3: mole ratio
Step 4: Calculate the mol of NaCl, then the mass
Step 5: find the percentage of NaCl in the sample:
Example: The content of saccharine (C
7
H
7
NO
3
S) in diet sweetener tablets can be determined by oxidising the sulfur to sulfate and precipitating it as barium sulfate
(BaSO
4
). A 0.607 g sample yields 0.3196 g barium sulfate. What was the percentage of saccharine in the sample?
Step 1: Calculate mol on barium sulfate.
Step 2: 1 mole of C
7
H
7
NO
3
S yields 1 mole of BaSO
4
(only 1 sulfur atom in each)
Mole ratio
Step 3: Calculate mass of C
7
H
7
NO
3
S

Step 4: Calculate the percentage of C
7
H
7
NO
3
S in the sample.
QUESTIONS: 17, 44, 46, 47 & 48.