5.5.1 Percentage by Volume
When referring to percentage composition of
gases or liquids, we mean percentage by moles, although it is usually referred
to as percentage by volume. Remember that each component of a mixture of gases
or miscible liquids occupies the total volume.
Example
Consider air to be by volume 20 percent oxygen and 80
percent nitrogen. Find the average molar mass of air.
Given: Air is (by moles) 20% O2 and
80% N2
RTF: Average molar mass
Solution
When solving percentage problems it is good practice to
consider 100 units of the substance in question. In this case, the substance is
air. Consider 100 moles of air. If we can find the mass of 100 moles of air,
and we divide this mass by 100 (mol), we get the average mass of 1 mole of air,
i.e. the average ‘molar mass’.
Step 1: Consider 100 mol of air:
This contains 80 mol of N2 and 20
mol of O2.
Step 2: Find the total mass:
Mass of 80 mol of N2 = 80
mol × 28gmol28gmol = 2240 g
Mass of 20 mol of O2 = 20
mol × 32gmol32gmol = 640 g
Total mass of 100 mol of air = 2240 + 640 = 2880 g
Step 3: Find the average molar mass: 2880 g / 100 mol = 28.8 g/mol
Average molar mass of air is 28.8 g.
Consider the
atmosphere on Mars to be by volume 95% carbon dioxide, 3% nitrogen gas and 2%
argon gas. Find the average molar mass of the air on Mars.
Given Molar mass in
g.mol−1: C = 12, O = 16, N = 14, Ar = 40
Give your answer to
three significant figures.
The average molar mass of air on Mars is …… g
Example
A
hydrocarbon (a compound which contains only hydrogen and carbon) is 80 percent
carbon by weight (i.e. by mass). Find its empirical (simplest) formula.
Given:
hydrocarbon, 80% C (and therefore 20% H).
RTF: Simplest formula.
Solution
Asking for
the simplest formula is really asking for the mole ratio. We are given the mass
ratio and we need to find the ratio by moles. The strategy is to assume we have
100 g of the hydrocarbon and to convert mass to moles.
100 g of
the hydrocarbon contain:
80 g C = 80
g × 1 mol12 g =
(20/3) moles
20 g H = 20
g × 1 mol1 g =
20 moles
Atoms of CAtoms of H=mols of Cmols of H=20/320=13
So the empirical (or simplest) formula is CH3.
A hydrocarbon is 75 percent carbon by weight.
Find its empirical (simplest) formula.
Given Molar mass in
g.mol-1: H = 1, C = 12
The empirical formula
is _____
CH
CH4
CH2
CH6
What is the % by mass
of each element in the compound C6H5NO2?
Given Molar mass in
g.mol−1: H = 1, N = 14, O = 16, C = 12
C
H
N
O
Carbon has the ability to form chains of carbon
atoms connected to other atoms, mainly hydrogen, oxygen and sometimes nitrogen.
These compounds are called organic
compounds.
Organic
compounds burn in oxygen. Any hydrogen in the compound burns into water vapor.
Any carbon burns into carbon dioxide. If we determine the mass of the water
vapor and carbon dioxide produced we can calculate how much hydrogen and carbon
were in the sample.
A typical
setup is shown below
Procedure for analyzing a sample of organic compound:
A sample of the compound to be analyzed is placed in a porcelain dish
inside a Pyrex tube with a large diameter. Dry air or oxygen is allowed into
the tube and the compound is heated to burning point. A pump sucks air gently
out of the free end of Trap 4 to allow air to go through the system. The sample
burns in the air passing above it. The products of combustion are sucked
through Traps 2, 3 and 4. Traps 2 and 3 should be carefully weighed just before
we start the experiment.
Trap 1 contains KOH (s). The purpose of
Trap 1 is to make sure that the air that passes over the heated sample is free
of CO2 and H2O vapor. Solid KOH
is deliquescent (i.e. can
absorb enough water to form a saturated solution). It is also a strong base,
which absorbs (acidic) CO2. Therefore air or oxygen that passes
through Trap 1 will be freed of any CO2 or H2O.
Trap 2 contains
concentrated H2SO4. The purpose of
Trap 2 is to absorb all the water vapor produced as a result of burning the
sample of the compound. Sulfuric acid is hygroscopic (absorbs
water from the air, e.g. table salt). Sometimes anhydrous CaCl2 is
placed instead of H2SO4, as it is also
deliquescent.Trap 2 is weighed again at the end of the experiment. The increase
in the weight of Trap 2 is the weight of the water vapor produced as a result
of burning the sample.
Trap 3 contains NaOH or KOH, solution or solid. The purpose of Trap 3
is to absorb CO2 produced as a result of burning the
sample of the compound. Trap 3 is weighed again at the end of the experiment.
The increase in the weight of Trap 3 is the weight of the CO2 produced
as a result of burning the sample.
Trap 4 contains NaOH or KOH, solution or solid. The purpose of Trap 4
is to prevent any CO2 from going backward to Trap 3. No CO2 can
get to Trap 3 from the outside air. The only CO2 that Trap 3
traps is that coming from the burnt compound samples.
The
apparatus is used to determine the percentage composition of organic compounds
containing C, H and O. The weight of oxygen in the sample can only be obtained
by subtraction, as the following problem shows.
Example
6.0g of an
organic compound are burned. The only products of burning are CO2 and
H2O. The increase in the weights of Traps 2 and 3 were 3.6 g
and 8.8 g respectively. Determine the empirical formula of the compound:
Given: Mass
of sample = 6.0 g
The products of combustion are only H2O and CO2
Increase of weight due to H2O produced = 3.60 g
Increase of weight due to CO2 produced = 8.80 g
RTF: Simplest formula
Solution
Since the
products of combustion are only H2O and CO2 then the
compound contains only C, H and possibly O. There is no way to tell if all the
oxygen for combustion came from the air, or if some of it came from the
compound itself.
Step 1:
Determine the mass of H in the compound:
Molar mass of H2O is (1 + 1 + 16)
= 18 g mol–1
18 g of H2O contain 2 g H
(1+1=2)
Therefore 3.6 g H2O contain 3.6 g H2O
× 2 g H18 g H2O =
0.40 g H
Step 2: Determine the mass of C in the compound:
Molar mass of CO2 is
(12 + 16 + 16) = 44 g mol–1
44 g of CO2 contain 12 g
C
Therefore 8.8 g CO2 contain
8.8 g CO2 × 12 g C44 g CO2 =
2.4 g C
Step 3: Determine the mass of O in the compound:
The sum of the masses of C and H in
the sample = 2.4 + 0.40 = 2.8 g
Therefore the rest of the sample must
be oxygen: 6.0 – 2.8 = 3.2 g
Step 4: Determine the number of moles of each element in the sample:
Number of moles of H = 0.4 g H × 1 mol1 g =
0.40 mol
Number of moles of C = 2.4 g H × 1 mol12 g =
0.20 mol
Number of moles of O = 3.2 g O × 1 mol16 g =
0.20 mol
Step 5: Determine the ratio of the atoms:
The ratio of the atoms is the ratio
of the moles. The simplest way of finding this ratio is to divide all by the
smallest, which is C or O.
The ratio of C to C is (obviously) 1
The ratio of O to C is 0.20/0.20 = 1
The ratio of H to C is 0.40/0.20 = 2
Therefore the simplest formula is CH2O.
4.40 g of an organic compound are burned. The
only products of burning are CO2 and H2O. The mass of water vapor produced was 3.60 g and the mass of
carbon dioxide produced 8.80 g.
Choose the correct
empirical formula of the compound.
Given Molar mass in
g.mol−1: H = 1, C = 12, O = 16
The empirical formula
is _____.
C2H2O
CHO
C2HO
CH4
C2H4O
