|
 9.
The molar gas volume in calculations - moles, gas volumes and Avogadro's Law
It
is better to know how to rearrange an equation than use a formula triangle
A summary of how to do basic molar gas
volume
calculations and rearrangement of the formula needed to solve problems
-
Avogadro's Law states that equal volumes of gases under the same conditions of temperature
and pressure contain the same number of molecules (irrespective of
the nature of the gaseous particles)
-
This means equal amounts of moles
of gases occupy the same volume under the same conditions of temperature
and pressure.
-
So the volumes have equal moles of
separate particles (molecules or individual atoms) in them.
-
Therefore one mole of any gas (formula mass in g), at the same temperature and pressure occupies the
same volume .
-
This is
24 dm3 (24 litres) or
24000 cm3, at room temperature of
20oC/293K and
normal pressure of 101.3 kPa/1 atmosphere (such conditions are often referred to
as RTP/NTP).
-
20oC (293K) is
often treated as room temperature, and the molar gas volume is
24.0 dm3
at normal pressure.
-
25oC (298K) is treated as
a standard temperature, at which the molar volume of the gas is
24.4 dm3
at normal pressure.
-
The molar volume for s.t.p is 22.4 dm3
(22.4 litres) at 0oC and 101 kPa1 atmosphere pressure.
-
Historically, s.t.p unfortunately
stands for standard temperature and pressure, but these days 25oC/298K
is usually considered the standard temperature for data e.g.
thermochemical and thermodynamic data at a higher academic level.
-
Unfortunately, on the internet, both
20oC and 25oC are referred to as 'room
temperature'.
-
Reminder about the Avogadro Number (Avogadro's Constant):
-
The Avogadro number is 6.02 x 1023 particles/mole
of defined chemical 'species'.
-
Therefore, e.g. at 20oC, 24.0 dm3 of a
gas contains 6.03 x 1023 molecules.
-
6.03 x 1023 / 24.0 = 2.5 x 1022
= number of molecules in 1 dm3 of gas.
-
2.5 x 1022 / 1000 = 2.5 x 1019 =
number of molecules in 1 cm3 of gas.
-
For more details see
Part 7
Introduction to the mole and Avogadro's number
-
 Some handy
(essential) relationships for
substance Z below:
-
It is much better to be able to rearrange an equation than use a
triangle.
-
moles Z = mass of Z
gas (g) / atomic or formula mass of gas Z (g/mol)
-
mass of Z in g =
moles of Z x atomic or formula mass
of Z
-
atomic or formula mass of
Z
= mass of Z / moles of Z
-
1 mole = formula
mass of Z
in g.
-
gas volume of Z
= moles of Z x volume
of 1 mole
-
rearranging this
equation gives ...
-
moles of Z
= gas volume of Z
/ volume of 1 mole
-
 moles
= V(dm3) / 24 (at RTP)
-
The latter form of the
equation can be used to calculate molecular mass from experimental data
because
-
moles = mass / molecular
mass = gas volume / volume of 1 mole
-
mass / molecular mass =
gas volume / volume of 1 mole
-
molecular mass = mass x
volume of 1 mole/volume
-
therefore at RTP: Mr
= mass(g) x 24 / V(dm3)
-
so, if you know the mass
of a gas and its volume, you can work out moles of gas and then work
out molecular mass.
-
This has been done
experimentally in the past, but these days, molecular mass is
readily done very accurately in a mass spectrometer.
-
Note (i): In the following examples, assume you are
dealing with room temperature and pressure i.e. 25oC and 1 atmosphere
pressure so the molar volume is 24dm3 or 24000cm3.
-
Note (ii):
-
Apart from solving the problems using the
mole concept (method (a) below, and reading any equations
involved in a 'molar way' ...
-
It is also possible to solve them without using the mole
concept (method (b) below). You still use the molar volume itself,
but you think of it as the volume occupied by the formula mass of the
gas in g and never think about moles!
TOP OF PAGE
and INDEXES
Quiz 9
on moles and the molar volume of a gas
Quiz 10. Calculations involving gas volume ratios
in reactions
Methods
of measuring how much gas is formed in a reaction
TOP OF PAGE
and INDEXES
Quiz 9
on moles and the molar volume of a gas
Quiz 10. Calculations involving gas volume ratios
in reactions
Calculations using the molar as volume
Examples explained and problems for you to solve
(answers at bottom of
page)
-
From now on the calculations involve
interpreting an equation in terms of moles e.g.
-
(e) If 10 g of calcium carbonate
(limestone) is thermally decomposed what volume of carbon dioxide is formed
at room temperature and pressure?
-
The equation is:
CaCO3(s)
===> CaO(s) + CO2(g)
-
Atomic masses: Ca = 40, C = 12,
O =16. Formula mass CaCO3 = 40 + 12 + (3 x 16) = 100
-
1 mol of calcium carbonate gives 1 mol of
carbon dioxide.
-
Therefore mol CaCO3 = mol CO2
= 10/100 = 0.10 mol
-
Therefore volume CO2 = mol CO2
x molar gas volume
-
Volume of CO2 = 0.10 x 24.0 =
2.4 dm3 (or 2.4 x 1000 = 2400 cm3)
-
AND
from now on, you have to the questions - ANSWERS at
bottom of page
-
AND
remember it is better to be able to rearrange a formula than use a triangle.
-
AND
do not assume all atomic masses quoted are needed to solve the problem.
(At RTP 1 mol gas = 24 dm3 or 24000 cm3)
-
Given the atomic masses:
Pb = 207, N = 14,
O = 16
-
and the formula of lead(II)
nitrate: Pb(NO3)2
-
If lead(II)
nitrate thermally decomposes on strong heating according to
the equation
-
lead(II) nitrate ===> lead(II) oxide +
nitrogen dioxide + oxygen
-
2Pb(NO3)2(s)
==> 2PbO(s) + 4NO2(g) + O2(g)
-
(a) Calculate
the mass of lead(II) oxide formed if 66.2 g is decomposed.
-
(b) Calculate
the volumes of nitrogen dioxide and oxygen formed in cm3.
-
Fully
worked out ANSWER
TOP OF PAGE
and INDEXES
Quiz 9
on moles and the molar volume of a gas
Quiz 10. Calculations involving gas volume ratios
in reactions
Learning objectives
for problem solving using the molar gas volume
Know that the volume for 1 mole of gas varies with temperature and
pressure.
Know the different units of temperature and pressure and be able to
convert from one to the other in the context of calculations of moles
and volumes of gases.
Know how to solve problems and do calculations based on the volume of 1
mole of gas (molar volume, usually taken as 24 dm3 at GCSE
level).
Know at RTP moles of gas = volume of gas in dm3 / 24 and be
able to rearrange this moles of gas' equation.
Know that Avogadro's Law states that equal volumes of gases under the same conditions of temperature
and pressure contain the same number of molecules.
Be able to describe and explain how to measure the volume of a gas formed
in a chemical reaction e.g. collection in gas syringe or recording loss of
mass from a system as a gas is produced and escapes into the atmosphere.
Know how to read equations in terms of (molar) mole ratios e.g. to deduce
volumes of gases formed from a given mass of reactant - spot the crucial
mole ratio in the equation.
TOP OF PAGE
and INDEXES
Quiz 9
on moles and the molar volume of a gas
Quiz 10. Calculations involving gas volume ratios
in reactions
WHERE NEXT?
Self-assessment Quizzes on reacting gas volume ratios and
Avogadro's Law
Quiz 9
on moles and the molar volume of a gas, Avogadro's Law
See also for gas calculations
Notes on
the gas laws and P V T gas calculations
Part 10.
Reacting gas volume
ratios, Avogadro's Law
& Gay-Lussac's Law
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Quiz 9
on moles and the molar volume of a gas
Quiz 10. Calculations involving gas volume ratios
in reactions
| OTHER CALCULATION PAGES
1.
What is relative atomic mass, relative isotopic mass, calculating relative atomic mass
2.
Calculating relative
formula/molecular mass of a compound or element molecule
3.
Law of Conservation of Mass and simple reacting mass calculations
4. Composition by percentage mass of elements
in a compound
5. Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
6a. Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination (see calculations section 14.)
6b.
Reacting masses, concentration of solution and volumetric titration calculations
(NOT using moles)
7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
8. Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
9.
Moles and the molar volume of a gas, Avogadro's Law
10. Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
reactants to products)
11. Molarity, volumes and solution
concentrations (and diagrams of apparatus)
(this page)
12.
How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
13. Electrolysis products calculations (negative cathode and positive anode products)
14. Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
14.1
% purity of a product 14.2a
% reaction yield
14.2b
atom economy 14.3
dilution of solutions
14.4
water of crystallisation
calculation
14.5
how
much of a reactant is needed? limiting reactant calculations
Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
Radioactivity and half-life calculations including
dating materials
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Worked out ANSWERS to the moles and gas volume
questions
(At RTP 1 mol gas = 24 dm3 or 24000 cm3)
-
Given the atomic masses:
Pb = 207, N = 14,
O = 16
-
and the formula of lead(II)
nitrate: Pb(NO3)2
-
If lead(II)
nitrate thermally decomposes on strong heating according to
the equation
-
lead(II) nitrate ===> lead(II) oxide +
nitrogen dioxide + oxygen
-
2Pb(NO3)2(s)
==> 2PbO(s) + 4NO2(g) + O2(g)
-
(a) Calculate
the mass of lead(II) oxide formed if 66.2 g is decomposed.
-
(b) Calculate
the volumes of nitrogen dioxide and oxygen formed in cm3.
-
The best way
to solve this problem is to think moles!
-
2 mol
===> 2 mol + 4 mol + 1 mol
-
Formula
masses: Pb(NO3)2 = 331 (watch
the arithmetic!) and PbO = 223
-
(a) for Pb(NO3)2
===> PbO: 2 mol ===> 2 mol (1 : 1 ratio)
-
66.2 / 331 =
0.20 mol lead nitrate
-
so 0.20 mol
PbO formed
-
mass = mol x
formula mass
-
mass PbO
formed = 0.20 x 223 =
44.6 g
-
(b) From (a)
0.20 mol Pb(NO3)2, so,
-
from the mol
ratio of the equation, this will yield 0.40 mol NO2
and 0.10 mol O2
-
The molar gas
volume at RTP = 240000 cm3
-
Volume of NO2
= 0.40 x 24000 =
3600 cm3
-
Volume of O2
= 0.10 x 24000 =
900 cm3
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Quiz 9
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Quiz 10. Calculations involving gas volume ratios
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