|
 9.
The molar gas volume in calculations, moles, gas volumes and Avogadro's Law
A summary of how to do basic molar gas
volume
calculations and rearrangement of the formula
-
Avogadro's Law states that equal volumes of gases under the same conditions of temperature
and pressure contain the same number of molecules.
-
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).
-
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.
-
The molar volume for s.t.p is 22.4 dm3
(22.4 litres) at 0oC and 1 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/thermodynamic data at a higher academic level.
-
Unfortunately, on the internet, both
20oC and 25oC are referred to as 'room
temperature'.
-
 Some handy relationships for
substance Z below:
-
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!
-
Methods
of measuring how much gas is formed in a reaction (volume can be
compared with theoretical prediction!)
-
(a)
You can collect the gases in a calibrated gas syringe.
-
You must make sure too much gas
isn't produced and too fast!
-
A gas syringe is more accurate
than collecting the gas in an inverted measuring cylinder under water shown
below, but its still only accurate to the nearest cm3.
-
You can collect any gas by this
method.
-
(b)
The gas is collected in a measuring cylinder filled with water and inverted
over a trough of water.
-
You can get a more accurate
result by using an inverted burette instead of a measuring cylinder.
-
However, this method is no good
if the gas is soluble in water!
-
Burettes are calibrated in 0.10
cm3 intervals. measuring cylinders to the nearest cm3
or worse!
-
In both methods the reaction is
carried out in conical flask fitted with a sealing rubber bung, but a tube
enabling the gas evolved to be collected in some suitable container.
-
(c)
 A
third method is to measure the mass loss due to gas loss by carrying out the reaction in a
flask set up on an accurate one-pan electronic balance.
-
You need to put a cotton wool
plug in the neck of the conical flask in case you lose any of the solution
in a spray as the gas bubbles up - effervescence can produce an aerosol.
-
This method can be used for any
reaction that produces a gas, but the gas is released into the laboratory,
ok if its harmless.
-
It is potentially the most
accurate method, BUT, the mass loss may be quite small especially hydrogen
[Mr(H2) = 2], better for the 'heavier' gas
carbon dioxide [MrCO2) = 44]
-
There is also a possible error if the gas
dissolves in water, oxygen has a very low solubility, but carbon dioxide is
much more soluble and will reduce the measured mass of gas evolved and
measured.
-
Molar
gas volume calculation
Example 9.1
-
(a) What is the volume of 3.5 g of hydrogen
(H2)? [Ar(H)
= 1]
-
common thinking: hydrogen exists as H2 molecules, so Mr(H2)
= 2, so 1 mole or formula mass in g = 2 g
-
method (i)
-
method (ii):
-
2 g occupies 24 dm3, so scaling up for the
volume
of hydrogen ...
-
3.5 g will have a volume of 3.5/2 x 24 =
42 dm3 (or 42000 cm3)
-
-
-
(b) What is the volume of 11.6 g of
butane gas at RTP?
-
The formula of butane is C4H10,
atomic masses are C = 12 and H = 1
-
Formula mass of butane = (4 x 12) + (10 x
1) = 58
-
moles butane = mass butane / formula mass
= 11.6/58 = 0.20 mol
-
volume = moles x molar volume
-
volume butane = 0.20 x 24 =
4.8 dm3
-
-
-
(c) What mass of chlorine (Cl2)
does 6.0 dm3 of the gas contain at rtp?
-
Atomic mass of Cl = 35.5, molecular mass
of chlorine = 2 x 35.5 = 71
-
1 mol gas = 24 dm3, so mol Cl2
= 6/24 = 0.25 mol
-
mass = moles x molecular mass = 0.25 x 71
= 17.75 g (4sf, 2dp)
-
-
-
(d) What mass of methane (CH4)
is contained in a 10 dm3 cylinder at three times normal
atmospheric pressure at room temperature.
-
Atomic masses: C = 12, H = 1, molar
gas volume = 24 dm3.
-
Formula mass of methane = 16 + (4 x 1) =
16
-
At normal pressure mol of CH4
would be 10/24 = 0.4166
-
However, in 3x normal pressure there will
be 3x as many methane particles in the same volume.
-
So the actual number of moles at this
higher pressure will be 3 x 0.4166 = 12.5 mol
-
mass CH4 = mol CH4
x formula mass of CH4 = 12.5 x 16 =
200 g (or 0.20
kg)
-
-
-
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)
-
-
-
Molar gas volume calculation
Example 9.2
-
Given the equation
-
MgCO3(s) + H2SO4(aq) ==>
MgSO4(aq)
+ H2O(l) + CO2(g)
-
This equation is read as 1 mole of
magnesium carbonate reacts with 1 mole of sulfuric acid to form 1 mole of
magnesium sulfate, 1 mole of water and 1 mole of carbon dioxide gas
-
What mass of magnesium carbonate is needed to make 6 dm3
of carbon dioxide?
-
The important mole ratio is 1 MgCO3
==> 1 CO2
-
method (a):
-
since 1 mole = 24 dm3, 6 dm3 is
equal to 6/24 = 0.25 mol of gas
-
From the equation, 1 mole of MgCO3 produces 1
mole of CO2, which occupies a volume of 24 dm3.
-
so 0.25 moles of MgCO3 is needed to make 0.25
mol of CO2
-
formula mass of MgCO3 = 24 + 12 +
(3 x 16) = 84,
-
so required mass of MgCO3
= mol x formula
mass = 0.25 x 84 =
21g
-
-
-
method (b):
-
converting the equation into the required reacting masses ..
-
formula masses: MgCO3 = 84 (from above), CO2 = 12 +
(2 x 16) = 44
-
MgCO3 : CO2 equation ratio is 1 :
1
-
so 84g of MgCO3 will form 44g of CO2
-
44g of CO2 will occupy 24dm3
-
so scaling down, 6 dm3 of CO2 will
have a mass of 44 x 8/24 = 11g
-
if 84g MgCO3 ==> 44g of CO2,
then ...
-
21g MgCO3
==> 11g of CO2 by solving the ratio, scaling down by
factor of 4
-
-
-
Molar gas volume calculation
Example 9.3
-
Molar gas volume calculation
Example 9.4
-
Given the equation ... (and Ar's
Ca = 40, H = 1, Cl = 35.5)
-
Ca(s) + 2HCl(aq) ==> CaCl2(aq)
+ H2(g)
-
The equation is read as 1 mole of calcium
atoms reacts with 2 moles of hydrochloric acid to form 1 mole of calcium
chloride salt and 1 mole of hydrogen gas molecules (NOT atoms).
-
What volume of hydrogen is formed when
...
-
(i) method (a):
-
3g Ca = 3/40 = 0.075 mol Ca
-
from 1 : 1 ratio in equation, 1 mol Ca produces 1 mol H2
-
so 0.075 mol Ca produces 0.075 mol H2
-
so volume H2
= 0.075 x 24 =
1.8 dm3
(or 1800 cm3)
-
-
-
(i) method (b):
-
from equation 1 Ca ==> 1 H2 means 40g
==> 2g
-
so scaling down: 3g Ca will produce 2 x 3/40 = 0.15g H2
-
2g H2 has a volume 24 dm3, so
scaling down ...
-
0.15g H2 has a
volume of (0.15/2) x 24 =
1.8 dm3
(or 1800 cm3)
-
-
-
(ii) method (a) only:
-
from equation: 2 moles HCl ==> 1 mole H2
(mole ratio 2:1)
-
so 0.25 mol HCl ==> 0.125 mol H2, volume 1
mole gas = 24 dm3
-
so volume H2
= 0.125 x 24 =
3 dm3
-
-
-
Molar
gas volume calculation
Example 9.5
-
Given the equation ... (and Ar's
Mg = 24, H = 1, Cl = 35.5)
-
Mg(s) + 2HCl(aq) ==> MgCl2(aq)
+ H2(g)
-
The equation is read as 1 mole of
magnesium reacts with 2 moles of hydrochloric acid to form 1 mole of
magnesium chloride salt and 1 mole of hydrogen gas molecules (NOT atoms).
-
How much magnesium is needed to make 300 cm3
of hydrogen gas?
-
method (a)
-
The important mole ratio is 1 Mg
==> 1 H2
-
300 cm3 = 300/24000 = 0.0125 mol H2
(since 1 mol of any gas = 24000 cm3)
-
from the equation 1 mole Mg ==> 1 mole H2
-
so 0.0125 mole Mg needed to make 0.0125 mol H2
-
so mass of Mg = mole Mg x Ar(Mg)
-
so mass Mg needed =
0.0125 x 24 =
0.30 g
-
-
-
method (b)
-
reaction ratio in equation is 1 Mg ==> 1 H2,
-
so reacting mass ratio is 24g Mg ==> 2g H2,
-
2g H2 has a volume of 24000 cm3
(volume of formula mass in g)
-
so scaling down: mass Mg needed
= 24 x (300/24000) =
0.30 g
-
-
Molar gas volume calculation
Example 9.6
-
A small teaspoon of sodium hydrogencarbonate
(baking
soda) weighs 4.2g.
-
Calculate the moles, mass and volume of carbon dioxide
formed when it is thermally decomposed in the oven.
-
Assume room
temperature for the purpose of the calculation.
-
2NaHCO3(s) ==> Na2CO3(s)
+ H2O(g) + CO2(g)
-
This equation is read as 2 moles of
sodium hydrogencarbonate decomposes to give 1 mole of sodium carbonate, 1
mole of water and 1 mole of carbon dioxide gas.
-
The important mole ratio is 2
NaHCO3 ==> 1 CO2
-
Formula mass of NaHCO3 is 23+1+12+(3x16) = 84 =
84g/mole
-
Formula mass of CO2 = 12+(2x16) = 44 = 44g/mole
(not needed by this method)
-
In the equation 2 moles of NaHCO3 give 1 mole
of CO2 (2:1 mole ratio in equation)
-
Moles NaHCO3 = 4.2/84 = 0.05 moles ==>
0.05/2 = 0.025 mol CO2 on decomposition.
-
Mass = moles x formula mass, so mass CO2 =
0.025 x 44 = 1.1g CO2
-
Volume = moles x molar volume = 0.025 x 24000 =
600 cm3
of CO2
-
-
-
Molar gas volume calculation Example 9.7
-
Molar gas volume calculation
Example 9.8
-
Molar gas volume calculation
Example 9.9
-
3.27g of metal dissolved in dilute
sulfuric acid to form 1.2 dm3 of hydrogen gas at RTP.
-
It was believed that the equation for the
reaction is:
-
1 mole of gas at RTP occupies a
volume of 24 dm3
-
Calculate the atomic mass of the metal.
-
moles of hydrogen = 1.2 / 24 = 0.05
-
According to the equation 1 mole of metal
produces 1 mole of hydrogen.
-
Therefore moles of metal M also equals
0.05
-
rearranging: moles = mass / atomic mass
-
gives atomic mass = mass / moles
-
therefore atomic mass of metal M =
3.27 / 0/05 = 65.4
-
-
-
Molar gas volume calculation
Example 9.9
-
Sodium hydrogencarbonate reacts with
acids to form a salt, water and carbon dioxide e.g.
-
sodium hydrogencarbonate +
sulfuric acid ===> sodium sulfate + water +
carbon dioxide
-
The balanced symbol equation:
-
Atomic masses: Na =23, H = 1,
S = 32, O = 16
-
(a) Calculate the formula masses of
sodium hydrogencarbonate, sodium sulfate and carbon dioxide.
-
(b)
What is the maximum volume of carbon dioxide that can be collected if 10.0 g
of sodium hydrogencarbonate is dissolved in excess sulfuric acid?
-
From the equation 2 mol of the carbonate
gives 2 mol of the gas.
-
Therefore mol NaHCO3 = mol CO2
= 10.0/84 = 0.119 mol
-
volume of CO2 = mol CO2
x molar gas volume
-
volume CO2 = 0.119 x 24 =
2.86 dm3 (3sf, 2.86 x 1000 = 2860 cm3)
-
-
-
(c) What mass of sodium sulfate is made
assuming a 100% yield (hardly so in practice!)
-
From the equation 2 mol of NaHCO3
produces 1 mol of Na2SO4
-
So mol of Na2SO4 =
mol NaHCO3/2 = 0.119/2 = 0.0595
-
mass Na2SO4 = mol
Na2SO4 x formula mass of Na2SO4
= 0.0595 x 142 = 8.45 g (3 sf)
-
-
-
Molar gas volume calculation
Example 9.10
-
Atomic masses: Na = 23, H = 1,
O = 16 and the volume of 1 mole of gas is 24.0 dm3.
-
Sodium reacts with excess water to form
sodium hydroxide and hydrogen gas.
-
If 4.6 g of sodium reacts with water,
what volume of hydrogen is formed at room temperature and pressure?
-
The equation is:
2Na(s) +
2H2O(l) ===> 2NaOH(aq) + H2(g)
-
From the equation 2 moles of sodium
produce 1 mole of hydrogen.
-
mol Na = 4.6/23 = 0.20
-
Therefore mol hydrogen produced = 0.20/2
= 0.10
-
So volume of hydrogen formed = 0.1 x 24 =
2.4 dm3
See also for gas calculations
Advanced
notes on gas law calculations, kinetic
model theory of an IDEAL GAS & non-ideal gases
Reacting gas volume
ratios, Avogadro's Law
& Gay-Lussac's Law Calculations
Above is typical periodic table used in GCSE science-chemistry specifications in
doing gas volume ratio calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
TOP OF PAGE
OTHER CALCULATION PAGES
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What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
-
Calculating relative
formula/molecular mass of a compound or element molecule
-
Law of Conservation of Mass and simple reacting mass calculations
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Composition by percentage mass of elements
in a compound
-
Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
-
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
-
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
-
Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
-
Moles and the molar volume of a gas, Avogadro's Law
(this page)
-
Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
reactants-products)
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Molarity, volumes and solution
concentrations (and diagrams of apparatus)
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How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
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Electrolysis products calculations (negative cathode and positive anode products)
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Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
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Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
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Gas calculations involving PVT relationships,
Boyle's and Charles Laws
-
Radioactivity & half-life calculations including
dating materials
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