Molar gas volume and
Avogadro's Law calculations
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Brown's Chemistry  GCSE/IGCSE/GCE (basic A level)
O Level Online Chemical Calculations
9.
The molar gas volume in calculations, application of Avogadro's Law
Quantitative chemistry
calculations Help for problem solving
in doing molar gas volume calculations. Practice revision questions on
involving moles and volumes of gases and Avogadro's Law, using
experiment data, making predictions. This page describes and
explains,
with fully worked out examples, how to calculate the volume of gas
formed from given masses of reactants. You need to know the formula
connecting moles, mass and formula mass AND know how to use the molar
volume in these gas volume calculation methods. Online practice exam
chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE
CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB
courses. These revision notes and practice questions on how to use the
molar gas volume and Avogadro's Law chemical calculations and worked
examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1)
chemistry science courses.
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See also 10.
Calculations using reacting gas volume
ratios  GayLussac's Law
AND
P V T gas volume, pressure and
temperature calculations
Selfassessment Quizzes
on the molar gas volume and Avogadro's Law calculations
type in answer
QUIZ or
multiple choice
QUIZ
Note on two acronyms and associated units:
STP or s.t.p. is a
largely defunct abbreviation for 'standard temperature and
pressure', 0^{o}C/273K temperature and a pressure of 1
atm/101.3 kPa (101325 Pa). (In bygone times 760 mm of Hg of a
'mercury' barometer/manometer)
NTP is a frequently
used abbreviation for 'normal laboratory temperature and
pressure', 25^{o}C/298K temperature and pressure
of 1 atm/101.3 kPa pressure (101325 Pa).

9.
The molar gas volume in calculations, moles, gas volumes and Avogadro's Law

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 24dm^{3} (24 litres) or 24000 cm^{3}, at room temperature
of 25^{o}C/298K and
normal pressure of 101.3 kPa/1 atmosphere (such conditions are often referred to
as RTP).

The molar volume for s.t.p is 22.4 dm^{3}
(22.4 litres) at 0^{o}C and 1atmosphere pressure.

Historically, s.t.p
unfortunately stands for standard temperature and pressure, but these days 25^{o}C/298K
is usually considered the standard temperature (RTP).

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(dm^{3}) / 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: M_{r}
= mass(g) x 24 / V(dm^{3})

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. 25^{o}C and 1 atmosphere
pressure so the molar volume is 24dm^{3} or 24000cm^{3}.

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 (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 cm^{3}.

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
cm^{3} intervals. measuring cylinders to the nearest cm^{3}
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 gas loss by carrying out the reaction in a
flask set up on an accurate onepan 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
[M_{r}(H_{2}) = 2], better for the 'heavier' gas
carbon dioxide [M_{r}CO_{2}) = 44]

Molar
gas volume calculation
Example 9.1

What is the volume of 3.5g of hydrogen? [A_{r}(H)
= 1]

common thinking: hydrogen exists as H_{2} molecules, so M_{r}(H_{2})
= 2, so 1 mole or formula mass in g = 2g

method (a)

method (b):

2g occupies 24 dm^{3}, so scaling up for the
volume
of hydrogen ...

3.5 g will have a volume of 3.5/2 x 24 =
42 dm^{3} (or 42000 cm^{3})



Molar gas volume calculation
Example 9.2

Given the equation

MgCO_{3(s)} + H_{2}SO_{4(aq)} ==>
MgSO_{4(aq)}
+ H_{2}O_{(l)} +CO_{2(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 dm^{3}
of carbon dioxide?

The important mole ratio is 1 MgCO_{3}
==> 1 CO_{2}

method (a):

since 1 mole = 24 dm^{3}, 6 dm^{3} is
equal to 6/24 = 0.25 mol of gas

From the equation, 1 mole of MgCO_{3} produces 1
mole of CO_{2}, which occupies a volume of 24 dm^{3}.

so 0.25 moles of MgCO_{3} is needed to make 0.25
mol of CO_{2}

formula mass of MgCO_{3} = 24 + 12 +
(3 x 16) = 84,

so required mass of MgCO_{3}
= mol x formula
mass = 0.25 x 84 = 21g



method (b):

converting the equation into the required reacting masses ..

formula masses: MgCO_{3} = 84 (from above), CO_{2} = 12 +
(2 x 16) = 44

MgCO_{3} : CO_{2} equation ratio is 1 :
1

so 84g of MgCO_{3} will form 44g of CO_{2}

44g of CO_{2} will occupy 24dm^{3}

so scaling down, 6 dm^{3} of CO_{2} will
have a mass of 44 x 8/24 = 11g

if 84g MgCO_{3} ==> 44g of CO_{2},
then ...

21g MgCO_{3}
==> 11g of CO_{2} 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 A_{r}'s
Ca = 40, H = 1, Cl = 35.5)

Ca_{(s)} + 2HCl_{(aq)} ==> CaCl_{2(aq)}
+ H_{2(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 H_{2}

so 0.075 mol Ca produces 0.075 mol H_{2}

so volume H_{2}
= 0.075 x 24 = 1.8 dm^{3}
(or 1800 cm^{3})



(i) method (b):

from equation 1 Ca ==> 1 H_{2} means 40g
==> 2g

so scaling down: 3g Ca will produce 2 x 3/40 = 0.15g H_{2}

2g H_{2} has a volume 24 dm^{3}, so
scaling down ...

0.15g H_{2} has a
volume of (0.15/2) x 24 = 1.8 dm^{3}
(or 1800 cm^{3})



(ii) method (a) only:

from equation: 2 moles HCl ==> 1 mole H_{2}
(mole ratio 2:1)

so 0.25 mol HCl ==> 0.125 mol H_{2}, volume 1
mole gas = 24 dm^{3}

so volume H_{2}
= 0.125 x 24 = 3 dm^{3}



Molar
gas volume calculation
Example 9.5

Given the equation ... (and A_{r}'s
Mg = 24, H = 1, Cl = 35.5)

Mg_{(s)} + 2HCl_{(aq)} ==> MgCl_{2(aq)}
+ H_{2(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 cm^{3}
of hydrogen gas?

method (a)

The important mole ratio is 1 Mg
==> 1 H_{2}

300 cm^{3} = 300/24000 = 0.0125 mol H_{2}
(since 1 mol of any gas = 24000 cm^{3})

from the equation 1 mole Mg ==> 1 mole H_{2}

so 0.0125 mole Mg needed to make 0.0125 mol H_{2}

so mass of Mg = mole Mg x A_{r}(Mg)

so mass Mg needed =
0.0125 x 24 = 0.3g



method (b)

reaction ratio in equation is 1 Mg ==> 1 H_{2},

so reacting mass ratio is 24g Mg ==> 2g H_{2},

2g H_{2} has a volume of 24000 cm^{3}
(volume of formula mass in g)

so scaling down: mass Mg needed
= 24 x (300/24000) = 0.3g


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.

2NaHCO_{3(s)} ==> Na_{2}CO_{3(s)}
+ H_{2}O_{(g)} + CO_{2(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
NaHCO_{3} ==> 1 CO_{2}

Formula mass of NaHCO_{3} is 23+1+12+(3x16) = 84 =
84g/mole

Formula mass of CO_{2} = 12+(2x16) = 44 = 44g/mole
(not needed by this method)

In the equation 2 moles of NaHCO_{3} give 1 mole
of CO_{2} (2:1 mole ratio in equation)

Moles NaHCO_{3} = 4.2/84 = 0.05 moles ==>
0.05/2 = 0.025 mol CO_{2} on decomposition.

Mass = moles x formula mass, so mass CO_{2} =
0.025 x 44 = 1.1g CO_{2}

Volume = moles x molar volume = 0.025 x 24000 =
600 cm^{3}
of CO_{2}



Molar gas volume calculation Example 9.7

Molar gas volume calculation
Example 9.8

See also QA7.1 on the mole
introduction page

Selfassessment Quizzes on reacting gas volume ratios and
Avogadro's Law
See also for gas calculations
Advanced
notes on gas law calculations, kinetic
model theory of an IDEAL GAS & nonideal gases
Reacting gas volume
ratios, Avogadro's Law
& GayLussac's Law Calculations
Above is typical periodic table used in GCSE sciencechemistry specifications in
doing gas volume ratio calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
OTHER CALCULATION PAGES

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Calculating relative
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Law of Conservation of Mass and simple reacting mass calculations

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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)

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and GayLussac's Law (ratio of gaseous
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