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Online Chemical Calculations
9.
The molar gas volume in calculations, application of Avogadro's Law
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. (updated Feb 23rd)
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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.

So the volumes have equal moles of
separate particles in them.

One mole of any gas (or the 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 or 1 atmosphere (both = 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

The latter form of the
equation can be used to calculate molecular mass from experimental data
because

moles = mass / molecular
mass

molecular mass = mass /
moles

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.

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!

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)}

What mass of magnesium carbonate is needed to make 6 dm^{3}
of carbon dioxide? [A_{r}'s: Mg = 24, C = 12, O = 16, H =1 and S
= 32]

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 need to make 0.25
mol of CO_{2}

formula mass of MgCO_{3} = 24 + 12 + 3x16 = 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 +
2x16 = 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)}

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)}

How much magnesium is needed to make 300 cm^{3}
of hydrogen gas?

method (a)

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

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
OTHER CALCULATION PAGES

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

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 GayLussac's Law (ratio of gaseous
reactantsproducts)

Molarity, volumes and solution
concentrations (and diagrams of apparatus)

How to
do volumetric titration calculations e.g. acidalkali titrations
(and diagrams of apparatus)

Electrolysis products calculations (negative cathode and positive anode products)

Other calculations
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy

Energy transfers in physical/chemical changes,
exothermic/endothermic reactions

Gas calculations involving PVT relationships,
Boyle's and Charles Laws

Radioactivity & halflife calculations including
dating materials
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