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 dm^{3} (24 litres) or
24000 cm^{3}, at room temperature of
20^{o}C/293K and
normal pressure of 101.3 kPa/1 atmosphere (such conditions are often referred to
as RTP/NTP).

20^{o}C (293K) is
often treated as room temperature, and the molar gas volume is
24.0 dm^{3}
at normal pressure.

25^{o}C (298K) is treated as
a standard temperature, at which the molar volume of the gas is
24.4 dm^{3}
at normal pressure.

The molar volume for s.t.p is 22.4 dm^{3}
(22.4 litres) at 0^{o}C and 101 kPa1 atmosphere 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 for data e.g.
thermochemical and thermodynamic data at a higher academic level.

Unfortunately, on the internet, both
20^{o}C and 25^{o}C are referred to as 'room
temperature'.

Reminder about the Avogadro Number:

The Avogadro number is 6.02 x 10^{23} particles/mole
of defined chemical 'species'.

Therefore, e.g. at 20^{o}C, 24.0 dm^{3} of a
gas contains 6.03 x 10^{23} molecules.

6.03 x 10^{23} / 24.0 = 2.5 x 10^{22}
= number of molecules in 1 dm^{3} of gas.

2.5 x 10^{22} / 1000 = 2.5 x 10^{19} =
number of molecules in 1 cm^{3} of gas.

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(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 in a reaction

In these
gas volume calculations assume the
molar gas volume is 24 dm^{3} (24 000 cm^{3}) at 'normal'
temperature and pressure.

Molar
gas volume calculation
Example 9.1

(a) What is the volume of 3.5 g of hydrogen
(H_{2})? [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 = 2 g

method (i)

method (ii):

2 g 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})



(b) What is the volume of 11.6 g of
butane gas at RTP?

The formula of butane is C_{4}H_{10},
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 dm^{3}



(c) What mass of chlorine (Cl_{2})
does 6.0 dm^{3} 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 dm^{3}, so mol Cl_{2}
= 6/24 = 0.25 mol

mass = moles x molecular mass = 0.25 x 71
= 17.75 g (4sf, 2dp)



(d) What mass of methane (CH_{4})
is contained in a 10 dm^{3} cylinder at three times normal
atmospheric pressure at room temperature.

Atomic masses: C = 12, H = 1, molar
gas volume = 24 dm^{3}.

Formula mass of methane = 16 + (4 x 1) =
16

At normal pressure mol of CH_{4}
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 CH_{4} = mol CH_{4}
x formula mass of CH_{4} = 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: CaCO_{3}(s)
===> CaO(s) + CO_{2}(g)

Atomic masses: Ca = 40, C = 12,
O =16. Formula mass CaCO_{3} = 40 + 12 + (3 x 16) = 100

1 mol of calcium carbonate gives 1 mol of
carbon dioxide.

Therefore mol CaCO_{3} = mol CO_{2}
= 10/100 = 0.10 mol

Therefore volume CO_{2} = mol CO_{2}
x molar gas volume

Volume of CO_{2} = 0.10 x 24.0 =
2.4 dm^{3} (or 2.4 x 1000 = 2400 cm^{3})



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.

Molar gas volume calculation
Example 9.2 (At RTP 1 mol gas = 24 dm^{3} or 24000 cm^{3})

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}

Fully
worked out ANSWERS at the bottom of the page

Molar gas volume calculation
Example 9.3 (At RTP 1 mol
gas = 24 dm^{3} or 24000 cm^{3})

6g of a hydrocarbon gas had a volume of 4.8
dm^{3}, calculate its molecular mass.

Fully
worked out ANSWERS at the bottom of the page


Molar gas volume calculation
Example 9.4 (At RTP 1 mol
gas = 24 dm^{3} or 24000 cm^{3})

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
...

Molar
gas volume calculation
Example 9.5 (At
RTP 1 mol gas = 24 dm^{3} or 24000 cm^{3})

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?

Fully
worked out ANSWERS at the bottom of the page
Molar gas volume calculation
Example 9.6 (At RTP 1 mol
gas = 24 dm^{3} or 24000 cm^{3})

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) }

atomic masses: Na = 23, H = 1,
C = 12, O = 16

Fully
worked out ANSWERS at the bottom of the page



Molar gas volume calculation Example 9.7
(At RTP 1 mol gas = 24 dm^{3} or 24000 cm^{3})

Molar gas volume calculation
Example 9.8 (At RTP 1 mol gas = 24
dm^{3} or 24000 cm^{3})

Molar gas volume calculation
Example 9.9 (At RTP 1 mol gas = 24
dm^{3} or 24000 cm^{3})

3.27g of metal dissolved in dilute
sulfuric acid to form 1.2 dm^{3} 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 dm^{3}

Calculate the atomic mass of the metal.

Fully worked out ANSWERS at the bottom of the
page



Molar gas volume calculation
Example 9.10

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?

(c) What mass of sodium sulfate is made
assuming a 100% yield (hardly so in practice!)

Molar gas volume calculation
Example 9.11

Atomic masses: Na = 23, H = 1,
O = 16 and the volume of 1 mole of gas is 24.0 dm^{3}.

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) +
2H_{2}O(l) ===> 2NaOH(aq) + H_{2}(g)

Fully worked out ANSWERS at the bottom of the
page
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 dm^{3} at GCSE
level).
Know at RTP moles of gas = volume of gas in dm^{3} / 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.
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Part 10.
Reacting gas volume
ratios, Avogadro's Law
& GayLussac's Law
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2.
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3.
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5. Empirical formula and formula mass of a compound from reacting masses
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(starting with reacting masses or % composition)
9.
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10. Reacting gas volume
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and GayLussac's Law (ratio of gaseous
reactants to products)
11. Molarity, volumes and solution
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(this page)
12.
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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
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14.2b
atom economy 14.3
dilution of solutions
14.4
water of crystallisation
calculation
14.5
how
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are unofficial. 
Worked out ANSWERS to the moles and gas volume
questions

Molar gas volume calculation
Example 9.2 (At RTP 1 mol
gas = 24 dm^{3} or 24000 cm^{3})

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 (At RTP 1 mol
gas = 24 dm^{3} or 24000 cm^{3})

Molar gas volume calculation
Example 9.4 (At RTP 1 mol
gas = 24 dm^{3} or 24000 cm^{3})

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 (At
RTP 1 mol gas = 24 dm^{3} or 24000 cm^{3})

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 =
0.30 g



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) =
0.30 g


Molar gas volume calculation
Example 9.6 (At RTP 1 mol
gas = 24 dm^{3} or 24000 cm^{3})

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
(At RTP 1 mol gas = 24 dm^{3} or 24000 cm^{3})

Molar gas volume calculation
Example 9.8 (At RTP 1 mol gas = 24
dm^{3} or 24000 cm^{3})

Molar gas volume calculation
Example 9.9 (At RTP 1 mol gas = 24
dm^{3} or 24000 cm^{3})

3.27g of metal dissolved in dilute
sulfuric acid to form 1.2 dm^{3} 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 dm^{3}

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 9.10
(At RTP 1 mol gas = 24 dm^{3} or 24000 cm^{3})

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.

M_{r}(NaHCO_{3})
= 23 + 1 + 12 + 3x16 = 84

M_{r}(Na_{2}SO_{4})
= 2x23 + 32 + 4x16 = 142

M_{r}(CO_{2}) =
12 + 2x16 = 44



(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 NaHCO_{3} = mol CO_{2}
= 10.0/84 = 0.119 mol

volume of CO_{2} = mol CO_{2}
x molar gas volume

volume CO_{2} = 0.119 x 24 =
2.86 dm^{3} (3sf, 2.86 x 1000 = 2860 cm^{3})



(c) What mass of sodium sulfate is made
assuming a 100% yield (hardly so in practice!)

From the equation 2 mol of NaHCO_{3}
produces 1 mol of Na_{2}SO_{4}

So mol of Na_{2}SO_{4} =
mol NaHCO_{3}/2 = 0.119/2 = 0.0595

mass Na_{2}SO_{4} = mol
Na_{2}SO_{4} x formula mass of Na_{2}SO_{4}
= 0.0595 x 142 = 8.45 g (3 sf)



Molar gas volume calculation
Example 9.11
(At RTP 1 mol gas = 24 dm^{3} or 24000 cm^{3})

Atomic masses: Na = 23, H = 1,
O = 16 and the volume of 1 mole of gas is 24.0 dm^{3}.

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) +
2H_{2}O(l) ===> 2NaOH(aq) + H_{2}(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 dm^{3}
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