UNIT V : THE MOLE CONCEPT (HONOURS)
CALCULATIONS BASED ON AVOGADRO'S HYPOTHESIS
One of Dalton's hypotheses incorrectly assumed that all molecules were made up of only one atom of each
element involved in the molecule. Avogadro's Hypothesis helped find how many atoms were actually present
in a molecule, allowing chemists to re–calculate the atomic masses of elements more accurately.
Experimentally, the following data is found.
1.00 L of NH3(g) reacts with exactly 1.00 L of HCl(g).
1.00 L of NH3(g) has a mass of 0.759 g
1.00 L of HCl(g) has a mass of 1.63 g .
Therefore
mass of HCl(g)
1.63 g
=
= 2.15
mass of NH 3 (g)
0.759 g
and:
mass of HCl(g) = 2.15 x mass of NH3(g).
According to
1 L of NH3(g) and 1 L of HCl(g) contain the same number of particles, so
Avogadro's Hypothesis,

that if:
1 L of HCl(g) is 2.15 times heavier than 1 L of NH3(g), and
1 L of HCl(g) has the same number of molecules as 1 L of NH3(g),
then each particle of HCl must be 2.15 times heavier than each particle of NH3.
Therefore you can find the mass of one gas particle relative to the mass of another gas particle by comparing
the masses of equal gas volumes.
EXAMPLE: 1 L of hydrogen gas at 0oC and a pressure of 101.3 kPa ("kilopascals") has a mass of 0.0893 g
and 1 L of oxygen gas under the same conditions has a mass of 1.43 g. Assigning a mass of
2.00 to a hydrogen particle, what is the mass of an oxygen particle relative to hydrogen?
Setting up the ratio
mass of oxygen gas
mass of oxygen particle
=
mass of hydrogen gas
mass of hydrogen particle
M
1.43 g
=
, which solves to give: M = 32.0
0.0893 g
2.00


Therefore a particle of oxygen has a mass of 32.0.
so that
 g. Under the same conditions, 5.68 L of nitrogen have a
EXAMPLE: 1.85 L of helium have amass of 0.400
mass of 8.60 g. Assigning a mass of 4.00 to helium, what is the mass of a nitrogen particle
relative to a helium particle?
The gases cannot be directly compared since they have different volumes. An additional
calculation is required: finding the amount of gas which exists in an arbitrarily selected
volume, say 1 L; that is, calculating the density of the gas.
density of helium =
0.400 g
= 0.216 g/L
1.85 L
density of nitrogen =
8.60 g
= 1.51 g/L
5.68 L

mass of nitrogen gas
mass of nitrogen particle
Now equal, 1 L, volumes are compared:
=
mass of helium gas
mass of helium particle
M
1.51g
=
, which solves to give: M = 28.0
0.216 g
4.00
particle is 28.0.
Therefore the mass of a nitrogen
EXERCISES:



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Hebden : Chemistry 11 Honours
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1. If 65.0 g of acetylene gas, at 0oC and 101.3 kPa pressure, occupy 56.0 L, what mass of acetylene
occupies 44.8 L?
2. If 7.5 g of hydrogen gas occupy 84 L, what is the volume occupied by 20.0 g of hydrogen?
3. If 100 g of oxygen gas occupy 72 L, what mass of oxygen occupies 50 L?
4. If 75 g of nitrogen gas, at 0oC and 101.3 kPa pressure, occupy 60 L, what is the volume occupied by
40 g of nitrogen?
5. At 25oC and 101.3 kPa pressure, 1.00 L of hydrogen chloride gas has a mass of 1.49 g, and 1.00 L of
oxygen gas has a mass of 1.31 g. What is the mass of a hydrogen chloride molecule relative to an
oxygen molecule having a mass of 32 units?
6. At 25oC and 101.3 kPa pressure, 3.50 L of sulphur dioxide have a mass of 9.14 g, and 3.50 L of
nitrous oxide have a mass of 4.29 g. What is the mass of a nitrous oxide molecule relative to a mass
of 64 units for a molecule of sulphur dioxide?
7. Under the same conditions, 7.00 L of fluorine have a mass of 11.9 g and 7.00 L of chlorine have a
mass of 22.2 g. If a fluorine molecule has a mass of 38 units, what is the mass of a chlorine
molecule?
8. Under the same conditions, 8.00 L of hydrogen bromide have a mass of 28.9 g and 14.0 L of xenon
gas have a mass of 81.9 g. If a xenon atom has a mass of 131 units, what is the mass of a hydrogen
bromide molecule?
9. Under the same conditions, 15 L of sulphur trioxide have a mass of 53.6 g and 4.0 L of carbon
monoxide have a mass of 5.0 g. If a carbon monoxide molecule has a mass of 28 units, what is the
mass of a sulphur trioxide molecule?
10. At 0oC and 101.3 kPa pressure, 10 L of hydrogen have a mass of 0.89 g and 10 L of oxygen have a
mass of 14.3 g. At 25oC and 202.6 kPa pressure, 1.0 L of argon has a mass of 1.64 g and 1.0 L of
hydrogen has a mass of 0.082 g. What is the mass of an argon atom relative to an oxygen molecule
having a mass of 32 units?
11. At 0oC and 101.3 kPa pressure, 3.00 L of helium gas have a mass of 0.536 g and 3.00 L of krypton
have a mass of 11.2 g. At 25oC and 152.0 kPa pressure, 2.00 L of krypton have a mass of 10.3 g and
2.00 L of neon have a mass of 2.47 g. What is the mass of a neon atom relative to a mass of 4.00
units for a helium atom?
DETERMINING EMPIRICAL FORMULAE FROM COMBUSTION OF HYDROCARBONS
Burning a compound that contains C, H (and possibly O), gives a reaction such as those shown below.
or
C2H6 + 7/2 O2
C2H6O + 3 O2
 2 CO2 + 3 H2O
 2 CO2 + 3 H2O
Notice that the products in both cases above are CO2 and H2O, and that there is the same amount of CO2
and H2O in both cases. When a compound containing hydrogen and carbon (a "hydrocarbon") is burned, a
huge excess of oxygen is always used, so it is impossible to differentiate between the burning of C2H6 and
C2H6O by trying to measure the amount of O2 used.
CASE I: If the compound does not contain any oxygen, only the amount of CO2 and H2O produced needs to
be known in order to find the empirical formula.
UNIT V : THE MOLE CONCEPT (HONOURS)
3
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EXAMPLE:
A sample of a hydrocarbon is burned to produce 30.8 g of CO2 and 25.2 g of H2O. What is the
empirical formula of the hydrocarbon?
1mol CO 2
1mol C
x
= 0.700 mol
44.0 g CO 2
1mol CO 2
1mol H2O
2 mol H
moles H = 25.2 g H2O x
x
= 2.80 mol
18.0 g H2O 1mol H2O
moles C = 30.8 g CO2 x
1
4
 formula is: CH4 .
Therefore, the empirical
EXAMPLE:

 to produce 72.4 mg of CO2 and 11.8 mg of H2O. What is
A sample of a hydrocarbon
is burned
the empirical formula of the hydrocarbon?
1mol CO 2
1mol C
x
= 0.00165 mol
44.0 g CO 2
1mol CO 2
1mol H2O
2 mol H
moles H = 0.0118 g H2O x
x
= 0.00131 mol
18.0 g H2O 1mol H2O
moles C = 0.0724 g CO2 x
1.26
5
1
4
 formula is: C5H4 .
Therefore, the empirical


CASE II: If a hydrocarbon might contain oxygen, then the original mass of compound burned must be known.
Once the moles (and mass) of carbon and hydrogen are found, the calculated masses of hydrogen
and carbon are subtracted from the original mass of compound to find the mass of oxygen.
EXAMPLE:
When 17.00 g of a compound containing C, H and O is burned, 32.5 g of CO2 and 20.0 g of
H2O are formed. What is the empirical formula of the compound?
moles C = 32.5 g CO2 x
1mol CO 2
1mol C
x
44.0 g CO 2
1mol CO 2
1mol H2O
2 mol H
x
18.0 g H2O 1mol H2O


Next, find the mass of C, H and O in the compound.
moles H = 20.0 g H2O x
mass C = 0.739 molx
mass H = 2.22 mol x
= 0.739 mol
= 2.22 mol
12.0 g
 g
= 8.87
1mol
1.0 g
= 2.22 g
1mol
mass O = mass compound – mass C – mass H = 17.00 g – 8.87 g – 2.22 g = 5.91 g
Now the moles of O can be found and the problem can be completed.
moles C = 0.739 mol
moles H = 2.22 mol
1mol
moles O = 5.91 g x
= 0.369 mol
16.0 g
2
6
1
and the empirical formula is: C2H6O .

EXERCISES:
12. A compound contains C, H and O. A 4.35 g sample of the compound is burned to produce 9.71 g of
CO2 and 1.70 g of H2O. What is the empirical formula of the compound?
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Hebden : Chemistry 11 Honours
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13. A compound contains C, H and O. A 10.0 g sample of the compound is burned to produce 16.6 g of
CO2 and 8.49 g of H2O. What is the empirical formula of the compound?
14. A 7.30 g sample of a hydrocarbon is burned to give 23.8 g of CO2 and 7.30 g of H2O. What is the
empirical formula of the compound?
15. A 5.00 g sample of a hydrocarbon is burned to give 16.9 g of CO2 and 3.46 g of H2O. What is the
empirical formula of the compound?
16. A compound is known to contain C, H and S. A 7.00 g sample contains 3.73 g of S and can be
burned to give 10.27 g of CO2 and 4.20 g of H2O. What is the empirical formula of the compound?
17. A compound is known to contain C, H and N. A 1.000 g sample contains 0.609 g of N and can be
burned to give 1.276 g of CO2 and 0.392 g of H2O. What is the empirical formula of the compound?
18. A compound contains only C, H and O. A 6.50 g sample is burned to give 13.0 g of CO2 and 5.32 g of
H2O. What is the empirical formula of the compound?
19. A compound contains C, H, N and O. A 16.800 g sample contains 5.680 g of C, 0.237 g of H and
3.31 g of N. What is the empirical formula of the compound?
20. A poisonous gas contains only Sn and H. When a 3.600 g sample of the gas is burned, 1.056 g of
H2O is formed. What is the empirical formula of the gas?
21. A compound contains C, H and O. A 5.90 g sample is burned to yield 11.18 g of CO2 and 3.66 g of
H2O. What is the empirical formula of the compound?
22. A compound is known to contain C and H, but the presence or absence of O has not been
determined. When 0.875 g of the compound is burned, 1.31 g of H2O and 2.67 g of CO2 are
produced. What is the empirical formula of the compound?
Finding the Molar Mass Using Avogadro's Hypothesis Calculations
Example: At the same temperature and pressure, a 5.0 L bulb can contain 8.00 g of CH4 or 22.0 g of
gas X. What is the molar mass of X?
22.0 g
molar mass of X
molar mass of X
=
=
8.00 g
molar mass of CH 4
16.0 g
which solves to give:


molar mass of X = 16.0 g x

22.0 g
= 44.0 g
8.00 g
COMBINED EXERCISES:
23. A gas has the empirical formula CHO. A 1.0 L bulb
at 30oC and 200 kPa is filled with O2 and
weighed. The bulb is then emptied and refilled with the unknown gas at the same temperature and
pressure. The amount of gas in the bulb was found to be 5.4 times heavier than the previous amount
of O2. What is the molecular formula of the gas?
24. A gas has an empirical formula CH2. A bulb is filled with the gas at 50oC and 101.3 kPa and its mass
measured. The bulb is emptied and refilled with O2 at the same temperature and pressure. The
mass of the gas is 1.75 times as much as the mass of the O2. What is the molecular formula of the
gas?
25. A gas has an empirical formula CHN. A 750 mL bulb is filled with the gas at 150oC and 122 kPa
pressure and its mass determined. The same bulb was then filled with CH4 at the same temperature
and pressure. The mass of the gas was about 5 times as much as that of the CH4. What is the
molecular formula of the gas?
UNIT V : THE MOLE CONCEPT (HONOURS)
5
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26. A compound contains C, H and O. A 5.000 g sample contains 3.215 g of C and 0.357 g of H. A
gaseous sample of the compound has a mass 5.25 times as much as O2 under the same conditions.
What is the molecular formula of the compound?
27. A gas has the empirical formula NH2. Under the same conditions, a sample bulb can contain either
0.138 g of the gas or 0.121 g of N2(g). What is the molecular formula of the gas?
28. A compound containing C and H is burned to give 23.7 g CO2 and 4.85 g H2O. A sample of the
compound, when vaporized, was found to have a mass 2.44 times as much as an equal volume of O2
at the same temperature and pressure. What is the molecular formula of the compound?
29. A compound contains B and H. When a 2.40 g sample was burned, 4.69 g of H2O were produced.
The gas is only 62.7% as heavy as CO2 under the same conditions. What is the molecular formula of
the compound?
30. A compound contains C, H and O. A 3.500 g sample is burned to give 8.21 g of CO2 and 1.26 g of
H2O. A 0.30 L sample of the gas has a mass of 2.01 g at STP. What is the molecular formula of the
compound?
ALTERNATE METHODS OF EXPRESSING CONCENTRATION
Although molarity is used extensively when expressing concentration, there are several other methods which
frequently are used under special circumstances.
Percentage by Weight and Percentage by Volume
These ways of measuring concentration generally are used when great accuracy is not required, although
the methods can give accurate values.
Definitions: % (w/w) = percentage by weight =
% (v/v) = percentage by volume =
Chemists prefer:
mass of solute
x 100%
mass of solution
volume of solute
x 100%
volume of solution

volume measurements for liquid solutes, and
mass measurements for solid solutes.

EXAMPLES: (a) How would you prepare 300.0 g of a solution which was 5.00 % (w/w) NaCl(aq)?
Analysis: Both the mass of NaCl and water used must be known. Since the solution’s
TOTAL mass is 300.0 g, you can find either the mass of NaCl or the mass of
water. The other mass is found by subtracting the first mass from 300.0 g.
mass of NaCl required = 5.00% of 300.0 g = 0.0500 x 300.0 g = 15.0 g
mass of water needed = 300.0 – 15.0 = 285.0 g
but the density of water is 1.000 g/mL and therefore: volume of water = 285.0 mL
Hence, put 15.0 g of salt into a container and add 285.0 mL of water.
(Note that it is not necessary to convert the mass of water into a volume of water but it
is quicker to pour out a volume of water than to have to determine a specific mass.)
(b) How would you prepare 300.0 mL of a solution which is 5.00 % (v/v) CH3COOH(aq)?
(CH3COOH, acetic acid, is a liquid)
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Hebden : Chemistry 11 Honours
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Analysis: Percent volume/volume problems have to be solved in a specific manner
which is similar but not identical to the method used with percent mass/mass
problems. A problem which arises is that when two liquids are combined
their combined volume is NOT NECESSARILY equal to the sum of their
individual volumes. Since the final volume must be 300.0 mL and 5.00 % of
that volume be made up of acetic acid, it will be necessary to calculate the
volume of acetic acid needed and then simply add water until the total
volume comes to EXACTLY 300.0 mL. This means you will not be sure
what exact volume of water has been added, but the added volume is not
important.
volume of CH3COOH required = 5.00 % of 300.0 mL = 0.0500 x 300.0 mL = 15.0 mL
Hence, dilute 15.0 mL of acetic acid to a total volume of 300.0 mL.
[Incidentally, the added volume will be CLOSE to (300 – 15 = ) 285 mL of water.]
(c) What mass of 5.00 % (w/w) NaCl solution is needed to obtain 2.50 g of NaCl?
Analysis: "5.00 % (w/w) NaCl)" can be translated as a conversion factor
5.00 g NaCl
100 g solution
This conversion factor is applied to the starting information to find the required mass.
mass solution
 = 2.50 g NaCl x
100 g solution
= 50.0 g
5.00 g NaCl
EXERCISES:

31. Recall that you do not know how much water is added when making up a % (v/v) solution. What
measurement(s) can be made in order to determine the exact amount of water added?
32. How would you make up the following solutions?
(a) 500.0 g of 8.00 % (w/w) KBr
(b) 1.00 L of 20.0 % (v/v) liquid acetone in water
(c) 250.0 mL of 4.00 % (v/v) liquid ethanol in acetone
(d) 2.00 kg of 2.50 % (w/w) NaOH
(e) 500.0 mL of 15.0 % (v/v) liquid ethylene glycol (antifreeze) in water
(f) 200.0 g of 1.50 % (w/w) sugar
33. What mass of a 2.00 % (w/w) NaCl solution is needed to obtain 15.0 g of NaCl?
34. What volume of 3.00 % (v/v) acetic acid solution, CH3COOH(aq), is needed to get 25.0 mL of acetic
acid?
35. If the density of acetic acid, CH3COOH, is 1.050 g/mL, what volume of 3.00 % (v/v) acetic acid is
needed to get 0.100 mol of acetic acid?
36. If the density of methanol, CH3OH, is 0.7917 g/mL, what volume of 25.0 % (v/v) methanol is needed
to obtain 50.0 g of methanol?
Parts per Million and Parts per Billion
These ways of measuring concentration are used when the solute is present in very small concentrations
("trace quantities") and are often used when analyzing solutions for the presence of impurities.
Definitions: parts per million (ppm) =

grams of solute
10 6 grams of solution
UNIT V : THE MOLE CONCEPT (HONOURS)
7
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parts per billion (ppb)
=
grams of solute
10 grams of solution
9
Because of the low concentrations involved, the density of the pure solvent is not changed by the
presence of the dissolved solute and hence the density of the solution is just the density of the solvent.
For aqueous solutions, which areof most interest to us, the density of pure water is 1.00 g/mL and the
definitions become:
# of ppm =
# of mg (solute)
L (water)
# of ppb =
# of g (solute)
L (water)

EXAMPLE: Sea water contains 6.0 x 10–6 ppm of gold. What is the molarity of the gold in sea water?
 6.0 x 10 -6 mg Au 10 -3 g Au
1mol Au
[Au] =
x
x
= 3.0 x 10–11 M
197.0 g Au
L solution
1mg Au
EXAMPLE: If a solution of sodium cyanide, NaCN, is 1.26 x 10–3 M, what is the concentration of NaCN
expressed in ppm?



The question requires the number of milligrams of NaCN in 1 L of solution:
# of ppm =
49.0 g NaCN
1mg NaCN
1.26 x 10 3 mol NaCN
x
x
= 61.7 mg/L = 61.7 ppm
1mol NaCN
L solution
10 3 g NaCN
EXERCISES:
 the definition:
37. Show that
is equivalent to the definition:
grams ofsolute

ppm =
10 6 g of solution
ppm =
mg solute
L solution

38. What is the concentration, in ppm, of the following?
provided the solvent is water.
(a) 3.5 x 10–5 M Pb
(b) 0.0010 M Hg
39. Find the molarity of the solutions having the following concentrations.

(a) 65 ppm of Br
(b) 3.0 ppb of Sn
(c) 2.50 ppm of Hg
(d) 1.0 ppb of dioxin, C12H4O2Cl4
40. What is the concentration, in ppb, of the following?
(a) 4.5 x 10–7 M Be
(b) 8.0 x 10–10 M Cd
41. What mass of fluoride ion is present in 2.50 kL of water containing 1.4 ppm of F–?
42. What volume of seawater containing 1.5 x 10–4 ppm of uranium must be processed to obtain
1.0 kg of uranium?
43. What is the value of the silver in 1.0 km3 (109 kL) of sea water having 2.0 x 10–4 ppm of Ag, if
silver is $8.00 per troy ounce (31.1 g = 1 troy ounce)?