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Calculations
on the mole concept
Francine
Taylor-Campbell, Contributor
This
week we will continue to look at questions
on the mole concept. You can never
have too much practice in this area.
Let
us try this question.
1.
(a) A compound with relative molecular
mass of 180 was found to contain 40%
carbon, 6.7% hydrogen and 53.3% oxygen.
Determine the molecular of this compound.
(b)
1.89g of zinc nitrate was heated according
to the equation
2Zn(NO3)2
(s) == 2ZnO (s) + 4NO2 (g) + O2 (g)
(i)
What is the percentage by mass of
oxygen in zinc nitrate?
(ii)
What volume of oxygen at r.t.p. is
given off when 1.89g of zinc nitrate
are heated?
ANSWERS
| 1. |
Carbon |
Hydrogen |
Oxygen |
| Mass
in 100g |
40 |
6.7 |
53.3 |
| Molar
mass (g/mol) |
12 |
1 |
16 |
| #
mol |
40/12 |
6.7/1 |
53.3/16 |
| |
3.33
|
6.7 |
3.33 |
| Ratio
of mols |
1
: |
2
: |
1 |
Empirical
formula of compound = CH2O
Therefore
the molecular formula = (CH2O)n
If
the molecular mass of the compound
= 180
Then
(CH2O)n = 180
(12
+ (1*2) + 16)n = 180 thus 30n = 180
; n=6
Molecular
formula = (CH2O)6
= C6H12O6
(b)
2Zn(NO3)2 ==
2ZnO + 4NO2 + O2
Molar
mass of Zn(NO3)2
= 65 + 14*2 + 16*6 = 189g
%
O = (mass of oxygen in zinc nitrate)/(Total
mass of zinc nitrate) = (96/189)*100
= 50.8%
(ii)
1.89g of Zn(NO3)2
represents 1.89/189 = 0.01 mol
According
to the equation 2 mol of Zn(NO3)2
gives off 1 mol O2
Therefore,
0.01 mol Zn(NO3)2
will produce 0.01/2 = 0.005 mol O2
At
r.t.p. 1 mole of any gas occupies
a volume of 24dm3
0.005
mol O2 has a volume of
0.005*24 = 0.12dm3
Question
2
2.
Ammonia and carbon dioxide react to
form water and a solid, urea, CON2H4.
In the reaction, 72dm3 of carbon dioxide
at r.t.p. are converted to urea.
(i)
Write the equation for the formation
of urea.
(ii)
Calculate the volume of ammonia at
r.t.p. which reacted.
(iii)
Calculate the mass of urea formed.
ANSWERS
(i)
2NH3 (g) + CO2
(g) == H2O (l) + CON2H4
(s)
(ii)
Based on the equation 2 mol NH3
react with 1 mol CO2 (2:1
ratio)
Hence
if 72 dm3 of CO2
is used then 2 x 72 dm3
of NH3 would react.
Volume
of ammonia = 144dm3
(iii)
1 mol of gas at r.t.p has a volume
of 24dm3
#
mol CO2 used = 72/24 =
3 mol
Using
the equation again 1 mol CO2
produces 1 mol CON2H4
(urea)
Thus,
3 mol CO2 produces 3 mol
urea
Molar
mass of urea = (12 + 16 + 14*2 + 1*4)
= 60g
3
mol urea has a mass of 60 x 3 = 180g.
NOW
ATTEMPT THIS QUESTION
3.
Give the equation for the reaction
between methane and steam.
Calculate
the maximum volume of hydrogen, measured
at s.t.p., which can be obtained from
16g of methane.
REMEMBER:
When working problems with moles,
it is always best to start by finding
the number of moles of the known substance,
whether by using its mass or volume.
Also, in most cases, an equation is
essential, as it helps to determine
in what mole ratio the reactants combine
or the products form.
In
next week's lesson we will look at
the mole and its application to solutions.
Francine
Taylor-Campbell is an independent
contributor. Send questions and comments
to kerry-ann.hepburn@gleanerjm.com
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