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Mole Concept Assignment Four – Answers
Question 1.
Under certain conditions, steam reacts with methane to produce carbon monoxide and hydrogen only.
Construct the equation for the reaction and calculate the total volume of gas that can be obtained when
500 cm3 of methane react completely with steam.
H2O(g) + CH4(g)  CO(g) + 3H2(g)
moles of gas = volume in dm 3  24
OR
moles of gas = volume in cm 3  24 000
moles of CH4 = 0.5  24 = 0.0208 mol
from the balanced chemical equation, 1 mol of CH4 produce 1 mol of CO + 3 mol H2
 0.0208 mol of CH4 produces 0.0208 mol of CO + (3  0.208) mol of H2 = 0.0832 mol of gas in total
total volume of gas produced = moles  24
total volume of gas produced = 0.0832  24 = 2.00 dm3 to 3 s.f.
Question 2.
a)
Hydrogen and carbon monoxide can be converted into methanol (CH3OH). Construct the equation
for this reaction.
2H2(g) + CO(g)  CH3OH(l)
b)
When 100 g of coal was completely burned in oxygen it produced 330 g of carbon dioxide. Calculate
the percentage by mass of carbon in this sample of coal.
moles = mass in grams  relative molecular mass
moles of CO2 = 330  (12 + 16 + 16)
= 330  44 = 7.50 mol
1 molecule of CO2 contains 1 atom of carbon  1 mol of CO2 contains 1 mol of carbon
 7.50 mol of CO2 contains 7.50 mol of carbon
mass in grams = moles  relative atomic mass
mass of carbon = 7.50  12 = 90.0 g
percentage carbon in the coal = (90  100)  100 = 90.0%
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Question 3.
a)
Write the equation, including state symbols, for the reaction between magnesium oxide and
hydrochloric acid.
MgO(s) + 2HCl(aq)  MgCl2(aq) + H2O(l)
b)
Tablets containing magnesium oxide can be used to relieve stomach pains caused by excess
hydrochloric acid. If the stomach contains the equivalent of 100 cm 3 of excess hydrochloric acid of
concentration 3 mol/dm 3, how many tablets each containing 3 g of magnesium oxide should be
taken?
moles of chemical in solution = concentration (in mol/dm3)  volume (in cm3)  10–3
moles of HCl in excess = 3  100  10–3 = 0.300 mol
from the balanced chemical equation, 2 mol of HCl react with 1 mol of MgO
 0.300 mol of HCl react with 1/2  0.300 = 0.150 mol of MgO
mass in grams = moles  relative molecular mass
mass of MgO = 0.150  (24 + 16)
= 0.150  40 = 6.00 g
number of tablets that should be taken = 6.00  mass of one tablet
= 6.00  3 = 2.00 tablets should be taken
Question 4.
The sugar glucose has an empirical formula of CH2O and a relative molecular mass, Mr, of 180.
a)
Calculate the molecular formula of glucose.
relative molecular mass of CH2O = 12 + 1 + 1 + 16 = 30
180  30 = 6
 glucose is composed of 6 units of CH2O
 the formula of glucose is C6H12O6
b)
If an aqueous solution of glucose is heated with alkaline copper(II) sulphate, a brick red precipitate is
formed. The precipitate is an oxide of copper. A 1.44 g sample of this oxide was found to contain
1.28 g of copper.
i)
Calculate the empirical formula of this oxide of copper.
1.44 g of compound contains 1.28 g of copper
 1.44 g of compound contains 1.44 – 1.28 = 0.16 g of oxygen
moles = mass in grams  relative atomic mass
moles of copper = 1.28  64 = 0.02 mol
moles of oxygen = 0.16  16 = 0.01 mol
divide through by the smallest answer, which is 0.01:
for copper – 0.02  0.01 = 2
for oxygen – 0.01  0.01 = 1
formula = Cu2O
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ii)
What is the oxidation state of the copper in this oxide?
2  Cu + oxidation state of oxygen = 0
2  Cu + (–2) = 0
2  Cu = +2
Cu = +1
iii)
Why does this reaction show that glucose is a reducing sugar?
The glucose has reduced the copper from +2 in the copper(II) sulphate to +1 in the copper(I)
oxide.
Question 5.
A 0.50 g sample of vegetable oil reacted with 60 cm 3 of hydrogen, measured at r.t.p. Assuming that one
mole of the oil reacts with one mole of hydrogen gas, calculate the relative molecular mass, Mr, of the oil.
moles of gas = volume in cm 3  24 000
moles of hydrogen = 60  24 000 = 0.0025 mol
1 mol of H2 reacts with 1 mol of oil  0.0025 mol of H2 reacts with 0.0025 mol of oil
 0.50 g of oil contains 0.0025 mol of oil
relative molecular mass = mass in grams  moles
relative molecular mass of the vegetable oil = 0.50  0.0025 = 200
Question 6.
Ammonia and carbon dioxide react together to form water and a solid, urea, CON 2H4. In the reaction, 72 dm3
of carbon dioxide at r.t.p. are converted to urea.
a)
Write the equation, including state symbols, for the formation of urea.
2NH3(g) + CO2(g)  CON2H4(s) + H2O(l)
b)
Calculate the volume of ammonia at r.t.p. which reacted.
moles of gas = volume in dm 3  24
moles of CO2 = 72  24 = 3.00 mol
from the balanced chemical equation, 1 mol of CO2 reacts with 2 mol of NH3
 3.00 mol of CO2 react with 2  3.00 = 6.00 mol of NH3
volume of gas in dm3 = moles  24
volume of NH3 = 6.00  24 = 144 dm3
c)
Calculate the mass of urea formed.
from the balanced chemical equation, 1 mol of CO2 produces 1 mol of CON2H4
therefore 3.00 mol of CO2 produces 3.00 mol of CON2H4
mass in grams = moles  relative molecular mass
mass of CON2H4 = 3.00  (12 + 16 + (2  14) + (4  1)) = 3.00  60
mass of CON2H4 = 180 g
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Question 7.
CFCs are compounds that contain only carbon, chlorine and fluorine. They are atmospheric pollutants and
destroy ozone in the upper atmosphere.
a)
“CFC11” Has the following composition by mass:
C = 8.7%
F = 13.8%
Cl = 77.5%
Calculate the empirical formula of CFC11.
Carbon
Fluorine
Chlorine
Divide percentage by Ar
8.7  12 = 0.725
13.8  19 = 0.726
77.5  35.5 = 2.18
Divide by smallest answer
0.725  0.725 = 1
0.726  0.725 = 1
2.18  0.725 = 3
Empirical formula = CFCl3
Check your answer – C is in group IV of the Periodic Table and makes 4 covalent bonds, while F
and Cl are in Group VII of the Periodic Table and each make 1 covalent bond.
b)
“CFC12” has the molecular formula CF2Cl2. It can be made by the reaction of hydrogen fluoride, HF,
with tetrachloromethane, CCl4.
CCl4 + 2HF  CCl2F2 + 2HCl
What is the maximum mass of CFC12 that can be made from 10.0 g of hydrogen fluoride?
moles = mass in grams  relative molecular mass
moles of HF = 10.0  (1 + 19) = 0.500 mol
from the balanced chemical equation, 2 mol of HF produce 1 mol of CCl2F2
 0.500 mol of HF produce 1/2  0.500 = 0.250 mol of CCl2F2
mass in grams = moles  relative molecular mass
mass of CCl2F2 = 0.250  (12 + (2  35.5) + (2  19))
= 0.250  121 = 30.25 g
= 30.3 g to 3 s.f.
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