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Worksheet Set B
Worksheet B001
(K. Nel - 2012)
A. Mass and Quantity
1. Describe briefly the difference between Ar and Mr
Ar is using for the monoatomic elements for instance the relative atomic mass of hydrogen
atom is 1 so it is for the atom. Mr is using for diatomic elements such that the relative molecular
mass of hydrogen molecule is 2 so it is for molecule.
2. What is the mole in chemistry? How does this relate to Ar and Mr?
If we work out the Ar or Mr of a substance, andthen weigh out that number of grams of the
substance, we can say how many atoms or molecules it contains. For instance, the Ar of carbon
is 12 and it contains 602 000 000 000 000 000 000 000(6,02x1023 ) carbon atoms. It is called
Avogadro’s number.
3. How many grams are there in the following
𝑚
We use 𝑛 = 𝑀𝑟 𝑜𝑟 𝐴𝑟
a. 1 mole of copper atoms = 64 gr
b. 18 moles of sulphur atoms = (18x32)=576 gr
c. 5.4 moles of hydrogen molecules (remember that hydrogen forms diatomic molecules
H2)=(5.4x2)=10.8 gr
d. 12 moles of sulphur molecules (S8)
= (12x8x32)=3072 gr
e. 2.5 moles of ozone (O3)
=(2.5x3x16)=120 gr
f.
8 moles of water
= ((2+16)x8)= 144 gr
g. 4 moles of hydrated copper sulphate (CuSO4.H2O). What is meant by “hydrated” copper
sulphate?
((64+32+64+2+16)x4)=712gr
h. 13 moles of ammonia
=((14+3)x13)=221 gr
i.
½ mole of sodium thiosulphate (Na3S2O3)
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=((69+64+48)/2)= 90.5 gr
j.
5 moles of iron (III)chloride
=((168+213)x5)=1905 gr
4. How many moles of atoms are there in
𝑚
We use 𝑛 = 𝑀𝑟 𝑜𝑟 𝐴𝑟
a. 16grams of sulphur
= 0.5 moles
b. 32g of oxygen
= 2 moles
c. 8g of hydrogen
=8 moles
d. 161g of sodium
=7 moles
e. 2.8kg of lithium
=121.739 moles
f.
558g of phosphorous
= 18 moles
5. How many moles of each atom can be found in one mole of the following:
a. Lead oxide, Pb3O4 = 3 moles of Pb, 4 moles of O
b. Acetic acid (also called ethanoic acid), CH3COOH= 2 moles of C, 4 moles of H, 2 moles of O
c. Ammonium nitrate =
d. Hydrated coppoer sulphate
e. Hydrated iron(II) sulphate, FeSO4.7H2O = 1 mole of Fe, 1 mole of , 11 moles of O, 14 moles
of H
B. Formulae and % Composition
1. The formula for carbon dioxide is CO2 . The Ar values are: C = 12 , O = 16.
Complete the following statements:
a. 1 mole of C (12.g) and 1 mole of O2 (32.g) combine to form 1. mole of CO2 (…44.g)
b. 48.0g of carbon and 128. g of oxygen combine to form 176 g of carbon dioxide.
c. When 1.2g of carbon reacts with oxygen, the increase in mass is ….g
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d. When 12 moles of carbon dioxide is decomposed into carbon and oxygen, 12.moles of
carbon and 12… moles of oxygen would be obtained.
e. The percentage by mass of carbon in carbon dioxide is …27.%
C. Practical data
1. Copper (II) oxide is converted to copper by passing a
stream of hydrogen through the apparatus while heating
the copper oxide as shown. Hydrogen is allowed to pass
through the tube for some time before it is lit. When all
the oxide had converted to copper, the test tube is
allowed to cool while maintaining the flow of hydrogen.
a. Why do we pass hydrogen through the tube for a
while before it is lit?
Because, fill the tube with hydrogen gas. It is like to prepare the environment for the reaction.
The experiment is repeated several times
and the following results are captured:
Experiment
Mass of
Mass of copper
Mass of oxygen
number
copper oxide [g]
produced [g]
lost [g]
1
0.62
0.55
0.07
2
0.90
0.80
0.10
3
1.12
1.00
0.12
4
1.69
1.50
0.19
5
1.80
1.60
0.2
b. Complete the table by calculating the missing figures.
c. Use graph paper to plot the mass of oxygen against the mass of copper, showing the
mass of oxygen on the y-axis and the mass of copper on the x-axis. Draw the line of best
fit through the origin and set of points.
d. From the graph, find the mass of oxygen that would combine with 0.70g of copper, as
well as the mass of oxygen that would combine with 1.20g of copper. Clearly indicate
these on the graph.
0.7 copper  0.9 oxygen
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1.20 copper  1.5 oxygen
e. Calculate the mass of oxygen that would combine with 1.20g of copper.
1.20 [g] copper  1.5 [g] oxygen
f.
How many moles of copper are there in 1.20g of copper?
𝑚
Ar=29 of copper 𝑛 = 𝐴𝑟 =
1.2
29
= 0.04
g. What is the simplest formula for this oxide of copper?
𝐶𝑢𝑂
h. What is another term for the simplest formula found in this way?
𝐶𝑢𝑂2
i.
Why was hydrogen kept flowing during the cooling phase?
Because the reaction doesn’t finished until the all the copper oxide is used.