S 4 chemistry / revision exercise / mole concept / 03-04 / P.1
Mole concept
1.
Find the number of moles of ions in
(a) 2 moles of Fe2(SO4) 3
From the formula, we know that 1 mole of Fe2(SO4) 3 contains 2 moles of Fe3
ions and 3 moles of SO42 ions.

=
=
=
22=4
32=6
4 + 6 = 10

=
=
=
0.2
0.2  3 = 0.6
0.2 + 0.6 = 0.8
No. of moles of Fe3 ions
No. of moles of SO42 ions
Total no. of moles of ions
(b) 0.2 moles of Al(NO3) 3
2.
No. of moles of Al3 ions
No. of moles of NO3 ions
Total no. of moles of ions
Given 1.6 g of methane (CH4), find
(a) number of moles of CH4
= 12 + 1  4 = 16 gmol1
= mass of CH4 /molar mass of CH4
= 1.6/16 = 0.10
(b) number of molecules of CH4
Molar mass of CH4
No. of moles of CH4
No. of molecules of CH4 =
=
=
no. of moles  Avogadro Number
0.10  6.02  1023
6.02  1022
(c) number of H atoms.
One CH4 molecule contain 4 H atoms,

3.
4.
no. of H atoms
=
=
6.02  1022  4
2.41  1023
Find the mass of
(a) 1 H2O molecule
18 / 6.02  1023
3.0  1023 g
(b) 1 Cu atom
63.5 / 6.02  1023
1.05  1022 g
(c) 1 Na ion
23 / 6.02  1023
3.8  1023 g
(d) 1 OH ion
17 / 6.02  1023
2.8  1023 g
(e) 1 neutron
1.0 / 6.02  1023
1.7  1024 g
(f) 1 electron
0.00055 / 6.02  1023
9.1  1028 g
Find the mass of
6.02  1022 lead atoms
3.01  1024 carbon dioxide
molecules
No. of moles of Pb atoms No. of moles of CO2
= 6.02  1022 / 6.02  1023 molecules
= 0.100
= 3.01  1024 / 6.02  1023
= 5.00
Mass of Pb atoms
Molar mass of CO2
= 0.100  207
= 12 + 16  2 = 44 g
= 20.7 g
Mass of CO2 molecules
= 5.00  44
= 220 g
3.01  1023 sulphate ions
No. of moles of SO42 ions
= 3.01  1023 / 6.02  1023
= 0.500
Molar mass of SO42
= 32 + 16  4 = 96 g
Mass of SO42 ions
= 0.500  96
= 48 g
S 4 chemistry / revision exercise / mole concept / 03-04 / P.2
5.
(a) How many molecules are there in 3.00 moles of oxygen molecules?
Number of oxygen molecules
=
=
3.00  6.02  1023
1.806  1024
(b) How many ions are there in 0.600 moles of potassium ions?
Number of potassium ions
=
=
6.
0.600  6.02  1023
3.612  1023
(Relative atomic masses: O = 16.0, K = 39.0)
Calculate the number of moles of atoms in
(a) 127 g of copper
Number of moles of copper
=
127 / 63.5
=
2 mol
(b) 12.8 g of sulphur.
Number of moles of sulphur
=
12.8 / 32.0
=
0.4 mol
(Relative atomic masses: Cu = 63.5, S = 32.0)
7.
How many atoms are there in
(a) 2.50 moles of oxygen atoms?
Number of oxygen atoms
=
=
2.50  6.02  1023
1.505  1024
(b) 6.00 g of magnesium atoms?
Number of mole of magnesium atoms
=
6.00 / 24.0 = 0.25 mol
Number of magnesium atoms
8.
=
0.25  6.02  1023
=
1.505  1023
What is the mass of 2.50 moles of magnesium atoms?
(Relative atomic mass: Mg = 24.0)
Mass of magnesium atoms
=
=
2.50  24.0
60.0 g
S 4 chemistry / revision exercise / mole concept / 03-04 / P.3
9.
A pure sample of calcium chloride CaCl2 was found to contain 7.10 g of Cl ions.
What mass of Ca ions does the sample contain?
Number of moles of Cl ions=
=
7.10 / 35.5
0.200
The formula CaCl2 shows that the ratio of Ca2 ions to Cl ions is 1 : 2,
hence no. of moles of Ca2 ions = 0.200 / 2 = 0.100
 Mass of Ca2 ions in the sample = 0.100  40
= 4.0 g
10. A metal M ionizes to give Mn ions. If atomic mass of M is 24, and 1.2 g of M ionize
to give 6.02  1022 electrons, calculate n (the charge on each ion of M).
No. of moles of electrons given
= no. of electrons / Avogadro Number
= 6.02  1022 / 6.02  1023 = 0.100
No. of moles of M ionized
= 1.2 / 24 = 0.050
 0.050 mole of M gives 0.100 mole of electrons on ionization.
Thus 1 mole of M gives 0.100 / 0.050 = 2 moles of electrons.
Since the charge on an ion of M is numerically equal to the number of moles of
electrons given by 1 mole of M, each ion of M carries 2 charges, i.e. n = 2.
11. Complete the table below:
Substance
Sulphuric acid
(H2SO4)
Molar mass of
substance
Mass of
substance
present
Number of
moles of
substance
present
Number of
molecules/
formula units
present
98 g mol1
58.8 g
0.6 mol
3.612  1023
Sodium
hydroxide
2.0 g
0.05 mol
40 g mol1
3.01  1022
(NaOH)
Potassium
carbonate
2.4 mol
331.2 g
138 g mol1
1.44  1024
(K2CO3)
(Relative atomic masses: C = 12.0, H = 1.0, K = 39.0, Na = 23.0, O = 16.0, S = 32.0)
12. If we breathed in 3913  1018 molecules of air pollutant, nitrogen dioxide (NO2),
how many grams of NO2 would we breathe in?
(Relative atomic masses: N = 14.0, O = 16.0)
Number of mole of nitrogen dioxide
= 3913  1018 / 6.02  1023
= 6.5  103 mol
Mass of nitrogen dioxide
=
=
13. (I)
6.5  103  (14.0 + 2  16.0)
0.299 g
How many moles of calcium fluoride (CaF2) are present in 16.5 g of it?
Formula mass of CaF2
= 40.0 + 19.0  2
= 78.0 g
Number of mole of CaF2
=
16.5 / 78.0
=
0.21 mol
S 4 chemistry / revision exercise / mole concept / 03-04 / P.4
(II) How many calcium and fluoride ions are present?
Number of Ca2+ ions
= 0.21  6.02  1023 = 1.26  1023
Number of F ions
= 0.42  6.02  1023 = 2.53  1023
(Relative atomic masses: Ca = 40.0, F = 19.0)
14. What mass of water contains the same number of molecules as 2.20 g of carbon
dioxide?
(Relative atomic masses: C = 12.0, H = 1.0, O = 16.0)
Number of mole of CO2 = 2.20 / (12.0 + 2  16.0) = 0.05 mol
Mass of water
=
=
0.05  (1.0  2 + 16.0)
0.9 g
15. Identity the substance which contains the greater number of molecules from each set:
(a) 2 moles of carbon dioxide molecules (CO2) or 8.40 g of sulphuric acid (H2SO4)
Number of CO2 molecules
= 2  6.02  1023
= 1.20  1024
(Greater number of molecules)
Number of mole of H2SO4
= 8.40 / (1.0  2 + 32.0 + 16.0  4) = 0.086 mol
Number of H2SO4 molecules
=
0.086  6.02  1023
=
5.16  1022
(b) 88.0 g 0f carbon dioxide (CO2) or 84.0 g of nitrogen (N2)
Number of CO2 molecules
= (88.0 / 44.0)  6.02  1023
= 1.20  1024
Number of mole of N2
=
=
(84.0 / 28.0)  6.02  1023
1.81  1024
(Greater number of molecules)
(c) 5  1024 ammonia molecules (NH3) or 8.00 g of sulphuric acid (H2SO4)
Number of NH3 molecules
=
5  1024
(Greater number of molecules)
Number of mole of H2SO4 molecules
=
=
[8.00 / (1.0  2 + 32.0 + 16.0  4)] 6.02  1023
4.91  1022
(Relative atomic masses: C = 12.0, H = 1.0, N = 14.0, O = 16.0, S = 32.0)
S 4 chemistry / revision exercise / mole concept / 03-04 / P.5
16. Calculate the molarity of each of the following solutions:
(a) 23.4 g of sodium chloride (NaCl) solid in 1.0 dm3 solution.
Number of mole of NaCl
= 23.4 / (23.0 + 35.5)
= 0.4 mol
Molarity of NaCl solution
= 0.4 / 1
= 0.4 M
(b) 17.0 g of silver nitrate (AgNO3) solid in 200.0 cm3 solution.
Number of mole of AgNO3
= 17.0 / (108.0 + 14.0 + 16.0  3)
= 0.1 mol
Molarity of AgNO3 solution
= 0.1 / (200.0 / 1000.0)
= 0.5 M
(Relative atomic masses: Ag = 108.0, Cl = 35.5, N = 14.0, Na = 23.0, O = 16.0)
17. Complete the table below:
Substance
KOH
CuSO4
(COOH) 2˙2H2O
Molar mass Concentration
(g mol-1)
(g dm-3)
Molarity
(M)
Mass of solute
required to
prepare 250.0 cm3
of solution (g)
5.60 g dm-3 ×
(250.0/1000) dm3
= 1.40 g
56.0 g mol-1
5.60 g dm-3
0.10 M
159.5 g mol-1
31.9 g dm3
0.20 M
7.975 g
126.0 g mol-1
30.24 g dm3
0.24 M
7.56 g
(Relative atomic masses: C = 12.0, Cu = 63.5, H = 1.0, N = 14.0, O = 16.0, S = 32.0)
18. Calculate the molarity of each of the following solutions
(a) 500.0 cm3 of 0.10 M sodium hydrogencarbonate (NaHCO3) solution.
Number of mole of NaHCO3
= 0.10  (500.0 / 1000.0) =
Mass of NaHCO3
0.05 mol
= 0.05  (23.0 + 1.0 + 12.0 + 16.0  3)
= 0.05  84
= 4.2 g
(b) 100.0 cm3 of 3.00 M potassium dichromate (K2Cr2O7) solution.
Number of mole of K2Cr2O7
= 3.00  (100.0 / 1000.0) =
Mass of K2Cr2O7
0.3 mol
= 0.3  (39.0  2 + 52.0  2 + 16.0  7)
= 0.3  294
= 88.2 g
(Relative atomic masses: C = 12.0, Cr = 52.0, H = 1.0, K = 39.0, Na = 23.0, O =
16.0)
S 4 chemistry / revision exercise / mole concept / 03-04 / P.6
19. 5.6 g of a metal M combine with 2.4 g of oxygen to form an oxide with the formula
M2O3. What is the atomic mass of M?
Let the atomic mass of M be Ar.
Mass of M in compound / Mass of oxygen in compound
= 2Ar / 16  3
= 5.6 / 2.4
= 56
Hence atomic mass of the metal M is 56.
20. A crystalline salt of formula M2S2O3 . 5H2O is found to contain 36.3% by mass of
water of crystallization. Calculate
(a) the formula mass of the hydrated salt
(b) the atomic mass of the metal M.
(a) 5(1  2 + 16) / formula mass of hydrated salt
= 36.3 / 100
= 248
Formula mass of hydrated salt = 248.
(b) Let the atomic mass of M be A.
Formula mass of hydrated salt
= 2Ar2 + 32  2 + 16  3 + 5(1  2 + 16) = 248
Ar = 23
Therefore the atomic mass of M is 23.
21. 10.0 g of hydrated iron(II) sulphate, Fe2SO4.nH2O, on strong heating, gave 4.53 g of
water. Find the value of n.
Relative mass (g)
Relative no. of mass
FeSO4
5.47
5.47 / 152 = 0.036
0.036 / 0.036 = 1
H2 O
4.53
4.53 / 18 = 0.252
0.252 / 0.036 = 7
Therefore, n = 7
22. A metal M forms two chlorides A and B which contains 55.9% and 65.5% by mass
of chloride respectively. The empirical formula of A is found to be MCl2. Determine
the empirical formula of B(without having to find the atomic mass of M).
If mass of M in chloride B is the same as that in chloride A (i.e. 44.1 g), mass of
chlorine in chloride B will be (by proportion) equal to
65.5  44.1 / 34.5 = 83.7 g
Therefore for the same mass (hence same no. of moles) of M,
Mass of chlorine in chloride B / Mass of chlorine in chloride A
= no. of moles of chlorine in chloride B / no. of moles of chlorine in chloride A
= 83.7 / 55.9 = 1.5
Since formula of chloride A is given to be MCl2, formula of chloride B should be
MCl3.