S 4 chemistry / revision exercise / mole concept / 03-04 / P.6

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.

\ No. of moles of Fe3+ ions = 2 ´ 2 = 4

No. of moles of SO42- ions = 3 ´ 2 = 6

Total no. of moles of ions = 4 + 6 = 10

(b)  0.2 moles of Al(NO3) 3

\ No. of moles of Al3+ ions = 0.2

No. of moles of NO3- ions = 0.2 ´ 3 = 0.6

Total no. of moles of ions = 0.2 + 0.6 = 0.8

2.  Given 1.6 g of methane (CH4), find

(a)  number of moles of CH4

Molar mass of CH4 = 12 + 1 ´ 4 = 16 gmol-1

No. of moles of CH4 = mass of CH4 /molar mass of CH4

= 1.6/16 = 0.10

(b)  number of molecules 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,

\ no. of H atoms = 6.02 ´ 1022 ´ 4

= 2.41 ´ 1023

3.  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

4.  Find the mass of

6.02 ´ 1022 lead atoms
No. of moles of Pb atoms
= 6.02 ´ 1022 / 6.02 ´ 1023
= 0.100
Mass of Pb atoms
= 0.100 ´ 207
= 20.7 g / 3.01 ´ 1024 carbon dioxide molecules
No. of moles of CO2 molecules
= 3.01 ´ 1024 / 6.02 ´ 1023
= 5.00
Molar mass of CO2
= 12 + 16 ´ 2 = 44 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

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

= 0.600 ´ 6.02 ´ 1023

= 3.612 ´ 1023

(Relative atomic masses: O = 16.0, K = 39.0)

6. 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

= 0.25 ´ 6.02 ´ 1023

= 1.505 ´ 1023

8. 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

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 / Molar mass of substance / Mass of substance present / Number of moles of substance present / Number of molecules/ formula units present
Sulphuric acid (H2SO4) / 98 g mol-1 / 58.8 g / 0.6 mol / 3.612 ´ 1023
Sodium hydroxide (NaOH) / 40 g mol-1 / 2.0 g / 0.05 mol / 3.01 ´ 1022
Potassium carbonate (K2CO3) / 138 g mol-1 / 331.2 g / 2.4 mol / 1.44 ´ 1024

(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

= 6.5 ´ 10-3 ´ (14.0 + 2 ´ 16.0)

= 0.299 g

13. (I) 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

(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)

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 / Molar mass
(g mol-1) / Concentration
(g dm-3) / Molarity
(M) / Mass of solute required to prepare 250.0 cm3 of solution (g)
KOH / 56.0 g mol-1 / 5.60 g dm-3 / 0.10 M / 5.60 g dm-3 × (250.0/1000) dm3
= 1.40 g
CuSO4 / 159.5 g mol-1 / 31.9 g dm-3 / 0.20 M / 7.975 g
(COOH) 2˙2H2O / 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) = 0.05 mol

Mass of NaHCO3

= 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) = 0.3 mol

Mass of K2Cr2O7

= 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)

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.

FeSO4 / H2O
Relative mass (g) / 5.47 / 4.53
Relative no. of mass / 5.47 / 152 = 0.036 / 4.53 / 18 = 0.252
0.036 / 0.036 = 1 / 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.