Calculations on Mole Concepts
1.
What is the mass of 0.2 mole of calcium carbonate ?
(R. a. m. : C = 12.0, O = 16.0, Ca = 40.1)
2.
Calculate the number of gold atoms in 20g of gold atom. (R.a.m. : Au = 197.0)
3.
It is given that the molar mass of water is 18.0g mol-1.
(a) What is the mass of 4 moles of water molecule?
(b) How many molecules are there?
(c) How many atoms are there?
4.
A magnesium chloride solution contains 10 g of magnesium chloride solid
(a) Calculate the number of moles of magnesium chloride in the solution.
(b) Calculate the number of magnesium ions in the solution.
(c) Calculate the number of chloride ions in the solution.
(d) Calculate the total number of ions in the solution.
(R.a.m.: Mg = 24.3, Cl = 35.5)
5.
What is the mass of carbon dioxide molecule?
(R.a.m. : C = 12.0, O = 16.0)
6.
(a) Find the mass in grams of 0.01 mole of zinc sulphide.
(b) Find the number of ions in 5.61 g of calcium oxide.
(c) Find the number of atoms in 32.05g of sulphur dioxide.
(d) There is 4.80 g of ammonium carbonate. Find the
(i) number of moles of the compound,
(ii) number of moles of ammonium ions,
(iii) number of moles of carbonic ions, and
(v) number of hydrogen atoms.
(R.a.m. : S = 32.1, Zn = 65.4, O = 16.0, Ca = 40.1)
7.
Find the volume occupied by 3.55 g of chlorine gas at room temperature and
pressure (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1; R.a.m. : Cl = 35.5)
8.
Find the number of molecules in 4.48 cm3 of carbon dioxide gas at standard
temperature and pressure.
9.
(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1; Avogrado constant = 6.02 1023
mol-1)
The molar volume of nitrogen gas is found to be 24.0 dm3 mol-1 at room
temperature and pressure. Find the density of nitrogen gas. (R.a.m. : N = 14.0)
10. 1.6 g of a gas occupies 1.2 dm3 at room temperature and pressure. What is the
relative molecular mass of the gas?
(Molar Volume of gas at R.T.P. = 24.0 dm3 mol-1)
11. (a) Find the volume of 0.6 g of hydrogen gas at room temperature and pressure.
(b) Calculate the number of molecules in 4.48 dm3 of hydrogen at standard
temperature and pressure.
P.1
(c) The molar volume of oxygen is 22.4 dm3 mol-1 at standard temperature and
pressure. Find the density of oxygen in g cm-3 at S.T.P..
(d) What mass of oxygen has the same number of moles as that in 3.2 g of
sulphur dioxide?
(R.a.m. : H = 1.0; O = 16.0, S = 32.1, molar volume at R.T.P. = 24.0 dm3 mol-1,
at S.T.P. = 22.4 dm3 mol-1)
12. A 500 cm3 sample of a gas in a sealed container at 700 mm Hg and 25oC is
heated to 100oC. What is the final pressure of the gas?
13. A reaction vessel of 500 cm3 is filled with oxygen at 25oC and the final pressure
exerted on it is 101 325 Nm-2. How many moles of oxygen are there?
(Ideal gas constant = 8.314 J K-1 mol-1)
14. A 5 dm3 vessel can withstand a maximum internal pressure of 50 atm. If 2 moles
of nitrogen gas is pumped into the vessel, what is the highest temperature it can
be safely heated to? (1 atm = 101325 Nm-2, R = 8.314 J K-1 mol-1)
15. (a) A reaction vessel is filled with a gas at 20oC and 5atm. If the vessel can
withstand a maximum internal pressure of 10 atm, what is the highest
temperature it can be safely heated to?
(b) A balloon is filled with helium at 25oC. The pressure exerted and the
volume of balloon are found to be 1.5 atm and 450 cm3 respectively. How
many moles of helium have been introduced into the balloon? ( 1 atm = 101325
Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
(c) 25.8 cm3 sample of a gas has a pressure of 690 mm Hg and a temperature of
17oC. What is the volume if the pressure is changed to 1.85 atm and the
temperature to 345 K? ( 1 atm = 760 mmHg)
16. A sample of gas occupying a volume of 50 cm3 at 1 atm and 25oC is found to
a mass of 0.0286 g. Find the relative molecular mass of the gas.
( Ideal gas constant = 8.314 J K-1mol-1; 1 atm = 101 325 Nm-2)
have
17. The density of a gas at 450 oC and 380 mmHg is 0.033 7 g dm-3. What is its
relative molecular mass? ( 1 atm = 760 mmHg = 101 325 Nm-2 ; ideal gas
constant = 8.314 J K-1mol-1)
18. 0.25 mole of nitrogen and 0.30 mole of oxygen are introduced into a vessel of 12
dm3 at 50oC. Calculate the partial pressures of nitrogen and oxygen and hence the
total pressure exerted by the gases.
( 1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
19. 4.0 g oxygen and 6.0 g of nitrogen are introduced into a 5 dm3 vessel at 27oC.
(a) What are the mole fractions of oxygen and nitrogen in the mixture?
(b) What is the final pressure of the system?
( 1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1; R.a.m. : N = 14.0, O =
16.0)
P.2
20. (a) 204 g of phosphorus vapour occupies a volume of 81.0 cm3 at 327oC and 1 atm.
Determine the relative molecular mass of phosphorus. ( 1 atm = 101325 Nm-2 ;
ideal gas constant = 8.314 J K-1 mol-1)
(b) A sample of gas has a mass of 12.0 g and occupies a volume of 4.16 dm3
measured at 97oC and 1.62 atm. Calculate the relative molecular mass of the
gas. ( 1 atm = 101325 Nm-2; ideal gas constant = 8.314 JK-1 mol-1)
(c) A sample of 0.037g magnesium reacted with hydrochloric acid to give 38.2
cm3 of hydrogen gas measured at 25oC and 740 mmHg. Use this information to
calculate the relative atomic mass of magnesium. (1 atm = 760 mmHg =
101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
21. (a) The valve between a 6 dm3 vessel containing gas A at a pressure of 7 atm
and an 8 dm3 vessel containing gas B at a pressure of 9 atm is opened.
Assuming that the temperature of the system remains constant and there is
no reaction between the gases, what is the final pressure of the system?
( 1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
(b) 2 g of helium, 3 g of nitrogen and 4 g of argon are introduced into 15 dm3
vessel at 100 oC.
(i) What are the mole fractions of helium, nitrogen and argon in the system?
(ii) Calculate the total pressure of the system, and hence the partial pressures
of helium, nitrogen and argon.
(1 atm = 101325 Nm-2; R = 8.314 J K-1mol-1 ; R.a.m. : He = 4.0, N = 14.0, Ar =
39.9)
22. What is the mass of copper formed at the cathode when a current of 0.25 A is
passed through a copper(II) sulphate solution for 1 hour ( R.a.m. : Cu = 63.5)?
23. Find the masses of products formed when a dilute sulphuric acid solution is
electrolysed with a current of 0.6 A for 90 minutes. (R.a.m. : H = 1.0, O = 16.0)
24. What mass of copper would be desposited by the quantity of electricity that
liberates 2.4 dm3 of oxygen measured at room temperature and pressure?
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1; R.a.m. : O = 16.0, Cu = 63.5)
25. (a) What current in amperes is required to deposit 6.35 g of copper in 50 minutes
from a copper(II) sulphate solution?
(1 F = 96500 ; R.a.m. : Cu = 63.5)
(b) What is the time required to pass 1 Faraday of electricity through an electrolyte
with a current of 0.35A? ( 1F = 96 500C )
(c) Calculate the mass of aluminium that would be deposited during the
electrolysis of a molten aluminum salt by a current of 10 A for 5 hours.
( 1F = 96500C ; R.a.m. : Al = 27.0)
(d) A current of 0.37A flowing for 15 minutes through an electrolyte liberates
0.20 g of metal X. what mass of X would be liberated by a current of 0.30 A
for minutes?
26. (a) 5 g of sulphur forms 10 g of an oxide on burning.What is the empirical
formula of the oxide?
(R.a.m. : O = 16.0, S = 32.1)
P.3
(b) 19.85 f of element M combines with 25.61 g of oxygen to form an oxide. If the
relative atomic mass of M is 331.0, find the empirical formula of the oxide.
(R.a.m. : O =16.0)
(c) Determine the empirical formula of copper(II) oxide using the following
results.
Experimental results:
Mass of test tube = 21.430 g
Mass of test tube + Mass of copper(II) oxide = 23.321g
Mass of test tube + Mass of copper = 22.940g
(R.a.m. : Cu = 63.5, O = 16.0)
27. Compound A contains carbon and hydrogen only. It is found that the compound
contains 75% carbon by mass. Determine its empirical formula. (Relative atomic
masses: C=12, H=1 )
28. The percentage by mass of phosphorus and chlorine in a sample of a phosphorus
chloride are 22.55% and 77.45% respectively. Find the empirical formula of the
chloride. (R.a.m. : P = 31.0, Cl = 35.5)
29. (a) Find the empirical formula of vitamin C if it consists of 40.9% carbon, 54.5%
oxygen and 4.6% hydrogen by mass. ( R.a.m.: C = 12.0, H = 1.0, O = 16.0)
(b) Each 325 mg tablet of aspirin consists of 195.0 mg carbon 14.6 mg hydrogen
and 115.4mg oxygen. Determine the empirical formula of aspirin.
(R.a.m. : C= 12.0, H = 1.0, O = 16.0)
30. A hydrogen was burnt completely in excess oxygen. It was found that 5.00 g of
the hydrocarbon gives 14.6 g of carbon dioxide and 9.0 g of water. Given that the
relative molecular mass of the hydrocarbon is 30.0, determine its molecular
formula. hydrocarbon.?
(R.a.m.* : H = 1.0, C = 12.0, O = 16.0)
31. Compound X is known to contain 44.44% carbon, 6.18% hydrogen and 49.38%
oxygen by mass. A typical analysis shows that it has a relative molecular mass of
162.0. Find its molecular formula.
(R.a.m.* : H = 1.0, C = 12.0, O = 16.0)
32. The chemical formula of hydrated copper(II) sulphate is known to be CuSO4.xH2O.
It is found that the percentage of water by mass in the compound is 36%. Find x.
(R.a.m. : H=1.0, O=16.0, S=32.1, Cu=63.5)
33. (a) Find Compound Z is the major component of a healthy drink. It contains
40.00% carbon, 6.67% hydrogen and 53.33% oxygen.
(i) Find the empirical formula of compound Z.
(ii) If the relative molecular mass of compound Z is 180, finds its molecular
formula.
(R.a.m. : C= 12.0, H = 1.0, O = 16.0)
(b) (NH4)2Sx contains 72.72% sulphur by mass is water. Find the value of x.
(R.a.m.: H = 1.0, N = 14.0, O = 16.0)
(c) In the compound MgSO4nH2O, 51.22% by mass is water. Find the value of n.
(R.a.m.: H = 1.0, O = 16.0, Mg = 24.3, S = 32.1)
P.4
34. The chemical formula of ethanoic acid is CH3COOH. Calculate the percentages
by mass of carbon, hydrogen and oxygen by mass respectively.
(R.a.m. : C=12.0, H=1.0, O=16.0 )
35. Calculate the mass of iron metal in a sample of 20g of hydrated iron (II) sulphate,
FeSO47H2O.
(R.a.m. : Fe = 55.8 ,
H=1.0, O=16.0 )
36. (a) Calculate percentages by mass of potassium, chromium and oxygen in
potassium chromate (VI), K2Cr2O7.
(R.a.m. : K = 39.1 . Cr = 52.0, O = 16.0)
(b) Find the mass of metal and water of crystallization in
(i) 100 g of Na2SO4·10H2O;
(ii) 70g of Fe2O3·8H2O.
(R.a.m.: H = 1.0, O = 16.0, Na = 23, S = 32.1, Fe = 55.8)
37. Give the chemical equations for the following reactions:
(a) Zinc + steam → zinc oxide + hydrogen
(b) Magnesium + silver nitrate → silver + magnesium nitrate
(c) Butane + oxygen → carbon dioxide + water
38. Calculate the mass of copper formed when 12.45g of copper(II) oxide is completely
reduced by hydrogen. (R.a.m. : H=1.0, O=16.0, Cu = 63.5 )
39. Sodium hydrogencarbonate decomposes according to the following equation.
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)
In order to obtain 240 cm3 of CO2 at room temperature and pressure, what is the
minimum amount of sodium hydrogencarbonate required?
(R.a.m. : H = 1.0, C =12.0, O = 16.0, Na = 23.0; molar volume of gas at R.T.P. =
24.0 dm3mol-1)
40. Calculate the volume of carbon dioxide formed when 20 cm3 of ethane and
70 cm3 of oxygen are exploded, assuming all volumes are measured at room
temperature and pressure.
41. 10 cm3 of a gaseous hydrocarbon was mixed with 80cm3 of oxygen which was in
excess. The mixture was exploded and then cooled. The volume left was 70cm3.
Upon passing the resulting gaseous mixture through concentrated sodium
hydroxide solution ( to absorb carbon dioxide), the volume of the residual gas
became 50 cm3. Find the molecular formula of the hydrocarbon.
42. (a) Find the volume of hydrogen produced at R.T.P. when 2.43 g of magnesium
reacts with excess hydrochloric acid. (R.a.m. : Mg = 24.3; molar volume of
gas at R.T.P. = 24.0 dm3mol-1.
(b) Find the minimum mass of chlorine required to produced 100 g of phosphorus
trichloride ( PCl3).
(c) 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen were exploded in a
closed vessel. After cooling, 110 cm3 of gases remained. After passing through
a solution of concentrated sodium hydroxide, the volume left was 50 cm3 .
P.5
Determine the molecular of the hydrocarbon.
(d) Calculate the volume of carbon dioxide formed when 5cm3 of methane burns in
excess oxygen, assuming all volumes are measured at room temperature and
pressure.
43. 25.0cm3 of sodium hydroxide solution was titrated against 0.067 M of sulphuric(VI)
acid using methyl orange as indicator. The indicator changed colour from yellow to
red when 22.5 cm3 of sulphuric(VI) acid had been added. Calculate the molarity of
the sodium hydroxide solution.
44. 2.52 g of a pure dibasic acid with formula mass of 126.0 was dissolved in water
and made up to 250.0 cm3 in a volumetric flask 25.0 cm3 of this solution was
found to neutralize 28.5 cm3 of sodium hydroxide solution.
(a) Calculate the molarity of the acid solution.
(b) If the dibasic acid is represented by H2X, write an equation for the reaction
between the acid and sodium hydroxide.
(c) Calculate the molarity of the sodium hydroxide solution
45. 0.186 g of sample of hydrate sodium carbonate, Na2CO3·nH2O, was dissolved in
100 cm3 of distilled water in conical flask. 0.10 M by hydrochloric acid was added
from a burette, 2 cm3 at a time. The pH value of the solution was measured by a pH
meter. The result was recorded and shown in the following figure. Calculate the
value of n in Na2CO3·nH2O.
P.6
46. 5 cm3 of 0.5M sulphuric(VI) acid was added to 25.0 cm3 of potassium hydroxide
solution. The mixture was then stirred and the highest temperature was recorded.
The experiment was repeated with different volumes of the sulphuric(VI) acid. he
laboratory set-up and the results were as follows:
(a) Plot the graph of temperature against volume of sulphuric(VI) acid added.
(b) Calculate the molarity of the potassium hydroxide solution.
(c) Explain why the temperature rose to a maximum and the fell.
47. When excess potassium iodide solution (KI) is added to 25.0 cm3 of acidified
potassium iodate solution (KIO3) of unknown concentration, the solution turns
brown. This brown solution requires 22.0 cm3 of 0.05 M sodium thiosulphate
solution to react completely with the iodine formed, using starch solution as
indicator. Find the molarity of the potassium iodate solution.
48. A piece of impure iron wire weighs 0.22 g. When it is dissolved in hydrochloric
acid, it is oxidized to iron(II) ions. The solution requires 36.5 cm3 of 0.02 M
acidified potassium manganate(VII) for complete reaction to form iron(III) ions.
What is the percentage purity of the iron wire?
49. (a) 5g of anhydrous sodium carbonate is added to 100 cm3 of 2 M hydrochloric
acid. What is the volume of gas evolved at room temperature and pressure ?
( R.a.m. : C = 12.0, O = 16.0, Na = 23.0; molar volume of gas at R.T.P. =
24.0 dm3 mol-1)
(b) 8.54g of impure hydrated iron(II) sulphate (formula mass of 392.14) was
dissolved in water and made up to 250 cm3. 25 cm3 of this solution required
20.76 cm3 of 0.0203 M acidified potassium manganate(VII) solution for
complete reaction. Determine the percentage purity of the hydrated iron(II)
ulphate.
P.7
Solutions :
1.
The chemical formula of calcium carbonate is CaCO3.
Molar mass of calcium carbonate
= (40.1 + 12.0 + 3 x 16.0) g mol-1
= 100.1 g mol-1
Mass of calcium carbonate
= Number of moles  Molar mass
= 0.2 mol  100.1 g mol-1
= 20.02 g
2.
Number of gold atoms in 20g of gold coin
=
20g
197.0 g mol
-1
 6.02  1023 mol-1
= 6.11  1023
3.
(a) Mass of water = Number of moles  Molar mass
= 4 mol  18.0 g mol-1 = 72.0 g
(b) There are 4 moles of water molecules.
Number of water molecules = Number of moles  Avogadro constant
= 4 mol  6.02  1023 mol-1 = 2.408  1024
(c) 1 water molecule has 3 atoms (including 2 hydrogen atoms and 1 oxygen
atoms).
1 mole of water molecule has 3 moles of atoms.
Thus, 4 moles of water molecules have 12 moles of atoms.
Number of atoms = 12 mol  6.02  1023 mol-1 = 7.224  1024
4.
(a) The chemical formula of magnesium chloride is MgCl2.
Molar mass of MgCl2 = (24.3 + 35.5  2) g mol-1 = 95.3 g mol-1
10g
Number of moles of MgCl2 = 95.3 g mol 1 = 0.105 mol
(b) 1 mole of MgCl2 contains 1 mole of Mg2+ and 2 moles of Cl- . Therefore,
0.105 mole of MgCl2 contains 0.105 mol  6.02  1023 mol-1.
Number of Mg2+ ions = Number of moles of Mg2+  Avogadro constant
= 0.105 mol  6.02  1023 mol-1
= 6.321  1022
(c) 0.105 mole of MgCl2 contains 0.21 mole of Cl- .
= Number of moles of Cl-  Avogadro constant
= 0.21 mol  6.02  1023 mol-1
= 1.264  1023
(d) Total number of ions = 6.321  1022 + 1.264  1023 = 1.896  1023
Number of Cl- ions
5.
The chemical formula of carbon dioxide is CO2.
Molar mass of CO2 = (12.0 + 16.0 x 2) g mol-1 = 44.0 g mol-1
Mass
∵ Number of mole
=
= Number of molecules
Molar Mass
∴
Mass of CO2 molecule
44.0 g mol -1
Mass of a CO2 molecule
=
1
6.02 10 23
= 7.31  10-23 g
P.8
Avogadro constant
=
44.0 g mol -1
6.02 x 10 23
6.
(a) Mass = No. of moles  Molar mass
Mass of ZnS = 0.01 mol  (65.0 + 32.1) g mol-1
= 0.01 mol  95.7 g mol-1 = 0.975g
(b) No. of moles of CaO = 5.61g / (40.1 + 16.0)g = 0.1 mol
1 CaO formula unit contains 1 Ca2+ ion and 1 O2- ion.
No. of moles of ions = 0.1 mole  2 = 0.2 mol
No of ions = 0.2 mol x 6.02  1023 mol-1 = 1.204  1023
(c) No. of moles of SO2 = 32.05 g / (32.1 + 16.0  2)g mol-1 = 0.5 mol
1 SO2 molecule contains 1 S atom and 2O atoms.
No. of mole of atoms = 0.5 mole  3 = 1.5 mol
No of atoms = 1.5 mol  6.02  1023 mol-1 = 9.03  1023
(d) Molar mass of (NH4)2CO3 = 96.0 g mol-1
(i) No. of mole of (NH4)2CO3 = 4.80 g / 96.0 g mol-1
(ii) ∵ 1 mole (NH4)2CO3 gives 2 moles NH4+.
No. of moles of (NH4)2CO3 = 0.05 mol  2 = 0.1 mol
(iii) ∵ 1 mole (NH4)2CO3 gives 1 mole CO32- .
No. of moles of CO32- = 0.05 mol
(iv) 1 (NH4)2CO3 formula unit contains 8H atoms.
No. of moles of H atoms = 0.05 mol  8 = 0.4 mol
No. of H atoms = 0.4 mol  6.02  1023 mol-1 = 2.408 1023
7.
Molar mass of chloride gas (Cl2) = 35.5  2 g mol-1 = 71.0 g mol-1
Number of moles of Cl2 =3.55g / 71 g mol-1 = 0.05 mol
Volume of Cl2 = Number of moles of Cl2  Molar volume
= 0.05 mol  24.0 dm3mol-1=1.2 dm3
8.
Molar volume of carbon dioxide gas at S.T.P. = 22.4 dm3 mol-1
Number of moles of CO2 = 4.48 cm3 / 22400 cm3 mol-1 = 2  10–4 mol
Number of CO2 molecules = 2  10-4 mol  6.02 x 1023 mol-1=1.204  1020
9.
Molar mass of nitrogen gas (N2) = (14.0 + 14.0) g mol-1= 28.0 g mol-1
∵ Density = mass /volume = molar mass / molar volume
∴ Density of N2 = 28 g mol-1 / 24 dm3 mol-1 = 1.167 g dm-3
10. Number of moles of the gas = 1.2 dm3 / 24 dm3 mol-1 = 0.05 mol
Molar mass of the gas = 1.6 g / 0.05 mol = 32 g mol-1
Relative molecular mass of the gas = 32 (no unit)
11. (a) No. of moles of H2 = 0.6 g / 1.0  2 g mol-1 = 0.3 mol
Volume = No. of moles  Molar volume = 0.3 mol  24.0 dm3 mol-1 = 7.2 dm3
(b) No. of moles of H2 = 4.48 g / 22.4 dm3 mol-1 = 0.2 mol
No of H2 molecules = 0.2 mol  6.02  1023 mol-1 = 1.204  1023
(c) Density = mass / volume = molar mass / molar volume
Molar mass of O2 = 16.0  2 g mol-1 =32.0 g mol-1
Molar volume of O2 = 22.4 dm3 mol-1 = 22 400 cm3 mol-1
Density = 32.0 g mol-1 / 22400 cm3 mol-1 = 1.43  10-3 g cm-3
(d) No. of moles of SO2 = 3.2 g / ( 32.1 + 16.0  2 ) g mol-1 = 0.05 mol
No of moles of O2 = 0.05 mol
P.9
Mass = No. of moles  Molar mass
Mass of O2 = 0.05 mol  16.0g mol-1 = 1.6 g
12. As the number of moles of the gas is fixed, PV/T should be a constant
P1V1/T1 = P2V2/T2
700 mmHg  500 cm3 / (273 + 25) K = P2  500 cm3 / (273 + 100) K
P2 = 876.17 mmHg.
Note: All temperature values used in gas laws are on the Kelvin scale.
13. PV = nRT
101 325 Nm-2  500  10-6 m3 = n  8.314 J K-1 mol-1  (273 + 25) K
n = 0.02 mol
There is 0.02 mole of oxygen in the reaction vessel.
14.
T = PV/nR
= 50  101325 Nm-2  5 10-3 / 2 mol  8.314JK-1mol-1 = 1523.4 K
The highest temperature it can be safely heated to is 1250.4oC.
15. (a) P1/T1 = P2/T2
5 atm / (273 + 50) K= 10 atm / T2
T2 = 586 K
(b) PV= nRT
1.5  101 325 Nm-2 
n = 0.0276 mol
450  10-6 m3 = n  8.314 J K-1 mol-1  (273+25) K
(c ) (690/760)atm  25.8cm3 / (273 + 17) K = 1.85 atm  V2 / 345 K
V = 15.06cm3
16. PV = m/M RT
101 325 Nm-2  50  10-6 m3 = ( 0.0286 g / M )  8.314 JK-1 mol-1  (273 + 25) K
∴ M = 13.99 g mol-1
Therefore, the relative molecular mass of the gas is 13. 99.
17. The unit of density of the gas has to be converted to g m-3 for the calculation.
0.0337 g dm-3 = 0.0337  103 gm-3 = 33.7 g m-3
PM = RT
M = RT/ P
= 33.7 g cm-3  8.314 JK-1 mol-1  (273 + 450)K
(380/760)  101325Nm-2
= 4.0 g mol-1
Therefore, the relative molecular mass of the gas is 4.0.
18. Let the partial pressure of nitrogen be PA. Using the ideal gas equation PV = nRT,
PA  12  10-3 m3 = 0.25 mol x 8.314 JK-1 mol-1  (273 + 50) K
PA = 55 946 Nm-2 ( or 0.552 atm)
Let the partial pressure of oxygen be PB.Using the ideal gas equation PV = nRT,
PB = 0.30 mol  8.314 JK-1 mol-1  (273 + 50) K
PB = 67 136 Nm-2 ( or 0.663 atm)
Total pressure = (55 946 + 67 136) Nm-2 = 123 082 Nm-2
P.10
Hence, the partial pressures of nitrogen and oxygen are 0.552 atm and 0.663 atm
respectively, and the total pressure of the mixture is 1.215 atm.
19. (a) Number of moles of oxygen = 0.125 mol
Number of moles of nitrogen = 0.214 mol
Total number of moles of gases = ( 0.125 + 0.214) mol = 0.339 mol
Mole fraction of oxygen = 0.15 mol / 0.339 mol = 0.369
Mole fraction of nitrogen = 0.214 mol / 0.339 mol = 0.631
(b) Let P be the total pressure of the system.
By ideal gas equation PV = nRT,
P  5  10-3 m3 = 0.339 mol  8.314 JK-1mol-1  (273 + 27)K
P = 169 107 Nm-2 ( or 1.67 atm)
20. (a) PV = m/M RT
101325 Nm-2  81.0 10-6m3 = (0.204g / M)  8.314J K-1mol-1  (273 + 327) K
M = 123.99 g mol-1
∴ The relative molecular mass of phosphorus is 123.99.
(b) PV= (m/M)RT
1.62 101325 Nm-2 4.16 x 10-3m3 = 12.0g/ M  8.314 J K-1mol-1 x (273+97)K
M = 54.06 g mol-1
∴ The relative molecular mass of the gas is 54.06.
(c) Mg(s) + 2 HCl(aq)
PV = nRT
MgCl2(aq) + H2(g)
740/760  101 325 Nm-2  38.2  10-6m3 = n  8.314 J K-1 mol-1  (273 + 25) K
n = 1.52  10-3 mol
∴ No. of moles of H2 produced = 1.52  10-3.
No. of mole of Mg reacted = No. of moles of H2 produced = 1.52  10-3 mol
Molar mass of Mg = Mass / No. of moles
= 0.037g / 1.52  10-3 mol-1= 24.3 g mol-1
∴ The relative atomic mass of Mg is 24.3.
21. (a) By Boyle’s law: P1V1 = P2V2
Partial pressure of gas A = 7 atm 
6 dm3
= 3.00 atm
(6  8) dm3
Partial pressure of gas B = 9 atm 
8 dm 3
= 5.14 atm
(6  8) dm 3
Final pressure = 3.00 atm + 5.14 atm = 8.14 atm
(b) (i) No. of moles of He = 2g / 4.0 g mol-1 = 0.05 mol
No. of moles of N2 = 3g / 14  2 g = 0.11 mol
No. of moles of Ar = 4g / 39.9 g mol-1 = 0.10 mol
Total no. of moles of gases = 0.05 mol + 0.11 mol + 0.10 mol = 0.71 mol
Mole faction of He = 0.05 mol / 0.71 mol = 0.704
Mole fraction of N2 = 0.11 mol / 0.71 mol = 0.155
Mole fraction of Ar = 0.1 mol / 0.71 mol = 0.141
P.11
(ii) Let the total pressure of the system be P.
PV = nRT
P  15  10-3m3 = 0.71 mol  8.314 JK-1mol-1  ( 273 + 100 ) K
P = 146 786 Nm-2
Partial pressure of He = 146 786 Nm-2  0.704 = 103 337 Nm-2
Partial pressure of N2 = 146 786 Nm-2  0.155 = 22 752 Nm-2
Partial pressure of Ar = 146 786 Nm-2  0.141 = 20 697 Nm-2
22. Cu2+(aq) + 2eCu(s)
2+
To discharge 1 mole of Cu , 2 moles of electrons (i.e. 2F) are required.
Number of moles of Cu formed = It/nF
= 0.25 A x (60x60)s1
2 x 96 500 C mol
= 4.66  10-3 mol
Mass of Cu formed = 4.66  10-3 mol  63.5g mol-1 = 0.296 g
23. When a dilute sulphuric acid is electrolysed, hydrogen is formed at the cathode and
oxygen is formed at the anode of the electrolytic cell.
At cathode: 2H(aq) + 2eH2(g)
To give 1 mole of hydrogen gas, 2 moles of electrons (i.e. 2F) are required.
Number of moles of H2 (g) formed= It / nF
=
0.6 A x (90x60)s
2x96500 C mol 1
= 0.016 8 mol
Mass of H2(g) formed = 0.016 8 mol  1.0  2 g mol-1 = 0.033 6g
At anode: 4OH-(aq)
O2(g) + 2H2O(l) + 4eTo give 1 mole of oxygen gas, 4 moles of electrons (i.e. 4F) are given out by the
hydroxide ions.
Number of moles of O2 (g) formed= It / nF
=
0.6 A x (90x60)s
4x96500 C mol 1
= 8.394 x 10-3 mol
Mass of O2(g) formed = 8.394  10-3 mol  16.0  2 g mol-1 = 0.2686g
24. 4OH-(aq)
O2(g) + 2H2O(l) + 4eTo give 1 mole of oxygen gas, 4 mole of electrons (i.e. 4 F)are given out by the
hydroxide ions.
2.4 dm 3
Number of moles of O2 given outm =
= 0.1 mol
24.0dm 3 mol 1
Number of moles of electrons given out = 0.1 mol  4 = 0.4 mol
Cu2+(aq) + 2eCu(s)
To discharge 1 mole of Cu2+, 2 moles of electrons (i.e. 2F) are required.
Number of moles of Cu formed = 0.4 mol / 2 = 0.2 mol
Mass of Cu deposited = 0.2 mol  63.5 g mol-1 = 12.7 g
P.12
25. (a) Q = It
96 500 C = 0.35 A  t
= 275 714 s
2+
(b) Cu (aq) + 2eCu(s)
No. of moles of Cu formed = It / 2F
I x 50 x 60s
6 . 3 5g
=
1
2 x 96 500 C mol 1
6 3 . 5g m o l
I = 6.43 A
(c) Al3+(l) + 3eAl(s)
No. of moles of Al(s) formed = It / 3F
=
10A x 5 x 60 x 60s
=
3 x 96 500Cmol 1
0.622 mol
Mass of Al formed = 0.622 mol  27.0g mol-1 = 16.794
(d) No. of moles of X formed = It / nF
Mass of X = 0.324 g Since molar mass, n and F are constants, It /( Mass of X)
0.30 A x 30 x 60s
0.37 A x 15 x 60s
is a constant.
=
Mass of X
0.20 g
26. (a) Mass of sulphur = 5 g
;
Mass of oxygen = (10 – 5) g
Sulphur
Oxygen
5
5
Number of moles
5 / 32.1 = 0.156
5 / 16.0 = 0.313
Divided by the smallest
0.156 / 0.156 = 1
0.313 / 0.156 = 2
Mass (g)
Simplest ratio
1
The empirical formula of the sulphur oxide is SO2.
2
(b)
M
O
Mass (g)
19.85
25.61
Number of moles
19.85 / 31.0
= 0.64
25.61 / 16.0
= 1.6
Divided by the smallest
0.64 / 0.64 = 1
1.6 / 0.64 = 2.5
Simplest ratio
2
The empirical formula of the oxide is M2O5.
5
(c) Mass of Cu = (22.940 - 21.430) = 1.51g
Mass of O = (23.321 - 22.940) = 0.381 g
Cu
O
Mass (g)
1.51
0.381
Number of moles
1.51/63.5
= 0.0238
0.381/16.0
= 0.0238
Divided by the smallest
0.0238/0.0238 =1
0.0238/0.0238 =1
Simplest ratio
1
1
The empirical formula of the oxide is CuO.
27. Let the empirical formula of the hydrocarbon be CxHy, and the mass of the
compound be 100 g.
Mass of carbon in the compound = 75 g
Mass of hydrogen in the compound=(100 –75) g = 25 g
P.13
Carbon
Hydrogen
75
25
Number of moles
75/12.0 = 6.25
25/1.0 = 25
Divided by the smallest
6.25/6.25 = 1
25/6.25 = 4
1
4
Mass (g)
Simplest ratio
Therefore, the empirical formula of the hydrocarbon is CH4.
28. Let the mass of phosphorus chloride be 100g. Then,
Mass of phosphorus in the compound = 22.55g
Mass of chloride in the compound = 77.45g
Phosphorus
Chloride
22.55
77.45
Mass (g)
Number of mole (mol)
Divided by the smallest
22.55/31.0 = 0.727 77.45/35.5 = 2.182
0.727/0.727 = 1
2.182/0.727 = 3
1
3
Simplest ratio
Therefore, the empirical formula of the phosphorus chloride is PCl3.
29. (a) Let the mass of vitamin C analyzed be 100g.
Mass (g)
Number of moles
Divided by the smallest
Simplest ratio
Carbon
Hydrogen
Oxygen
40.9
4.6
54.5
40.9/12.0 = 3.41
4.6/1.0 = 4.60
54.5/16.0 = 3.41
3.41/3.41 = 1
4.61/3.41 =1.35
3.41/3.41 = 1
3
4
3
The empirical formula of vitamin C is C3H4O3.
(b) In order to facilitate calculation, the masses of the elements are multiplied by
1000 first.
Mass (g)
Carbon
Hydrogen
Oxygen
195.0
14.6
115.4
Number of moles (mol) 195.0/12.0 =16.25 14.6/7.21 = 2.02
7.21/7.21 = 1
Divided by the smallest 16.25/7.21 = 2.25 14.6/7.21 = 2.02
7.21/7.21 = 1
Simplest ratio
9
8
4
The empirical formula of aspirin is C9H8O4.
30. Let the empirical formula of the hydrocarbon be CxHy.
Mass of carbon in the hydrocarbon = 14.6g  12.0 / 44.0 = 4.0g
Mass of hydrogen in the hydrocarbon = 9.0g  2.0 / 18.0 = 1.0g
Carbon
Hydrogen
4.0
1.0
Number of moles
4.0 / 12.0 = 0.333
1.0 / 1.0 = 1
Divided by the smallest
0.333 / 0.333 = 1
1 / 0.333 = 3
1
3
Mass (g)
Simplest ratio
Therefore, the empirical formula of the hydrocarbon is CH3. The molecular formula
of the hydrocarbon is (CH3)n.
Relative molecular mass of (CH3)n = 30.0
P.14
n  (12.0 + 1.0  3) = 30.0 ;
n= 2
Therefore, the molecular formula of the hydrocarbon is C2H6.
31. Let the empirical formula of the hydrocarbon be CxHyOz. Mass of carbon in the
compound = 44.44g
Mass of hydrogen in the compound = 6.18g
Mass of oxygen in the compound = 49.38g
Carbon
Hydrogen
Oxygen
44.44
6.18
49.38
44.44/12.0 = 3.70
6.18/1.0 = 6.18
49.38/16.0 = 3.09
3.70/3.09 = 1.2
6.18/3.09 = 2
3.09/3.09 = 1
6
10
5
Mass (g)
Number of moles
Divided by the smallest
Simplest ratio
The empirical formula of compound X is C6H10O5.
The molecular formula of compound X is (C6H10O5)n.
Relative molecular mass of (C6H10O5)n = 162.0
n  (12.0  6 + 1.0  10 + 16.0  5) = 162.0
;
n=1
Therefore, the molecular formula of compound is C6H10O5.
32. Let formular mass of CuSO4xH2O
= 63.5 + 32.1 + 16.0 x 4 + (1.0  2 + 16.0)x = 159.6 + 18x
Relative mass of water of crystallization = 18x / (159.6 + 18x) = 36 / 100
1800x = 5745.6 + 648x
x = 4.99  5
Therefore, the chemical formula of hydrated copper(II) sulphate is CuSO4 5H2O
33. (a) (i)
Let the mass of compound Z be 100g.
Carbon
Hydrogen
Oxygen
40.00
6.67
53.33
40.00/12.0 = 3.33
6.67/1.0 = 6.67
53.33/16.0=3.33
3.33/3.33 =1
6.67/3.33 =2
3.33/3.33 =1
1
2
1
Mass (g)
Number of moles
Divided by the smallest
Simplest ratio
The empirical formula of compound Z is CH2O.
(ii) Let the molecular formula of compound Z be (CH2O)n.
n  (12.0 + 1.0  2 + 16.0) = 180
30n = 180
;
n=6
The molecular formula of Z is C6H12O6.
(b)
Mass (g)
Number of moles
Divided by the smallest
NH4+ unit
S
27.28
72.72
27.28 / 18.0 = 1.52
72.72 / 32.1 = 2.27
1.52 / 1.52 = 1
2.27 / 1.52 = 1.49
2
3
Simplest ratio
Since the chemical formula of (NH4)2Sx is (NH4)2S3, the value of x is 3.
P.15
(c)
MgSO4
H2O
48.78
51.22
48.78 / 120.4 = 0.405
51.22 / 18.0 = 2.846
1.52 / 1.52 = 1
2.846 / 0.405 = 7
1
7
Mass (g)
Number of moles
Divided by the smallest
Simplest mole ratio
Since the chemical formula of MgSO4nH2O is MgSO47H2O , the value of x
is 7.
34. Relative molecular mass of CH3COOH = 12.0  2 + 1.0  4 + 16.0  2 = 60.0
% by mass of C = 12.0  2 / 60.0  100%= 40.00%
% by mass of H = 1.0  4 / 60.0  100% = 6.67%
% by mass of O = 16.0  2 / 60.0  100% = 53.33%
The percentage by mass of carbon, hydrogen and oxygen are 40.00%, 6.67% and
53.33% respectively.
35. Relative molecular mass of FeSO4·7H2O =
55.8 + 32.1 + 16.0  4 + (1.0  2 + 16.0)  7 = 277.9
% by mass of Fe = 55.8 / 277.9  100% = 20.08 %
Mass of Fe = 20 g  20.08 % = 4.02 g
36. (a) Molar mass of K2Cr2O7 = (39.1  2 + 52.0 + 16.0  7) g mol-1 = 294.2 g mol-1
% by mass of K = 39.1  2 g mol-1 / 294.2 g mol-1  100% = 26.58 %
% by mass of Cr = 52.0  2 g mol-1 / 294.2g mol-1  100% =35.25 %
% by mass of O = 16.0  7 g mol-1 / 294.2g mol-1  100% = 38.07 %
(b) (i) Molar mass of Na2SO4·10H2O = 322.1 g mol-1
Mass of Na = 23.0  2 g mol-1 / 322.1 g mol-1  100 g = 14.28 g
Mass of H2O = 18.0  10 g mol-1 / 322.1 g mol-1  100 g = 14.28 g
(ii) Molar mass of Fe2O3·8H2O = 303.6 g mol-1
Mass of Fe = 55.8  2 g mol-1 / 303.6g mol-1  70g = 25.73 g
Mass of H2O = 18.0  8 g mol-1 / 303.6g mol-1  70g = 33.20 g
37. (a) Zn(s) + H2O(g) → ZnO(s) + H2(g)
(b) Mg(s) + 2 AgNO3(aq) → 2 Ag(s) + Mg(NO3)2(aq)
(c) 2 C4H10(g) + 13 O2(g) →
38. CuO(s) + H2(g)
→
8 CO2(g) + 10 H2O(l)
Cu(s) + H2O(l)
As the mole ratio of Cu : CuO is 1 : 1, the number of moles of Cu formed is the
same as the number of moles of CuO reduced.
Number of moles of CuO reduced = 12.45 / (63.5 + 13.0) g mol-1 = 0.157 mol
Number of mole of Cu formed = 0.157 mol
Mass of Cu / 63.5 g mol-1 = 0.157
Mass of Cu = 0.157 mol  63.5 g mol-1= 9.97 g
Therefore, the mass of copper formed in the reaction is 9.97 g.
P.16
39. Number of moles of CO2 formed = 240 cm3/ 24000 cm3 mol-1 = 0.01 mol
From the equation, 2 moles of NaHCO3(s) will form 1 mole of CO2(g).
Number of moles of NaHCO3 required = 0.01  2 = 0.02 mol
Mass of NaHCO3 required = 0.02 mol  (23.0 + 1.0 + 16.0  3) g mol-1
= 0.02 mol  84.0g mol-1 = 1.68 g
Therefore, the minimum amount of sodium hydrogencarbonate required is 1.68 g.
40. Number of moles of CO2 formed :
C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
2 mol : 7 mol
:
2 volumes : 7 volumes :
4 mol
:
4 volumes :
6 mol
liquid
(by Avogadro’s law)
It can be judged from the equation that the mole ratio of CO2 : C2H6 is 4 :2, and the
volume ratio of CO2 : C2H6 should also be 4:2.
Let x be the volume of CO2(g) formed :
x / 20cm3 = 4 / 2
;
x = 40 cm3
Therefore, the volume of CO2 formed is 40 cm3.
41. Let the molecular formula of the hydrocarbon be CxHy.
Volume of hydrogen reacted = 10 cm3
Volume of O2(g) unreacted = 50 cm3
Volume of O2(g) reacted = 30 cm3
Volume of CO2(g) formed = 20 cm3
CxHy
+ (x + y/4) O2
→
CO2
+
1 volume :
(x + y/4) volumes
:
y/2 H2O
x volumes
Volume of CO2 (g) / volume of CxHy(g) = 20 cm3 / 10cm3 = 2
Volume of O2(g) / volume of CxHy(g) = (x + y/4) / 1 = 30/ 10
(2 + y/4) = 3 ;
y=2
Molecular formula is C2H4.
;
x =2
42. (a) No. of moles of H2 = No. of moles of Mg
Volume of H2/ 24.0 dm3 mol-1 = 2.43 g / 24.3 g mol-1
Volume of H2 = 2.4 dm3
(b) 1/3  no. of moles of Cl2 = 1/2  no. of moles of PCl3
1/3  mass of Cl2 / (35.5  2) g mol-1 = 1/2  100g / (31.0 + 35.5 + 3 ) g mol-1
Mass of Cl2 = 77.45 g
(c) Volume of CxHy used = 20 cm3
Volume of CO2 formed = 60 cm3
Volume of O2 used = 100 cm3
Volume of CxHy : volume of CO2 = 1 : x = 20 : 60 ;
x=3
Volume of CxHy : volume of O2 = 1 : x + y/4 = 20 : 100
3 + y/4 = 5
;
y=8
3
(d) Volume of CxHy used = 20 cm
It can be judged from the equation that the mole ratio of CO2 : CH4 is 1:1, the
volume ratio of CO2 : CH4 should also be 1:1.
x / 5 = 1/1
;
x=5
The volume of carbon dioxide gas is 5 cm3.
P.17
43. Number of moles of NaOH(aq) / Number of moles of H2SO4(aq) = 2 / 1
½  Number of moles of NaOH(aq) = Number of moles of H2SO4 (aq)
= 0.067 mol dm-3  22.5 x 10-3 dm3 = 1.508  10-3 mol
Number of moles of NaOH(aq) = 2  1.508  10-3 mol = 3.016  10-3 mol
Molarity of NaOH(aq) = 3.012  10-3 mol / 25.0  10-3 mol = 0.1221 mol dm-3
The molarity of NaOH is 0.121M
44.. (a) Number of moles of acid = 2.52 g / 126.0 g mol-1 = 0.02 mol
Molarity of acid solution = 0.02 mol / 250  10-3 dm3 = 0.08M
(b) H2X(aq) + 2NaOH(aq) → Na2X(aq) + 2NaOH(l)
(c) Number of moles of H2X = ½  number of moles of NaOH
Molarity of NaOH = 0.14M
45. There is a sudden drop in the pH value of the solution (from pH 3 to pH 8) with the
end point at 30.0 cm3.
Na2CO3·nH2O(s) + 2 HCl(aq) → 2NaCl(aq) + CO2(g) + (n+1)H2O(l)
Number of moles of Na2CO3·nH2O = ½  0.1 mol dm-3  30  10-3 dm3
106.0 + 18.0n = 124.0
; n=1
The formula is Na2CO3·H2O
46. (a)
(b) From the graph, it is found that the end point of the titration is reached when
20 cm3 of H2SO4 is added.
Number of moles of = 0.5 mol dm-3  20 / 1000 dm3 = 0.01 mol
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)
2 mol
1 mol
Mole of KOH(aq) : H2SO4 = 2 : 1
Number of moles of KOH(aq) = 2  0.01 mol = 0.02 mol
Molarity of KOH(aq) = 0.02 mol / (25  10-3 dm3) = 0.8 M
(c) Neutralization is an exothermic reaction. When more and more sulphuric(VI)
acid was added and reacted with potassium hydroxide, the temperature rose.
The temperature rose to a maximum value at which the equivalence point of the
reaction was reached. After that, any excess sulphuric (VI) acid added cooled
down the reacting solution, causing the temperature to drop.
P.18
47. IO3-(aq) + 5I- + 6H+(aq) →
2-
3I2(aq)
-
+
3H2O(l)
2-
I2(aq) + 2S2O3 (aq) → 2I (aq) + S4O6 (aq)
… (1)
… (2)
From (1), Number of moles of IO3-(aq) = 1/3  number of moles of I2(aq)
From(2), Number of moles of I2(aq) = 1/2  number of moles of S2O32-(aq)
Number of moles of IO3-(aq) = 1/6  number of moles of S2O32-(aq)
Molarity of IO3-(aq)  25.0 / 1000 dm3 = 1/6  0.05 mol dm-3  22.0 / 1000 dm3
Molarity of IO3-(aq) = 7.33  10-3 M
48. MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) →
Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Number of moles of Fe2+(aq) = 5  number of moles of MnO4-(aq) =
5  0.02 mol dm-3  36.5 x 10-3 dm3 = 3.65  10-3 mol
Number of moles of Fe dissolved = number of mole of Fe2+ formed
= 3.65  10-3 mol
Mass of Fe = 3.65  10-3 mol  55.8 g mol-1 = 0.204 g
Percentage purity of Fe = 0.204g / 0.22g  100 % = 92.73 %
49. (a) Na2CO3 (s) + 2 HCl (aq) →
2NaCl(aq) +
H2O(l) +
CO2(g)
No. of moles of Na2CO3 used = 5g / (23.0  2 +12 +16.0  3)g mol-1
= 0.0472 mol
No. of moles of HCl used = 2M  100/ 1000 dm3 = 0.2 mol
Since HCl is in excess, Na2CO3 is the limiting agent.
No. of moles of CO2 produced = no. of moles of Na2CO3 used = 0.0472 mol
Volume of CO2 produced = 0.0472 mol  24.0 dm3 mol-1= 1.133 dm3
(b) No. of moles of MnO4- = 0.0203M  20.76 / 1000 dm3
= 4.214 x 10-4 mol
No. of moles of Fe2+ = 5  no. of moles of MnO4- = 2.107  10-3 mol
No. of mole of Fe2+ in 25.0 cm3 solution = 2.107  10-3 mol
No. of mole of Fe2+ in 250.0 cm3 solution = 0.02107 mol
Molar mass of hydrated FeSO4 = 392.14 g mol –1
Mass of hydrated FeSO4 = 0.02107 mol  392.14 g mol–1= 8.26 g
5 purity of FeSO4 = 8.26 g / 8.54 g  100 % = 96.72 %
P.19