Exercise Page: 43
1. A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.
Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g
To calculate percentage composition of the compound:
Percentage of boron = mass of boron / mass of the compound x 100
= 0.096g / 0.24g x 100 = 40%
Percentage of oxygen = 100 – percentage of boron
= 100 – 40 = 60%
2. When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
11.00g of carbon dioxide is formed when 3.00g carbon is burnt in 8.00g of oxygen.
Carbon and oxygen are combined in the ratio 3:8 to give carbon dioxide using up all the carbon and
Hence, for 3g of carbon and 50g of oxygen, 8g of oxygen is used and 11g of carbon is formed, the
left oxygen is unused i.e., 50-8=42g of oxygen is unused.
This depicts the law of definite proportions – The combining elements in compounds are present in
definite proportions by mass.
3. What are polyatomic ions? Give examples.
Polyatomic ions are ions that contain more than one atom but they behave as a single unit
Example: CO32- , H2PO4–
4. Write the chemical formula of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
The following are the chemical formula of the above-mentioned list:
(a) Magnesium chloride – MgCl2
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO3)2
(d) Aluminium chloride – AlCl3
(e) Calcium carbonate – CaCO3
5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
The following are the names of the elements present in the following compounds:
(a) Quick lime – Calcium and oxygen (CaO)
(b) Hydrogen bromide – Hydrogen and bromine (HBr)
(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO3)
(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K2SO4)
6. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
Listed below is the molar mass of the following substances:
(a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g
(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8 x 32 = 256g
(c) Molar mass of Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g
(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g
(e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+
3×16 = 63g
7. What is the mass of –
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms((Atomic mass of aluminium =27)?
(c) 10 moles of sodium sulphite (Na2SO3)?
The mass of the above mentioned list is as follows:
(a) Atomic mass of nitrogen atoms = 14u
Mass of 1 mole of nitrogen atoms= Atomic mass of nitrogen atoms
Therefore, mass of 1 mole of nitrogen atom is 14g
(b) Atomic mass of aluminium =27u
Mass of 1 mole of aluminium atoms = 27g
1 mole of aluminium atoms = 27g, 4 moles of aluminium atoms = 4 x 27 = 108g
(c) Mass of 1 mole of sodium sulphite Na2SO3 = Molecular mass of sodium sulphite = 2 x Mass of Na + Mass of S + 3 x Mass of O = (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g
Therefore, mass of 10 moles of Na2SO3 = 10 x 126 = 1260g
8. Convert into mole.
(a) 12g of oxygen gas
(b) 20g of water
(c) 22g of carbon dioxide
Conversion of the above-mentioned molecules into moles is as follows:
(a) Given: Mass of oxygen gas=12g
Molar mass of oxygen gas = 2 Mass of Oxygen = 2 x 16 = 32g
Number of moles = Mass given / molar mass of oxygen gas = 12/32 = 0.375 moles
(b) Given: Mass of water = 20g
Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g
Number of moles = Mass given / molar mass of water
= 20/18 = 1.11 moles
(c) Given: Mass of carbon dioxide = 22g
Molar mass of carbon dioxide = Mass of C + 2 x Mass of Oxygen = 12 + 2x 16 = 12+32=44g
Number of moles = Mass given/ molar mass of carbon dioxide = 22/44 = 0.5 moles
9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
The mass is as follows:
(a) Mass of 1 mole of oxygen atoms = 16u, hence it weighs 16g
Mass of 0.2 moles of oxygen atoms = 0.2 x 16 = 3.2u
(b) Mass of 1 mole of water molecules = 18u, hence it weighs 18g
Mass of 0.5 moles of water molecules = 0.5 x 18 = 9u
10. Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.
To calculate molecular mass of sulphur:
Molecular mass of Sulphur (S8) = 8xMass of Sulphur = 8×32 = 256g
Mass given = 16g
Number of moles = mass given/ molar mass of sulphur
= 16/256 = 0.0625 moles
To calculate the number of molecules of sulphur in 16g of solid sulphur:
Number of molecules = Number of moles x Avogadro number
= 0.0625 x 6.022 x 10²³ molecules
= 3.763 x 1022 molecules
11. Calculate the number of aluminium ions present in 0.051g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)
To calculate the number of aluminium ions in 0.051g of aluminium oxide:
1 mole of aluminium oxide = 6.022 x 1023 molecules of aluminium oxide
1 mole of aluminium oxide (Al2O3) = 2 x Mass of aluminium + 3 x Mass of Oxygen
= (2x 27) + (3 x16) = 54 +48 = 102g
1 mole of aluminium oxide = 102g = 6.022 x 1023 molecules of aluminium oxide
Therefore, 0.051g of aluminium oxide has = 0.051 x 6.022 x 1023 / 102
= 3.011 x 1020 molecules of aluminium oxide
One molecule of aluminium oxide has 2 aluminium ions, hence number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 1020 molecules of aluminium oxide
= 6.022 x 1020
Er. Aman Blogger, Educator