Given, Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g
\( Percentage \ of \ boron = \) \( mass \ of \ boron \over mass \ of \ the \ compound \) \( × 100 \) \( = \) \( 0.096 \over 0.24 \) × 100 = 40%
Percentage of oxygen = 100 – percentage of boron
= 100 – 40 = 60%
When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. Carbon and oxygen are combined in the ratio 3:8 to give 11 carbon dioxide . Hence, for 3g of carbon and 50g of oxygen, 8g of oxygen is used and 11g of carbon is formed, and 42g of oxygen is unused. This is govern by law of definite proportions.
Polyatomic ions are ions that contain more than one atom but they behave as a single unit. Example: OH \(^- \) , H\(_2 \)PO\( _4^– \)
The chemical formula are:
(a) Magnesium chloride – MgCl\(_2\)
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO3)\(_2\)
(d) Aluminium chloride – AlCl\(_3\)
(e) Calcium carbonate – CaCO\(_3\)
The names of the elements present are:
(a) Quick lime – Calcium and oxygen
(b) Hydrogen bromide – Hydrogen and bromine
(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen
(d) Potassium sulphate – Sulphur, Oxygen, Potassium
We know that molar of a substance is the sum of the atomic masses of all the elements.
(a)So, molar mass of Ethyne C\(_2\)H\(_2\)= 2 × Mass of C+2 ×
Mass of H = (2 × 12) + (2 × 1) = 24 + 2 = 26g
(b) Molar mass of Sulphur molecule S8 = 8 × Mass of S = 8 × 32 = 256g
(c) Molar mass of Phosphorus molecule, P4 = 4 × Mass of P = 4 × 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 × Mass of O = 1 + 14 + 3 × 16 = 63g
(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 × 27 = 108g
(c) Mass of 1 mole of sodium sulphite Na\(_2\)SO\(_3\) = Molecular mass of sodium sulphite = 2 × Mass of Na + Mass of S + 3 × Mass of O = (2 × 23) + 32 +(3 × 16) = 46 + 32 + 48 = 126g
Therefore, mass of 10 moles of Na\(_2\)SO\(_3\) = 10 × 126 = 1260g
(a) Given, Mass of oxygen gas = 12g
Molar mass of oxygen gas = 2 × Mass of Oxygen = 2 × 16 = 32g
\( Number \ of \ moles \ = \) \( Mass \ given \over molar \ mass \ of \ oxygen \ gas \) \( = \) \( 12 \over 32 \) \( = 0.375 \ moles \)
(b) Mass of water = 20g
Molar mass of water = 2 × Mass of Hydrogen + Mass of Oxygen = 2 × 1 + 16 = 18g
\( Number \ of \ moles \ = \) \( Mass \ given \over molar \ mass \ of \ water \) \( = \) \( 20 \over 18 \) \( = 1.11 \ moles \)
(c) Mass of carbon dioxide = 22g
Molar mass of carbon dioxide = Mass of C + 2 × Mass of Oxygen = 12 + 2 × 16 = 12 + 32 = 44g
\( Number \ of \ moles = \) \( Mass \ given \over molar \ mass of \ carbon \ dioxide \) \( = \) \( 22 \over 44 \) \( = 0.5 \ moles \)
(a) Mass of 1 mole of oxygen atoms = 16u
∴ Mass of 0.2 moles of oxygen atoms = 0.2 × 16 = 3.2u
(b) Mass of 1 mole of water molecules = 18u
∴ Mass of 0.5 moles of water molecules = 0.5 × 18 = 9u
(a) Mass of 1 mole of oxygen atoms = 16u
∴ Mass of 0.2 moles of oxygen atoms = 0.2 × 16 = 3.2u
(b) Mass of 1 mole of water molecules = 18u
∴ Mass of 0.5 moles of water molecules = 0.5 × 18 = 9u
Molecular mass of Sulphur (S\(_8\)) = 8 × Mass of Sulphur = 8 × 32 = 256g
Mass of solid sulpur = 16g
\( Number \ of \ moles = \) \( mass \ given \over molar \ mass \ of \ sulphur \) = \( 16 \over 256 \) = 0.0625 moles
Number of molecules = Number of moles × Avogadro number
\( = 0.0625 × 6.022 x 10²³ = 3.763 × 10^{22} \) molecules
1 mole of aluminium oxide = 6.022 × 1023 molecules of aluminium oxide
1 mole of aluminium oxide (Al(S\(_2\))O(S\(_3\))) = 2 × Mass of aluminium + 3 × Mass of Oxygen
= (2 × 27) + (3 × 16) = 54 + 48 = 102g
1 mole of aluminium oxide \( = 102g = 6.022 × 10^{23} \) molecules of aluminium oxide
Therefore, 0.051g of aluminium oxide has \( = \) \( 0.051 × 6.022 × 10^{23} \over 102 \)
\( = 3.011 × 10^{20} \) 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 \( = 2 ×3.011 × 10^{20} \) molecules of aluminium oxide
\( = 6.022 × 10^{20} \)
Sodium carbonate + acetic acid ? Sodium acetate + carbon dioxide + water
(5.3g) + (6g) -> (8.2g) + (2.2g) + (0.9g)
11g = 11g
=> LHS = RHS
i.e., Mass of reactant = Mass of product
which satisfies the law of conservation of mass.
Given, ratio of H:O by mass to form water = 1:8
i.e., 1g of hydrogen required 8g of oxygen to react completely.
To find, mass of oxygen required to completely react with 3g of hydrogen.
So, 3g of hydrogen will require \( (8 × 3) \)g of oxygen
∴: 24g of oxygen is required to completely react with 3g of hydrogen.
The postulate of Dalton’s Atomic theory which is a result of the law of conservation of mass is, “Atoms can neither be created nor destroyed”.
The postulate of Dalton’s atomic theory that can explain the law of definite proportions is, " The relative number and kinds of atoms are equal in given compounds."
One atomic mass is equal to 1/12th the mass of one atom of carbon-12 i.e., 1amu = \( 1 \over 12 \) th mass of one atom of carbon.
We cannot see atoms with naked eye beacuse their size is very small, i.e., in nanometers( \( 1m = 10^9 nm \) ).
The formulae are:
(i) sodium oxide -> Na\( _2 \)O
(ii) aluminium chloride -> AlCl\( _3 \)
(iii) sodium sulphide -> Na\( _2 \)S
(iv) magnesium hydroxide -> Mg (OH)\( _2 \)
The names of the compounds are:
(i) Al\( _2 \)(SO\( _4 \))\( _3 \) –->Aluminium sulphate
(ii) CaCl\( _2 \) -> Calcium chloride
(iii) K\( _2 \)SO\( _4 \) -> Potassium sulphate
(iv) KNO\( _3 \) -> Potassium nitrate
(v) CaCO\( _3 \) -> Calcium carbonate
Chemical formula of a chemical compound is the symbolic representation of its composition. Eg. chemical formula of water is H\( _2 \)O.
The number of atoms present are:
(i) H\( _2 \)S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence total 3 atoms.
(ii) PO\( _4 \)\(^{3-} \) ion has 1 atom of phosphorus and 4 atoms of oxygen hence total 5 atoms.
We know that molecular mass of a substance is the sum of the atomic masses of all the atoms in the molecule of the substance.
i.e., molecular mass of H\( _2 \) = atomic mass of H + atomic mass of H = 2 × atoms atomic mass of H = 2 × 1 = 2u
Similarly,
The molecular mass of O\( _2 \) = 2 × atoms atomic mass of O = 2 × 16 = 32u
The molecular mass of Cl\( _2 \) = 2 × atoms atomic mass of Cl = 2 × 35.5 = 71u
The molecular mass of CO\( _2 \) = atomic mass of C + 2 × atomic mass of O = 12 + ( 2 × 16) = 44u
The molecular mass of CH\( _4 \) = atomic mass of C + 4 × atomic mass of H = 12 + ( 4 × 1) = 16u
The molecular mass of C\( _2 \)H\( _6 \) = 2 × atomic mass of C + 6 × atomic mass of H = (2 × 12) + (6 × 1) = 24 + 6 = 30u
The molecular mass of C\( _2 \)H\( _4 \) = 2 × atomic mass of C + 4 × atomic mass of H = (2 × 12) + (4 × 1) = 24 + 4 = 28u
The molecular mass of NH\( _3 \) = atomic mass of N + 3 × atomic mass of H = 14 + (3 × 1) = 17u
The molecular mass of CH\( _3 \)OH = atomic mass of C + 3 × atomic mass of H + atomic mass of O + atomic mass of H = 12 + (3 × 1) + 16 + 1 = 12 + 3 + 17 = 32u
Given, atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, O = 16u
We know that formula unit mass of a substance is the sum of the atomic masses of all the atoms in the formula unit of the substance.
So, the formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65 + 16 = 81u
The formula unit mass of Na\( _2 \)O = 2 × Atomic mass of Na + Atomic mass of O = (2 × 23) +16 = 46 + 16 = 62u
The formula unit mass of K\( _2 \)CO\( _3 \) = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O = (2 × 39) + 12 + (3 × 16) = 78 + 12 + 48 = 138u
Given, 1 mole of carbon weighs = 12g
1 mole of carbon atoms \( = 6.022 × 10^{23} \)atom
i.e., \( 6.022 × 10^{23} \)atom = 12g
So, mass of 1 atom of carbon \( = \) \( 12 \over 6.022 × 10^{23} \) \( = 1.99 × 10^{-23} \)
Given, Atomic mass of Na=23u, and of Fe= 56u
Now, 23g of Na contains \( = 6.022 × 10^{23} \) atoms
1g of Na contains = \( 6.022 × 10^{23} \over 23 \)atoms
100g of Na contains = \( 6.022 × 10^{23} \over 23 \) × 100 \( = 2.6182 × 10^{24} \) atoms
Now, 56g of Fe contains = 6.022 × 1023 atoms
1g of Fe contains = \( 6.022 × 10^{23} \over 56 \)atoms
100g of Fe contains = \( 6.022 × 10^{23} \over 56 \) × 100 \(1.075 × 10^{24} \) atoms
So, 100g of Na has more atoms.