The Language of Chemistry

a molecule

Reason — The formula of a compound employs symbols to denote the molecule of a compound.

Al₂O₃

Reason — The formula of aluminium oxide can be determined as below:

$\text{Al}^{3+} \space \text{O}^{2-} \\[1em] \overset{\phantom{3}{3}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{2}{2}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{O}} \\[1em] \text{Al}_2\text{O}_3$

two

Reason — The valency of lead in lead oxide can be determined from the formula PbO as below:

$\text{PbO} = \underset{\phantom{1}{1}}{\text{Pb}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{Pb}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]$

But valency of O is 2. Multiplying by 2, we get:

$\overset{{1 \times 2}}{\text{Pb}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{Pb}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]$

Therefore, valency of Lead is 2.

Phosphorus

Reason

Atomic mass of P = 31 μ

Molecular form = P₄ ⟶ Molecular mass = 31 × 4 = 124 μ

124 μ is 4 times the atomic mass

Hence, phosphorus has molecular mass four times its atomic mass.

The basic and acidic radicals in the compounds are given in the table below:

The formula and valency are given in the table below:

(a) H → One atom of Hydrogen.

(b) H₂ → One molecule of Hydrogen.

(c) 2H → Two atoms of Hydrogen.

(d) 2H₂ → Two molecules of Hydrogen.

(a) Ca₃(PO₄)₂ → Calcium Phosphate

(b) K₂CO₃ → Potassium Carbonate

(c) K₂MnO₄ → Potassium Manganate

(d) Mn₃(BO₃)₂ → Manganese (II) borate

(e) Mg(HCO₃)₂ → Magnesium Hydrogen Carbonate

(f) Na₄Fe(CN)₆ → Sodium Ferrocyanide

(g) Ba(ClO₃)₂ → Barium Chlorate

(h) Ag₂SO₃ → Silver Sulphite

(i) (CH₃COO)₂Pb → Lead Acetate

(j) Na₂SiO₃ → Sodium Silicate

(a) Chemical formula of Aluminium Sulphate:

$\text{Al}^{3+} \space \text{SO}_4^{2-} \\[1em] \overset{\phantom{3}{3}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{2}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{SO}_4} \\[1em] \text{Al}_2\text{(SO}_4)_3$

∴ Chemical formula of Aluminium Sulphate is Al₂(SO₄)₃

(b) Chemical formula of Ammonium Sulphate:

$\text{NH}_4^{+} \space \text{SO}_4^{2-} \\[1em] \overset{\phantom{3}{1}}{\text{NH}_4} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{2}}{\text{NH}_4} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{1}}{\text{SO}_4} \\[1em] \text{(NH}_4)_2\text{SO}_4$

∴ Chemical formula of Ammonium Sulphate is (NH₄)₂SO₄

(c) Chemical formula of Zinc Sulphate:

$\text{Zn}^{2+} \space \text{SO}_4^{2-} \\[1em] \text{Dividing by H.C.F. it becomes} \\[1em] \text{Zn}^{1+} \space \text{SO}_4^{1-}\\[1em] \overset{\phantom{22}{1}}{\text{Zn}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{1}}{\text{Zn}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{1}}{\text{SO}_4} \\[1em] \text{ZnSO}_4$

∴ Chemical formula of Zinc Sulphate is ZnSO₄

Formula of the compound formed by the combination of A and B:

$\overset{\phantom{3}{3}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{B}} \Rightarrow \underset{\phantom{2}{2}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{B}} \\[1em] \text{A}_2\text{B}_3$

∴ Formula of the compound is A₂B₃

(a) KClO → Potassium hypochlorite

(b) KClO₂ → Potassium Chlorite

(c) KClO₃ → Potassium Chlorate

(d) KClO₄ → Potassium Perchlorate

Given, formula of sulphate is M₂(SO₄)₃.

$\text{M}_2(\text{SO}_4)_3 = \underset{\phantom{2}{2}}{\text{M}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{3}{\text{SO}_4} \Rightarrow \overset{\phantom{3}{3}}{\text{M}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{SO}_4} \\[0.5em]$

∴ Valency of M is 3.

(a) Formula of Chloride:

$\text{M}^{3+} \space \text{Cl}^{1-} \\[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{Cl}} \\[1em] \text{M}\text{Cl}_3$

∴ Formula of chloride of element M is MCl₃

(b) Formula of Oxide:

$\text{M}^{3+} \space \text{O}^{2-} \\[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{2}{2}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{O}} \\[1em] \text{M}_2\text{O}_3$

∴ Formula of oxide of element M is M₂O₃

(c) Formula of Phosphate:

$\text{M}^{3+} \space \text{PO}_4^{3-} \\[1em] \text{Dividing by H.C.F. it becomes} \\[1em] \text{M}^{1+} \space \text{PO}_4^{1-} \\[1em] \overset{\phantom{3}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{PO}_4} \Rightarrow \underset{\phantom{2}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{1}}{\text{PO}_4} \\[1em] \text{MPO}_4$

∴ Formula of phosphate of element M is MPO₄

(d) Formula of Acetate:

$\text{M}^{3+} \space \text{CH}_3\text{COO}^{1-} \\[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{\phantom{CHCOOO}1}{\text{CH}_3\text{COO}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{CH}_3\text{COO}} \\[1em] \text{M}\text{(CH}_3\text{COO})_3$

∴ Formula of acetate of element M is M(CH₃COO)₃

A symbol is a short form that stands for the atom of a specific element or the abbreviation used for the name of an element.

A symbol conveys the following information:

1. It represents the name of the element.
2. It represents one atom of the element.
3. It represents a definite mass of the element (equal to atomic mass expressed in grams).

When the first letter of more than one element is the same the elements are denoted by two letters. Sulphur, Sodium and Silicon all have the first letter as S. Therefore, only sulphur is denoted by S, Silicon is denoted by two letter Si, Sodium is denoted by two letter Na taken from its latin name Natrium.

Full form of IUPAC is International Union of Pure and Applied Chemistry.

The elements represented by the following symbols are:

Au → Gold

Pb → Lead

Sn → Tin

Hg → Mercury

When writing symbols, we need to be careful about the case of the letters in the symbol. CO means the compound Carbon Monoxide not Cobalt.

Valency is the combining capacity of an atom or of a radical. The valency of an element or of a radical is the number of hydrogen atoms that will combine with or displace one atom of that element or radical.
Variable valency is the ability of certain elements to have more than one valency or different combining capacities. An atom of an element can sometimes lose more electrons than are present in its valence shell, i.e., there is a loss of electrons from the penultimate shell too. Such element is said to exhibit variable valency.

If an element exhibits two different positive valencies, then the suffix "ous" is used for the lower valency and the suffix "ic" is used for the higher valency. Modern chemists use Roman numerals in place of these trivial names.
For example, Iron can exhibit a valency of +2 or +3. The ion with valency +2 is named as Ferrous ion (Fe²⁺) and with valency +3 is named as Ferric ion (Fe³⁺). Its oxides will be named as Ferrous oxide or Iron (II) oxide [FeO] and Ferric oxide or Iron (III) oxide [Fe₂O₃].

(a) A chemical formula also known as molecular formula employs symbols to denote the molecule of an element or of a compound.

(b) The chemical/molecular formula of a compound has quantitative significance. It represents:

1. both the molecule and the molecular mass of the compound.
2. the respective numbers of different atoms present in one molecule of the compound.
3. the ratios of the respective masses of the elements present in the compound.

For example, the formula CO₂ means that:

1. the molecular formula of carbon dioxide is CO₂.
2. each molecule contains one carbon atom joined by chemical bonds with two oxygen atoms.
3. the molecular mass of carbon dioxide is 44, given that the atomic mass of carbon is 12 and that of oxygen is 16.

(a) An Acid radical is the radical that remains after an acidic molecule loses a hydrogen ion (H⁺). Acid radicals typically have a negative charge. They are also called electronegative radicals or anions.

(b) A Basic radical is the radical that remains after a base molecule loses a hydroxyl ion (OH^-). Basic radicals typically have a positive charge. They are also called electropositive radicals or cations.

(a) Sodium sulphate
     Formula: Na₂SO₄
   ∴ It contains 2 atoms of Sodium (Na), 1 atom of Sulphur (S) and 4 atoms of Oxygen (O).

(b) Quick lime
     Formula: CaO
   ∴ It contains 1 atom of Calcium (Ca) and 1 atom of Oxygen (O).

(c) Baking soda
     Formula: NaHCO₃
   ∴ It contains 1 atom of Sodium (Na), 1 atom of Hydrogen (H), 1 atom of Carbon (C) and 3 atoms of Oxygen (O).

(d) Ammonia
     Formula: NH₃
   ∴ It contains 1 atom of Nitrogen (N) and 3 atoms of Hydrogen (H).

(e) Ammonium dichromate
     Formula: (NH₄)₂Cr₂O₇
   ∴ It contains 2 atoms of Nitrogen (N), 8 atoms of Hydrogen (H), 2 atoms of Chromium (Cr) and 7 atoms of Oxygen (O).

(a) Chemical formula of Oxide of A:

$\text{A}^{4+} \space \text{O}^{2-} \\[1em] \text{Dividing by H.C.F. it becomes} \\[1em] \text{A}^{2+} \space \text{O}^{1-}\\[1em] \overset{{2}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{O}} \Rightarrow \underset{\phantom{2}{1}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{O}} \\[1em] \text{AO}_2$

∴ Chemical formula of Oxide of A is AO₂

(b) Chemical formula of Nitrate of A:

$\text{A}^{4+} \space \text{NO}_3^{1-} \\[1em] \overset{{4}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{NO}_3} \Rightarrow \underset{\phantom{2}{1}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{4}}{\text{NO}_3} \\[1em] \text{A(NO}_3)_4$

∴ Chemical formula of Nitrate of A is A(NO₃)₄

(c) Chemical formula of Phosphate of A:

$\text{A}^{4+} \space \text{PO}_4^{3-} \\[1em] \overset{{4}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{3}{\text{PO}_4} \Rightarrow \underset{\phantom{2}{3}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{4}}{\text{PO}_4} \\[1em] \text{A}_3(\text{PO}_4)_4$

∴ Chemical formula of Phosphate of A is A₃(PO₄)₄

(a) CH₃COONa (Sodium acetate)
     Cation: Na⁺ (Sodium ion)
     Anion: CH₃COO^- (Acetate ion)

(b) NH₄Cl (Ammonium chloride)
     Cation: NH₄⁺ (Ammonium ion)
     Anion: Cl^- (Chloride ion)

(c) PbCl₂ (Lead(II) chloride)
     Cation: Pb²⁺ (Lead(II) ion)
     Anion: Cl^- (Chloride ion)

(d) MgO (Magnesium oxide)
     Cation: Mg²⁺ (Magnesium ion)
     Anion: O^2- (Oxide ion)

Aluminium is trivalent cation with a valency of 3 → Al³⁺

Sulphate is divalent anion with a valency of 2 → SO₄^2-

Formula:

$\text{Al}^{3+} \space \text{SO}_4^{2-} \\[1em] \overset{3}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{2}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \underset{3}{\text{SO}_4} \\[1em] \text{Al}_2(\text{SO}_4)_3$

Hence, the formula is : Al₂(SO₄)₃

(a) PO₄^3-

P = 1 atom

O = 4 atoms

Total atoms = 1 + 4 = 5 atoms

(b) P₂O₅

P = 2 atoms

O = 5 atoms

Total atoms = 2 + 5 = 7 atoms

(c) H₂SO₄

H = 2 atoms

S = 1 atom

O = 4 atoms

Total atoms = 2 + 1 + 4 = 7 atoms

(d) SO₄^-2

S = 1 atom

O = 4 atoms

Total atoms = 1 + 4 = 5 atoms

A chemical equation is the symbolic representation of a chemical reaction using the symbols and formulae of the substances involved in the reaction.

An equation must be balanced in order to comply with the "Law of Conservation of Matter", which states that matter is neither created nor destroyed in the course of a chemical reaction. An unbalanced equation would imply that atoms have been created or destroyed.

The given equation tells us:

1. About the reactants involved and the products formed as a result of the reaction and their state. Zinc (Zn) and Hydrochloric acid (HCl) are the reactants. Zinc is in solid state and HCl is in aqueous solution state. The product Zinc Chloride (ZnCl₂) is also in aqueous solution state and Hydrogen is in gaseous state.
2. About the number of molecules of each substance taking part and formed in the reaction. Here one molecule of Zinc and two molecules of Hydrochloric acid react to give one molecule of Zinc chloride and one molecule of Hydrogen gas.
3. About chemical composition of respective molecules. For example, one molecule of Zinc chloride contains one atom of Zinc and two atoms of Chlorine.
4. About molecular mass; that 65 parts by weight of Zinc reacts with 73 parts by weight of Hydrochloric acid to produce 136 parts by weight of Zinc chloride and 2 parts by weight of Hydrogen.
   $\underset{65}{\text{Zn}}\text{(s)} + \underset{73}{\text{2HCl}}\text{(aq)} \longrightarrow \underset{136}{\text{ZnCl}_2}\text{(aq)} + \underset{2}{\text{H}_2}\uparrow$
5. 65 g of Zinc on treatment with 73 g of HCl, will produce 22.4 litres of hydrogen gas at S.T.P.
6. It also proves the law of conservation of mass. According to the above equation, 138 gram of reactants are producing 138 gram of products.

The equation of the given reaction does not tell us:

1. the time taken for the completion of the reaction.
2. whether heat is given out or absorbed during the reaction.
3. the respective concentrations of the reactants and the products.
4. the rate at which the reaction proceeds.
5. whether the reaction is completed or it is not completed.

The balanced chemical equations for the word equations are given below:

(a) C + O₂ ⟶ CO₂

(b) N₂ + O₂ ⟶ 2NO

(c) 3Ca + N₂ ⟶ Ca₃N₂

(d) CaO + CO₂ ⟶ CaCO₃

(e) Mg + H₂SO₄ ⟶ MgSO₄ + H₂↑

(f) 2Na + 2H₂O ⟶ 2NaOH + H₂↑

(a) 3Fe + 4H₂O ⟶ Fe₃O₄ + 4H₂

(b) 3Ca + N₂ ⟶ Ca₃N₂

(c) Zn + 2KOH ⟶ K₂ZnO₂ + H₂

(d) Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

(e) 3PbO + 2NH₃ ⟶ 3Pb + 3H₂O + N₂

(f) 2Pb₃O₄ ⟶ 6PbO + O₂

(g) 2PbS + 3O₂ ⟶ 2PbO + 2SO₂

(h) S + 2H₂SO₄ ⟶ 3SO₂ + 2H₂O

(i) S + 6HNO₃ ⟶ H₂SO₄ + 6NO₂ + 2H₂O

(j) MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

(k) C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂

(l) 2KOH + Cl₂ ⟶ KCl + KClO + H₂O

(m) 2NO₂ + H₂O ⟶ HNO₂ + HNO₃

(n) Pb₃O₄ + 8HCl ⟶ 3PbCl₂ + 4H₂O + Cl₂

(o) 2H₂O + 2Cl₂ ⟶ 4HCl + O₂

(p) 2NaHCO₃ ⟶ Na₂CO₃ + H₂O + CO₂

(q) 2HNO₃ + H₂S ⟶ 2NO₂ + 2H₂O + S

(r) P + 5HNO₃ ⟶ 5NO₂ + H₂O + H₃PO₄

(s) Zn + 4HNO₃ ⟶ Zn(NO₃)₂ + 2H₂O + 2NO₂

(a) NaCl + AgNO₃ ⟶ AgCl + NaNO₃

(b) The equation is balanced.

(c) Calculating the weights of reactants and products from the equation:

$\begin{matrix} \text{NaCl} & + & \text{AgNO}_3 & \longrightarrow & \text{AgCl} & + & \text{NaNO}_3 \\ 23 + 35.5 & + & 108 + 14 + 3(16) & & 108 + 35.5 & + & 23 + 14 + 3(16) \\ \Rightarrow 58.5 & + & 170 & & 143.5 & + & 85 \\ \Rightarrow & 228.5 & & & & 228.5 & \end{matrix}$

Thus, weights of reactants = weights of products = 228.5 g

(d) The equation satisfies the law of conservation of mass. It states that the total mass of the substances on either side of the equation is the same.

$\text{(a) }\underset{65}{\text{Zn}} + \underset{98}{\text{H}_2\text{SO}_4} \longrightarrow \underset{161}{\text{Zn}\text{SO}_4} + \underset{2}{\text{H}_2} \\[1em]$

The above balanced equation conveys the following information:

1. One molecule of zinc reacts with one molecule of sulphuric acid to produce one molecule of zinc Sulphate and one molecule of hydrogen.
2. 65 g of zinc reacts with 98 g of sulphuric acid to produce 161 g of zinc Sulphate and 2 g of hydrogen.
3. 65 g of zinc reacts with 98 g of sulphuric acid to produce 22.4 litres of hydrogen gas at S.T.P.
4. It also proves the law of conservation of mass. According to the above equation, 163 gram of reactants are producing 163 gram of products.

$\text{(b) }\underset{24}{\text{Mg}} + \underset{73}{\text{2HCl}} \longrightarrow \underset{95}{\text{MgCl}_2} + \underset{2}{\text{H}_2} \\[1em]$

The above balanced equation conveys the following information:

1. One molecule of magnesium reacts with two molecules of hydrochloric acid to produce one molecule of magnesium chloride and one molecule of hydrogen.
2. 24 g of magnesium reacts with 73 g of hydrochloric acid to produce 95 g of magnesium chloride and 2 g of hydrogen.
3. 24 g of magnesium reacts with 73 g of hydrochloric acid to produce 22.4 litres of hydrogen gas at S.T.P.
4. It also proves the law of conservation of mass. According to the above equation, 97 gram of reactants are producing 97 gram of products.

(a) Polyatomic ions are ions composed of two or more atoms that are covalently bonded together and carry a net electrical charge. Examples — Nitrate ion (NO₃^-) and Sulphate ion (SO₄^2-).

(b) Every equation must fulfill the "Law of Conservation of Matter". It states that matter is neither created nor destroyed in the course of a chemical reaction. Thus, the total mass of the substances on either side of the equation must be the same.

By looking at the chemical formula, we understand the ratio in which different atoms are united to form the molecule.

The molecular formula of a compound has quantitative significance. It represents:

1. both the molecule and the molecular mass of the compound.
2. the respective numbers of different atoms present in one molecule of the compound.
3. the ratios of the respective masses of the elements present in the compound.

For example, the formula CO₂ means that:

1. the molecular formula of carbon dioxide is CO₂
2. each molecule contains one carbon atom joined by chemical bonds with two oxygen atoms;
3. the molecular mass of carbon dioxide is 44, given that atomic mass of carbon is 12 and that of oxygen is 16.

A radical is an atom or a group of atoms of the same or of different elements that behaves as a single unit with a positive or a negative charge.

An Acid radical is the radical that remains after an acidic molecule loses a hydrogen ion (H⁺). Acid radicals typically have a negative charge. They are also called electronegative radicals or anions.

A Basic radical is the radical that remains after a base molecule loses a hydroxyl ion (OH^-). Basic radicals typically have a positive charge. They are also called electropositive radicals or cations.

For example, in the compound ammonium carbonate (NH₄)₂CO₃, ammonium (NH₄⁺) is a basic radical with combining power 1 and carbonate (CO₃^2-) is an acidic radical with combining power 2.

Berzelius

Reason — Berzelius suggested that the initial letter of an element written in capitals should represent that particular element. This method suggested by him laid the basis of the IUPAC system of chemical symbols and formulae.

one

Reason — Hydrogen carbonate radical is represented as HCO₃^-. Thus, it shows that one carbon atom is present in a hydrogen carbonate radical.

Fe₂(SO₄)₃

Reason — Chemical formula of iron (III) sulphate:

$\text{Fe}^{3+} \space \text{SO}_4^{2-} \\[1em] \overset{\phantom{22}{3}}{\text{Fe}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{2}}{\text{Fe}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{3}}{\text{SO}_4} \\[1em] \text{Fe}_2\text{(SO}_4)_3$

∴ Chemical formula of iron (III) sulphate is Fe₂(SO₄)₃

1 : 8

Reason — Hydrogen-to-oxygen mass ratio in water can be found as below:

Molecualr formula of water = H₂O

∴ Hydrogen-to-oxygen mass ratio in water

   = $\dfrac{2}{16}$ = $\dfrac{1}{8}$

   = 1 : 8

Ca(HCO₃)₂

Reason — Chemical formula of calcium hydrogen carbonate:

$\text{Ca}^{2+} \space \text{HCO}_3^{1-} \\[1em] \overset{\phantom{22}{2}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{HCO}_3} \Rightarrow \underset{\phantom{2}{1}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{HCO}_3} \\[1em] \text{Ca}\text{(HCO}_3)_2$

∴ Chemical formula of calcium hydrogen carbonate is Ca(HCO₃)₂

C, Ca, Cu, Cd

Reason — Carbon is denoted by the first letter of its name C, Calcium by the first two letters of its name Ca, Cadmium by first and third letters Cd and Copper is denoted by the first two letters of its latin name Cuprum so Cu.

Mercury

Reason — The term "Hydrargyrum" is derived from the Greek words "hydr-" meaning water and "argyros" meaning silver. In Latin, "Hydrargyrum" is used to refer to the element mercury, which is a silvery, liquid metal.

2 and 3

Reason — Given the formula Zn₃(PO₄)₂ :

$\text{Zn}_3(\text{PO}_4)_2 = \underset{\phantom{2}{3}}{\text{Zn}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{PO}_4} \Rightarrow \overset{\phantom{3}{2}}{\text{Zn}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{3}{\text{PO}_4} \\[0.5em]$

∴ Valency of Zinc is 2 and Phosphate is 3.

2

Reason — Chemical formula of calcium phosphate:

$\text{Ca}^{2+} \space \text{PO}_4^{3-} \\[1em] \overset{\phantom{22}{2}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \space \overset{3}{\text{PO}_4} \Rightarrow \underset{\phantom{2}{3}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{PO}_4} \\[1em] \text{Ca}_3\text{(PO}_4)_2$

∴ Chemical formula of calcium phosphate is Ca₃(PO₄)₂ hence x = 2.

4

Reason — Given the formula C₂H₄ :

$\text{C}_2\text{H}_4 = \underset{\phantom{2}{2}}{\text{C}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{4}{\text{H}} \Rightarrow \overset{\phantom{3}{4}}{\text{C}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{H}} \\[0.5em]$

∴ Valency of Carbon is 4.

Ca₃N₂

Reason — Calcium (Ca) forms a +2 ion (Ca⁺²) and nitride (N^-3) has a –3 charge.

$\text{Ca}^{2+} \space \text{N}^{3-} \\[1em] \overset{\phantom{22}{2}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \space \overset{3}{\text{N}} \Rightarrow \underset{\phantom{2}{3}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{N}} \\[1em] \text{Ca}_3\text{N}_2$

It is perfectly balanced. Therefore, Ca₃N₂ is correct.

Law of conservation of mass

Reason — Law of conservation of mass states that the total mass of the substances on either side of the equation is the same.

1/12th the mass of carbon atom (C-12)

Reason — Atomic mass is expressed in atomic mass units [a.m.u.] or "μ". Atomic mass unit is defined as 1/12 the mass of carbon atom C-12. (The mass of an atom of carbon-12 isotope was given the atomic mass of 12 units, i.e., 12 amu or simply 12 μ ).

Only Q

Reason — The valency of elements in compound magnesium nitride (Mg₃N₂) is,

$\text{Mg}_3\text{N}_2 = \underset{\phantom{2}{3}}{\text{Mg}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{N}} \Rightarrow \overset{\phantom{3}{2}}{\text{Mg}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{3}{\text{N}} \\[0.5em]$

Hence, Mg is divalent and N is trivalent.

Both P and R

Reason — Certain elements exhibit more than one valency and they show variable valency.

For example, copper shows valency of 1 and 2 and forms compounds like, cuprous [copper(I)] oxide and cupric [copper(II)] oxide, respectively.

Hence, both P and R are true.

Only Q

Reason — Given element 'a' is having atomic mass 32. The figure shows there are eight 'a' atoms forming a octatomic molecule. So when we calculate the molecular mass it will be

32 x 8 = 256

Hence, element 'a' is octatomic molecule with molecular mass 256.

A is true but R is false.

Explanation — The atomic mass of sodium is 23 amu. Hence, the assertion (A) is true.

An atom of sodium is 23 times heavier than one-twelfth the mass of a carbon-12 atom, not 23 times heavier than an atom of carbon with mass 12 amu. Hence, reason (R) is false.

A is true but R is false.

Explanation — An atom is the smallest particle of an element and can take part in a chemical reaction. Hence, the assertion (A) is true.

Atoms may or may not exist independently.
For example:

Noble gas atoms like He, Ne can exist freely.

But atoms like H, O, and N do not exist alone under normal conditions; they exist as diatomic molecules (H₂, O₂, N₂). Hence, reason (R) is false.

Both A and R are true and R is the correct explanation of A.

Explanation — An equation must be balanced in order to comply with the "Law of Conservation of Matter", which states that matter is neither created nor destroyed in the course of a chemical reaction. Hence, the assertion (A) is true.
An unbalanced equation would imply that atoms have been created or destroyed. Hence, both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Both A and R are true and R is the correct explanation of A.

Explanation — Phosphorus can show variable valency (like 3 and 5), so we name its compounds with prefixes like phosphorus trichloride (PCl₃).
Aluminium, however, shows a fixed valency of 3, so AlCl₃ is simply called aluminium chloride without needing a prefix. Hence, both assertion (A) and reason (R) are true, and reason (R) correctly explains assertion (A).

Both A and R are true but R is not the correct explanation of A.

Explanation — In a chemical reaction, the total mass of the substances on either side of the equation is the same, this is referred to be as law of conservation of mass. Hence, the assertion (A) is true.
In a chemical reaction the products are formed by the rearrangement of atoms or elements in the reactants. Hence, the reason (R) is true but reason (R) is not the correct explanation for assertion (A).

A is true but R is false.

Explanation — The molecular formula of Magnesium phosphate is Mg₃(PO₄)₂. Hence, the given assertion (A) is true.

Given the formula Mg₃(PO₄)₂ :

$\text{Mg}_3(\text{PO}_4)_2 = \underset{\phantom{2}{3}}{\text{Mg}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{PO}_4} \Rightarrow \overset{\phantom{3}{2}}{\text{Mg}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{3}{\text{PO}_4} \\[0.5em]$

∴ Valency of magnesium (Mg) is 2 and Phosphate is 3. Hence, the reason (R) is false.

Both A and R are true but R is not the correct explanation of A.

Explanation — A chemical equation represents molecules and it is necessary for the chemical equation to follow the law of conservation of mass. Hence, both assertion (A) and reason (R) are true.
However, balancing the equation is what ensures the law of conservation of mass. Therefore, reason (R) is not the correct explanation of assertion (A).

A is false but R is true.

Explanation — Chemical combination may take place between two or more elements. Hence, the assertion (A) is false.
A compound is a substance formed by the chemical combination of two or more elements in a fixed proportion. So, a single substance is formed in a chemical combination. Hence, the reason (R) is true.

(i) Molecular mass of Na₂SO₄.10H₂O
     = 2 x 23 + 32 + 4 x 16 + 10(2 x 1 + 16)
     = 46 + 32 + 64 + 180
     = 322 amu

(ii) Molecular mass of (NH₄)₂CO₃
     = 2(14 + 4 x 1) + 12 + 3 x 16
     = 36 + 12 + 48
     = 96 amu

(iii) Molecular mass of (NH₂)₂CO
     = 2(14 + 2 x 1) + 12 + 16
     = 32 + 12 + 16
     = 60 amu

(iv) Molecular mass of Mg₃N₂
     = 3 x 24 + 2 x 14
     = 72 + 28
     = 100 amu

(a) The relative molecular mass of CHCl₃
     = 12 + 1 + 3 x 35.5
     = 12 + 1 + 106.5
     = 119.5 amu

(b) The relative molecular mass of (NH₄)₂Cr₂O₇
     = 2(14 + 4 x 1) + 2 x 52 + 7 x 16
     = 36 + 104 + 112
     = 252 amu

(c) The relative molecular mass of CuSO₄.5H₂O
     = 63.5 + 32 + 4 x 16 + 5(2 x 1 + 16)
     = 63.5 + 32 + 64 + 90
     = 249.5 amu

(d) The relative molecular mass of (NH₄)₂SO₄
     = 2(14 + 4 x 1) + 32 + 4 x 16
     = 36 + 32 + 64
     = 132 amu

(e) The relative molecular mass of CH₃COONa
     = 12 + 3 + 12 + 16 + 16 + 23
     = 82 amu

(f) The relative molecular mass of KClO₃
     = 39 + 35.5 + 3 x 16
     = 39 + 35.5 + 48
     = 122.5 amu

(g) The relative molecular mass of (NH₄)₂PtCl₆
     = 2(14 + 4 x 1) + 195 + 6 x 35.5
     = 36 + 195 + 213
     = 444 amu

Relative molecular mass of MgSO₄.7H₂O
   = 24 + 32 + 4 x 16 + 7(2 x 1 + 16)
   = 24 + 32 + 64 + 126
   = 246 amu

246 g of Epsom salt contains 126 g of water of crystallisation

∴ 100 g of Epsom salt contains $\dfrac{126 \times 100}{246}$

= $\dfrac{12600}{246}$ = 51.22 g of water of crystallisation.

∴ % mass of water in Epsom salt, MgSO₄.7H₂O = 51.22%

(a) Relative molecular mass of Ca(H₂PO₄)₂
     = 40 + 2(2 x 1 + 31 + 4 x 16)
     = 40 + 2(2 + 31 + 64)
     = 40 + 194
     = 234 amu

Wt. of P in Ca(H₂PO₄)₂ = 2 x 31 = 62 g

% of P = $\dfrac{\text{Wt. of P}}{\text{Total Wt. of Ca(H}_2\text{PO}_4)_2}\times 100$

= $\dfrac{62}{234}$ x 100 = 26.5%

∴ Phosphorus in Calcium hydrogen phosphate is 26.5%

(b) Relative molecular mass of Calcium phosphate, Ca₃(PO₄)₂
     = 3 x 40 + 2(31 + 4 x 16)
     = 120 + 2(31 + 64)
     = 120 + 190
     = 310 amu

Wt. of P in Ca₃(PO₄)₂ = 2 x 31 = 62 g

% of P = $\dfrac{\text{Wt. of P}}{\text{Total Wt. of Ca}_3\text{(PO}_4)_2}\times 100$

= $\dfrac{62}{310}$ x 100 = 20%

∴ Phosphorus in Calcium phosphate is 20%

Relative molecular mass of KClO₃
   = 39 + 35.5 + 3 x 16
   = 39 + 35.5 + 48
   = 122.5 amu

122.5 g of KClO₃ contains 39 g of Potassium

∴ 100 g of KClO₃ contains $\dfrac{39 \times 100}{122.5}$ g of Potassium

= $\dfrac{39000}{1225}$ = 31.83 g of Potassium

122.5 g of KClO₃ contains 35.5 g of Chlorine

∴ 100 g of KClO₃ contains $\dfrac{35.5 \times 100}{122.5}$ g of Chlorine

= $\dfrac{35500}{1225}$ = 28.98 g of Chlorine

122.5 g of KClO₃ contains 48 g of Oxygen

∴ 100 g of KClO₃ contains $\dfrac{48 \times 100}{122.5}$ g of Oxygen

= $\dfrac{48000}{1225}$ = 39.18 g of Oxygen

∴ In KClO₃ : K = 31.83%, Cl = 28.98% and O = 39.18%

Relative molecular mass of CON₂H₄
   = 12 + 16 + 2 x 14 + 4 x 1
   = 12 + 16 + 28 + 4
   = 60 amu

Wt. of C in CON₂H₄ = 12 g

% of C = $\dfrac{\text{Wt. of C}}{\text{Total Wt. of CON}_2\text{H}_4}\times 100$

= $\dfrac{12}{60}$ x 100 = 20%

∴ Carbon in Urea is 20%

Relative molecular mass of Ferric oxide (Fe₂O₃)
  = 56 x 2 + 16 x 3
  = 112 + 48
  = 160 amu

Since 160 g of Ferric oxide (Fe₂O₃ ) contains 112 g of iron

∴ 100 g of contains Ferric oxide (Fe₂O₃ ) contains $\dfrac{112}{160}$ x 100 = 70%

∴ Percentage of Iron in Ferric oxide (Fe₂O₃ ) is 70%

Molecular formula of Urea is CON₂H₄

Relative molecular mass of CON₂H₄
   = 12 + 16 + 2 x 14 + 4 x 1
   = 12 + 16 + 28 + 4
   = 60 amu

Wt. of N in CON₂H₄ = 28 g

% of N = $\dfrac{\text{Wt. of N}}{\text{Total Wt. of CON}_2\text{H}_4}\times 100$

= $\dfrac{28}{60}$ x 100 = 46.67%

Wt. of N in 50 kg CON₂H₄

= $\dfrac{46.67}{100}$ x 50 = 23.33 kg

∴ Amount of Nitrogen in one bag (50 kg) of Urea is 23.33 kg.

The relative molecular mass of (NH₄)₂ PtCl₆
     = 2(14 + 4 x 1) + 195 + 6 x 35.5
     = 36 + 195 + 213
     = 444 amu

Wt. of Pt in [(NH₄)₂ PtCl₆] = 195 g

∴ Percentage of platinum in (NH₄)₂PtCl₆ is

    = $\dfrac{195}{444}$ x 100

    = 43.91 %

∴ Rounding to the nearest whole number, the percentage of platinum in ammonium chloride platinate (NH₄)₂ PtCl₆ is 44%.

Relative molecular mass of Calcium phosphate, Ca₃(PO₄)₂
     = 3 x 40 + 2(31 + 4 x 16)
     = 120 + 2(31 + 64)
     = 120 + 190
     = 310 amu

310 g of Ca₃(PO₄)₂ contains 120 g of Calcium

∴ 100 g of Ca₃(PO₄)₂ contains $\dfrac{120 \times 100}{310}$ g of Calcium

= $\dfrac{12000}{310}$ = 38.70 % of Calcium.

310 g of Ca₃(PO₄)₂ contains 62 g of Phosphorus

∴ 100 g of Ca₃(PO₄)₂ contains $\dfrac{62 \times 100}{310}$ g of Phosphorus

= $\dfrac{6200}{310}$ = 20 % of Phosphorus.

310 g of Ca₃(PO₄)₂ contains 128 g of Oxygen.

∴ 100 g of Ca₃(PO₄)₂ contains $\dfrac{128 \times 100}{310}$ g of Oxygen.

= $\dfrac{12800}{310}$ = 41.3 % of Oxygen.

∴ In Ca₃(PO₄)₂ : Ca = 38.70%, P = 20% and O = 41.3%

(a) Valency of fluorine in CaF₂ is -1.

(b) Valency of sulphur in SF₆ is +6.

(c) Valency of phosphorus in PH₃ is +3.

(d) Valency of carbon in CH₄ is +4.

(e) Valency of manganese in MnO₂ is +4.

(f) Valency of copper in Cu₂O is +1.

(g) Valency of magnesium in Mg₃N₂ is +2.

(h) Valency of nitrogen in the given compounds :

1. N₂O₃ = +3
2. N₂O₅ = +5
3. NO₂ = +4
4. NO = +2

An equation must be balanced in order to comply with the "Law of Conservation of Matter", which states that matter is neither created nor destroyed in the course of a chemical reaction. An unbalanced equation would imply that atoms have been created or destroyed.

For example, consider the following unbalanced equation:

KNO₃ ⟶ KNO₂ + O₂

In this equation, there are 3 oxygen atoms on the left side, but 4 oxygen atoms on the right side. This means that the equation is not balanced and does not satisfy the law of conservation of mass.

The balanced form of the equation is:

2KNO₃ ⟶ 2KNO₂ + O₂

Now there are 6 oxygen atoms on both the sides. This means that the law of conservation of mass is satisfied and the equation correctly represents the reaction.

Atomic mass unit is defined as 1⁄12 the mass of an carbon atom C-12.

A Basic radical is the radical that remains after a base molecule loses a hydroxyl ion (OH^-). Basic radicals typically have a positive charge. They are also called electropositive radicals or cations.

(a) Al₂O₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 3H₂O

(b) 8NH₃ + 3Cl₂ ⟶ 6NH₄Cl + N₂

(c) C₅H₁₂ (pentane) + 8O₂ ⟶ 5CO₂ + 6H₂O

(d) 2C₄H₁₀ (butane) + 13O₂ ⟶ 8CO₂ + 10H₂O

(e) 2FeSO₄ + H₂SO₄ + Cl₂ ⟶ Fe₂(SO₄)₃ + 2HCl

(f) 2NaCl + MnO₂ + 3H₂SO₄ ⟶ 2NaHSO₄ + MnSO₄ + 2H₂O + Cl₂

(g) Na₂Cr₂O₇ + 14HCl ⟶ 2NaCl + 2CrCl₃ + 7H₂O + 3Cl₂

(h) 2KMnO₄ + 3H₂SO₄ + 5H₂S ⟶ K₂SO₄ + 2MnSO₄ + 8H₂O + 5S

The atomic mass of calcium being 40 amu signifies that, on average, a calcium atom weighs 40 times the mass of one twelfth of a carbon-12 atom.

(a) 2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

(b) 2KHCO₃ + H₂SO₄ ⟶ K₂SO₄ + 2CO₂ + 2H₂O

(c) Fe + H₂SO₄ ⟶ FeSO₄ + H₂

(d) Cl₂ + SO₂ + 2H₂O ⟶ H₂SO₄ + 2HCl

(e) 2AgNO₃ ⟶ 2Ag + 2NO₂ + O₂

(f) 3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 2NO + 4H₂O

(g) 4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

(h) BaCl₂ + H₂SO₄ ⟶ BaSO₄ + 2HCl

(i) 2ZnS + 3O₂ ⟶ 2ZnO + 2SO₂

(j) Al₄C₃ + 12H₂O ⟶ 4Al(OH)₃ + 3CH₄

(k) 4FeS₂ + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂

(l) 2KMnO₄ + 16HCl ⟶ 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

(m) Al₂(SO₄)₃ + 8NaOH ⟶ 3Na₂SO₄ + 2NaAlO₂ + 4H₂O

(n) 2Al + 2NaOH + 2H₂O ⟶ 2NaAlO₂ + 3H₂

(o) 2K₂Cr₂O₇ + 8H₂SO₄ ⟶ 2K₂SO₄ + 2Cr₂(SO₄)₃ + 8H₂O + 3O₂

(p) K₂Cr₂O₇ + 14HCl ⟶ 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

(q) S + 6HNO₃ ⟶ H₂SO₄ + 6NO₂ + 2H₂O

(r) 2NaCl + MnO₂ + 3H₂SO₄ ⟶ 2NaHSO₄ + MnSO₄ + 2H₂O + Cl₂

(a) Valency of X is +3 and Y is -1. Formula of the compound formed between X and Y is :

$\text{X}^{3+} \space \text{Y}^{1-} \\[1em] \overset{\phantom{22}{3}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{Y}} \Rightarrow \underset{\phantom{2}{1}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \underset{{3}}{\text{Y}} \\[1em] \text{X}\text{Y}_3$

∴ Chemical formula of the compound formed between X and Y is XY₃

(b) Valency of X is +3 and Z is -2. Formula of the compound formed between X and Z is :

$\text{X}^{3+} \space \text{Z}^{2-} \\[1em] \overset{\phantom{22}{3}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{Z}} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \underset{{3}}{\text{Z}} \\[1em] \text{X}_2\text{Z}_3$

∴ Chemical formula of the compound formed between X and Z is X₂Z₃

(a) 6 Carbon and 12 Hydrogen atoms are present.

(b) Its molecular formula is C₆H₁₂.
As the simplest ration between C and H is 1:2, hence the empirical formula is CH₂

(c) Relative molecular mass of C₆H₁₂
   = (6 x 12) + (1 x 12)
   = 72 + 12
   = 84 g

84 g of C₆H₁₂ contains 72 g of Carbon

∴ 100 g of C₆H₁₂ contains $\dfrac{72 \times 100}{84}$ g of Carbon

= $\dfrac{7200}{84}$ = 85.7 g of Carbon.

84 g of C₆H₁₂ contains 12 g of Hydrogen

∴ 100 g of C₆H₁₂ contains $\dfrac{12 \times 100}{84}$ g of Hydrogen

= $\dfrac{1200}{84}$ = 14.3 of Hydrogen.

∴ In C₆H₁₂, C = 85.7% and H = 14.3%

(a) Dalton used symbol $\bigcirc$ for oxygen and ${\bigcirc}\mathllap{\bullet\phantom{\space}}$ for hydrogen.

(b) Symbol represents one atom of an element.

(c) Symbolic expression for a molecule is called molecular formula .

(d) Sodium chloride has two radicals. Sodium is a basic radical while chloride is a acidic radical.

(e) Valency of phosphorus in PCl₃ is 3 and in PCl₅ is 5 .

(f) Valency of Iron in FeCl₂ is 2 and in FeCl₃ it is 3 .

(g) Formula of iron (III) carbonate is Fe₂(CO₃)₃ .

The completed table is given below:

(a) Atoms in H₂SO₄ are

H ⟶ 2

S ⟶ 1

O ⟶ 4

(b) Atoms in HClO₄

H ⟶ 1

Cl ⟶ 1

O ⟶ 4

(c) Atoms in K₂Cr₂O₇

K ⟶ 2

Cr ⟶ 2

O ⟶ 7

(d) Atoms in KMnO₄

K ⟶ 1

Mn ⟶ 1

O ⟶ 4

(e) Atoms in K₄[Fe(CN)₆]

K ⟶ 4

Fe ⟶ 1

C ⟶ 6

N ⟶ 6

(f) Atoms in Na₂CrO₄

Na ⟶ 2

Cr ⟶ 1

O ⟶ 4

(g) Atoms in Mn₃(BO₄)₂

Mn ⟶ 3

B ⟶ 2

O ⟶ 8

(h) Atoms in HNO₂

H ⟶ 1

N ⟶ 1

O ⟶ 2

(a) A molecular formula represents the molecule of an element or of a compound.

(b) Molecular formula of water is H₂O.

(c) A molecule of sulphur is octatomic.

(d) CO represents Carbon Monoxide and Co represents cobalt.

(e) Formula of iron (III) oxide is Fe₂O₃.

1. Ratio of C and H atoms in Benzene (C₆H₆) = 6 : 6
   Simplest Ratio = 1 : 1
   ∴ Empirical formula of Benzene (C₆H₆) = CH
2. Ratio of C, H and O atoms in Glucose (C₆H₁₂O₆) = 6 : 12 : 6
   Simplest Ratio = 1 : 2 : 1
   ∴ Empirical formula of Glucose (C₆H₁₂O₆) = CH₂O
3. Ratio of C and H atoms in Acetylene (C₂H₂) = 2 : 2
   Simplest Ratio = 1 : 1
   ∴ Empirical formula of Acetylene (C₂H₂) = CH
4. Ratio of C, H and O atoms in Acetic acid (CH₃COOH) = 2 : 4 : 2
   Simplest Ratio = 1 : 2 : 1
   ∴ Empirical formula of Acetic acid (CH₃COOH) = CH₂O