Samacheer Class 11 Chemistry - Basic Concepts of Chemistry and Chemical Calculations

(c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\)
Converting grams to moles calculator is as simple as multiplying the total mass by the moles per unit of mass.

(c) The ratio between the number of molecules in A to number of molecules in B is 2: 1
No. of moles of oxygen = 8 g/32 g = 0.25 moles of oxygen
No. of moles of sulphur dioxide = 8 g/64 g
= 0.125 moles of sulphur dioxide
Ratio between the no. of molecules = 0.25: 0.125 = 2: 1

(d) 1 moles of HCl (g)
H 2(g) + Cl 2(g) → 2HCl (g)
Content
CH 4
0 2
CO 2
Stoichiometric coefficient
1
1
2
No. of moles of reactants allowed to react at 273 K and 1 atm pressure
22.4 L (1 mol)
11.2 L (0.5 mol)
_
No. of moles of a reactant reacted and product formed
0.5
0.5
_
Amount of HCl formed = 1 mol

Relative atomic mass is defined as the ratio of the average atomic mass factor to the unified atomic mass unit.
Relative atomic mass (A r ) = \(\frac{\text { Average mass of the atom }}{\text { Unified atomic mass }}\)

The mole is defined as the amount of a substance which contains 6.023 x 10 23 particles such as atoms, molecules, or ions. It is represented by the symbol.

Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine. Equivalent mass has no unit but gram equivalent mass has the unit g eq-1.
Gram equivalent mass = \(\frac{\text { Molar mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}{\text {Equivalence factor }\left(\mathrm{eq} \mathrm{mol}^{-1}\right)}\)

Oxidation
Reduction
1. Reactions involving the addition of oxygen
Reactions involving removal of hydrogen
2. Reactions involving loss of an electron
Reactions involving gain of electron
3. Reaction in which oxidation number of the element increases.
Reaction in which oxidation number of the element decreases.

Oxidation:
According to the classical concept, oxidation is a process of addition of oxygen or removal of hydrogen.
Removal of hydrogen
2H 2 S + O 2 → H 2 O + 2S
Addition of oxygen
C + O 2 → CO 2
According to the electronic concept, loss of electrons is called oxidation reaction.
Ca → Ca 2+ + 2e –
During oxidation, oxidation number increases.
Dining oxidation, reducing agent gets oxidised.
Reduction:
Reduction is a process of removal of oxygen or addition of hydrogen.
Addition of hydrogen
Ca + H 2 → CaH 2
Removal of oxygen
Zn O + C → Zn + CO
According to the electronic concept, gain of electrons is called a reduction reaction.
Zn 2+ + 2e – → Zn
During reduction, oxidation number decreases.
During reduction, oxidising agent gets reduced.

(i) urea [CO(NH 2 ) 2 ]:
Molar mass of urea = (4 × Atomic mass of hydrogen) + ( 1 × Atomic mass of carbon) + ( 2 × Atomic mass of nitrogen) + ( 1 × Atomic mass of oxygen)
= (4 × 1.008) +(1 × 12) + (2 × 14)+ ( 1 × 16)
= 4.032 + 12 + 28 + 16 = 60.032 g mol -1
(ii) acetone[CH 3 COCH 3 ]
Molar mass of Acetone = (6 × Atomic mass of hydrogen) + ( 3 × Atomic mass of carbon) + (1 × Atomic mass of oxygen) = (6 × 1.008) + (3 × 12) + ( 1 × 16)
= 6.048 + 36 + 16 = 52.024 g mol -1.
(iii) boric acid [H 3 BO 3 ]:
Molar mass of Boric acid = (3 × Atomic mass of hydrogen) + ( 1 × Atomic mass of boron) + ( 3 × Atomic mass of oxygen)
= (3 × 1.008) + (3 × 11) + ( 1 × 16)
= 3.024 + 33 + 16 = 52.024 g mol -1.
(iv) sulphuric acid [H 2 SO 4 ] = (2 × Atomic mass of hydrogen) + ( 1 × Atomic mass of sulphur) + ( 4 × Atomic mass of oxygen)
= (2 × 1.008) +(1 × 32) + (4 × 16)
= 2.016 + 32 +64 = 98.016 g mol -1.

Given:
The density of C0 2 at 273 K and 1 atm pressure = 1.965 kgm -3
Molar mass of CO 2 =?
At 273 K and 1 atm pressure, 1 mole of CO 2 occupies a volume of 22.4 L
Mass of 1 mole of CO 2 = \(\frac{1.965 \mathrm{Kg}}{1 \mathrm{~m}^{3}}\) × 22.4 L
= \(\frac{1.965 \times 10^{3} \mathrm{~g} \times 22.4 \times 10^{-3} \mathrm{~m}^{3}}{1 \mathrm{~m}^{3}}\)
= 44.01 g
Molar mass of CO 2 = 44 gmol -1

Compound
Given no. of moles
No. of oxygen atoms
Ethanol – C 2 H 5 OH
1
1 × 6.022 × 10 23
Formic acid -HCOOH
1
2 × 6.022 × 10 23
Water – H 2 O
1
1 × 6.022 × 10 23
Formic acid

there in 0.320 kg
Given:
mass of 1 atom = 6.645 × 10 -23 g
∴ mass of 1 mole of atom = 6.645 × 10 -23 23 g × 6.022 × 10 23 = 40 g
∴ Number of moles of element in 0.320 kg = \(\frac{1 \mathrm{~mol}}{40 \mathrm{~g}}\) × 0.320 kg
= \(\frac{1 \mathrm{~mol} \times 320 \mathrm{~g}}{40 \mathrm{~g}}\) = 8 mol

Molecular mass:
* Relative molecular mass is defined as the ratio of the mass of the molecule to the unified atomic mass unit.
* It can be calculated by adding the relative atomic masses of its constituent atoms.
* For carbon monoxide (CO) Molecular mass = Atomic mass of carbon + Atomic mass of oxygen 12 + 16 = 28 u.
Molar mass:
* It is defined as the mass of one mole of a substance.
* The molar mass of a compound is equal to the sum of the relative atomic masses of its constituent expressed in g mol- 1.
* For carbon monoxide (CO) 12 + 16 = 28 g mol -1 Both molecular mass and molar mass are numerically the same but the units are different.

Compound
Molecular formula
Empirical formula
Fructose
C 6 H 12 O 6
CH 2 O
Caffeine
C 8 H 10 N 4 O 2
C 4 H 5 N 2 O

Given:
2Al + Fe 2 O 3 → Al 2 O 3 + 2Fe
Molar mass of Al 2 O 3 formed = 6mol × 102 g mol -1 = 612 g
[Al 2 O 3 = (2 × 27) + (3 × 16) = 54 + 48 = 102]
Excess reagent = Fe 2 O 3.
Amount of excess reagent left at the end of the reaction = 1 mol × 160 g mol -1 = 160 g
[Fe 2 O 3 = (2 × 56) + (3 × 16) = 112 + 48 = 160 g] = 160 g

C 2 H 6 + \(\frac{7}{2}\)O 2 → 2CO 2 + 3H 2 O
⇒ 2C 2 H 6 + 7O 2 → 4CO 2 + 6H 2 O
∴ To produce 4 moles of CO 2, 2 moles of ethane is required
To produce 1 mole (44g) of CO 2 required number of moles of ethane
= \(\frac{1}{2}\) mole of ethane
= 0.5 mole of ethane

Empirical formula = C 6 H 6 O
η = Molar mass / Calculated empirical formula mass = \(\frac{2 \times \text { vapour density }}{94}\)
= \(\frac{2 \times 47}{94}\) = 1
∴ Molecular formula(C 6 H 6 O) × 1 = C 6 H 6 O

∴ Empirical formula = Na 2 SH 20 O 14
η = Molar mass / Calculated empirical formula mass
= \(\frac{322}{322}\) = 1
[Na 2 SH 20 O 14 = (2 × 23) + (1 × 32) + (20 × 1) + (14 × 16)
= 46 + 32 + 20 + 224 = 322]
Molecular formula = Na 2 SH 20 O 14
Since all the hydrogen in the compound present as water
∴ Molecular formula is Na 2 SH 20 O 14

K 2 Cr 2 O 7 + 6KI + H 2 SO 4 → K 2 SO 4 + Cr 2 (SO 4 ) 3 + I 2 + H 2 O
K 2 Cr 2 O 7 + 6KI + H 2 SO 4 → K 2 SO 4 + Cr 2 (SO 4 ) 3 + 3I 2 + H 2 O
K 2 Cr 2 O 7 + 6KI + 7H 2 SO 4 → 4K 2 SO 4 + Cr 2 (SO 4 ) + I 2 + 7H 2 O
(ii) KMnO 4 + Na 2 SO 3 → MnO 2 + Na 2 SO 4 + KOH
⇒ 2KMnO 4 + 3Na 2 SO 3 → MnO 2 + Na 2 SO 4 + KOH
⇒ 2KMnO 4 + 3Na 2 SO 3 → 2MnO 2 + 3Na 2 SO 4 + KOH
⇒ 2KMnO 4 + 3Na 2 SO 3 → MnO 2 + Na 2 SO 4 + 2KOH
(iii) Cu + HNO 3 → Cu(NO 3 ) 2 + NO 2 + H 2 O
Cu + 2HNO 3 → Cu(NO 3 ) 2 + NO 2 + H 2 O
Cu + 2HNO 3 + 2HNO 3 → Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O
Cu + 4HNO 3 → Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O
(iv) KMnO 4 + H 2 C 2 O 4 + H 2 SO 4 → K 2 SO + MnSO 4 + CO 2 + H 2 O
KMnO 4 + 5H 2 C 2 O 4 + H 2 SO 4 → K 2 SO + MnSO 4 + CO 2 + H 2 O
2KMnO 4 + 5H 2 C 2 O 4 + H 2 SO 4 → 2MnSO 4 + 10CO 2 + H 2 O
2KMnO 4 + 5H 2 C 2 O 4 + 3H 2 SO 4 → K 2 SO + 2MnSO 4 + 10CO 2 + 8H 2 O

(ii) C 2 O 4 2- + Cr 2 O 7 2- →
Cr 3+ + CO 2 (in acid medium)
(iii) Na 2 S 2 O 3 + I 2 → Na 2 S 4 O 6 + NaI 2 (in acid medium)
S 2 O 3 2- → S 4 O 6 2- ………….(1)
Half reaction ⇒ I 2 → I – …………….(2)
(iv) Zn + NO 3 – → Zn +2 + NO
Half reactions are