☰ Contents · Chemistry

The law of equivalents

Lessons 26 · 1 lessons · I.R. Asqarov, K. G‘opirov, N.X. To‘xtaboyev. Chemistry Grade 8, revised 4th edition. “Yangiyul Poligraph Service”, Tashkent, 2019
26

The law of equivalents

Textbook: pp. 111–115
GoalKnow the concepts of equivalent and equivalent mass and the law of equivalents, and use them in simple calculations.
New words
equivalent · ekvivalentequivalent mass · ekvivalent massalaw of equivalents · ekvivalentlar qonunivalency · valentlik
Explanation

Substances combine in fixed quantitative ratios. An equivalent is the amount that combines with, or takes the place of, 1 mol of hydrogen atoms (1 g). The mass of one equivalent is called the equivalent mass: E(H) = 1 g/mol, E(O) = 16 : 2 = 8 g/mol. For an element E = Aᵣ : valency, for example E(Mg) = 24 : 2 = 12, E(Al) = 27 : 3 = 9. For compounds: E(oxide) = M : (valency · number of atoms of the element), E(acid) = M : n(H), E(base) = M : n(OH), E(salt) = M : (valency of the metal · number of metal atoms). The law of equivalents: the masses of reacting substances are proportional to their equivalent masses, m₁ : m₂ = E₁ : E₂. Many elements show different valencies in different compounds, so their equivalent changes too (iron: 28 in FeCl₂, 56 : 3 ≈ 18.7 in FeCl₃).

Worked examples
12 g Mg combine with 8 g O (MgO). m(Mg) : m(O) = E(Mg) : 8, so E(Mg) = 12·8 : 8 = 12 — in agreement with 24 : 2.
E(H₂SO₄) = 98 : 2 = 49, E(NaOH) = 40 : 1 = 40. So 49 g of H₂SO₄ neutralize 40 g of NaOH; 98 g of H₂SO₄ need 80 g of NaOH.
Class activity

“Equivalent cards”: write 8 elements on cards (Na, Mg, Al, Ca, O, Cl, S(VI), C(IV)); write E on the back and check with a friend. Only calculate; no substances are used.

Practice
1
What are E(H) and E(O)?
2
The equivalent mass of Ca (valency II, Aᵣ = 40)?
3
The equivalent mass of Al₂O₃ (M = 102): M : (3·2) = ?
4
Why does the equivalent mass of iron differ in FeCl₂ and FeCl₃?