☰ Contents · Chemistry (General chemistry)

Equivalent and the law of equivalents

Lessons 7 · 1 lessons · S. Masharipov, A. Mutalibov, E. Murodov, H. Islomova. General chemistry, Grade 11, 1st edition. G‘afur G‘ulom Publishing and Printing House, Tashkent, 2018
7

Equivalent

Textbook: pp. 39–45
GoalExplain equivalent and equivalent mass, calculate the equivalent mass of elements, oxides, acids, bases and salts, and solve problems with the law of equivalents.
New words
equivalent · ekvivalentequivalent mass · ekvivalent massavalence · valentliklaw of equivalents · ekvivalentlar qonuni
Explanation

Substances react with one another in equal-valued — equivalent — amounts. The equivalent mass of an element (the book says «equivalent weight») is the mass of the element that combines with 1 part by mass of hydrogen or 8 parts by mass of oxygen: E = A/V, where A is the atomic mass and V the valence; hence E(H) = 1, E(O) = 8, E(Cl) = 35.5. If one element shows different valences in different compounds, its equivalent mass differs too (sulfur: 32/4 = 8 in SO₂ and 32/6 ≈ 5.33 in SO₃). For compounds: oxide E = M/(n·V) (n is the element’s subscript, V its valence); acid E = M/n(H) (the number of H replaceable by a metal); base E = M/n(OH); salt E = M/(n·V) (by the metal); ion E = M/z. Law of equivalents: the masses of reacting substances are proportional to their equivalent masses, m₁/m₂ = E₁/E₂. For gases an equivalent volume is also used: under normal conditions it is 11.2 L for hydrogen and 5.6 L for oxygen, since 1 g of H₂ and 8 g of O₂ occupy these volumes. Equivalent mass is expressed in g/mol (the book writes g/equiv). Note: «equivalent» and «equivalent weight» are older terms; current international (IUPAC) practice uses the mole and the reaction equation instead, but they still appear in school and test problems.

Worked examples
Equivalent mass of Al(OH)₃: M = 78 g/mol, 3 OH groups, E = 78/3 = 26. For the salt Na₂CO₃: M = 106, metal subscript 2, valence 1, E = 106/(2·1) = 53.
6 g of a metal combined with oxygen to give 10 g of oxide. The oxygen mass is 10 – 6 = 4 g. From m(Me)/m(O) = E(Me)/8, E(Me) = 6·8/4 = 12 (if divalent, A = 24 — magnesium).
Class activity

Filling a table: the class works in groups to calculate the equivalent masses of the acids HCl, H₂SO₄, H₃PO₄ and the bases NaOH, Ca(OH)₂ and enters the results in a common table on the board.

Practice
1
Find the equivalent mass of the acid H₂S (M = 34 g/mol).
2
What is the equivalent mass of the salt K₂SO₄ (M = 174 g/mol)?
3
18 g of a metal reacted with acid and released 22.4 L (n.c.) of hydrogen. Find the equivalent mass of the metal (equivalent volume of hydrogen is 11.2 L).
4
Why is the equivalent mass of sulfur different in SO₂ and SO₃?