Get the equivalent weight of a substance as its molar mass divided by its n-factor. Enter a formula or the molar mass directly.
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From the article Equivalent Weight — The Concept Students Get Wrong.
Find the equivalent weight of H₂SO₄.
M = 2 × 1.008 + 32.06 + 4 × 15.999 = 2.016 + 32.06 + 63.996 = 98.072 g mol⁻¹
Both hydrogens are replaceable, so n-factor = 2.
E = 98.072 ÷ 2 = 49.04 g/eq
Consequence: a 1 M solution of H₂SO₄ is 2 N. Molarity and normality are not interchangeable words.
6.3 g of oxalic acid dihydrate, H₂C₂O₄·2H₂O, is dissolved to make 250 mL of solution. Find its normality.
M = (2 × 1.008) + (2 × 12.011) + (4 × 15.999) + 2 × 18.015
= 2.016 + 24.022 + 63.996 + 36.030 = 126.064 g mol⁻¹
Oxalic acid is dibasic, so n-factor = 2 and E = 126.064 ÷ 2 = 63.03 g/eq
Equivalents = 6.3 ÷ 63.03 = 0.09995 ≈ 0.100 eq
N = 0.100 ÷ 0.250 L = 0.40 N
The waters of crystallisation are part of the molar mass. Leaving them out gives E = 45.02 and a normality of 0.56 N — a 40 % error, and the classic reason a titration "does not match the book value".
Worked in full in Equivalent Weight — The Concept Students Get Wrong.
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