Equivalent Mass of a Metal: Hydrogen Displacement and Oxide Methods
In Class 11 stoichiometry and in many practical books, you are asked to find the equivalent mass of a metal from an experiment. The data is usually given as the mass of a metal and either the volume of hydrogen it displaces or the mass of its oxide. Once you know the idea, every problem is the same short calculation. This article covers the three classical methods and the correct way to handle the gas data.
What is equivalent mass of a metal?
The equivalent mass of a metal is the number of grams of the metal that combine with, or displace, 1.008 g of hydrogen, 8 g of oxygen or 35.5 g of chlorine. These three numbers come from the atomic masses (H = 1.008, O = 16, Cl = 35.5): one equivalent of oxygen is 16 / 2 = 8 g because oxygen has valency 2.
For a known metal, equivalent mass = atomic mass / valency (n-factor). For example, magnesium: 24.305 / 2 = 12.15 g per equivalent. The experimental methods below find this number from measurements.
Oxide formation: E = (mass of metal × 8) / (mass of oxygen combined)
Chloride formation: E = (mass of metal × 35.5) / (mass of chlorine combined)
The number 11200 is the volume in mL occupied at STP (using 22.4 L/mol) by 1.008 g of hydrogen, which is half a mole of H2 gas: 22400 / 2 = 11200 mL. Some books use 22.7 L/mol for STP defined as 1 bar. Then the number is 11350 mL. Check which value your textbook uses, and use the same one throughout.
Worked examples
Example 1: hydrogen displacement at STP. 0.12 g of a metal displaces 112 mL of hydrogen at STP from dilute acid. Find the equivalent mass.
E = (0.12 × 11200) / 112 = 1344 / 112 = 12.0 g/equiv.
Cross-check: the metal is magnesium, whose true value is 24.305 / 2 = 12.15. The result is within experimental error.
Example 2: oxide method. 0.48 g of a metal gives 0.80 g of its oxide on heating in air. Find the equivalent mass.
Mass of oxygen combined = 0.80 − 0.48 = 0.32 g.
E = (0.48 × 8) / 0.32 = 3.84 / 0.32 = 12.0 g/equiv.
This is again magnesium. Both methods give the same answer for the same metal, which is the law of equivalent proportions at work.
Example 3: hydrogen collected over water. 0.300 g of a metal reacts with dilute acid and gives 140 mL of hydrogen collected over water at 27 °C. The atmospheric pressure is 760 mm Hg, and the aqueous tension at 27 °C is taken as 27 mm Hg. Find the equivalent mass.
Step 1: pressure of dry hydrogen = 760 − 27 = 733 mm Hg.
Step 2: use P1V1/T1 = P2V2/T2 to go to STP (760 mm Hg, 273.15 K). T1 = 300.15 K.
V2 = (733 × 140 × 273.15) / (760 × 300.15) = 122.9 mL.
Step 3: E = (0.300 × 11200) / 122.9 = 27.3 g/equiv.
Check by moles: moles of H2 from pV = nRT: n = (733/760 × 0.140) / (0.08206 × 300.15) = 5.48 × 10−3 mol, mass of H2 = 0.01105 g. E = 0.300 / 0.01105 × 1.008 = 27.4. The two routes agree to the accuracy of the data. This value is close to 55.85 / 2 = 27.9, so the metal is consistent with iron giving Fe2+.
If you forget the aqueous-tension correction, you use 760 instead of 733 and the answer is about 3.7 % off.
Example 4: from E to atomic mass. A metal has equivalent mass 9.0 and forms an oxide with valency 3. Find its atomic mass.
Atomic mass = E × valency = 9.0 × 3 = 27.0. This is aluminium (the true value is 26.98).
Why the answers are the same by every route
Equivalent mass is just the mass that supplies one mole of electrons in a reaction. One mole of H+ picks up one mole of electrons to give half a mole of H2. One mole of oxygen atoms picks up two moles of electrons, so 8 g (half a mole of O atoms) takes one mole of electrons. So when you equate a metal with 1.008 g of hydrogen or 8 g of oxygen, you equate the same amount of electron transfer. That is why the hydrogen method, the oxide method and the chloride method all agree.
Common mistakes
- Using 22400 instead of 11200: that gives the mass per mole of hydrogen molecules and so doubles the answer.
- Forgetting to subtract aqueous tension when the gas is collected over water.
- Using the volume at room temperature directly without converting to STP.
- Using the mass of the oxide in place of the mass of oxygen. Always subtract the mass of the metal first.
- Giving atomic mass when the question asks for equivalent mass, or the reverse. The valency links them.
- Using 273 K in one line and 273.15 K in the next. Pick one value and keep it, and state it.
Quick summary
| Method | You measure | Equivalent mass |
|---|---|---|
| Hydrogen displacement | Mass of metal, V of H2 at STP (mL) | mass × 11200 / V |
| Oxide | Mass of metal, mass of oxide | mass of metal × 8 / (oxide − metal) |
| Chloride | Mass of metal, mass of chloride | mass of metal × 35.5 / (chloride − metal) |
| Known metal | Atomic mass and valency | atomic mass / valency |
Exam patterns for this topic differ between boards and between years, so follow your own syllabus and last year's papers.
Check atomic mass divided by valency for a known metal, then compare it with your experimental value.
Open the Equivalent Weight Calculator →Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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