Q1.
A student burns 10 g of ethanol in an open beaker. After the reaction, no residue is left in the beaker. Does this mean the Law of Conservation of Mass is violated? Explain.
Answer
No, the law is not violated. Nothing is left in the beaker because every product of the combustion is a gas, and in an open beaker the gases float away. The matter has moved, not disappeared.
Ethanol burns in the oxygen of the air:
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
Now do the mass accounting. Take H = 1 u, C = 12 u, O = 16 u.
Molecular mass of ethanol C₂H₅OH = (12 × 2) + (1 × 6) + 16 = 46 u
Oxygen used: 3 O₂ = 3 × 32 u = 96 u, so for 10 g of ethanol
Mass of O₂ = 10 g × 96 ÷ 46 = 20.87 g
Carbon dioxide formed: 2 CO₂ = 88 u → 10 g × 88 ÷ 46 = 19.13 g
Water formed: 3 H₂O = 54 u → 10 g × 54 ÷ 46 = 11.74 g
Oxygen used: 3 O₂ = 3 × 32 u = 96 u, so for 10 g of ethanol
Mass of O₂ = 10 g × 96 ÷ 46 = 20.87 g
Carbon dioxide formed: 2 CO₂ = 88 u → 10 g × 88 ÷ 46 = 19.13 g
Water formed: 3 H₂O = 54 u → 10 g × 54 ÷ 46 = 11.74 g
Total mass of reactants = 10 g + 20.87 g = 30.87 g
Total mass of products = 19.13 g + 11.74 g = 30.87 g
Mass of reactants = Mass of products ✓
Total mass of products = 19.13 g + 11.74 g = 30.87 g
Mass of reactants = Mass of products ✓
So the balance would have shown a fall — but only because 30.87 g of gas has drifted out of the beaker while 20.87 g of oxygen quietly came in from the air. Neither of those crossings was weighed.
Why it happens: a conservation law is a statement about a closed system. Burning in an open beaker is about as open as a system gets — matter enters as oxygen and leaves as carbon dioxide and steam. To test the law you must seal the system, exactly as set-up 2 of Activity 9.2 does.
Check it yourself: burn a candle inside a closed jar standing on a balance. The reading stays constant while the flame burns, even though wax is disappearing — because the CO₂ and water vapour are trapped inside.