Exercises — Law of conservation of mass (Lavoisier) — proof, examples

WHAT we know: sealed flask ⇒ closed system ⇒ ${\text{mass}}_{\text{reactants}}={\text{mass}}_{\text{products}}$. WHY we just add: every hydrogen and oxygen atom ends up inside a water molecule; none escapes. $m_{\text{water}}=m_{{\text{H}}_2}+m_{{\text{O}}_2}=4.0+32.0=\boxed{36.0\text{ g}}$

The masses are equal, so as far as the balance can tell, nothing left or entered — behaviour of a closed system. Equal masses = conservation is visibly obeyed. (Strictly, an open system could coincidentally show equal masses if equal masses of gas entered and left, but for a single simple reaction, equal in = equal out points to a closed system.)

WHAT: oxygen is the only other reactant, so total reactant mass $=24.0+m_{{\text{O}}_2}$. WHY: closed chamber ⇒ $M_R=M_P$. Rearrange to isolate the unknown: $m_{{\text{O}}_2}=m_{\text{MgO}}-m_{\text{Mg}}=40.0-24.0=\boxed{16.0\text{ g}}$ WHAT IT LOOKS LIKE: see the mass-bar figure below — the oxygen bar is exactly the gap between the metal you started with and the oxide you ended with.

WHAT: stoppered flask = closed system. Nothing enters or leaves. WHY: forming a solid inside the flask only rearranges atoms; it does not change total mass. $m_{\text{after}}=m_{\text{before}}=\boxed{210.5\text{ g}}$ The solid looks like "new stuff appeared", but its atoms came from the dissolved reactants.

(a) The $\uparrow$ symbol means a gas leaves — this is an open system, so the balance reading for the crucible is allowed to drop. No law is violated; we just stopped weighing part of the products. (b) The escaped ${\text{CO}}_2$ is the missing mass: $m_{{\text{CO}}_2}=m_{{\text{CaCO}}_3}-m_{\text{CaO}}=20.0-11.2=\boxed{8.8\text{ g}}$ Check: solid left $+$ gas gone $=11.2+8.8=20.0\text{ g}$ — the full original mass is still accounted for once we count the invisible gas.

WHAT is odd: in Exercise 3.1 an open system got lighter; here it got heavier. Both are fine — an open system's mass can move either way depending on whether gas leaves or joins. WHY heavier: oxygen atoms from the air joined the iron. That oxygen was never on the balance before, so its mass is added: $m_{{\text{O}}_2}=m_{\text{rust}}-m_{\text{Fe}}=80.0-56.0=\boxed{24.0\text{ g}}$ The universe's mass is still conserved; only the lump's mass rose because matter entered the (open) system.

(a) Sealed vessel, closed system: $m_{{\text{CO}}_2}=m_{\text{C}}+m_{{\text{O}}_2}=12.0+32.0=44.0\text{ g}$ (b) Second reaction is also mass-conserving. Reactants are the $44.0\text{ g}$ of ${\text{CO}}_2$ plus $12.0\text{ g}$ of fresh carbon: $m_{\text{CO}}=m_{{\text{CO}}_2}+m_{\text{C}}=44.0+12.0=\boxed{56.0\text{ g}}$ WHY it's just addition again: every atom of the ${\text{CO}}_2$ and every atom of the added carbon ends up in $\text{CO}$; nothing leaves during the reaction. Two conservation statements chained together.

WHAT disappeared: only the visible solid wax — $2.0\text{ g}$ of it turned into invisible gas and vapour, but that gas is still inside the sealed jar. WHY the total is unchanged: the jar is closed. Atoms left the solid wax and joined gas molecules, all trapped: $m_{\text{jar,after}}=m_{\text{jar,before}}=\boxed{305.0\text{ g}}$ The "$2.0\text{ g}$ lost" is a red herring: it left the solid, not the jar.

(a) The classical conservation-of-mass law is a chemistry law: it assumes nuclei stay intact and only bonds rearrange. In a nuclear reaction, nuclei themselves change, and a tiny mass defect converts to energy. Mass alone is not conserved here — but total mass-energy is. See Mass–Energy Equivalence (E=mc²). (b) The energy comes from Einstein's relation. Here $E=m{c}^2$ answers the question "how much energy hides in this lost mass?" — that is exactly what $m{c}^2$ measures. $E=m{c}^2=(7.2\times 10^{-6})\times (3.00\times 10^8{)}^2$ $E=7.2\times 10^{-6}\times 9.00\times 10^{16}=\boxed{6.48\times 10^{11}\text{J}}$ That enormous number from a mere $7.2$ milligrams is why the mass change is undetectable in ordinary chemistry: chemical energy changes are billions of times smaller, so the corresponding mass change is far below any balance's precision.

(a) No reaction = the ultimate degenerate case: $N_i^{\text{before}}=N_i^{\text{after}}$ trivially, so mass is unchanged. Conservation holds even when nothing happens — it is the null case. (b) Chemically, mass is unchanged: the atoms are all still inside, only their bonds rearranged. Strictly, releasing energy $E$ implies a mass loss of $E/c^2$ (by Mass–Energy Equivalence (E=mc²)) — but if the heat stays trapped in the box, even that energy stays inside, so the box's mass does not change at all. For any energy that escaped, the mass drop would be $E/c^2$, which for chemical heats is around $10^{-10}\text{ g}$ — utterly immeasurable. So to every balance ever built: mass is conserved.

Ask these in order for any conservation problem: Is the system open or closed? ::: Closed → masses balance exactly. Open → account for gas entering or leaving. If open and mass dropped, where did it go? ::: Out as escaping gas (e.g. CO₂↑). If open and mass rose, where did it come from? ::: In from the surroundings (e.g. O₂ absorbed during rusting). When does mass genuinely change? ::: Only in nuclear reactions, via the mass defect and E=mc². Why can't ordinary chemistry show a mass change? ::: The energy change is so small the mass shift (E/c²) is below any balance's precision.