Checking Mass Conservation in a Balanced Equation

Introduction

Balanced equations conserve each element, so they also conserve total mass. We can check this numerically by adding the relative masses of all reactant formula units and comparing them with the products. This is especially useful when a reaction includes a gas that might escape from an open container.

Core explanation

Relative formula mass, often written Mᵣ, is calculated by summing relative atomic masses for the atoms in a formula. For a simple calculation use H = 1, C = 12 and O = 16. H₂ has Mᵣ = 2, O₂ has Mᵣ = 32 and H₂O has Mᵣ = 18. In 2H₂ + O₂ → 2H₂O, the reactant total is 2 × 2 + 1 × 32 = 36 relative mass units. The product total is 2 × 18 = 36. The figures agree because hydrogen and oxygen atoms were conserved.

For calcium carbonate decomposition, CaCO₃ → CaO + CO₂, use Ca = 40, C = 12 and O = 16. CaCO₃ has Mᵣ = 40 + 12 + 48 = 100. CaO has Mᵣ = 56, and CO₂ has Mᵣ = 44. Thus 100 = 56 + 44. If the carbon dioxide escapes from an open apparatus, the remaining solid mass falls, but the mass of all products together still equals the starting mass in the ideal reaction.

Coefficients matter in the mass calculation. For 2Mg + O₂ → 2MgO, using Mg = 24 and O = 16, left total is 2 × 24 + 32 = 80, and right total is 2 × (24 + 16) = 80. Comparing Mᵣ(Mg) directly with Mᵣ(MgO) without the coefficients would mix one unit of each substance rather than the balanced reaction quantities.

A matching total mass is a useful check, but it does not replace an element-by-element audit. Two different mistakes could in principle cancel in a total mass sum, especially with rounded atomic masses. Formula errors are best caught by checking identities and each element count first; the mass comparison then illustrates the same conservation in numerical form.

These relative sums can also be read as gram amounts for the corresponding mole quantities. In the magnesium equation, 2 mol Mg has approximate mass 48 g and 1 mol O₂ has mass 32 g, yielding 2 mol MgO with mass 80 g. The exact values depend on the atomic masses chosen, but the equality follows from conservation.

Step-by-step reasoning

1. Confirm the chemical equation is balanced by counting every element. 2. Calculate the relative mass of each intact reactant and product formula. 3. Multiply each formula mass by its coefficient and sum each side. 4. Compare totals, inspect any discrepancy and distinguish total-system mass from the mass left in an open vessel.

Visual explanation

Imagine a scale with a left tray holding two 2-unit H₂ packets and one 32-unit O₂ packet. The right tray holds two 18-unit H₂O packets. Both trays total 36 units, although the packets have been reorganised.

Real-world analogy

Suppose a shipping box contains a 56-g tool and a 44-g accessory. Together they weigh 100 g, matching an original 100-g kit even if the accessory is shipped separately. Looking only at the remaining tool would misleadingly suggest 44 g vanished, like ignoring released CO₂ from heated calcium carbonate.

Real-world example

Heating limestone converts CaCO₃ into CaO and CO₂. The calculated 100:56:44 mass relationship is useful for understanding why the solid residue is lighter. It also shows why an open-vessel measurement must include the escaped gas to demonstrate total mass conservation directly.

Why?

Why does equal atom count imply equal total mass? Each element's atoms carry the same mass before and after an ordinary chemical reaction. Rearranging bonds changes substances but does not change the number or identity of those atoms, so summing their masses gives equal totals.

Common misconception

“If the solid loses mass, the reaction violates conservation.” The missing mass may be in a gas product. Include every reactant and product in the system boundary before judging whether total mass is conserved.

Worked example

Audit 2CO + O₂ → 2CO₂ using C = 12 and O = 16. CO has Mᵣ 28, O₂ has 32 and CO₂ has 44. Reactants total 2 × 28 + 32 = 88. Products total 2 × 44 = 88. Atom check also gives C 2 and O 4 on both sides, so the numerical mass agreement is expected.

Quick check

1. Why does CaCO₃ → CaO + CO₂ give 100 = 56 + 44 with the stated atomic masses? Answer: One formula unit of CaCO₃ has the same atoms as one CaO plus one CO₂; their relative masses therefore sum equally.

Exam focus

Show formula-mass calculations and multiply by coefficients. Use all products, including gases. Round atomic masses consistently and explain that a numerical mass check supports, but does not replace, the element-by-element audit.

Advanced insight

If an equation conserves each element, total mass conservation follows by multiplying each element's count equality by that element's atomic mass and adding the equalities. This is why a correctly balanced ordinary chemical equation must pass a mass audit for any consistent set of atomic masses, apart from negligible nuclear mass-energy effects outside normal chemistry.

Summary

Multiply each substance's relative formula mass by its balanced coefficient and add the terms on each side. Equal totals illustrate conservation of mass. Count atoms first, include escaping gases and keep the system boundary clear when comparing with a measured mass change.

Practice questions

1. Check 2H₂ + O₂ → 2H₂O by relative mass, using H = 1 and O = 16. Answer: Left 2 × 2 + 32 = 36; right 2 × 18 = 36. 2. For CaCO₃ → CaO + CO₂, what relative mass of CO₂ accompanies 56 units of CaO? Answer: 44 units, completing the 100-unit starting mass of CaCO₃ with the stated masses. 3. Why must the coefficient of MgO be included in a mass audit of 2Mg + O₂ → 2MgO? Answer: Two MgO formula units form, so their combined mass is twice the mass of one formula unit.