4.3.2.3,4.3.2.4 - Balancing from masses and limiting reactants

Converting masses into simple mole ratios (Higher)

Masses alone cannot give the balancing numbers because different substances have different molar masses.

Illustrative complete-reaction data: 13.0 g zinc and 3.20 g oxygen form 16.20 g zinc oxide. The mass ratio is 13.0:3.20:16.20, while the coefficient ratio relates particles rather than grams. Use $A_r(\ce{Zn})=65$ and $A_r(\ce{O})=16$, giving molar masses 65, 32 and $81\,\mathrm{g\,mol}^{-1}$, respectively.

Converting each mass separately gives:

Calculate each amount using $n=m/M$, then divide every amount by the smallest:

The equation is $\ce{2Zn + O2 -> 2ZnO}$. Dividing all amounts by the same value preserves their ratio while revealing small whole numbers.

Illustrative lithium chloride data: 2.10 g lithium reacts with 10.65 g chlorine to form 12.75 g lithium chloride. Use molar masses 7, 71 and $42.5\,\mathrm{g\,mol}^{-1}$:

The same coefficient ratio can arise from very different mass ratios because each substance has a different molar mass.

If a simplified ratio is 1:1.5:1, multiply every entry by 2 to obtain 2:3:2. Do not round 1.5 to 2: that changes the ratio. Experimental values may be slightly away from a whole-number ratio because measurements have uncertainty. Use precision and any supplied uncertainty to judge a plausible simple ratio, then check atom counts. Do not force poor or incomplete data to fit a desired equation.

Finding balancing coefficients from data (Higher)

Once the chemical formulae are known, the simplest mole ratio gives the coefficients placed in front of them.

Illustrative experiment: 5.40 g aluminium and 4.80 g oxygen form 10.2 g aluminium oxide. The formulae are $\ce{Al}$, $\ce{O2}$ and $\ce{Al2O3}$; use $A_r(\ce{Al})=27$ and $A_r(\ce{O})=16$. The amounts are:

Dividing by 0.100 gives 2:1.5:1. Multiplying the entire ratio by 2 gives 4:3:2, so:

Check four aluminium atoms and six oxygen atoms on each side. The formula $\ce{Al2O3}$ stays fixed; the measured data determine the number of its formula units, not a new identity for aluminium oxide.

Illustrative calcium chloride data: 2.00 g calcium and 3.55 g chlorine form 5.55 g calcium chloride. With molar masses 40, 71 and $111\,\mathrm{g\,mol}^{-1}$:

One calcium atom and two chlorine atoms appear on each side. A coefficient of one is left unwritten; it still counts in the ratio.

The stated masses must refer to quantities that actually react or are formed. An initial reactant mass can include an unreacted excess, which cannot be used directly as the reacted amount. If a collected gas escaped, its measured mass may also understate its production. The assumption of a complete, correctly measured reaction is essential to the method.

Equations can be rearranged to recover missing data. From $n=m/M$, multiplying by $M$ gives $m=nM$; dividing that expression by $n$ gives $M=m/n$. Keep units attached to each physical quantity: grams divided by grams per mole gives moles. If a product mass is inferred by conservation, include all products and any gases.

Limiting and excess reactants (Higher)

A reaction stops producing its intended product when one required reactant has been used up, even if another reactant remains.

The limiting reactant is completely consumed and limits the amount of product. A reactant in excess is supplied in more than the amount needed to react with the limiting reactant. Deliberately using an excess can help ensure that all of the other reactant reacts.