Reacting masses, limiting reagent and yield

The three-step method

Every reacting-mass question follows the same route, and writing the three steps out is what stops you getting lost:

1. convert what the question gives into moles;
2. use the ratio in the balanced equation to get moles of the other substance;
3. convert those moles into what the question asks for.

The equation ratio applies to moles only. Applying it to masses is the commonest structural error in the chapter, and it gives a wrong answer that looks entirely reasonable.

Balance the equation first

Reading the ratio off an unbalanced equation produces a wrong ratio, and everything after it inherits the error. For

the ratio $\ce{Mg}:\ce{MgO}$ is $2:2$, that is $1:1$. From the unbalanced version it would look like $1:1$ as well — but for $\ce{Mg}:\ce{O2}$ it is $2:1$, not $1:1$.

Limiting reagent

When the question gives amounts of two reactants, one of them runs out first and decides how much product forms. To find it: convert both to moles, then divide each by its coefficient in the equation. The smaller result is the limiting reagent.

Percentage yield and purity

Yield is below $100\%$ in real experiments because some product is lost in transfer or filtering, the reaction may be reversible and not go to completion, or side reactions produce something else.

Empirical and molecular formula

From percentages: divide each percentage by that element's $A_{r}$, then divide all the answers by the smallest. That gives the empirical formula — the simplest whole-number ratio.

To get the molecular formula, divide the true $M_{r}$ by the empirical formula mass, and multiply every subscript by that number.