Mass Relations from Equations

The relative numbers of reactant and product molecules (or the relative numbers of moles) are indicated by the coefficients of a balanced chemical equation. Using molar masses, we can compute the relative masses of reactants and products in a chemical reaction.

Example 1

Ammonia reacts with oxygen according to the equation

Thus 4 moles of NH₃ react with 3 moles of O₂ to form 2 moles of N₂ and 6 moles of H₂O. Suppose that the reaction consumes 10.00 g NH₃. This corresponds

From the mole ratios, the reaction must involve

Using the molar masses, we can compute the mass of O₂ reacting and the masses of N₂ and H₂O produced:

Notice that mass is conserved (as required) in the reaction:

Often a problem gives the mass of one reactant or product and you are to determine the masses of the other substances involved. In problems of this kind, you may assume that sufficient amounts of the other reactants are present. If the masses of two reactants are given, you must determine which one (if either) is in excess. The best approach is to convert the given masses to moles and compare the mole ratio to that determined by the balanced equation.

Example 2

A solution containing 2.00 g of Hg(NO₃)₂ was added to a solution containing 2.00 g of Na₂S. Calculate the mass of products formed according to the reaction

We begin by calculating the number of moles of each reactant:

The equation indicates that equimolar quantities of reactants are required. Hence the given amount of Hg₂(NO₃)₂ requires only 6.16 × 10^-3 mol of Na₂S. Thus 1.94 × 10-2 mol (1.51 g) of Na₂S remains unreacted. Hg₂(NO₃)₂ is the limiting reagent and we expect to form 6.16 × 10^-3 mol HgS and 2(6.16 × 10^-3) = 1.23 × 10^-2 mol NaNO₃:

In summary, the original 4.00 g of reagents has been transformed to 1.43 g HgS and 1.05 g NaNO₃ with 1.51 g Na₂S unreacted. Note that the sum of the final masses is 3.99 g, within round-off error of the original 4.00 g.