1: molar mass, grams ↔ moles, mole ratio · 2: mass to mass, which reactant is limiting · 3: mass of product from a limiting reagent, % yield. Answers within 1.5% count.
Atoms are far too small to count one by one, so chemists count them in moles. One mole is $6.022 \times 10^{23}$ particles (the Avogadro constant). The molar mass $M$ is the mass of one mole in grams; it equals the relative formula mass. For water, $M = 2 \times 1 + 16 = 18\ \text{g/mol}$, so 18 g of water contains one mole of molecules.
Use $n = \dfrac{m}{M}$: moles equal mass divided by molar mass. So 36 g of water is $36 \div 18 = 2$ mol. Going back, $m = n \times M$. You need moles because the equation’s ratio is a ratio of particles, not of masses: 2 g of hydrogen does not react with 1 g of oxygen.
When two reactants are mixed, the one that runs out first is the limiting reagent; it alone decides how much product forms, and the other is in excess. To find it, change each mass to moles and divide by its coefficient: the smaller number is limiting. The mass of product calculated from the limiting reagent is the theoretical yield. In a real experiment you usually get less, so the percentage yield = actual ÷ theoretical × 100.
In the limiting reagent lab choose $2\,\text{H}_2 + \text{O}_2$, set 6 molecules of hydrogen and 2 of oxygen, and press React: oxygen runs out after 2 sets and 2 hydrogen molecules are left. Then type masses in grams below the picture and compare. Next, open the Mole map and change the units to see which arrows you use, and finish with Level 1 of the game.
What mass of water forms when 8 g of hydrogen burns? $2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$
| Step | Working |
|---|---|
| 1. Moles of what you have | $n(\text{H}_2) = 8 \div 2 = 4$ mol |
| 2. Mole ratio | $\text{H}_2\text{O} : \text{H}_2 = 2 : 2$, so $n(\text{H}_2\text{O}) = 4$ mol |
| 3. Mass of what you want | $m = 4 \times 18 = 72$ g |
12 g of magnesium burns in 16 g of oxygen. $2\,\text{Mg} + \text{O}_2 \rightarrow 2\,\text{MgO}$
| Step | Working |
|---|---|
| 1. Moles of each reactant | $n(\text{Mg}) = 12 \div 24 = 0.5$ mol; $n(\text{O}_2) = 16 \div 32 = 0.5$ mol |
| 2. Divide by the coefficients | Mg: $0.5 \div 2 = 0.25$; O2: $0.5 \div 1 = 0.5$ → Mg is limiting |
| 3. Product from the limiting reagent | $n(\text{MgO}) = 0.5 \times \tfrac{2}{2} = 0.5$ mol; $m = 0.5 \times 40 = 20$ g |
| 4. Excess left over | O2 used $= 0.25$ mol (8 g), so $16 - 8 = 8$ g of oxygen is left. Check: $12 + 8 = 20$ g ✔ |
Heating 50 g of calcium carbonate gives 22.4 g of calcium oxide. $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$
| Step | Working |
|---|---|
| 1. Theoretical yield | $n(\text{CaCO}_3) = 50 \div 100 = 0.5$ mol → $n(\text{CaO}) = 0.5$ mol → $m = 0.5 \times 56 = 28$ g |
| 2. Percentage yield | $22.4 \div 28 \times 100 = 80\%$ |
Some product is lost when it is filtered, poured or dried; the reaction may be reversible and not go to completion; side reactions can make other products; and the reactants may be impure. A yield above 100% usually means the product is still wet or contains impurities.
Stoichiometry is the calculation of the amounts of reactants and products in a chemical reaction. It uses the balanced equation, whose coefficients give the mole ratio, together with molar masses to convert between grams and moles.
Key takeaway: the balanced equation gives the mole ratio; molar mass converts grams and moles.Convert the mass of each reactant to moles (divide by its molar mass), then divide each number of moles by that reactant’s coefficient in the balanced equation. The reactant with the smallest result is the limiting reagent; it is used up first and decides how much product forms.
Key takeaway: moles divided by coefficient: the smallest value is limiting.Find the limiting reagent, convert it to moles, use the mole ratio from the balanced equation to get the moles of product, then multiply by the product’s molar mass. The result is the maximum mass of product the reaction can give.
Key takeaway: theoretical yield comes from the limiting reagent.Percentage yield = actual yield ÷ theoretical yield × 100. For example, if the theoretical yield is 28 g and you collect 22.4 g, the percentage yield is 80%.
Key takeaway: actual divided by theoretical, times 100.Because the coefficients in an equation count particles, not grams. Different substances have different molar masses, so equal masses contain different numbers of particles. Only after converting to moles can you use the mole ratio.
Key takeaway: the equation’s ratio is a particle ratio, so use moles.Add up the relative atomic masses of all the atoms in the formula, multiplying each by its subscript. For calcium hydroxide, Ca(OH)2: 40 + 2 × (16 + 1) = 74 g/mol.
Key takeaway: sum the atomic masses, multiplied by the subscripts.Use the limiting reagent to work out how many moles of the excess reactant react (mole ratio), subtract that from the moles you started with, and convert the remainder to grams. The lab above shows this for any two masses you type.
Key takeaway: start minus used, worked out from the limiting reagent.School exam papers usually give rounded relative atomic masses (C = 12, Cl = 35.5), while data books and university courses use more precise IUPAC values (C = 12.011, Cl = 35.45). The method is identical; answers differ only slightly, and the practice game accepts answers within 1.5%.
Key takeaway: same method; only the last digits of the answers change.