Stoichiometry and the Mole Concept
The central tool
n = m / M, where M is the molar mass in g/mol. One mole contains 6.02x10^23 particles, and one mole of any ideal gas occupies 22.4 L at STP (0 °C, 1 atm) or about 24.5 L at 25 °C, 1 atm. Solution concentration is c = n / V in mol/L, and dilution obeys c1V1 = c2V2 because the number of moles does not change.
The procedure that earns marks
- Write and balance the equation - mole ratios come only from the balanced equation.
- Convert every given quantity to moles.
- Divide each amount by its coefficient to find the limiting reagent.
- Use the limiting reagent's ratio to find moles of product.
- Convert back to mass, volume or concentration, applying percent yield if asked: yield = actual / theoretical x 100%.
For empirical formulas, divide percentage masses by atomic masses, then divide through by the smallest ratio. The molecular formula is the empirical formula multiplied by M(molecular)/M(empirical).
Worked example
10.0 g of CaCO3 is heated: CaCO3(s) -> CaO(s) + CO2(g). M(CaCO3) = 100.1 g/mol, so n = 0.0999 mol; the 1:1 ratio gives 0.0999 mol CaO, and M(CaO) = 56.1 g/mol, so the theoretical mass is 5.60 g. If only 4.76 g is collected, the percent yield is 4.76/5.60 x 100% = 85%.
Traps
Using the excess reagent to compute product. Forgetting to balance before using ratios. Mixing up the molar volume conditions (22.4 L only at STP). Reporting percent yield above 100% without noticing that the product is wet or impure. In titration problems, forgetting that c1V1 = c2V2 works only when the acid and base react 1:1.