Stoichiometry

Key Ideas and Formulas

  • Mole (n): amount of substance in mol
  • Molar mass (M): mass in g·mol–1 from periodic table

  • \( n = \dfrac{m}{M} \) (moles from molar mass)
  • \( n = C \times V \) for solutions (V is in L, C is in mol·L–1 or M)

  • Balanced chemical equations provide mole ratios
  • Tip: try to represent all values in stoichiometry equations as fractions
    • Always start with what you’re given (with units!), multiply by conversion fractions, and cross units that cancel

Reacting Masses

    Steps:

  1. Write and balance the chemical equation
  2. Convert the known mass or volume to moles
  3. Use mole ratios to find the moles of the unknown substance
  4. Convert to the required units (mass, moles, volume)

Example 1 (mass → mass):

How many grams of CO2 are produced when 8.50 g propane burns completely?


$$ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} $$ $$ \text{Molar masses: } \quad M_{\text{C}_3\text{H}_8} = 44.10 \, \text{g/mol}, \quad M_{\text{CO}_2} = 44.01 \, \text{g/mol} $$ $$ 8.50 \, \text{g C}_3\text{H}_8 \times \frac{1 \, \text{mol C}_3\text{H}_8}{44.10 \, \text{g C}_3\text{H}_8} \times \frac{3 \, \text{mol CO}_2}{1 \, \text{mol C}_3\text{H}_8} \times \frac{44.01 \, \text{g CO}_2}{1 \, \text{mol CO}_2} = 25.4 \, \text{g CO}_2 $$

Example 2 (solution → mass):

What mass of CaCO3 precipitate forms when 25.0 mL of 0.200 M CaCl2 is mixed with excess Na2CO3?


$$ \text{CaCl}_2(aq) + \text{Na}_2\text{CO}_3(aq) \rightarrow \text{CaCO}_3(s) + 2\text{NaCl}(aq) $$ $$ 25.0\,\text{mL} \times \frac{1.00\,L}{1000\,\text{mL}} \times \frac{0.200\,\text{mol CaCl}_2}{1.00\,L} \times \frac{1\,\text{mol CaCO}_3}{1\,\text{mol CaCl}_2} \times \frac{100.09\,g\,\text{CaCO}_3}{1\,\text{mol CaCO}_3} = 0.501\,g $$

Limiting Reactant

In a reaction, one reactant will likely run out before the other reactant is used up fully. This will limit the amount of product produced. In order to figure out which reactant is the limiting reactant, and the maximum yield of products, follow the steps below.


    Steps:

  1. Write and balance the chemical equation
  2. Convert each reactant’s given amount (mass or volume) to moles
  3. Divide each reactant's number of moles by its coefficient in the equation. The smallest result is the limiting reactant
  4. Use the moles of the limiting reactant and molar ratios to compute the moles/mass of products

Example:

For the reaction: \(2\text{Al} + 3\text{Cl}_2 → 2\text{AlCl}_3 \), if you have 3.00 g Al and 6.50 g Cl2, which is limiting and how much AlCl3​ can be produced?


$$ \text{Molar masses: } \quad M_{\text{Al}} = 26.98 \, \text{g/mol}, \quad M_{\text{Cl}_2} = 70.90 \, \text{g/mol}, \quad M_{\text{AlCl}_3} = 133.33 \, \text{g/mol} $$
Limiting reactant calculation: $$ 3.00 \, \text{g Al} \times \frac{1 \, \text{mol Al}}{26.98 \, \text{g Al}} \times \frac12 = 0.0556 \, \text{mol Al} $$ $$ \boxed{6.50 \, \text{g Cl}_2 \times \frac{1 \, \text{mol Cl}_2}{70.90 \, \text{g Cl}_2} \times \frac13 = 0.0306 \, \text{mol Cl}_2} $$ $$ \text{Therefore, } \text{Cl}_2 \text { is the limiting reactant.} $$

Product formed from limiting reactant:

$$ 6.50 \, \text{g Cl}_2 \times \frac{1 \, \text{mol Cl}_2}{70.90 \, \text{g Cl}_2} \times \frac{2 \, \text{mol AlCl}_3}{3 \, \text{mol Cl}_2} \times \frac{133.33 \, \text{g AlCl}_3}{1 \, \text{mol AlCl}_3} = 8.15 \, \text{g AlCl}_3 $$

Theoretical Yield, Actual Yield, and Percent Yield

    Definitions:

  • Theoretical Yield: the maximum amount of product that can be produced from the limiting reactant
  • Actual Yield: the amount of product actually produced in an experiment (usually less than theoretical yield)
  • Percent Yield: the efficiency of an experimental reaction, calculated as:

      • \( \text{Percent Yield} = \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \)

Example (continuing the AlCl3 case):

If you actually obtain 7.20 g AlCl3, what is the percent yield?


$$ \text{Percent Yield} = \frac{7.20 \, \text{g AlCl}_3}{8.15 \, \text{g AlCl}_3} \times 100\% = 88.3\% $$

Titration Calculations

    Key Relations/Notes:

  • \( n = C \times V \)
    • n → moles
    • C → concentration
    • V → volume
  • Use mole ratios from balanced equations to relate moles of acid and base
  • Equivalence Point: when the moles of an acid and the moles of a base react in a stoichiometric ratio determined by the balanced chemical equation

    Steps:

  1. Write and balance the chemical equation
  2. Convert all volumes to litres
  3. Calculate moles of the substance with known concentration: \( n_1 = C_1 \times V_1 \)
  4. Use the mole ratio from the balanced equation to find moles of the substance with unknown concentration
  5. Calculate the unknown concentration: \( C_2 = \dfrac{n_2}{V_2} \)

Example (acid-base, 1:1 ratio):

25.00 mL of HCl of unknown concentration is titrated with 0.1000 M NaOH. The equivalence point is at 18.62 mL NaOH. Find the concentration of HCl.


$$ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} $$ $$ C_{\text{HCl}} = 18.62\,\text{mL} \times \frac{1.000\,L}{1000\,\text{mL}} \times \frac{0.1000\,\text{mol NaOH}}{1.000 \, L} \times \frac{1\,\text{mol HCl}}{1\,\text{mol NaOH}} \times \frac{1}{0.02500\,L} = 0.07448\,\text{mol L}^{-1} $$

Finding the Empirical Formula From Mass or Percentage

    Steps:

  1. Convert each given mass to moles using molar masses
    • If given the % composition, first assume 100 g to turn % → grams
    • If given a combustion analysis, determine the moles of C and H from the mass/moles of CO2 and H2O
  2. Divide the moles of all elements by the smallest mole amount to get a ratio
  3. If the resulting values are not integers, multiply all values by either 2, 3, 4… to reach whole numbers
  4. Write the empirical formula using the ratio of moles for each element

Example (combustion analysis):

1.050 g of an unknown hydrocarbon produces 3.306 g CO2 and 1.353 g H2O. Find the empirical formula.


Moles of C from CO2:

$$ 3.306 \, \text{g CO}_2 \times \frac{1 \, \text{mol CO}_2}{44.01 \, \text{g CO}_2} \times \frac{1 \, \text{mol C}}{1 \, \text{mol CO}_2} = 0.07512 \, \text{mol C} $$

Moles of H from H2O:

$$ 1.353 \, \text{g H}_2\text{O} \times \frac{1 \, \text{mol H}_2\text{O}}{18.02 \, \text{g H}_2\text{O}} \times \frac{2 \, \text{mol H}}{1 \, \text{mol H}_2\text{O}} = 0.1502 \, \text{mol H} $$

Divide by the smallest amount of moles (0.07512):

$$ \text{C:} \quad \frac{0.07512 \, \text{mol C}}{0.07512} = 1.00 $$ $$ \text{H:} \quad \frac{0.1502 \, \text{mol H}}{0.07512} = 2.00 $$ $$ \text{Empirical formula: } \text{CH}_2 $$

Note: We know oxygen is not present in the molecule because the question stated the molecule to be a hydrocarbon. The oxygen in the products comes from the oxygen added during combustion. If oxygen was present in the molecule (CxHyOz), find the mass of oxygen by subtracting the mass of C and H from the given mass of the molecule, then convert that into moles using the molar mass of oxygen.