Stoichiometry Lecture Notes

Stoichiometry

• Stoichiometry is the numerical relationship between the relative quantities of substances in a reaction or compound. The term originates from "stoikheion" (part) and "metron" (measure).
• Lavoisier (1794) established the principle of mass conservation.

Chemical Equations

• Chemical reactions involve mixing two or more species to produce new substances.

• Chemical equations describe chemical reactions in a numeric fashion.

• Reactants are placed on the left-hand side, and products on the right-hand side, separated by a forward arrow ($\rightarrow$) or an equilibrium arrow ($\rightleftharpoons$), depending on the nature of the reaction.

  2H2(g) + O2(g) \rightarrow 2H_2O(l)

  • $H2$ and $O2$ are reactants.
  • $H_2O$ is the product.

Net Ionic Equation

• A net ionic equation is a simplified form of a chemical equation that includes only the ions involved in the chemical reaction that form a product, such as a precipitate.

• It helps understand what drives a reaction.

  3KOH(aq) + Fe(NO3)3(aq) \rightarrow 3KNO3(aq) + Fe(OH)3(s)

  Net Ionic Equation:

  3OH^-(aq) + Fe^{3+}(aq) \rightarrow Fe(OH)_3(s)

• Spectator ions are not included in the net ionic equation.

Chemical Equations II

• The Law of conservation of mass requires the same number of each type of atom on each side of the arrow.

• Stoichiometric coefficients are used to balance an equation to meet this condition.

• Physical states can also be specified in chemical equations:

  • (g) for gas
  • (l) for liquid
  • (s) for solid
  • (aq) for aqueous solution (substance dissolved in water)

  2H2(g) + O2(g) \rightarrow 2H_2O(l)

Balancing Reactions

• Example:

  Ca3(PO4)2 + SiO2 + C \rightarrow P4 + CaSiO3 + CO

  Method:

  1. Balance the heaviest atoms first (heavy $\rightarrow$ light), ignoring H and O initially.

     1 Ca3(PO4)2 + 6 SiO2 + 10 C \rightarrow 1 P4 + 6 CaSiO3 + 10 CO

  2. Balance H (not applicable in this example).

  3. Balance O.

  4. If there is a fractional coefficient, multiply the entire equation by the denominator to obtain integer coefficients throughout.

Molecular Mass

• Molecular mass is the sum of atomic masses in a molecular formula (also known as molecular weight, MW).

  C6H{12} = (6 \times 12.01) + (12 \times 1.01) = 84.18 Da

• Determined using a mass spectrometer, which provides $M+$.

  C{23}H{46}O_2

Moles, n, and $N_A$

• Mole is the number of atoms in 12.000g of $^{12}C$, which is 6.022 \times 10^{23} (Avogadro’s Constant, $N_A$).
• Mole: SI unit for the amount of substance; a mole is simply $N_A$ entities of that substance.
• Molar mass ($M$) = mass of 1 mole. Units: g/mol
• Number of moles ($n$) = \frac{m}{M} (where $m$ is mass). Unit abbreviation: mol
• Note: Molar mass = MW with units of g/mol
• n = \frac{m}{M}

Worked Problems: Moles

• Calculating the number of moles:

  • 10.0 g of $O_2$: \frac{10.0 \text{ g}}{32.00 \text{ g/mol}} = 0.313 \text{ mol}
  • 25.000 kg of $CH_4$: \frac{25000 \text{ g}}{16.04 \text{ g/mol}} = 1559 \text{ mol}
  • 1.0 g of $C{12}H{22}O_{11}$: \frac{1.0 \text{ g}}{342.30 \text{ g/mol}} = 0.0029 \text{ mol}

• Calculating mass from the number of moles:

  • 0.250 mol of $H_2$: 2.02 \frac{\text{g}}{\text{mol}} \times 0.250 \text{ mol} = 0.505 \text{ g}
  • 150.0 mol of $NaCl$: 58.44 \frac{\text{g}}{\text{mol}} \times 150.0 \text{ mol} = 8766 \text{ g}

Mass Percent

• Mass percent is often used for solids at high concentrations.
• Example: A commonly used stainless steel (18/10) alloy in cutlery contains:
  • 72% Fe
  • 18% Cr
  • 10% Ni
• This means in 100g of the alloy, there are 72g of Fe, 10g of Ni, and 18g of Cr. This is often easier than using mole ratios.

Mass Percent in Compounds

• Mass percent is also used to express the percentage of an element (A) in a compound (cmpd):

  \text{mass } % A = \frac{\text{moles of A in cmpd } \times \text{ atomic mass of A}}{\text{molar mass of cmpd}} \times 100

• Examples:

  • What is the %C mass in ethanol ($C2H5OH$)? $Mr(C2H_5OH) = 46.1 \text{ g/mol}$

    \%C = \frac{(2 \times 12.01)}{46.1} \times 100 = 52.1\%

  • Mass % C in $C{12}H{22}O_{11}$: \frac{(12 \times 12.0)}{342.30} \times 100 = 42.1 \%

  • Mass % O in $C{12}H{22}O_{11}$: \frac{(11 \times 16.0)}{342.30} \times 100 = 51.4 \%

  • Mass % H in $C{12}H{22}O_{11}$: \frac{(22 \times 1.0)}{342.30} \times 100 = 6.5 \%

Worked Problems: Mass Percent

• Example: You have 53g of $Na2CO3$.

  • What class of compound is this? Ionic (group I metal).

  • What is the name of the compound? Sodium carbonate (Type 1 ionic cmpd).

  • How many moles of the compound do you have?

    n = \frac{53 \text{ g}}{106.0 \text{ g/mol}} = 0.500 \text{ moles}

  • How many atoms of sodium do you have?

    n \times 2 \times 6.022 \times 10^{23} = 6.02 \times 10^{22}

  • What is the %Na (mass) in this compound?

    \frac{(2 \times 23.0 \text{ g/mol})}{106.0 \text{ g/mol}} \times 100 = 43.4\%

Stoichiometry in Action

• The critical link between substances involved in a chemical reaction is the mole-to-mole ratio.

• Consider the following:

  2C8H{18}(l) + 25O2(g) \rightarrow 16CO2(g) + 18H_2O(g)

  • 2 moles of liquid octane react with 25 moles of oxygen gas to produce 16 moles of carbon dioxide gas and 18 moles of steam ($H_2O$).

• Sometimes it is difficult to work with moles directly, so we work with something more practical.

What Limits a Sundae

• Example Reaction Ratio:
  • 2 scoops ice cream + 1 cherry + 50 mL syrup $\rightarrow$ 1 Sundae
• This is a ratio: 2 scoops : 1 cherry : 50 mL

Fixed Proportions

• Because the mole-to-mole ratio is constant, reactions occur in fixed proportions.

• Example 1:

  • 1NaOH + 1HCl \rightarrow 1NaCl + 1H_2O

  • At 5 moles NaOH, the exact proportions are:

    5NaOH + 5HCl \rightarrow 5NaCl + 5H_2O

  • However, if there is an excess of HCl, e.g.,

    5NaOH + 10HCl \rightarrow 5NaCl + 5H_2O

    • Only 5 HCl molecules will react $\rightarrow$ 5 HCl molecules in excess (left over).

    5NaOH + 10HCl \rightarrow 5NaCl + 5H_2O + 5 HCl \text{ excess}

• Example 2:

  2Mg + 1O_2 \rightarrow 2MgO

  • If there is an excess of $O_2$, e.g.,

    2Mg + 3O_2 \rightarrow 2MgO

    • Then the reaction would be:

      2Mg + 3O2 \rightarrow 2MgO + 2O2 \text{ excess}

Limiting Reagent

• All reactions eventually use up a reactant and seem to stop.
• The reactant that is consumed first is called the limiting reagent (or limiting reactant) because it limits the amount of product formed.
• Any reagent not completely consumed during the reactions is said to be in excess and is called an excess reagent.
• The calculated amount of product is always based on the limiting reagent.

Limiting Reagent II

• A limiting reagent in a reaction is the reactant with the lowest number of moles when we consider the stoichiometry.

• The amount of product(s) is determined by the stoichiometry and the number of moles of the limiting reagent.

  C2H4 + H2 \rightarrow C2H_6

  • (M 28.06) (M 2.02) (M 30.08)

  • 0.50g 9.9g x g ?

  • 0.018 mol 4.9 mol 0.018 mol

  • Limiting reagent: $C2H4$ ($m = n \times M$)

    0.018 \times 30.08 \approx 0.54g

Worked Problems: Limiting Reagent

• Two 'easy' limiting reagent problems.

• Example 1a:

  2Mg + O_2 \rightarrow 2MgO

  • 24.30amu 2 x 16.00 24.30 + 16.00
  • 32.00 amu 40.30 amu
  • We react 48.60g of Mg in air (unlimited $O_2$), what mass of MgO will form?
    Ans: 80.60 g; Mg is the limiting reagent

• Example 1b:

  2Mg + O_2 \rightarrow 2MgO

  • 24.30 amu 2 x 16.00 24.30 + 16.00
  • 32.00 amu 40.30 amu
  • We react 48.60 g of Mg in an atmosphere containing 16.0 g of $O_2$, what is the theoretical maximum mass of MgO that will form?

Ans: 40.30 g; $O_2$ is the limiting reagent. There will be 24.30 g of unreacted Mg.
2.0 \text{ mol} + 0.5 \text{ mol} \rightarrow 2 \times 0.5 \text{ mol}

N.B. DO NOT use the stoichiometric coefficient to calculate the molar mass

Worked Problem 2

• 2 ICl3 + 3 H2O \rightarrow 1 ICl + 1 HIO_3 + 5 HCl
• (M 233.3) (M 18.02) (M 162.4) (M 175.9) (M 35.55)
• 735g 97.7g x g ?
• 3.150 mol 5.422 mol x? mol

1. Ratio the moles $\rightarrow$ find limiting reagent

   \frac{3.15}{2} < \frac{5.422}{3} \implies 1.575 < 1.807

2. Divide all the reaction coefficients by the coefficient of the limiting reagent – normalized eqn.

3. Calculate the amount of product using the normalized equation and moles of limiting reagent.

• Product (1/2 * 3.150 = 1.575 mol)

  1.575 \times 175.9 = 277g

% Yield

• In most experiments, the amount of a product isolated falls short of the maximum amount calculated.

• The actual yield of the desired product is simply how much is isolated in the laboratory.

• The theoretical yield of the product is what would be obtained if no losses occurred – calculated yield.

• The percentage yield is the actual yield calculated as a percentage of the theoretical yield.

• The calculation may be done in either grams or moles, but both yields must be in the same units.

• The actual yield cannot be more than the theoretical yield.

  \% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

Equations and Problems

• Balancing equations: Must be able to balance efficiently. Try some more examples: BLB 3e: 3.15 – 3.15
• Limiting reagent problems: Must be OK at solving these. Try some more
• Problems: BLB 3e:
  • Balance equations 3.25 – 3.28
  • Moles, grams, atoms/molecules 3.44 – 3.49
  • Equation and mole calculations 3.52 – 3.57
  • Limiting reagent