Stoichiometry

Chapter 6: Stoichiometry

6.1 Introduction

• Conservation of Matter:

  • Matter is neither created nor destroyed in chemical or physical changes.

• Chemical Reactions:

  • Bonds in reactant(s) break and new bonds form to produce product(s) without altering elemental composition.

• Chemical Equations:

  • Represent chemical reactions.
  • Example:
  • H2 + O2 → H2O
  • States of matter: (s, l, g, aq).

• Stoichiometry:

  • Refers to the relationship between the amounts of reactants and products in a chemical equation.

6.2 The Chemical Equation and Law of Conservation of Mass

• Antoine Lavoisier:

  • Discovered the principle that during a chemical reaction, the total mass remains unchanged.
  • Regarded as the father of modern chemistry.

• Law of Conservation of Mass:

  - Total mass of products equals total mass of reactants in a chemical reaction.

  • Atoms are neither created nor destroyed, thus:
  • Number and type of reactant atoms must equal those in products.

• Balanced vs Unbalanced Reactions:

  • Unbalanced: Example.
  • Balanced: Example:
  • A + B → C + D, where a, b, c, and d are coefficients representing the amount of reactants/products involved.

• Example:

  • Reaction of methane (CH4) with oxygen (O2) produces carbon dioxide (CO2) and water (H2O).

6.3 Reaction Stoichiometry

• Stoichiometry Defined:

  • Mathematical representation of a chemical reaction using coefficients.

• Example:

  • Making a s’more requires 2 Graham crackers, 1 marshmallow, and 3 chocolates.

• Stoichiometric Coefficients:

  • Used to determine amounts needed of all other components from one known quantity.

• Practical Example:

  • Given 6 chocolates:
  • Calculate needed ingredients:
    • For Graham crackers:
      $6 ext{ chocolates} imes \frac{2 ext{ Graham crackers}}{3 ext{ chocolates}} = 4 ext{ Graham crackers}$
    • For marshmallows:
      $6 ext{ chocolates} imes \frac{1 ext{ marshmallow}}{3 ext{ chocolates}} = 2 ext{ marshmallows}$
    • Total s’mores:
      $6 ext{ chocolates} imes \frac{1 ext{ s’more}}{3 ext{ chocolates}} = 2 ext{ s’mores}$

6.3.1 Mole-to-Mole Conversions

• Understanding Mole Ratios:

  • General representation of a reaction:
  • aA + bB
    ightarrow cC + dD
  • If moles of one species are known, moles of others can be determined.

• Chemical Equation: Glucose breakdown example:

  • C6H{12}O6 + 6 O2
    ightarrow 6 CO2 + 6 H2O
  • For 4.75 moles of glucose:
  • Moles of CO2 produced:
    $4.75 ext{ mol } C<em>6H</em>{12}O<em>6 imes \frac{6 ext{ mol } CO</em>2}{1 ext{ mol } C<em>6H</em>{12}O<em>6} = 28.5 ext{ mol } CO</em>2$
  • Moles of glucose needed with 4.75 moles of O2:
    $4.75 ext{ mol } O<em>2 imes \frac{1 ext{ mol } C</em>6H<em>{12}O</em>6}{6 ext{ mol } O<em>2} = 0.792 ext{ mol } C</em>6H<em>{12}O</em>6$

6.3.2 Molar Mass Conversions

• Atomic Mass:

  • Example:
  • 1 atom of He = 4.003 amu.
  • Conversion:
    • 1 mole of He = 6.022 x 10²³ atoms = 4.003 g.

• Molecular Mass Calculation:

  • SO2 example:
  • $1S = 32.07 ext{ amu}$
  • $2O = 2 imes 16.00 ext{ amu}$
  • Total: $SO_2 = 64.07 ext{ amu}$

• Moles of Methane (CH4):

  • Molar mass calculation:
  • $ext{Molar mass of CH}_4 = 12.01 ext{ g} + 4(1.008 ext{ g}) = 16.04 ext{ g}$
  • Calculated moles in given mass:
    $6.07 ext{ g CH}<em>4 imes \frac{1 ext{ mol CH}</em>4}{16.04 ext{ g CH}<em>4} = 0.378 ext{ mol CH}</em>4$

6.3.3 Mass-to-Mass Conversions

• General Concept:

  • Given mass can determine mass of all other species.

• Chemical Reaction Example:

  • SO2(g) + O2(g)
    ightarrow SO_3(g)

• Steps:

  • Write balanced equation.
  • Convert grams to moles.

• Example Calculation: How many grams of SO3 from 3.47 g of O2 (with molar masses provided)? Reactions yield the following:

  1. Convert grams to moles.
  2. Use stoichiometry.
  3. Convert moles SO3 back to grams.

• Result: 17.4 g of SO3.

6.4 The Limiting Reactant

• Limiting Reactant:

  • The reactant that limits the amount of product formation.
  • Always completely consumed during the reaction.

• Excess Reactant:

  • The reactant that remains after the limiting reactant is completely consumed.

Identifying the Limiting Reactant

• Example:
• Equation: N2(g) + 3H2(g)
  ightarrow 2NH_3(g)

1. Balance the equation.
2. Determine initial number of moles of reactants.
3. Calculate required moles to balance with stoichiometric ratios.
4. Compare with initial moles to identify limiting reactant.

Method 1: Comparing Reactants to Each Other

• Mole ratio: (aA + bB
  ightarrow cC + dD)

1. Perform balance.
2. Convert grams to moles for comparison.
3. Determine how many moles A & B are needed for the reaction.
4. Identify if A or B is limiting.

Method 2: Comparing Reactants to A Specific Product

1. Balance the equation and find moles of A & B.
2. Calculate potential product from both reactants.
3. The reactant yielding the smaller amount of product is limiting.

6.5 Percent Yield

• Theoretical Yield:

  • The calculated maximum amount of product formed under ideal conditions.

• Actual Yield:

  • The amount of product obtained in a real-world experiment.

• Percent Yield Formula:
  $ext{Percent Yield} = \frac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100$

6.6 Mass Percent Composition and Chemical Subscripts

• Mass Percent:

  • Ratio of the mass of an element in a compound to the total compound mass, multiplied by 100.
  • Example: with MgCl2:
  • 1 mol Mg = 24.32 g, 1 mol Cl = 35.45 g
  • $95.21 ext{ g/mol for MgCl2}$

• Calculating Mass Percents:

  • $ext{Mass percent of Mg} = \frac{1 ext{ mol Mg} imes 24.32 ext{ g}}{1 ext{ mol MgCl2} imes 95.21 ext{ g}} imes 100$

• Total mass percent will always sum to about 100% when rounded.

6.7 Empirical and Molecular Formulas

• Empirical Formulas:
  • Smallest whole-number ratio of elements in a compound.
  • Different compounds can share the same empirical formula, e.g., formaldehyde (CH2O) and glucose (C6H12O6) share CH2O as the empirical formula.

6.7.1 Determining Empirical Formulas

1. Gather mass (grams) from mass percent data.
2. Convert mass to moles using molar mass.
3. Divide mole quantities by the smallest mole number to find ratios.
4. Round to nearest whole number if close to integers.

Process to Determine Molecular Formulas

5. Use the whole-number ratio as subscripts for empirical formula.
6. Ratios may be scaled up if initial mole numbers are not close to integers.

• Example:
  • For a compound with given composition, empirical formula determination led to C6.50H8.93O1.00, which rounds to C13H18O2 as empirical.

6.7.2 Determining Molecular Formulas

• Need molar mass to convert empirical formulas to molecular formulas.
  • The ratio of EF molar mass is compared to the compound's molar mass to achieve the molecular formula.

6.7.3 Determining Empirical Formulas from Combustion Data

• Combustion Analysis:
  • Used to derive percent compositions; compounds combust in oxygen producing CO2 and H2O.
• Example: Calculating the empirical formula from measured combustion data in CO2 and H2O to reveal compound composition.