Moles and Mass Lecture Notes

Moles and Mass

Key Concepts

• Mole: A unit for measuring the amount of substance in chemistry.
• Molar Mass (M): The mass of one mole of a substance (in grams per mole, g/mol).
• Balanced Chemical Equation: Indicates the ratio of moles of reactants and products involved in a chemical reaction.

Learning Objectives

• Understand usage of mole ratios for calculating amounts of species in reactions.
• Grasp concepts of limiting reactants and percentage yield.

Chemical Equation Information

• A balanced equation indicates:
  • Relative number of molecules of reactants and products.
  • Example equation:
    $C<em>2H</em>5OH + 3O<em>2 \rightarrow 2CO</em>2 + 3H_2O$
  • Here, 1 mole of ethanol reacts with 3 moles of oxygen, producing 2 moles of carbon dioxide and 3 moles of water.

Mole-Mole Relationship

• Balanced equations predict moles of products from given reactants.
• Example: From the reaction $2H<em>2O \rightarrow 2H</em>2 + O_2$
  • 2 moles of water yield 2 moles of hydrogen and 1 mole of oxygen.

The Mole Ratio

• The mole ratio allows conversion between moles of different substances in a balanced equation.
• For example, from the equation $Na<em>2SiF</em>6 + 4Na \rightarrow Si + 6NaF$
  • From 4 moles of Na, 6 moles of NaF are produced.

Moles to Mass Conversion

• Relationship to convert moles to mass:
  • Formula:
    $n = \frac{m}{M}$
  • Where:
    • $n$ = number of moles
    • $m$ = mass (grams)
    • $M$ = molar mass (g/mol)
• Example: For 10.0 grams of water (H2O), calculate moles:
  • Molar mass (H2O) = 18.02 g/mol
    $n = \frac{10.0 \text{ g}}{18.02 \text{ g/mol}} \approx 0.555 \text{ mol}$

Converting Moles to Mass

• Using the formula:
  $m = M \times n$
• Example: For 2.5 moles of oxygen (O2):
  • Molar mass of O2 = 32.00 g/mol
    $m = 2.5 \text{ mol} \times 32.00 \text{ g/mol} = 80.0 \text{ grams}$

Stoichiometry, Limiting Reagents, and Percentage Yield

• Stoichiometry: Refers to mole-to-mole ratios linking substances in a chemical equation. Equations must be balanced.
• Limiting Reagents: The reactant that is consumed first, limiting the amount of product formed.
  • E.g. For the reaction
    $3C<em>2H</em>4 + 3H<em>2O \rightarrow 3C</em>2H_5OH$
  • If $C2H4$ runs out first, it's the limiting reagent.
• Percentage Yield: Calculated yield as a percentage of the theoretical yield.
  • Formula:
    \text{% yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%
  • Actual yield is the amount obtained from an experiment hence must be less than or equal to theoretical yield.

Example Problems

• Practice Question: Given the equation
  $C<em>2H</em>5OH + 3O<em>2 \rightarrow 2CO</em>2 + 3H_2O$
  How many moles of O2 are needed to produce 4.8 moles of CO2?

  • Answer Choices:
  • a) 3.2 mol O2
  • b) 4.8 mol O2
  • c) 7.2 mol O2
  • d) 1.5 mol O2

• Example Calculation: For the reaction of Sn with HF to produce SnF2:

  • Reaction:
    $Sn + 2HF \rightarrow SnF<em>2 + H</em>2$
  • Find grams of tin(II) fluoride from 55.0 g of HF with excess tin available.
  • Answer Choices:
    • a) 431 g
    • b) 215 g
    • c) 72.6 g
    • d) 1.37 g