By: JASON CAMERON
Key Concepts
Moles
- Avogadro’s Constant (NA): 6.022 x 10^23 particles/mol
- Number of particles (N) = Number of moles (n) * NA
- Moles (n) = Mass (m) / Molar Mass (M)
Molar Mass
- Mass of 1 mole of a substance (g/mol)
- Elements: Molar mass = Relative atomic mass
- Compounds: Molar mass = Sum of molar masses of elements
Percent Composition
- % Composition = (Mass of element / Total mass of compound) * 100%
Dimensional Analysis
- Convert units using conversion factors
- Given unit * (Desired unit / Given unit) = Desired unit
Hydrates
- Compounds with water molecules trapped in their crystal structure
- Determine the formula of a hydrate by finding the mole ratio of water to anhydrous compound
Limiting Reagents
- Reactant that is completely consumed first in a reaction
- Limits the amount of product formed
- Identify by comparing mole ratios of reactants to the balanced equation
Excess Reagents
- Reactants that are not completely consumed in a reaction
- Leftover after the limiting reagent is used up
Percent Yield
- % Yield = (Actual yield / Theoretical yield) * 100%
- Actual yield: Amount of product obtained experimentally
- Theoretical yield: Maximum amount of product calculated from stoichiometry
Sample Problems
Mole Calculations
- Calculate the number of moles in 1.2 x 10^24 atoms of carbon.
- n = N / NA = 1.2 x 10^24 / 6.022 x 10^23 = 2.0 mol
- Calculate the mass of 0.417 mol of sodium chloride (NaCl).
- Molar mass of NaCl = 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol
- Mass = n * M = 0.417 mol * 58.44 g/mol = 24.4 g
Percent Composition
- Calculate the percent composition of water (H2O).
- Molar mass of H2O = 2 * 1.01 (H) + 16.00 (O) = 18.02 g/mol
- % H = (2 * 1.01 / 18.02) * 100% = 11.2%
- % O = (16.00 / 18.02) * 100% = 88.8%
Limiting Reagents
- Identify the limiting reagent when 10.0 g of hydrogen (H2) reacts with 50.0 g of oxygen (O2) to produce water (H2O).
- Balanced equation: 2H2 + O2 → 2H2O
- Moles of H2 = 10.0 g / 2.02 g/mol = 4.95 mol
- Moles of O2 = 50.0 g / 32.00 g/mol = 1.56 mol
- Mole ratio H2:O2 = 4.95 mol / 1.56 mol = 3.17
- Stoichiometric ratio H2:O2 = 2:1
- Since the mole ratio of H2 to O2 is greater than the stoichiometric ratio, H2 is the excess reagent and O2 is the limiting reagent.
Percent Yield
- Calculate the percent yield if 5.00 g of product is obtained in an experiment with a theoretical yield of 6.00 g.
- % Yield = (5.00 g / 6.00 g) * 100% = 83.3%