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The Mole and Molar Mass

The Mole and Molar Mass. Moles are the currency of chemistry 1 mole = 6.02 x 10 23 particles = Avogadro’s number Molar mass of an element = mass of one mole of that element Molar mass = atomic mass of element in grams - Average mass of one atom of C = 12.01 amu

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The Mole and Molar Mass

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  1. The Mole and Molar Mass • Moles are the currency of chemistry • 1 mole = 6.02 x 1023 particles = Avogadro’s number • Molar mass of an element = mass of one mole of that element • Molar mass = atomic mass of element in grams - Average mass of one atom of C = 12.01 amu - Average mass of one mole of C = 12.01 g • Molar mass of a compound = sum of molar masses of each atom in the compound • Example: What is the molar mass of MgCl2? Molar mass of Mg = 24.31 g, Molar mass of Cl = 35.45 g Molar mass of MgCl2 = 24.31 g + (2 x 35.45 g) = 95.21 g • *Note: use as many sig. figs. as given in the atomic masses (don’t round molar masses to 1/10 of a gram)

  2. Calculations using Molar Mass • Molar mass is used to convert between moles and grams of a substance • Chemical equations are balanced with atoms, and moles are an extension of that (they also balance) • Quantities of reactants and products are measured by mass (grams) in the lab, but we use moles to balance equations • Example 1: How many grams of H2O are in 0.100 mol H2O? Molar mass of H2O = (2 x 1.008 g) + 16.00 g = 18.02 g Grams H2O = 0.100 mol x (18.02 g/1 mol) = 1.80 g H2O • Example 2: How many moles are in 0.250 g of HCl? Molar mass of HCl = 1.008 g + 35.45 g = 36.46 g Mol HCl = 0.250 g x (1 mol/36.46 g) = 6.86 x 10-3 mol HCl

  3. Percent Composition • Molecular formula tells you how many moles of each type of atom are in one mole of a compound • Using molar mass, you can calculate the % by mass of each element in a compound • Example: Calculate the % composition by mass of CO2 Molar mass of CO2 = 12.01 g + (2 x 16.00 g) = 44.01 g %C = (12.01 g/44.01 g) x 100% = 27.29% %O = (2 x 16.00 g/44.01 g) x 100% = 72.71%

  4. Empirical Formulas • Empirical formulas give the lowest whole-number ratio of atoms in a compound • Molecular formula = multiple of empirical formula • Example: empirical formula for glucose = CH2O molecular formula = C6H12O6 • Can calculate empirical formula from masses or from % composition by mass • Example: Calculate empirical formula for compound with 8.0 g C and 2.0 g H 8.0 g C x (1 mol/12.01 g) = 0.6661 mol C 2.0 g H x (1 mol/1.008 g) = 1.984 mol H 0.6661 mol C/0.6661 mol = 1.000 C = 1 C 1.984 mol H/0.6661 mol = 2.979 H = 3 H’s Empirical formula = CH3

  5. Mole Relationships and Mole Calculations • Atoms, moles and mass are conserved in a chemical reaction (shown in balanced equation) • Using chemical formulas, molar masses and mole ratios (from balanced chemical equation), amounts of products and reactants can be calculated • Example: For C3H8 + 5O2 3CO2 + 4H2O How many moles of O2 are needed to produce 2.0 moles of H2O? • Have moles H2O, want moles O2 • 5 mol O2 = 4 mol H2O • 2.0 mol H2O x (5 mol O2 /4 mol H2O) = 2.5 mol O2

  6. Mole Relationships and Mass Calculations Example: C3H8 + 5O2 3CO2 + 4H2O How many grams of CO2 are produced when 10.0 g of C3H8 are consumed? 1. Have grams C3H8, want grams CO2 (g C3H8  mol C3H8  mol CO2  g CO2) 2. Molar mass C3H8 = (3 x 12.01 g) + (8 x 1.008 g) = 44.094 g Mol C3H8 = 10.0g x (1 mol/44.094 g) = 0.2268 mol 3. Mol CO2 = 0.2268 mol C3H8 x (3 mol CO2/1 mol C3H8) = 0.6804 mol 4. Molar mass CO2 = 12.01 g + (2 x 16.00 g) = 44.01 g Grams CO2 = 0.6804 mol CO2 x (44.01 g/1 mol) = 29.9 g

  7. Percent Yield • Theoretical (or ideal) yield is calculated from balanced chemical equation and molar masses • Actual yield = what you got (usually < ideal) • % yield = (actual yield/theoretical yield) x 100% Example: 2Fe + 3S  Fe2S3 What is the % yield if 1.00 g Fe makes 1.50 g Fe2S3? 1. Calculate theoretical yield: (g Fe  mol Fe  mol Fe2S3  g Fe2S3) 1.00 g Fe x (1 mol/55.85 g) = 0.0179 mol Fe 0.0179 mol Fe x (1 mol Fe2S3/2 mol Fe) = 0.00900 mol Fe2S3 Molar mass Fe2S3 = (2 x 55.85 g) + (3 x 32.06 g) = 207.88 g 0.00900 mol Fe2S3 x (207.88 g/1 mol) = 1.87 g Fe2S3 2. % yield = (1.50 g Fe2S3 /1.871 g Fe2S3) x 100% = 80.2%

  8. Limiting Reactant • Limiting Reactant = reactant that is completely used up • Example: If 1.0 mol of Fe are reacted with 3.0 mol of S, which is the limiting reactant? - Since 2.0 mol Fe are required to react with 3.0 mol of S, then 1.0 mol of Fe can only react with 1.5 mol of S - So, all the Fe is use up and 1.5 mol of S is leftover - Fe is the limiting reactant

  9. Equilibrium Constants • For reversible reactions at equilibrium: - Rate of forward reaction = rate of reverse reaction - Amounts of reactants and products are constant (but not necessarily equal) • Ratio of products to reactants = equilibrium constant (Keq) • For a reaction aA + bB  cC + dD Keq = [C]c [D]d/[A]a [B]b, where [A] = concentration of A • Example: for 2H2 + O2 2H2O Keq = [H2O]2/[H2]2 [O2] • If Keq > 1, equilibrium favors products If Keq < 1, equilibrium favors reactants

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