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Chapter 3

Chapter 3. Calculations involving Chemical Formulae and Equations. Contents. Mass and Moles of a Substance Molecular weight Moles Determining Chemical Formulae Mass % from formula Elemental Analysis: gives % C, H, O Determining formula from elemental analysis Stoichiometry

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Chapter 3

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  1. Chapter 3 Calculations involving Chemical Formulae and Equations

  2. Contents • Mass and Moles of a Substance • Molecular weight • Moles • Determining Chemical Formulae • Mass % from formula • Elemental Analysis: gives % C, H, O • Determining formula from elemental analysis • Stoichiometry • Amounts of substances in a reaction • Limiting reagents

  3. Atomic and Molecular Weights • Molecular Weight (Formula weight): Summed masses of all atoms in the formula unit of the compound. • How? • For the general formula AaBbCcDd... fm = formula mass = aMA + bMB + cMC + dMD + ... • E.g. determine the formula mass (formula weight) of the following: • NaOH, H2O, CaCO3, C2H6O • Percentage composition from formula (mass % contribution of each element to the total mass of the compound). For AaBbCc • E.g. determine the mass % composition of each element in: NaOH, H2O, CaCO3, C2H6O

  4. The Mole • Since individual atoms are very small, they weigh very little. It is not convenient deal with masses that small, since we can typically measure gram amounts in the lab. • Mole: Number of atoms (or molecules) as there are in 12.00 g Carbon-12; 1 mol = 6.02x1023 and is called Avogadro's number. • Mole is used to indicate a certain number of particles (like a dozen or bushel might be used). 1 mol of • water occupies approximately 18 mL and has 6.02x1023 molecules. • gold occupies approximately 10 mL and has 6.02x1023 atoms.

  5. Molar mass • 1 mol = FM = 6.02x1023 particles • The definition of a mol allows us to perform a variety of conversions. • Let n = # mol, n’ = # atoms (molecules), m = mass and FM = formula mass then we can convert to: • m from n, FM E.g. Determine the mass of NaCl needed to have 5.0 mol of it. • n from m, FM E.g. Determine the number of mol of NaCl in 15.00 g of it. • n’ from n E.g. Determine the number of molecules in 3.222 mol of NaCl • n’ from m, FM E.g. Determine the number of atoms in 4.32 g of NaCl • m from n’, FM E.g. Determine the mass of one formula unit of NaCl.

  6. EMPIRICAL FORMULA Empirical Formula: Simplest formula where all coefficients are integers. • In earlier example Fe2O3, Fe4O6 Fe6O9 Fe8O12 possible formulas for iron oxide. Its empirical formula Fe2O3. • Often obtained from % composition data. • Assume 100 g of compound- • Determine mol of each element. • Divide by smallest # of moles. • Multiply by smallest whole number to make all coefficients whole numbers. • E.g. Determine the empirical formula of a compound with the following % compositions: Mass%O = 34.7% Mass%C = 52.1% Mass%H = 13.1%

  7. COMBUSTION ANALYSIS • Experimental mass % also determined. For organic compounds the sample is combusted. Analyzes the amount of C,H, and O in compounds. • C converted to CO2 and mass measured. • H converted to H2O and mass measured • E.g.: Combustion analysis of a 1.621 g sample of ethanol, which contains only C,H,O, was performed. Calculate the mass % composition of each element in ethanol. • Results: Mass CO2 = 3.095 g Mass of H2O = 1.902 g.

  8. MOLECULAR FORMULA • Measure molecular mass • Multiply empirical mass by integer to obtain molecular mass. • Multiply all the co-efficients by this integer. • E.g. Determine the molecular mass of a compound with the empirical formula NO2 and a molecular mass of 92.00 g/mol.

  9. STOICHIOMETRY The relative amounts of reactants and products in a reaction are given by the ratio of stoichiometric coefficients. (Conservation of mass). • E.g. For the reaction : • 2Na(s) + Cl2(g)  2NaCl(s) • 2 mol Na = 1 mol of Cl2 = 2 mol of NaCl. • E.g. 2 Determine # mol of Cl2 needed to react with 4.2 mol of Na. How many mol of NaCl will form? • Mol of Cl2 • Mol NaCl: • Summary: For aA + bB  cC

  10. STOICHIOMETRY2 • Mass of one reactant can be related to the mass of another reactant or product. (Law of definite proportions). E.g. Determine the mass of Na needed to react with 34.45 g of Cl2 and the maximum mass of NaCl that could be produced. • Write reaction and express the moles of one compared to the moles of the other. • Substitute for mol in terms of mass and formula mass in each part. Let n = mol; FM = formula mass then E.g. 1: Determine the mass of oxygen consumed when it reacts with 10 g CH3CHO. E.g. 2. Calculate the mass of oxygen needed to react with 100 g Al to for Al2O3.

  11. LIMITING REAGENTS • Limiting reagent: the reactant which limits the maximum amount of product that can be produced. It will be completely consumed in the reaction before any other reactants. • Limiting reagent must be known before theoretical yield can be determined. E.g. 1 Determine limiting reagent if 3.00 moles of Al react with 2.15 moles of O2 to form Al2O3. Strategy: • Determine mol of Al2O3 that could be produced from Al • Determine mol of Al2O3 that could be produced from O2 • Reactant producing smallest amount of Al2O3 is limiting reagent. E.g.2 Calculate the theoretical yield when 20 g Al react with 25 g O2. Strategy: • Determine expected yield from each reactant. • Compare. If mass of Al2O3 from produced from Al is less than from O2, then Al is limiting. Otherwise O2 is limiting.

  12. YIELDS IN CHEMICAL REACTIONS • Theoretical yield: the maximum amount of product that can be obtained from given amounts of reactants. • Actual Yield : the actual amount of product obtained from a reaction. • Theoretical yield calculated from amount of reactants and stoichiometry • Actual yield is the amount of product actually obtained. • % yield is the yield as a percent of the theoretical yield: • E.g. The reaction for the synthesis of acetic acid from methanol is shown below. What is the % yield if 15.0 g of methanol were combined with a stoichiometric amount of carbon monoxide to form19.1 g of product? CH3OH(l) + CO(g)  HC2H3O2(l)

  13. Summary • Moles are used to describe measurable quantities of a substance (1 mol = 6.02x1023 particles). • Reactions occur in well defined proportions and are represented by balanced chemical equations. • The ratio of stoichiometric coefficients tells how much of one compound will react if we know the amount of the other: aA + bB  cC • Limiting reagent is one that limits the amount of product. Use it to determine the “theoretical yield”. • Empirical formula: simplest formula; all integers. • Molecular formula: actual formula of a compound.

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