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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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chapter 3

Chapter 3

Calculations involving Chemical Formulae and Equations

  • 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
atomic and molecular weights
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
the mole
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.
molar mass
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.

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%

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.

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.

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
  • 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.

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.


    • 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.


    • 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.
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)

  • 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.