Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Prentice Hall © 2003 Chapter 3

Chapter 3Chapter 3Stoichiometry: Calculations with Stoichiometry: Calculations with

Chemical Formulas and EquationsChemical Formulas and Equations

FAYETTEVILLE STATE UNIVERSITYCOLLEGE OF BASIC AND APPLIED

SCIENCESDEPARTMENT OF NATURAL

SCIENCES

CHEM 140

Prentice Hall © 2003 Chapter 3

• Lavoisier: mass is conserved in a chemical reaction.

• Chemical equations: descriptions of chemical reactions.

• Two parts to an equation: reactants and products:

2H2 + O2 2H2O

Chemical EquationsChemical Equations

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• The chemical equation for the formation of water can be visualized as two hydrogen molecules reacting with one oxygen molecule to form two water molecules:

2H2 + O2 2H2O

Chemical EquationsChemical Equations

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2Na + 2H2O 2NaOH + H2

2K + 2H2O 2KOH + H2

Chemical EquationsChemical Equations

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• Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products.

Chemical EquationsChemical Equations

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Chemical EquationsChemical Equations

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• Law of conservation of mass: matter cannot be lost in any chemical reactions.

Chemical EquationsChemical Equations

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Combination and Decomposition Reactions• Combination reactions have fewer products than reactants:

2Mg(s) + O2(g) 2MgO(s)• The Mg has combined with O2 to form MgO.• Decomposition reactions have fewer reactants than products:

2NaN3(s) 2Na(s) + 3N2(g) (the reaction that occurs in an air bag)

• The NaN3 has decomposed into Na and N2 gas.

Some Simple Patterns of Some Simple Patterns of Chemical ReactivityChemical Reactivity

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Combination and Decomposition Reactions

Some Simple Patterns of Some Simple Patterns of Chemical ReactivityChemical Reactivity

Prentice Hall © 2003 Chapter 3

Combination and Decomposition Reactions

Some Simple Patterns of Some Simple Patterns of Chemical ReactivityChemical Reactivity

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Combustion in Air

Some Simple Patterns of Some Simple Patterns of Chemical ReactivityChemical Reactivity

Combustion is the burning of a substance in oxygen from air:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

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Formula and Molecular Weights• Formula weights (FW): sum of AW for atoms in formula.

FW (H2SO4) = 2AW(H) + AW(S) + 4AW(O)= 2(1.0 amu) + (32.0 amu) + 4(16.0)

= 98.0 amu• Molecular weight (MW) is the weight of the molecular

formula.MW(C6H12O6) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)

Formula WeightsFormula Weights

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Percentage Composition from Formulas

• Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:

Formula WeightsFormula Weights

( )( )100

Compound ofFW AWElement of Atoms

Element % ↔=

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Mole: convenient measure chemical quantities.• 1 mole of something = 6.0221367 1023 of that thing.• Experimentally, 1 mole of 12C has a mass of 12 g.

Molar Mass• Molar mass: mass in grams of 1 mole of substance (units

g/mol, g.mol-1).• Mass of 1 mole of 12C = 12 g.

The MoleThe Mole

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The MoleThe Mole

Prentice Hall © 2003 Chapter 3

The MoleThe Mole

Prentice Hall © 2003 Chapter 3

The MoleThe Mole

This photograph shows one mole of solid (NaCl), liquid (H2O), and gas (N2).

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Interconverting Masses, Moles, and Number of Particles

• Molar mass: sum of the molar masses of the atoms:molar mass of N2 = 2 (molar mass of N).

• Molar masses for elements are found on the periodic table.

• Formula weights are numerically equal to the molar mass.

The MoleThe Mole

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• Start with mass % of elements (i.e. empirical data) and calculate a formula, or

• Start with the formula and calculate the mass % elements.

Empirical Formulas from Empirical Formulas from AnalysesAnalyses

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Prentice Hall © 2003 Chapter 3

Molecular Formula from Empirical Formula

• Once we know the empirical formula, we need the MW to find the molecular formula.

• Subscripts in the molecular formula are always whole-number multiples of subscripts in the empirical formula

Empirical Formulas from Empirical Formulas from AnalysesAnalyses

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Combustion Analysis• Empirical formulas are determined by combustion

analysis:

Empirical Formulas from Empirical Formulas from AnalysesAnalyses

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• Balanced chemical equation gives number of molecules that react to form products.

• Interpretation: ratio of number of moles of reactant required to give the ratio of number of moles of product.

• These ratios are called stoichiometric ratios. NB: Stoichiometric ratios are ideal proportions

• Real ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles).

Quantitative Information Quantitative Information from Balanced Equationsfrom Balanced Equations

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Prentice Hall © 2003 Chapter 3

• If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess).

• Limiting Reactant: one reactant that is consumed

Limiting ReactantsLimiting Reactants

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Limiting ReactantsLimiting Reactants

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Theoretical Yields• The amount of product predicted from stoichiometry

taking into account limiting reagents is called the theoretical yield.

• The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:

Limiting ReactantsLimiting Reactants

100yield lTheoretica

yield ActualYield % ↔=