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

CHEMISTRY The Central Science

9th Edition

David P. White


• 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


• 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



2Na + 2H2O 2NaOH + H2

2K + 2H2O 2KOH + H2



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





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



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

test state one product is formed)

2Mg(s) + O2(g) 2MgO(s)• The Mg has combined with O2 to form MgO.

• Decomposition reactions have fewer reactants than products:(on test state one reactant)

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


Combination and Decomposition Reactions



Combination and Decomposition Reactions



Combustion in Air


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

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


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


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 %


Mole: convenient measure of 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

The MoleThe Mole


The MoleThe Mole


The MoleThe Mole

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


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


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

Do 50 and 52 as an example

Empirical Formulas from Empirical Formulas from AnalysesAnalyses


Combustion Analysis• Empirical formulas are determined by combustion

analysis:



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



• A 1.540 g sample burns in oxygen to produce 2.257 g of carbon dioxide and 0.9241 grams of water. The sample only contains carbon, hydrogen and oxygen.

• Give all mass percents• What is the simplest formula?• If the molar mass is between 50 and 70 grams per mole,

what is the molecular formula?


• The insecticide DDD contains only carbon, hydrogen and chlorine. When 3.200g is burned, 6.162 g of carbon dioxide and 0.9008 g of water are formed. What is the simplest formula of DDD?


• 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. 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


• 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




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:


100yield lTheoretica

yield ActualYield %


How many grams of aluminum sulfide can form from the reaction of 9.00g of aluminum with 8.00g of sulfur?


Chromium (III) hydroxide will dissolve in concentrated sodium hydroxide solution according to the following equation.

NaOH + Cr(OH)3 NaCr(OH)4

This process is one step on making high purity chromium chemicals. If you begin with 66.0g Cr(OH)3

and obtain 38.4g of NaCr(OH)4 , what is your percent

yield?


End of Chapter 3:End of Chapter 3:Stoichiometry: Calculations with Stoichiometry: Calculations with

Chemical Formulas and Chemical Formulas and EquationsEquations

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Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations