Prentice Hall © 2003Chapter 3 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations CHEMISTRY The Central Science 9th Edition David P. White

Prentice Hall © 2003Chapter 3 Lavoisier: mass is conserved in a chemical reaction. Chemical equations: descriptions of chemical reactions. Two parts to an equation: reactants and products: 2H 2 + O 2  2H 2 O Chemical Equations

Prentice Hall © 2003Chapter 3 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: 2H 2 + O 2  2H 2 O Chemical Equations

Prentice Hall © 2003Chapter 3 2Na + 2H 2 O  2NaOH + H 2 2K + 2H 2 O  2KOH + H 2 Chemical Equations

Prentice Hall © 2003Chapter 3 Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products. Chemical Equations

Prentice Hall © 2003Chapter 3 Chemical Equations

Prentice Hall © 2003Chapter 3 Law of conservation of mass: matter cannot be lost in any chemical reactions. Chemical Equations

Prentice Hall © 2003Chapter 3 Combination and Decomposition Reactions Combination reactions have fewer products than reactants: (on test state one product is formed) 2Mg(s) + O 2 (g)  2MgO(s) The Mg has combined with O 2 to form MgO. Decomposition reactions have fewer reactants than products:(on test state one reactant) 2NaN 3 (s)  2Na(s) + 3N 2 (g) (the reaction that occurs in an air bag) The NaN 3 has decomposed into Na and N 2 gas. Some Simple Patterns of Chemical Reactivity

Prentice Hall © 2003Chapter 3 Combination and Decomposition Reactions Some Simple Patterns of Chemical Reactivity

Prentice Hall © 2003Chapter 3 Combination and Decomposition Reactions Some Simple Patterns of Chemical Reactivity

Prentice Hall © 2003Chapter 3 Combustion in Air Some Simple Patterns of Chemical Reactivity Combustion is the burning of a substance in oxygen from air: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l)

Prentice Hall © 2003Chapter 3 Formula and Molecular Weights Formula weights (FW): sum of AW for atoms in formula. FW (H 2 SO 4 ) = 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(C 6 H 12 O 6 ) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) Formula Weights

Prentice Hall © 2003Chapter 3 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 Weights

Prentice Hall © 2003Chapter 3 Mole: convenient measure of chemical quantities. 1 mole of something =  of that thing. Experimentally, 1 mole of 12 C 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 12 C = 12 g. The Mole

Prentice Hall © 2003Chapter 3 The Mole

Prentice Hall © 2003Chapter 3 The Mole This photograph shows one mole of solid (NaCl), liquid (H 2 O), and gas (N 2 ).

Prentice Hall © 2003Chapter 3 Interconverting Masses, Moles, and Number of Particles Molar mass: sum of the molar masses of the atoms: molar mass of N 2 = 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 Mole

Prentice Hall © 2003Chapter 3 Start with mass % of elements (i.e. empirical data) and calculate a formula Do 50 and 52 as an example Empirical Formulas from Analyses

Prentice Hall © 2003Chapter 3 Combustion Analysis Empirical formulas are determined by combustion analysis: Empirical Formulas from Analyses

Prentice Hall © 2003Chapter 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 Analyses

Prentice Hall © 2003Chapter 3 A g sample burns in oxygen to produce g of carbon dioxide and 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?

Prentice Hall © 2003Chapter 3 The insecticide DDD contains only carbon, hydrogen and chlorine. When 3.200g is burned, g of carbon dioxide and g of water are formed. What is the simplest formula of DDD?

Prentice Hall © 2003Chapter 3 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 from Balanced Equations

Prentice Hall © 2003Chapter 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 Reactants

Prentice Hall © 2003Chapter 3 Limiting Reactants

Prentice Hall © 2003Chapter 3 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 Reactants

Prentice Hall © 2003Chapter 3 How many grams of aluminum sulfide can form from the reaction of 9.00g of aluminum with 8.00g of sulfur?

Prentice Hall © 2003Chapter 3 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?

Prentice Hall © 2003Chapter 3 End of Chapter 3: Stoichiometry: Calculations with Chemical Formulas and Equations