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Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

CHEMISTRY The Central Science 9th Edition. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations. David P. White. Chemical Equations. Lavoisier: mass is conserved in a chemical reaction. Chemical equations: descriptions of chemical reactions.

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Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

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  1. CHEMISTRYThe Central Science 9th Edition Chapter 3Stoichiometry: Calculations with Chemical Formulas and Equations David P. White Chapter 3

  2. Chemical Equations • 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 Chapter 3

  3. Chemical 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 Chapter 3

  4. Chemical Equations 2Na + 2H2O  2NaOH + H2 2K + 2H2O  2KOH + H2 Chapter 3

  5. Chemical Equations • Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products. Chapter 3

  6. Chemical Equations Chapter 3

  7. Chemical Equations • Law of conservation of mass: matter cannot be lost in any chemical reactions. Chapter 3

  8. Some Simple Patterns of Chemical Reactivity 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. Chapter 3

  9. Some Simple Patterns of Chemical Reactivity Combination and Decomposition Reactions Chapter 3

  10. Some Simple Patterns of Chemical Reactivity Combination and Decomposition Reactions Chapter 3

  11. Some Simple Patterns of Chemical Reactivity Combustion in Air Combustion is the burning of a substance in oxygen from air: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Chapter 3

  12. Formula Weights 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) Chapter 3

  13. Formula 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: Chapter 3

  14. The Mole • 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. Chapter 3

  15. The Mole

  16. The Mole Chapter 3

  17. The Mole This photograph shows one mole of solid (NaCl), liquid (H2O), and gas (N2). Chapter 3

  18. The Mole • 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. Chapter 3

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

  20. Empirical Formulas from Analyses • Combustion Analysis • Empirical formulas are determined by combustion analysis: Chapter 3

  21. Empirical Formulas from Analyses • 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 Chapter 3

  22. 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? Chapter 3

  23. 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? Chapter 3

  24. Quantitative Information from Balanced Equations • 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). Chapter 3

  25. Limiting Reactants • 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 Chapter 3

  26. Limiting Reactants Chapter 3

  27. Limiting 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: Chapter 3

  28. How many grams of aluminum sulfide can form from the reaction of 9.00g of aluminum with 8.00g of sulfur? Chapter 3

  29. 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? Chapter 3

  30. End of Chapter 3:Stoichiometry: Calculations with Chemical Formulas and Equations Chapter 3

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