1 / 32

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.

kalli
Download Presentation

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related