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Chapter 3 (short) Chemical Reactions and Reaction Stoichiometry

Learn about balancing equations, types of reactions like combination, decomposition, combustion, and calculations involving moles and molar mass. Understand formula mass, molecular mass, and percent composition.

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Chapter 3 (short) Chemical Reactions and Reaction Stoichiometry

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  1. Lecture Presentation Chapter 3(short)Chemical Reactions andReaction Stoichiometry James F. Kirby Quinnipiac University Hamden, CT

  2. Balancing Equations • From the Law of Conservation of Mass: same number of atoms for each element on both sides. • Balance equation by changing coefficients, NOT subscripts.

  3. Other Symbols in Chemical Equations Δ CH4(g) + 2O2(g) CO2(g) + 2H2O(g) The states of matter for the reactants and products are often written in parentheses to the right of each formula or symbol. • (g) = gas; (l) = liquid; (s) = solid; • (aq) = dissolve in aqueous (water) solution

  4. Other Symbols in Chemical Equations Δ CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Other symbols can be used to represent conditions during the chemical reaction. One example is the use of Δ over the reaction arrow, which means heat is needed for the reaction to take place.

  5. Simple Patterns of Chemical Reactivity • Types of reactions, which can be predicted at this point • Combination reactions • Decomposition reactions • Combustion reactions

  6. Combination Reactions • In a combination reaction, two or more substances react to form one product.

  7. Combination Reaction Predictions: Metal and Nonmetal • You should be able to predict the product of a combination reaction between a metal and a nonmetal, like the one below. (Hint: Remember common charges for Groups!)

  8. Decomposition Reactions • In a decomposition reaction one substance breaks down into two or more substances. • In the air bag, solid sodium azide releases nitrogen gas quickly.

  9. Decomposition Reaction Predictions: Heating a Metal Carbonate • Metal carbonates decompose when heated to give off carbon dioxide and a metal oxide. • Balancing these equations is based on the charge of the metal. Δ CaCO3(s)CaO(s) + CO2(g)

  10. Combustion Reactions • Combustion reactions are rapid reactions that produce a flame. • Combustion reactions most often involve oxygen in the air as a reactant.

  11. Combustion Reaction Predictions • When burning compounds with C and H in them, the products are CO2 and H2O. C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

  12. Formula Mass • A formula massis the sum of the atomic masses for the atoms in a chemical formula. The formula mass of sulfuric acid, H2SO4, would be • 2(1.0 u) + 32.1 u + 4(16.0 u) = 98.1 u

  13. Molecular Mass • For a molecule, the formula mass is also called molecular mass. • A molecular massis the sum of the atomic masses of the atoms in a molecule. • For glucose, which has a molecular formula of C6H12O6, the molecular mass is 6(12.0 u) + 12(1.0 u) + 6(16.0 u) = 180.0 u

  14. (number of atoms)(atomic mass) × 100 % Element = (Formula mass of the compound) Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

  15. (6)(12.0 u) %C = (180.0 u) = 40.0% Percent Composition So the percentage of carbon in glucose is:

  16. Mole: Unit for Amount • The mole is the amount of substance containing as many units as there are atoms in exactly 12 grams of 12C atoms (Always state explicitly the identity of the unit) 1 mole of 12C contains 6.022 × 1023 atoms 12C 1 gram = 6.022 × 1023 u Avogadro’s number (NA) = number of units per mole NA = 6.022 × 1023 units.mole-1

  17. Molar Mass • A molar mass is the mass of 1 mole of a substance (i.e., g/mol). • The molar mass of an element is the mass per mole of its atoms • The formula mass (in u) will be the same number as the molar mass (in g/mol).

  18. Mole Relationships • One mole of atoms, ions, or molecules contains Avogadro’s number of those particles. • The number of atoms of an element in a mole of compound is the subscript in the formula (number of atoms of that element in the formula) times Avogadro’s number.

  19. Converting Amounts • Moles provide a bridge from the molecular scale to the real-world scale. • Using equalities, we can convert from mass to moles and from moles to atoms. • How many atoms in 3 g of copper (Cu)? • 3 g Cu x (1 mol Cu/63.5 g Cu) x (6.02 x 1023 atoms/1 mol Cu) = 3 x 1022 atoms

  20. Determining Empirical Formulas One can determine the empirical formula from the percent composition by following these three steps.

  21. Determining Empirical Formulas—an Example The compound para-aminobenzoic acid (PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

  22. 1 mol 12.01 g 1 mol 1.01 g 1 mol 14.01 g 1 mol 16.00 g Determining Empirical Formulas—an Example Assuming 100.00 g of para-aminobenzoicacid and converting to moles: C: 61.31 g × = 5.105 mol C H: 5.14 g × = 5.09 mol H N: 10.21 g × = 0.7288 mol N O: 23.33 g × = 1.456 mol O

  23. 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 0.7288 mol 0.7288 mol 1.458 mol 0.7288 mol Determining Empirical Formulas—an Example Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005 ≈ 7 H: = 6.984 ≈ 7 N: = 1.000 O: = 2.001 ≈ 2

  24. Determining Empirical Formulas—an Example These are the subscripts for the empirical formula: C7H7NO2

  25. Determining a Molecular Formula • Remember, the number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula. • If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula.

  26. Determining a Molecular Formula—an Example • The empirical formula of a compound was found to be CH. It has a molar mass of 78 g/mol. What is its molecular formula? • Solution: Molar mass CH = 13 g/mol number multiple = 78 g/mol /13 g/mol = 6 The molecular formula is C6H6.

  27. Quantitative Information from a Balanced Equation • The coefficients in the balanced equation show • Relative numbers of moles of reactants and products, which can be converted to mass

  28. Stoichiometric Calculations To convert from moles (or grams) of A to moles (grams) of B we use the mole ratio from the balanced equation.

  29. An Example of a Stoichiometric Calculation C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) How many grams of water can be produced from 1.00 g of glucose?

  30. An Example of a Stoichiometric Calculation The next calculation is to convert moles of glucose to moles of water using the mole ratio from the coefficients in the balanced equation.

  31. An Example of a Stoichiometric Calculation The last step is to convert moles of water to grams of water.

  32. Limiting Reactants • The limiting reactant is the reactant that will run out of first (in this case, the H2).

  33. Limiting Reactants In the example below, the O2 would be the excess reagent.

  34. Limiting Reactants • The limiting reactant is used in all calculations to determine amounts of products and reactants that are used in a reaction.

  35. Theoretical Yield • The theoretical yield is the maximum amount of product that can be made.(calculated through the stoichiometryproblem) • The actual yieldis the amount one actually produces and measures.

  36. actual yield theoretical yield Percent yield = × 100 Percent Yield One finds the percent yield by comparing the actual yield to the theoretical yield

  37. Percentage Yield Calculation When excess NH3and 90.4 g CuO are reacted, the theoretical yield is 72.2 g Cu. 2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + H2O(g) The actual yield is 58.3 g Cu. What is the percent yield? = 80.7%

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