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Ch. 3 Stoichiometry: Calculations with Chemical Formulas

Ch. 3 Stoichiometry: Calculations with Chemical Formulas. Law of Conservation of Mass. Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged. Stoichiometry. The quantitative nature of chemical formulas and chemical reactions. Reactants.

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Ch. 3 Stoichiometry: Calculations with Chemical Formulas

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  1. Ch. 3 Stoichiometry: Calculations with Chemical Formulas

  2. Law of Conservation of Mass • Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged.

  3. Stoichiometry • The quantitative nature of chemical formulas and chemical reactions

  4. Reactants • The chemical formulas on the left of the arrow that represent the starting substances 2H2 + O2 2H2O Reactants

  5. Products • The substances that are produced in the reaction and appear to the right of the arrow 2H2 + O2 2H2O Products

  6. Because atoms are neither created nor destroyed in any reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  7. Balancing Chemical Equations CH4 + O2 CO2 + H2O C=1 C=1 H=4 H=2 O=2 O=3

  8. Balancing Chemical Equations CH4 + O2 CO2 + 2H2O C=1 C=1 H=4 H=2 X 2 =4 O=2 O=3

  9. Balancing Chemical Equations CH4 + O2 CO2 + 2H2O C=1 C=1 H=4 H=2 x 2 = 4 O=2 O= 2 + 2x1 = 4

  10. Balancing Chemical Equations CH4 + 2O2 CO2 + 2H2O C=1 C=1 H=4 H=2 x 2 = 4 O=2 x 2 = 4 O= 2 + 2x1 = 4

  11. Combustion Reactions • Rapid reactions that produce a flame. • Most combustion reactions involve O2 as a reactant • Form CO2 and H2O as products

  12. Combustion Reactions C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) C= 3 C=1 X 3 = 3 H=8 H=2 X 4 = 8 O= 2 X 5 = 10 O=(2 X3)+(1X4)=10

  13. Combination Reactions (synthesis) • 2 or more substances react to form 1 product.

  14. Combination Reactions (synthesis) 2Mg(s) + O2(g) 2MgO(s) Mg=1 x 2=2 Mg= 1 x 2=2 O= 2 O=1 x 2 = 2

  15. Decomposition Reaction • 1 substance undergoes a reaction to produce 2 or more substances

  16. Decomposition Reaction CaCO3 (s)  CaO (s) + CO2(g) Ca=1 Ca=1 C=1 C=1 O=3 O=1+2=3

  17. 3 Methods of Measuring • Counting • Mass • Volume

  18. Example 1 • If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?

  19. Example 1 • Count: 1 dozen apples = 12 apples • Mass: 1 dozen apples = 2.0 kg apples • Volume: 1 dozen apples = 0.20 bushels apples Conversion Factors: • 1 dozen2.0 k.g1 dozen 12 apples 1 dozen 0.20 bushels

  20. Example 1 • 0.50 bushel x 1 dozen x 2.0 kg = 0.20 bushel 1 dozen = 5.0 kg

  21. Avogadro’s Number • Named after the Italian scientist Amedo Avogadro di Quaregna • 6.02 x 10 23

  22. Mole (mol) • 1 mol = 6.02 x 10 23 representative particles • Representative particles: atoms, molecules ions, or formula units (ionic compound)

  23. Mole (mol) • Moles= representative x 1 mol particles 6.02 x 10 23

  24. Example 2 (atoms  mol) • How many moles is 2.80 x 10 24 atoms of silicon?

  25. Example 2 • 2.80 x 10 24 atoms Si x 1 mol Si 6.02 x 10 23 atoms Si = 4.65 mol Si

  26. Example 3 (mol  molecule) • How many molecules of water is 0.360 moles?

  27. Example 3 • 0.360 mol H2O x 6.02 x 10 23 molecules H2O 1 mol H2O =2.17 molecules H2O

  28. The Mass of a Mole of an Element • The atomic mass of an element expressed in grams = 1 mol of that element = molar mass Molar mass S Molar mass Hg Molar mass C Molar mass Fe

  29. 6.02 x 10 23 atoms S 6.02 x 10 23 atoms Hg 6.02 x 10 23 atoms C 6.02 x 10 23 atoms Fe

  30. Example 4 (mol  gram) • If you have 4.5 mols of sodium, how much does it weigh?

  31. Example 4 • .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na 1 mol Na

  32. Example 5 (grams  atoms) • If you have 34.3 g of Iron, how many atoms are present?

  33. Example 5 • 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms 55.8 g Fe 1 mol Fe =3.70 x 10 23 atoms Fe

  34. The Mass of a Mole of a Compound • To find the mass of a mole of a compound you must know the formula of the compound • H2O  H= 1 g x 2 O= 16 g 18 g = 1 mole = 6.02 x 10 23 molecules

  35. Example 6 (gram  mol) • What is the mass of 1 mole of sodium hydrogen carbonate?

  36. Example 6 • Sodium Hydrogen Carbonate = NaHCO3 • Na=23 g • H=1 g • C=12 g • O=16 g x3 • 84 g NaHCO3 = 1 mol NaHCO3

  37. Mole-Volume Relationship • Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical

  38. Avogadro’s Hypothesis • States that equal volumes of gases at the same temperature and pressure contain the same number of particles • Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible

  39. Standard Temperature and Pressure (STP) • Volume of a gas changes depending on temperature and pressure • STP= 0oC (273 K) 101.3 kPa (1 atm)

  40. Standard Temperature and Pressure (STP) • At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume

  41. Conversion Factors • AT STP • 1 mol gas22.4 L gas 22.4 L gas 1 mol gas

  42. Example 7 • At STP, what volume does 1.25 mol He occupy?

  43. Example 7 • 1.25 mol He x 22.4 L He = 28.0 L He 1 mol He

  44. Example 8 • If a tank contains 100. L of O2 gas, how many moles are present?

  45. Example 8 • 100. L O2 X 1 mol O2 = 4.46 mol O2 22.4 L O2

  46. Calculating Molar Mass from Density • The density of a gas at STP is measured in g/L • This value can be sued to determine the molar mass of gas present

  47. Example 9 • A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.

  48. Example 9 • 1 mol gas x 22.4 L gas X 3.58 g gas = 1 mol gas 1 L gas Molar Mass= 80.2 g

  49. Percent Composition • The relative amounts of the elements in a compound • These percentages must equal 100

  50. Percent Composition • %element = mass of element x 100 mass of compound

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