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Ch. 10 Chemical Quantities

Ch. 10 Chemical Quantities. 3 Methods of Measuring. Counting Mass Volume. 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?. Example 1. Count: 1 dozen apples = 12 apples Mass: 1 dozen apples = 2.0 kg apples

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Ch. 10 Chemical Quantities

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  1. Ch. 10 Chemical Quantities

  2. 3 Methods of Measuring • Counting • Mass • Volume

  3. 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?

  4. 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

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

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

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

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

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

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

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

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

  13. 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

  14. 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

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

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

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

  18. 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

  19. 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

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

  21. 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

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

  23. 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

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

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

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

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

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

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

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

  31. 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

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

  33. 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

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

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

  36. Example 10 • Find the percentage of each element present in Al2 (CO3)3

  37. Example 10 • Al2(CO3)3 • Al= 27 g x 2 = 54 g / 234 g x 100=23% • C= 12 g x 3 = 36 g/ 234 g x 100= 15% • O = 16 g x 9 = 144 g / 234 g x 100=62% 234 gAl2(CO3)3

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