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Chapter 3

Chapter 3. Atomic masses Average atomic mass Ex. What is the avg. atomic mass of a sample that is 69.09% 62.93 amu and 30.91% 64.93 amu? 0.6909(62.93amu) + 0.3091(64.93amu) = 63.55 amu. Mole Molar mass What is the mass of 3 moles of KOH? 3 n KOH x 56 g/n = 168 g = 200 g s.f.

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Chapter 3

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  1. Chapter 3 • Atomic masses • Average atomic mass • Ex. What is the avg. atomic mass of a sample that is 69.09% 62.93 amu and 30.91% 64.93 amu? • 0.6909(62.93amu) + 0.3091(64.93amu) = 63.55 amu

  2. Mole • Molar mass • What is the mass of 3 moles of KOH? • 3 n KOH x 56 g/n = 168 g = 200 g s.f. • How many atoms are in 15 g of K? • 15 g x 1n/39.1g x 6.022x1023atoms/n = 2.3x1023 atoms

  3. Percent composition • Ex. • What is the percent comp. of H2O? • %H = 2.02 g / 18.02 g x 100 = 11.2 % • %O = 16.0 g / 18.02 g x 100 = 88.8 %

  4. Determining Formulas • Empirical Formula • Steps • Mass • Moles • Divide • Ex. 8 g of O and 32 g of S • 8 g O x 1n/16g = 0.5 n O / 0.5 n = 1 O • 32 g S x 1n/32g = 1.0 n S / 0.5 n = 2 S • S2O

  5. Molecular Formula • Ex. • S2O has a molar mass of 160 g/n, what’s the molecular formula • 160 g/n / 80 g/n = 2 so S2O becomes • S4O2 • If given percentages assume its out of 100 g and change the sign from % to g.

  6. Chemical Equations • Reactants  Products • Balancing • Ex.s • H2 + O2 H2O • 2H2 + O2 2H2O • C2H6 + O2 CO2 + H2O • C2H6 + 7/2O2 2CO2 + 3H2O • 2C2H6 + 7O2 4CO2 + 6H2O

  7. Stoichiometry • Mole ratio • 2H2 + O2 2H2O • 2n H2 / 1n O2 • 2n H2 / 2n H2O • 1n O2/ 2n H2O

  8. How much NaCl is formed from 23 g of Na and excess Cl2? • 2Na + Cl2 2NaCl • 23g Na x 1n/23g x 2n NaCl / 2n Na = 1n NaCl • If we wanted to go to mass we would have multiplied by the molar mass of NaCl.

  9. How much oxygen is needed to burn 15 g of CH4? • CH4 + O2→ CO2 + H2O • CH4 + 2O2→ CO2 + 2H2O • 15g CH4 x 1n / 16.05 g x 2n O2 / 1n CH4 • 1.9 n O2 if we want mass then use molar mass

  10. Limiting Reactant • Cheeseburgers • 24 buns, 20 patties, 48 slices of cheese • With reactions we have to prove which reactant runs out by doing a stoichiometry problem.

  11. How much water is produced from 10. g of H2 and 10. g of O2? • 2H2 + O2 2H2O • 10g H2 x 1n/2g x 1n O2/ 2n H2 x 32 g/n = 80 g O2 • O2 is the limiting reactant need 80g have 10g! • 10.g O2 x 1n/32g x 2n H2O/1n O2 x 18g/n= 11g H2O

  12. If we want to know the remaining reactant we just start with the limiting reactant and calculate the amount of reactant used. • 10.g O2 x 1n/32g x 2n H2/1n O2 x 2g/n= 1.25g H2 10 g – 1.25 g = 8.75 g of H2 left over • Percent Yield • actual / theoretical x 100 • If a lab using the previous example is conducted and only 5.0 g of water is produced what’s the percent yield? • 5.0 g / 11 g x 100 = 46 %

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