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Stoichiometry

Stoichiometry. Chapter 3. Atomic Mass. 1961: Atomic Mass is based on 12 C 12 C is assigned a mass of EXACTLY 12 AMU Most Accurate Way to compare masses is to use a mass spectrometer Results in ratio of masses like 13 C/ 12 C Apply ratio to mass of 12 C to get atomic mass.

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Stoichiometry

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  1. Stoichiometry Chapter 3

  2. Atomic Mass • 1961: Atomic Mass is based on 12C • 12C is assigned a mass of EXACTLY 12 AMU • Most Accurate Way to compare masses is to use a mass spectrometer • Results in ratio of masses like 13C/12C • Apply ratio to mass of 12C to get atomic mass. • Atomic Mass on Periodic Table is a weighted average of the isotopes • 98.89% 12C, 1.11% 13C, negligible amounts of 14C • (.9889)(12) + (.011)(13.0034) = 12.01amu

  3. The Mole • Mole: the number equal to the number of carbon atoms in exactly 12 grams of pure 12C • Avogadro’s number: 6.022x1023 units = 1 mole • 1 mole of an element weighs exactly the atomic mass of the element in grams • 1 mole of a molecule weighs exactly the atomic mass of the molecule in grams

  4. Calculating Number of Atoms • A silicon chip for a computer has a mass of 5.68 mg. How many atoms of silicon are in the chip? • Convert milligrams to grams • Convert grams to moles • Convert moles to atoms

  5. Percent Composition of Compounds • 2 common ways to describe the composition of a compound: • Number of constituent atoms • Percent composition by mass • Example: % composition of ethanol (C2H5OH) • 2 mol C x 12.01 g/mol = 24.02g C • 6 mol H x 1.008 g/mol = 6.048 g H • 1 mol O x 16.00 g/mol – 16.00 g O • % C = 24.02g/(24.02g+6.048g+16.00g) x 100 = 52.14% C • %H = 6.048g/46.07g x 100 = 13.13% H • %O = 16.00g/46.07g x 100 = 34.73% O

  6. Determining Formula of Compounds • You have 0.1156g of a new compound composed of C, H, and N • You decomposed or reacted it with water to get CO2, and H2O, which was collected and weighed. • 0.1638g CO2 • 0.1676g H2O • How can you calculate the formula of the compound? • How much carbon was in the original compound? • How much nitrogen was in the original compound?

  7. Determining Formula of Compounds • Use % by mass of the product compounds: • C: 1 mol x 12.01 g/mol = 12.01g C • O: 2 mole x16.00g/mol = 32.00g O • Molar mass CO2 = 44.01 g/mol • 0.1638 g CO2 x 12.01gC/44.01gCO2 = 0.004470 g C • 0.00470g CO2/0.1156g compound x 100% = 38.67% C • 0.1676 g H2O x 2.016 g H/18.02 g H2O = 0.001875 g H • 0.001875 g H /0.1156g compound x 100% = 16.22% H • Rest (100 – 38.67 – 16.22) is N (45.11%)

  8. Calculate Empirical Formula • From % by mass, calculate Empirical Formula • ASSUME YOU HAVE 100 g OF COMPOUND • 38.67 g C x 1 mol/12.01 g = 3.220 mol C • 16.22 g H x 1 mol/1.008 g = 16.09 mol H • 45.11 g N x 1 mol/14.01 g = 3.219 mol N • Find smallest WHOLE-NUMBER ratios of elements: • 3.220 mol C/3.219 mol N = 1.000 = 1:1 • 16.09 mol O/3.220 mol C = 4.997 = 5:1 • Formula would be: CH5N • Or any whole number multiple of those elements

  9. Actual Molecular Formula • Must determine Actual molecular mass experimentally • Suppose compound is known to have mass of 62.12 g/mol • Actual mass/Empirical mass gives ratio • 62.12g/mol / 31.06 G/mol = 2.0 • Actual formula is (CH5N)x2 = C2H10N2 • Solve Questions 69-75 (one per group), page 119-120 in textbook.

  10. Limiting Reactants • Some chemical processes are carried out with excess amounts of one or more reactants. • Chemically we learn about this in equilibrium • The excess reactants are usually the cheaper ones. • How do you handle questions where you have quantities of reactants that are not stoichiometrically balanced? • Concept of limiting reactant.

  11. Limiting Reactants • How to solve the problem: • Solve the problem once for each reactant that MIGHT be limiting. • Determine which solution has the lowest number of product moles • That is your limiting reactant and maximum amount of product. • Any other reactant above the stoichiometric amount of the limiting reactant is “excess”

  12. Limiting Reactant • Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithim carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00kg of lithium hydroxide? • Step 1: Write the unbalanced equation: • LiOH(s) + CO2(g)  Li2CO3(s) + H2O(l) • Balance the equation: • 2LiOH(s) + CO2(g)  Li2CO3(s) + H2O(l)

  13. Limiting Reactant • 2LiOH(s) + CO2(g)  Li2CO3(s) + H2O(l) • Step 2: convert LiOH to moles Step 3: Determine amount of CO2 that reacts with a Given amount of LiOH Step 4: Calculate moles of CO2 that reacts with a given mass of LiOH

  14. Limiting Reagent • Step 5: Calculate mass of CO2 using molar mass

  15. Limiting Reactant • Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products are solid copper and water vapor. If a sample of 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How manuy grams of N2 will be formed? • Balanced equation: • 2NH3(g) + 3CuO(s) N2(g) + 3 Cu(s) + 3H2O(g)

  16. Limiting Reactant • Compute moles of NH3 and CuO • 18.1g NH3 x 1mol NH3/17.03gNH3 = 1.06mol NH3 • 90.4g CuO x 1 mol CuO/79.55g CuO = 1.14 mol CuO • Use Mole ratio of CuO and NH3 to determine limiting reactant: • 1.06 mol NH3 x 3mol CuO/2mol NH3 = 1.59 mol CuO • Because it takes 1.59 mole of CuO to react with 1.06 mole of NH3, but we only have 1.14 mole of CuO, CuO is the limiting reactant. • We will run out of CuO before we run out of NH3 • Verify: • Balanced Equation: Mol CuO/Mol NH3 = 3/2 = 1.5 • Amount Present: Mole CuO/Mol NH3 = 1.14/1.06 = 1.08

  17. Limiting Reactant • Calculate amount of N2 formed based on moles of CuO (limiting factor) present: • 1 mol N2/3 mol CuO x 1.14 mol CuO = 0.380 mol N2

  18. % Yield • Theoretical yield is the amount of product formed based on actual amounts and balanced equation stoichiometric calculations • Actual yield is the measured amount of product formed • % yield is actual yield/theoretical yield * 100 • Assignment: Questions 97a, 98c, 99-103 • One for each group

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