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Stoichiometry Review

Stoichiometry Review. Spring Final Exam. Molar Mass the mass of one mole of a substance. (207.2 g). Pb :. Pb O 2. 1. = 207.2 g. 239.3 g. O :. = 32.0 g. 2. (16.0 g). (1.0 g). H :. 1. = 1.0 g. H NO 3. 63.0 g. N:. = 14.0 g. 1. (14.0 g). O:. = 48.0 g. 3. (16.0 g).

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Stoichiometry Review

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  1. Stoichiometry Review Spring Final Exam

  2. Molar Massthe mass of one mole of a substance (207.2 g) Pb: PbO2 1 = 207.2 g 239.3 g O: = 32.0 g 2 (16.0 g) (1.0 g) H: 1 = 1.0 g HNO3 63.0 g N: = 14.0 g 1 (14.0 g) O: = 48.0 g 3 (16.0 g)

  3. g element x 100 % of element = molar mass of compound : : : : : : percentage composition: the mass % of each element in a compound (see calcs above) Find % composition. 207.2 g Pb = 86.6% Pb PbO2 239.2 g 32.0 g O = 13.4% O 239.2 g 42.0 g N = 28.2% N 149.0 g (NH4)3PO4 12.0 g H = 8.1% H 149.0 g 31.2 g P = 20.8% P 149.0 g 64.0 g O = 43.0% O 149.0 g

  4. Finding an Empirical Formula from Experimental Data 1. Find # of g of each element. 2. Convert each g to mol. 3. Divide each “# of mol” by the smallest “# of mol.” 4. Use ratio to find formula. A compound is 45.5% yttrium and 54.5% chlorine. Find its empirical formula. YCl3

  5. To find molecular formula… A. Find empirical formula. B. Find molar mass of empirical formula. C. Find x= MM molecular MM empirical D. Multiply all parts of empirical formula by x. (How many empiricals “fit into” the molecular?)

  6. A carbon/hydrogen compound is 7.7% H and has a molar mass of 78 g. Find its molecular formula. emp. form.  CH 78 g mmemp = 13 g C6H6 = 6 13 g

  7. NO2 92 g 46 g A compound has 26.33 g nitrogen, 60.20 g oxygen, and molar mass 92 g. Find molecular formula. mmemp = 46 g N2O4 = 2

  8. 1 mol = 6.02 x 1023 particles 1 mol = molar mass (in g) Mass (g) 1 mol = 22.4 L MOLE (mol) Volume (L or dm3) 1 mol = 22.4 dm3 Particle (at. or m’c) Island Diagram: a. Diagram has four islands. b. “Mass Island” for elements or compounds c. “Particle Island” for atoms or molecules d. “Volume Island”: for gases only 1 mol @ STP = 22.4 L = 22.4 dm3

  9. ( ) ( ) ( ) What mass is 1.29 mol ? Iron (II) nitrate Fe2+ NO31– Fe(NO3)2 179.8 g 1.29 mol = 232 g 1 mol How many molecules is 415 L at STP? sulfur dioxide sulfur dioxide SO2 1 mol 6.02 x 1023 m’c 415 L 1 mol 22.4 L = 1.12 x 1025 m’c

  10. ( ) ( ) ( ( ) ) aluminumsulfate aluminum sulfate What mass is 6.29 x 1024 m’cules ? Al3+ SO42– Al2(SO4)3 342.3 g 1 mol 6.29 x 1024 m’c 1 mol 6.02 x 1023 m’c = 3580 g At STP, how many g is 87.3 L of nitrogen gas? N2 1 mol 28.0 g 87.3 L = 109 g 22.4 L 1 mol

  11. Chemical Reactions --Chemical Equations-- • During a chem. rxn.; atoms are rearranged (NOT created or destroyed!) • Chemical equations must be balanced to show the relative amounts of all substances. • Balanced means: each side of the equations has the same # of atoms of each element. • CH4 + O2 —> H2O + CO2 • CH4+ 2O2 —> 2H2O + CO2 Unbalanced Balanced

  12. Balanced? Cellulose reacts with oxygen gas to form carbon dioxide gas & liquid water. C6H10O5 + 6O2 6CO2 +5H2O • 6 C6 • 10 H10 • 17 O17 Balanced!!!

  13. Balanced? Nitroglycerin decomposes to form nitrogen gas, oxygen gas, carbon dioxide gas & water vapor 2 C3H5(NO3)3 3N2 + O2 + 6CO2 +5H2O • 6 C 6 • 10 H 10 • 6 N 6 • 18 O 19 Not Balanced!

  14. Balancing Chemical Equat. • When balancing equations, you may change coefficients as much as you need to, but you may never change subscripts because you can’t change what substances are involved.

  15. H2(g) + O2 (g)  H2O (l) Example: • Atom Inventory: Reactants Products 2 H 2 2 O 1 • Add coefficients to balance 3. Double check to make sure it is all BALANCED! 2 2 4 4 2

  16. Balancing Chemical Equat. • Balancing Chemical Equations practice

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