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Stoichiometry

Learn about stoichiometry concepts such as molar mass, percent composition, moles, conversions, empirical formulas, and molecular formulas. Understand how to calculate molar mass, determine percent composition, convert between mass and moles, find empirical formulas, and determine molecular formulas.

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Stoichiometry

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  1. Stoichiometry Molar mass, Percent composition, Moles, Conversions, Empirical formulas, Molecular formulas

  2. The Mole • SI unit for amount • A counting unit for measuring large quantities of small items • Atoms • molecules • Particles 1 mole = 6.022 x 10 23 atoms = molar mass

  3. Molar Mass • The amount of mass (grams) in one mole of a substance • For elements • Average atomic mass  periodic table • For compounds • Must add up each element’s mass • Subscripts are factored in by multiplication • Units are g/mol • May be called formula mass or molecular mass

  4. Steps to solve • Find elements on periodic table • Write down the mass • Multiply the mass by the subscript • Add together

  5. Sample: Determine Molar Mass Calculations are FUN! 40.08 g/mol 2. 55.847 g/mol • Ca • Fe • NaCl Na: 22.989 g/mol Cl: 35.453 g/mol 58.443 g/mol

  6. Molar Mass (cont.) 4. Fe2O3 5. Na2O • Fe: 55.847 x (2) = 111.694 g/mol • O: 15.999x (3)= 47.997 g/mol • 159.691 g/mol • Na: 22.989 x(2) =  45.978 g/mol • O: 15.999= 15.999 g/mol • 61.977 g/mol

  7. Percent Composition • Steps to solve • Find the total mass of the COMPOUND • Take the mass of each element and divide it by the total mass • Multiply by 100

  8. Sample: Calculate the percent Composition C: 12.011 44.009 O: 31.998 44.009 CO2 C: 12.011 g/mol O: (2)15.999 g/mol = 31.998 g/mol 44.009 g/mol .2730 100 27.3 % 100 72.7% .727

  9. Sample 2 H2SO4 H: (2)1.008 g/mol = 2.016 g/mol S: 32.064 g/mol O: (4)15.999 g/mol = 63.996 g/mol 98.076 g/mol H: 2.016 98.076 S: 32.064 98.076 O: 63.996 98.076 .0206 100 2.06 % .327 100 32.7 % .653 100 65.3%

  10. Composition of Hydrates Hydrate: crystal contains water within • Water can fit into the salt crystal • Happens in fixed ratios • Each salt that forms a hydrate has a DIFFERENT water ratio • Each salt crystal is unique Anhydrous: water has been removed • Achieved through drying • Steps to solve • Calculate the total mass of the compound INCLUDING THE water • Divide the water mass by total mass • Multiply by 100

  11. That was nothing. Time for Coco! Sample Problem: Hydrate • What percentage of water is found in CuSO4 •5H2O • Cu: 63.54 • S: 32.064 • O: (4)15.999= 63.996 • H2O: (5)18.01= 90.05 • 249.65 90.05 249.65 100 .361 36.1% H2O

  12. The Mole conversion Mass (grams) Particles Ions Atoms Molecules Formula units ÷ by molar mass X by 6.022 x 10 23 Mole ÷ by 6.022 x 10 23 X by Molar mass ÷ by 22.4 L x by 22.4 L Volume (Liters) Gases at STP

  13. Stoichiometric Conversions: Mass to Moles • Steps to convert • Identify what you start with • Determine what units you end in • Might have to do side calculations • Set up conversion factor • Evaluate • Do the Math! • You can relate any unit to another by this method • Relate one unit to another • conversion factors

  14. Example • How many moles are in 28 grams of CO2? 28 grams CO2 .636 _______ moles CO2 C: 12.011g/mol O: (2)15.999 g/mol = 31.998 g/mol 44.009 g/mol

  15. Empirical Formula • When the elements are in the smallest whole number ratio within compound Steps to solve • When given a compound • Divide out by the common multiple Steps to solve • When given percentages • Change percent sign to grams (NO MATH) • Convert masses to moles • Using conversions • Re-divide ALL moles amounts by the smallest mole amount • Multiply if not whole numbers • Write compound with subscripts

  16. Example: What’s my Empirical Formula Piece of Cake! • C6H6 • C8H18 • C2H6O2 • X39Y13 • CH • C4H9 • CH3O • X3Y

  17. Example: with percent A compound is found to contain 36.48% Sodium, 25.41% Sulfur, and 38.11% Oxygen. Find its empirical formula. Na2SO3

  18. Molecular Formula Multiple of an empirical formula So many compounds with the same Empirical formula

  19. Steps to solve • Complete an empirical formula process if needed • Find the mass of the EMPIRICAL FORMULA • Divide the Empirical formula mass by the molecular mass • Molecular mass is normally given in the problem • Answer is the common multiple • Multiple the SUBSCRIPTS by the common multiple

  20. All done! Time to go sledding! Yippee!! Example • A compound with an empirical formula of C4H4O and a molecular mass of 136 grams. What is the molecular formula of this compound? • C: 4 (12.011) = 48.044 • H:4 (1.009 ) = 4.036 • O: 15.999 • 68.079 C4H4O x 2 = C8H8O2

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