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Stoichiometry

Stoichiometry. AP Chemistry Mr. Martin. Topics. Law of Conservation of Matter Balancing Chem Eq Mass Relationships in rxn’s Limiting Reagents Theoretical, Act and % yld Empirical and Molecular formulas. A. Balancing Equations.

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Stoichiometry

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  1. Stoichiometry AP Chemistry Mr. Martin

  2. Topics • Law of Conservation of Matter • Balancing Chem Eq • Mass Relationships in rxn’s • Limiting Reagents • Theoretical, Act and % yld • Empirical and Molecular formulas

  3. A. Balancing Equations • Law of Conservation of Mass – matter is neither created or destroyed – during chem rxn’s some matter is converted to energy – why we must bal equations • Stoichiometry – the relationship among the substances in chem rxn’s.

  4. Chemical Equations • < TARGET="display">

  5. Balancing Equations • < TARGET="display">

  6. Steps to Balance • < TARGET="display">

  7. Practice Problems • Cu + O2 ----> Cu2O • CaCl2 + AgNO3 ----> AgCl + Ca(NO3)2 • H2 + O2 ----> H2O • Mg + P4 ---> Mg3P2 • H2SO4 + NaOH ----> H2O + Na2SO4

  8. Solutions • 4Cu + O2 ----> 2Cu2O • CaCl2 + 2AgNO3 ----> 2AgCl + Ca(NO3)2 • 2H2 + O2 ----> 2H2O • 6Mg + P4 ---> 2Mg3P2 • H2SO4 + 2NaOH ----> 2H2O + Na2SO4

  9. Additional Practice • Fe2O3 + CO -> Fe3O4 + CO2 • Fe3O4 + CO -> FeO + CO2 • C12H22O11(s) + O2(g) -> CO2(g) + H2O • Fe(s) + O2(g) -> Fe2O3 • Ca(s) + H2O(l) -> Ca(OH)2(aq) + H2(g)

  10. B. Atomic and Molecular Weights • Atomic Mass Scale uses atomic mass units (AMU’S) relative mass scale originally based on hydrogen = 1 amu • Now 1 amu = 1/12 of the carbon-12 isotope • Average atomic masses (atomic weight) elements are a mixture of isotopes and therefore the mass of an element is a weighted average of the naturally occurring isotopes.

  11. Calculation of Atomic Mass • AM = sum (abundance) x (isotope mass) of all the naturally occurring isotopes Ex: Ne-20 19.992 amu 90% abund Ne-22 21.991 amu 10 % abund (0.9 x 19.992) + (0.1 x 21.991) = 20.192 amu’s atomic mass

  12. Formula Weight • Sum of the average atomic masses of each atom in it’s chem ormula • EX C12H22O11 C (12) 12 = 144 amu’s H (22) 1 = 22 amu’s O (11) 16 = 176 amu’s 342 amu’s formula weight

  13. Percent Composition by Weight • What is the % comp of Methane CH4?

  14. A number 6.02 x 10^23 – like a dozen = 12 , a pair = 2, or a gross = 144 A mol is the amount of matter that contains as many objects (atoms, molecules or formula units) as the # of atoms in exactly 12g of the carbon-12 isotope. The Mole

  15. One mole of Carbon = 6.02x10^23 atoms of carbon One mole of H2O is equal to 6.02 x 10^23 H2O molecules. One mole of HCl is equal to 6.02 x 10^23 formula units of HCl. Molar Mass = 12.01 g of Carbon Molecular Weight = H2O H (2) 1 = 2 0 (1) 16 = 16 18 g/mol Formula Weight = HCl 36.5g/mol The Mole and Molar Mass

  16. Conversion Problems • Gram moles  atoms, fu’s, molecules • How many moles are in 200g of sulfuric acid. ( 2.04 mole of H2SO4)

  17. Conversion Problems • How many formula units are in 10g of sulfuric acid? ( 6.14 x 10^22 fu’s)

  18. Empirical and Molecular Formulas • Empirical Formulas – gives the lowest whole # ratio of atoms (moles) in a compound. • EF’s are derived from exp info most notably % comp.

  19. Example Problems • An analysis of sodium dichromate gives the following mass percentages, 17.5% Na, 39.7 % Cr, and 42.8% O. Calc the emp. formula?

  20. Example Problem Mol. Formula • Molecular formula – is a multiple of the emp. formula. n = exp mm/ef weight • The % composition of acetaldehyde is 54% C, 9.2% H, and 36.3% O. Its molar mass is 88 amu’s. What is its molecular formula.

  21. A method of analysis where a sample cont. C,H, O undergoes complete combustion. The gas stream goes through two tubes one absorbs CO2, the other H2O. Combustion Analysis

  22. Combustion Analysis cont. • The change of mass in the tubes gives the mass of CO2, and H2O. • A 4.24mg Acedic Acid sample is completely burned yielding 6.24mg of CO2 and 2.54mg of H2O. What is the mass % of each element?

  23. Quantitive Information from Bal. Eq grms   mol   mol   grms Ex. A (moles to moles) How many moles of H2O will be produced if 17.5 moles of H2 combine with excess O2? (17.5) 

  24. Example Problems Cont. • How many grams of KClO3 must decompose to produce 185.0 moles of O2? (1.51 x 10^4 g)

  25. Example Prob Cont. • How many grams of silver chloride can be produced from the reaction of 17.5 g of silver nitrate with excess sodium chloride solution? (14.4g)

  26. Example Problems • How many mg of aluminum nitrate would be required to completely react with 1.75 kg of barium hydroxide? (1.45 x 10^6mg)

  27. Limiting Reagent & Percent Yld • The reactant that is entirely consumed when a reaction goes to completion is the limiting reagent. • The reactant that is not completely consumed is said to be in excess. • The amount of product that is produced is determined by the limiting reactant.

  28. Ex. Limiting Reagent Prob. • In a lab test reaction 20.0g of CH3CHO and 10.0g of O2 were put in a reaction vessel. (a) How many grams of acedic acid can be produced (b) How many grams of excess reactant remain after the reaction is complete.

  29. Theoretical and Percent Yld. • The max amount of product that can be obtained is the theoretical yield. ( based on stoichiometric calc.) • The actual yield is the mass obtained by the actual reaction. • % yield = actual yld/theorical yld x 100

  30. Problem • The amount of acedic acid produced from the previous reaction was 23.8g. What is the percent yield of the reaction. (87.2%)

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