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What is the empirical formula for the following unknown sample?

What is the empirical formula for the following unknown sample?. Sample is a carbohydrate ( only C, H & O). 50.00 g of sample yields: 73.29 g CO 2 30.00 g H 2 O Molar masses CO 2 : 44.01 g/mol H 2 O: 18.02 g/mol. Some Compounds with Empirical Formula

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What is the empirical formula for the following unknown sample?

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  1. What is the empirical formula for the following unknown sample? Sample is a carbohydrate (only C, H & O). 50.00 g of sample yields: 73.29 g CO2 30.00 g H2O Molar masses CO2: 44.01 g/mol H2O: 18.02 g/mol

  2. Some Compounds with Empirical Formula CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O) Molecular M Formula (g/mol) Name Use or Function CH2O 30.03 Formaldehyde Disinfectant; Biological preservative C2H4O2 60.05 Acetic acid Acetate polymers; vinegar ( 5% solution) C3H6O3 90.08 Lactic acid Causes milk to sour; forms in muscle during exercise C4H8O4 120.10 Erythrose Forms during sugar metabolism C5H10O5 150.13 Ribose Component of many nucleic acids and vitamin B2 C6H12O6 180.16 Glucose Major nutrient for energy in cells

  3. Ascorbic Acid ( Vitamin C ) - I Contains C , H , and O • Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O • Calculate it’s Empirical formula! • C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C • H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O) = 2.92 x 10-4 g H • Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O

  4. Vitamin C Combustion - II • C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C • H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H • O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O • Divide each by 2.21 x 10-4 • C = 1.00 Multiply each by 3 = 3.00 = 3.0 • H = 1.32 = 3.96 = 4.0 • O = 1.00= 3.00 = 3.0 C3H4O3

  5. Vitamin C

  6. Empirical v. Molecular Formula Compound EF MF n _ formaldehydeCH2O CH2O 1 acetic acidCH2O C2H4O2 2 glucoseCH2O C6H12O6 6 All 3 compounds are: 40.00 % C (by mass) 6.714 % H 53.27 % O n x (empirical formula) = molecular formula

  7. How is molecular formula determined? Must measure the mass of a molecule using Mass Spectrometry

  8. What is the molecular formula for the following unknown sample? Elemental analysis shows: 75.40 % C 4.43 % H 20.10 % O Mass spectrometry gives: molecular weight = 318 amu n = (molecular wt) / (formula wt)

  9. Chemical Equation C2H5OH+3O22CO2+3H2O Check that the equation is balanced. 1 mole of ethanolreacts with ► 3 moles of oxygen 1 mole of ethanolreacts to produce ► 2 moles of carbon dioxide ► 3 moles of water

  10. __ NH3 + __ O2→ __ NO + __ H2O • Balance the chemical equation • For every 1 mole of NH3 reacted, how many moles of NO are produced? • How many moles of O2 react with 2.0 moles of NH3

  11. Mole ratios:relating moles within chemical equations 1 5 3 4 • If we had 2 moles of C3H8 • - how many moles of CO2 would be produced? • - how many moles of H2O would be produced? • If we had 0.5 moles of C3H8 • - how many moles of O2 would be needed?

  12. Solving Mass  Moles Reaction Problems grams molar mass moles mass Mm = mole Generic Reaction:aA + bB→cC + dD mole ratio d a mass A (g) moles A moles D mass D (g)

  13. Stoichiometry Problem • If you burned 4.4 grams of propane in a plentiful supply of oxygen • - how many grams of O2 would be consumed? • - how many grams of carbon dioxide would be produced? 1 5 3 4

  14. Ethanol combusts to form CO2 & water. __C2H5OH + __O2 __CO2 + __H2O • Balance the chemical equation • Identify as many mole ratios as you can • How many moles of oxygen (O2) react with 15.0 moles of ethanol • How many grams of O2 react with 15.0 moles of ethanol • How many g of CO2 are formed when 1.00 kg of ethanol is burned with 1.00 kg of O2

  15. Sardine & Swiss on Rye Sandwich 20 sardines 12 slices 20 slices

  16. An Ice Cream Sundae Analogy for Limiting Reactions Fig. 3.10

  17. Ethanol combusts to form CO2 & water. __C2H5OH + __O2 __CO2 + __H2O • Balance the chemical equation • Identify as many mole ratios as you can • How many moles of oxygen (O2) react with 15.0 moles of ethanol • How many grams of O2 react with 15.0 moles of ethanol • How many g of CO2 are formed when 1.00 kg of ethanol is burned with 1.00 kg of O2

  18. Limiting Reactant (or reagent) The limiting reactant is the reactant in a chemical reaction which limits the amount of products that can be formed. The limiting reactant in a chemical reaction is present in insufficient quantity to consume the other reactant(s). This situation arises when reactants are mixed in non-stoichiometric ratios.

  19. Limiting Reactant Example 1 76.15 g/mol 32.00 g/mol

  20. Limiting Reactant Example 3 4NH3 + 5O2→ 4NO + 6H2O Add: 14 mol 20 mol Could make 16 mol NO Could make 14 mol NO NH3 is the limiting reagent. (Use this as basis for all further calculations)

  21. Using Stoichiometry Stoichiometry is used to answer two fundamental questions in chemical analysis: What is the theoretical yield? What is the limiting reactant? REMEMBER: stoichiometry shows molar ratios not mass ratios

  22. Percent yield actual yield % yield = actual yield: observed yield of product theoretical yield: calculated assuming 100% conversion of the LIMITING REAGENT Both yields can be in moles or grams x 100 theoretical yield

  23. Theoretical Yield: Which Reactant is Limiting? 1) calculate moles (or mass) of product formed by complete reaction of each reactant. 2) the reactant that yields the least product is the limiting reactant (or limiting reagent). 3) the theoretical yield for a reaction is the maximum amount of product that could be generated by complete consumption of the limiting reagent.

  24. Limiting Reactant Example 2 • When 66.6 g of O2 gas is mixed with 27.8 g of NH3 gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is produced by the following reaction: • 16.04 17.03 32.00 27.03 g/mol • 2CH4 + 2NH3 + 3O2→ 2HCN + 6H2O • What is the % yield of HCN in this reaction? • How many grams of NH3 remain?

  25. Mass to moles 66.6 g of O2→ 2.08 mol O2 27.8 g of NH3 → 1.63 mol NH3 25.1 g of CH4→ 1.56 mol CH4 Which reactant is limiting? 2.08 mol O2 can yield 1.39 mol HCN (or 37.5 g) 1.63 mol NH3 can yield 1.63 mol HCN (or 44.1 g) 1.56 mol CH4 can yield 1.56 mol HCN (or 42.2 g) Conclusion? O2 is the limiting reagent.

  26. % yield = actual yield O2 is the limiting reagent. Thus, theoretical yield is based on 100% consumption of O2. 2.08 mol O2can yield 1.39 mol (or 37.5 g) HCN x 100 theoretical yield % yield = 36.4 g HCN x 100 = 97.1% 37.5 g HCN moles could also be used

  27. 2. How many grams of NH3 remain? 36.4 g (or 1.35 mol) of HCN gas is produced 2CH4 + 2NH3 + 3O2→ 2HCN + 6H2O Since the reaction stoichiometry is 1:1, 1.35 mol of NH3 is consumed: 1.63 mol NH3initially present – 1.35 mol NH3 consumed 0.28 mol NH3 remaining 0.28 mol NH3 x (17.03 g NH3/mol) = 4.8 g NH3 remain

  28. 4 Types of Redox Reactions • A 8.50 g sample of NCl3 • 2) Decomposition reaction • 3) Combustion reaction • 4) Displacement reaction (or single-replacement) 2NI3(s)→ N2(g)+ 3I2(g) CH4(g) + 2O2(g)→CO2(g)+ 2H2O(g) Cu(s) + Zn2+(aq)→Cu2+(aq) + Zn(s)

  29. Chemical Equations identifies all reactants & products shows molar ratios between all chemical species in the reaction (“stoichiometry”) proper stoichiometry is represented ONLY when the chemical equation is balanced

  30. Chapter 4 • 3 general classes of chemical reactions: • Precipitation reactions • Ex: geology, heavy metal analysis • Solid formation from ionic compounds • 2) Acid/base reactions • Ex: many biochemical reactions • Protontransfer in polar covalent compounds • 3) Oxidation/Reduction (“redox”) reactions • Ex: batteries, metabolic energy production • Electrontransfer inionic & molecular compounds

  31. Fundamental rxn. similarities • Precipitation • Acid/base • Redox Textbook Classifying Chemical Reactions • Similarities in written equations • Combination reactions (CR) • often redox • Decomposition reactions (DR) • often redox • Single replacement (SR) • often redox • Double displacement (DD) • Acid/base, precipitation, redox Lab

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