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The Molecular Formula of a Compound

The Molecular Formula of a Compound. Is the sf of a cmpd the same as its molecular formula (mf)? Sometimes. One sf can represent many different molecules. The following cmpds all have sf = CH 2 O: name mf mm multiple (g/mol) of sf

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The Molecular Formula of a Compound

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  1. The Molecular Formula of a Compound Is the sf of a cmpd the same as its molecular formula (mf)? Sometimes.

  2. One sf can represent many different molecules The following cmpds all have sf = CH2O: name mf mm multiple (g/mol) of sf ________________________________________ formaldehyde CH2O 30 1 acetic acid C2H4O2 60 2 lactic acid C3H6O3 90 3 glucose C6H12O6 180 6

  3. Sample Problem #1 The sf of ribose, a carbohydrate, is CH2O. In a separate experiment using a mass spectrometer, the molar mass of ribose was determined to be 150 g/mol. What is the mf of ribose? How to solve this? What is mm of the sf of ribose? 30 g/mol By what do we multiply 30 to get 150? 5 mf = 5*sf = C5H10O5 for ribose.

  4. Sample Problem #2 The sf of a gas is C2H5. The density of the gas is 2.33 g/L at SATP. Find the mf of the gas. What is mm of C2H5, the sf of the gas? sf = 29 g/mol What piece of info do you need to remember about gases and mole concept? (Vol. of 1.00 mol of any gas = 24.8L @SATP) 2.33 g/L = x g/24.8 L  1.00 mol of the gas = 57.8 g/24.8 L (SATP) or 58 g/mol

  5. What is relationship of mf (58 g/mol) to the sf (29 g/mol)? mf = 2*sf . . . and the molecular formula is C4H10 (butane).

  6. Try this: • a) A gaseous cmpd is made up of 21.9% S, 78.1% F by mass. Find the sf of the cmpd. b) 60.9 g of the gas occupies 10.0L at SATP. Determine the mf of the gas.

  7. A Question . . . If you burn a sample—CaHb—or—CaHbOc—in (excess) O2, what will be the products? CO2 and H2O. Where will all the C in the sample end up? In the CO2. Where will all the H in the sample end up? In the H2O.

  8. Where will any O in the sample (ie. if the sample contains O) end up? We don’t know for sure. Could be in the water or in the CO2. We use this idea to determine the simplest formula of hydrocarbons, and hydro- carbons that also contain Oxygen. Take a look . . .

  9. Finding Simplest and Molecular Formulae by Experiment The Carbon-Hydrogen Combustion Analyzer stream of O2 •    furnace H2O absorber CO2 absorber CaHbOc What is the furnace for? To combust the sample. NB. Water absorber must be situated before the CO2 absorber. Why? Because CO2 absorber will also absorb water.

  10. Think about the C-H analyzer apparatus . . . All of C in sample ends up in CO2; all of the H ends up in H2O. What data must be recorded before beginning the C-H analysis of a compound? • mass of sample • mass of “empty” H2O absorber • mass of “empty” CO2 absorber

  11. What data must be recorded at the end of a C-H analysis? • mass of “full” H2O absorber • mass of “full” CO2 absorber How do we know how much O2 to add to the sample? We don’t—we use excess O2 in order to ensure complete combustion.

  12. But what will the data from a C-H analysis give us? mass of CO2 is used to get mass of C in sample using % composition. mass of H2O is used to get mass of H in sample using % composition

  13. So we know mass C and mass H in sample. What will this buy us? Convert mass of each element to mols— find mol : mol ratio—and finally . . . empirical formula of the compound. What about molecular formula? As before, more information required to find molecular formula.

  14. Sample Problem #3 A 1.00 g sample of a pure compound containing only C and H was combusted in a C-H analyzer. 0.692g of H2O and 3.338g of CO2 were produced. • Find mass C and mass H in sample. • Find sf of sample. Get busy . . .

  15. mass C in sample = (%C in CO2)*mass CO2 mass C = (12.01g/44.01g) * 3.338 g = 0.9109 g C in sample combusted mass H in sample = (%H in H2O)*mass H2O mass H = (2.02g/18.02g) * 0.6919 g = 0.07756 g H in sample combust.

  16. in 1.000 g sample, we have 0.9109 g C, convert to moles to get 0.9109 g/12.01 g·mol-1 = 0.07584 mol C 0.07756 g H, convert to moles to get 0.07756 g/1.01 g·mol-1 = 0.07694 mol H ratio of C : H is 0.07584 mol C : 0.07694 mol H sf is CH. How to check for reasonableness?

  17. Note: We did not need to find % composition in order to get sf. We converted each element to mols and got mol : mol ratio from there.

  18. Homework • p 275 PP 41 – 50 (do even #s) • Do the C-H combustion analyzer handout.

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