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Composition of Unknown Compounds

Composition of Unknown Compounds. How do we determine the chemical formula of new & unknown compounds? http://www.youtube.com/watch?v=6ZJqN8KKrBU. %Comp is simply the % of each element by mass in the compound %Comp = element molar mass compound molar mass

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Composition of Unknown Compounds

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  1. Composition of Unknown Compounds How do we determine the chemical formula of new & unknown compounds? http://www.youtube.com/watch?v=6ZJqN8KKrBU

  2. %Comp is simply the % of each element by mass in the compound %Comp = element molar mass compound molar mass Experimentally, %Comp is determined by a process called “combustion analysis” unknown substance is heated to very high temperatures; chemical behaviour is observed each element has unique behaviour at high temperatures ∴ we can predict what percent of the sample belongs to each element Step 1:Percentage Composition (%Comp)

  3. %Comp by Combustion Analysis

  4. Example 1: • If we have a 50g sample of jelly beans, with 20g of red jelly beans and 30g of green jelly beans, what is the percent composition of each colour? • Red: 20/50 x 100 = 40% • Green: 30/50 x 100 = 60%

  5. Example 2: Iron(iii) Oxide; Fe2O3, mmtotal= 159.69 g/mol % Fe = (2x55.85)/159.69 x100 = 69.9% % O = (3x16.00)/159.69 x100 = 30.1 % 6 • Iron(ii) Oxide; FeO, mmtotal= 71.84 g/mol • % Fe = 55.85/71.84 x100 = 77.7% • % O = 16.00/71.84 x100 = 22.3 %

  6. Step 2:Empirical Formula (Molar Ratio) 7 • Molecular formulae of many compounds is known, but how? • Knowing %Comp of unknown sample, we can calculate the molar ratio of elements • when identifying a new compound chemists first develop an empirical formula before determining the true molecular formula & structural arrangements • Empirical formula is simply a molar ratio of atoms within in the same molecule

  7. Empirical Formula (Molar Ratio) 8

  8. Empirical Formula: The Calculation! 9 • Given % composition assume a mass of 100g of the unknown compound • Convert this to a MOLAR amount of each elements using element’s molar mass (n=m/mm) • Create a ratio by dividing each molar value by the lowest total amount of moles (n)

  9. Example Calculation: 10 If we have an unknown sample with 5.88% H and 94.12% O, what is the Empirical Formula? • Given % composition assume a mass of 100g of the unknown compound Therefore; mH = 5.88g mO=94.12g • Convert this to a MOLAR amount of each elements using element’s molar mass (n=m/mm) nH = 5.88g/1.01g/mol = 5.82 mol nO=94.12g/16.00g/mol = 5.88mol • Create a ratio by dividing each molar value by the lowest total amount of moles (n) Hydrogren = 5.82/5.82 = 1 & Oxygen = 5.88/5.82 = 1 Therefore; the empirical formula is 1H: 1O or HO

  10. Step 3: True Molecular Formula 11 • to find the MOLECULAR MASS of an unknown molecule, chemists use a device called a MASS SPECTROMETER (see pg#166) • https://www.youtube.com/watch?v=tOGM2gOHKPc • https://www.youtube.com/watch?v=J-wao0O0_qM • the molecular formula is then determined by analyzing the factor difference between the molecule’s empirical molecular mass and the true molecular mass and the true molecular mass

  11. Example: 12 • If the Empirical Formula of our unknown is HO, and the Mass Spectrometer tells us the molar mass is 34.02g/mol, what is the true molar mass? • Factor Difference = True Mass/Empirical Mass = 34.02/17.01 = 2 THEREFORE, True Molecular Formula is 2 x HO = H2O2

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