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Aircraft Accident Report Mole Airlines Flight 1023 DC 6-02 O’Hare Airport, Chicago October 23, 2010. National Transportation Safety Board. SHS Task Force Report Number:60-21-023-AA. Mole Airlines Flight 1023. SHS Task Forces October 2011. Flight leaving O’Hare headed for Boston
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Aircraft Accident ReportMole AirlinesFlight 1023 DC 6-02 O’Hare Airport, ChicagoOctober 23, 2010 National Transportation Safety Board SHS Task Force Report Number:60-21-023-AA
Mole Airlines Flight 1023 SHS Task Forces October 2011
Flight leaving O’Hare headed for Boston • 6:02 pm EST plane crashed outside of Detroit, MI • Evidence of a Pre-Crash Explosion Found • At the site of the explosion a material has been found. • Subsequent chemical analysis of the material shows it to be: Carbon 37.01% Hydrogen 2.22% Nitrogen 18.5% Oxygen 42.27%
Mangled passengers are found in and around the crash. • They are not recognizable and their dental records are not available. • They must be identified by the substances found in their belongings or in their bodies. • Upon further investigation one passenger showed a time of death approximated at one hour prior to the crash. Perhaps they were murdered?
Split into 12 task forces (2 chemist teams) • Work on analysis chemical substances from each Toe Tag. • You should be able to determine the EMPIRICAL FORMULA from the % compositions that were found. • Compare the formulas to known compounds on the data table to name the chemical. • Use passenger list to make a tentative identification. • Submit reports to your supervisor.
C:37.01% H: 2.22% N: 18.5% O: 42.27% • How to find the Empirical Formula…. • Assume 100g & Divide % composition by molecular mass • 37.01g C / 12.01 g/mol = 3.083 mol C • 2.22g H / 1.01 g/mol = 2.20 mol H • 18.5g N / 14.0 g/mol = 1.32 mol N • 42.27g O / 16.00 g/mol = 2.642 mol O Atomic Mass of Carbon
How to find the Empirical Formula…. • Divide all answers by the smallest answer (1.32 mol) • 3.083 mol C / 1.32 mol = 2.33 • 2.20 mol H / 1.32 mol = 1.67 • 1.32 mol N / 1.32 mol = 1.00 • 2.642 mol O / 1.32 mol = 2.00
Write the number of moles as a subscript in a chemical formula 3.083 mol C / 1.32 mol = 2.33 Carbon 2.20 mol H / 1.32 mol = 1.67 Hydrogen 1.32 mol N / 1.32 mol = 1.00 Nitrogen 2.642 mol O / 1.32 mol = 2.00 Oxygen • C2.33H1.67N1.00O2.00 • Need to get whole numbers as subscripts
Multiply these answers by 2, 3, or 4 to get whole numbers • 2 x 2.33 = 4.66 Not close enough • 3 x 2.33 = 6.99 Very close to the WHOLE NUMBER 7 • 4 x 2.33 = 9.32 Not close enough • USE 3 • 3 x 2.33 = 7 There are 7 C • 3 x 1.67 = 5 There are 5 H • 3 x 1.00 = 3 There are 3 N • 3 x 2.00 = 6 There are 6 O • Write out the Empirical Formula using the whole numbers calculated. • Formula= C7H5N3O6
Formula= C7H5N3O6 • Look up Formula on Table to Identify • Trinitrotoluene • Description: Explosive (TNT=dynamite) • What conclusions can be draw from the evidence?