Aircraft Accident Report Mole Airlines Flight 1023 DC 6-02 O’Hare Airport, Chicago October 23,...
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Transcript of Aircraft Accident Report Mole Airlines Flight 1023 DC 6-02 O’Hare Airport, Chicago October 23,...
Aircraft Accident ReportMole 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 ForcesOctober 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 Hydrogen1.32 mol N / 1.32 mol = 1.00 Nitrogen2.642 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?