National Transportation Safety Board Aircraft Accident Report Mole Airlines Flight 1023 DC 6-02 O’Hare Airport, Chicago October 23, 2010 SHS Task Force Report Number:60-21-023-AA

SHS Task Forces October 2011 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.

How to find the Empirical Formula…. 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

Look up Formula on Table to Identify Formula= C7H5N3O6 Look up Formula on Table to Identify Trinitrotoluene Description: Explosive (TNT=dynamite) What conclusions can be draw from the evidence?