Part
5: STRUCTURE DETERMINATION
a. Use the molecular weight of X = 120.19 g/mol and the % of C and H to calculate the formula: (2 marks)
C: 120.19 x 89.94% g/mol C = 108.098 g/mol C. Now divide by atomic weight of C, 108.098 / 12.011 = 9
H: 120.19 x 10.06% g/mol H = 12.09 g/mol H. Now divide by atomic weight of H, 12.09 / 1.008 = 12
Hence the molecular formula is C_{9}H_{12} where the simplest ratio of C to H, the empirical formula is C_{3}H_{4}.
b. C_{9}H_{12} (see above) (2 marks)
c. For a hydrocarbon, the formula for the index of hydrogen deficiency, IHD = 0.5 ( 2c+2h), therefore, the IHD = 0.5 (2 x 9 + 2  12) = 4 (2 marks)
d. What information do we have ? (structure 5 marks, name 2 marks)

Common errors:
Did not know that an empirical formula was the simplest ratio of the elements. (High school / 1st yr knowledge)
Not checking molecular formula against the molecular weight provided.
Incorrectly counting or not checking the number of H and or C types in X
Drawing structures for X that did not fit the molecular formula of C_{9}H_{12}, the IHD and the IR data.
Note that IHD defines the number of rings + pi bonds present for a given molecular formula.
Could not name X correctly : incorrect numbering, incorrect alphabetisation, wrong root name (e.g. wrong root, mixing up phenyl with benzyl).