Menthol, from the oil of mint, has a characteristic cool taste. the compound contains only c, h, and o. if 95.6 mg of menthol burns completely in o2, and gives 269 mg of co2 and 110 mg of h2o, what is the empirical formula of menthol?

            Empirical Formula  =  C₁₀H₂₀O₁

Solution:

Data Given:

                      Mass of Menthol  =  95.6 mg  =  0.0956 g

                      Mass of CO₂  =  269 mg  =  0.269 g

                      Mass of H₂O  =  110 mg  =  0.110 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.269 ÷ 0.0956) × (12 ÷ 44) × 100

                      %C  =  (2.8138) × (12 ÷ 44) × 100

                      %C  =  2.8138 × 0.2727 × 100

                      %C  =  76.74 %

                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.110 ÷ 0.0956) × (2.02 ÷ 18.02) × 100

                      %H  =  (1.1506) × (2.02 ÷ 18.02) × 100

                      %H  =  1.1506 × 0.1120 × 100

                      %H  =  12.89 %

                      %O  =  100% - (%C + %H)

                      %O  =  100% - (76.74% + 12.89%)

                      %O  =  100% - 89.63%

                      %O  =  10.37 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 76.74 ÷ 12.01

                      Moles of C  =  6.3896 mol

                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 12.89 ÷ 1.01

                      Moles of H  =  12.7623 mol

                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 10.37 ÷ 16.0

                      Moles of O  =  0.6481 mol

Step 3: Find out mole ratio and simplify it;

               C                                        H                                     O

            6.3896                              12.7623                            0.6481

     6.3896/0.6481                  12.7623/0.6481                 0.6481/0.6481

               9.85                                   19.69                                   1

             ≈ 10                                      ≈ 20                                     1

Result:

         Empirical Formula  =  C₁₀H₂₀O₁

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.