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Dennis_Churaev [7]
3 years ago
7

Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol i

s combusted, producing 0.2546 mg of co2 and 0.1043 mg of h2o. what is the empirical formula for menthol? express your answer as a chemical formula.
Chemistry
1 answer:
Ket [755]3 years ago
8 0

Answer:

            Empirical Formula  =  C₁₀H₂₀O

Solution:

Data Given:

                      Mass of Menthol  =  9.045 × 10⁻² mg  =  9.045 × 10⁻⁵ g

                      Mass of CO₂  =  0.2546 mg  =  0.0002546 g

                      Mass of H₂O  =  0.1043 mg  =  0.0001043 g

Step 1: Calculate %age of Elements as;

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

                      %C  =  (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100

                      %C  =  (2.814) × (12 ÷ 44) × 100

                      %C  =  2.814 × 0.2727 × 100

                      %C  =  76.73 %


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

                      %H  =  (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100

                      %H  =  (1.153) × (2.02 ÷ 18.02) × 100

                      %H  =  1.153 × 0.1120 × 100

                     %H  =  12.91 %


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

                      %O  =  100% - (76.73% + 12.91%)

                      %O  =  100% - 89.64%

                     %O  =  10.36 %

Step 2: Calculate Moles of each Element;

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

                      Moles of C  = 76.73 ÷ 12.01

                     Moles of C  =  6.3888 mol


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

                      Moles of H  = 12.91 ÷ 1.01

                      Moles of H  =  12.7821 mol


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

                      Moles of O  = 10.36 ÷ 16.0

                      Moles of O  =  0.6475 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     O

            6.3888                              12.7821                            0.6475

     6.3888/0.6475                  12.7821/0.6475                 0.6475/0.6475

               9.86                                   19.74                                   1

             ≈ 10                                      ≈ 20                                     1

Result:

         Empirical Formula  =  C₁₀H₂₀O₁

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