1. 1) The female silkworm moth produces a sex attractant called bombykol, which

Question

1.       1)                    The female silkworm moth produces a sex attractant called bombykol, which is composed of  80.61% carbon, 12.68%                       hydrogen, and 6.71% oxygen.

(a) What is the empirical formula of bombykol?

(b) If the molecular weight of bombykol is 238.4 amu, what is the molecular formula?

(c) If 5.698 g of bombykol is burned in excess oxygen to form 16.83 g of CO2 and 6.452 g of H2O, how many grams of carbon were present in the original sample?

2.       2)             A 2.20 g sample of an unknown acid (empirical formula = C3H4O3) is dissolved in 1.0 L of water. A titration required 25.0                  mLs of 0.500 M NaOH to react completely with all the acid present. Assuming the unknown acid has one acidic proton per               molecule, what is the molecular formula of the unknown acid?

Explanation / Answer

moles of C in 100g = 80.61 / 12 = 6.72 moles
moles of H = 12.68 / 1.0079 = 12.58 moles
moles of O = 6.71 / 16 = 0.419 moles

ratio of C : H : O = 6.72 : 12.58 : .419 dividing by the smallest number we get

C : H : O = 16.02 : 30.00 : 1.00

empirical formula is C16H30O
empirical formula weight is 238.4 g/mol which is the molecular weight so the empirical formula is also the molecular formula C16H30O

mass of CO2 = 16.83 ; moles of CO2 = 16.83 / 44 = 0.383 moles
= moles of C in the sample = 0.383 x 12 g of C = 4.59g of C in the sample

or 4.59 / 5.698 % = 80.55 % .. same as given in the data..

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