Question

The following data were obtained from experiments to find the molecular formula of lidocaine, a local anesthetic, which contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 3.54-g sample of lidocaine with excess O2 formed 9.32 g CO2 and 3.00 g H2O. Another sample of mass 2.35 g was found to contain 0.281 g of N. The molar mass of lidocaine was found to be 234. Find the molar formula of lidocaine.

Answer 1

Answer:

C₁₄H₂₂N₂O    

Explanation:

Let's call the molecular formula CxHyNzOw. For the combustion reaction, all the carbon will form CO₂ and all the hydrogen will form H₂O

CxHyNzOw + O₂ → CO₂ + H₂O

So, the molar mass of H is 1 g/mol, of C is 12 g/mol, of O is 16 g/mol, and of N is 14 g/mol. The molar masses of the compounds are

CO₂ = 12 + 2x16 = 44 g/mol

H₂O = 2x1 + 16 = 18 g/mol

The number of moles is the mass divided by the molar mass, so:

CO₂ = 9.32/44 = 0.2118 mol

H₂O = 3/18 = 0.1667 mol

In 1 mol of CO₂, there is 1 mol of C, so the number of moles of C is 0.2118 mol, and in 1 mol of H₂O there is 2 mol of H, so the number of moles of H is 2x0.1667 = 0.3334 mol, which is the number of moles in the lidocaine. The masses are:

C = 0.2118x12 = 2.5416 g

H = 0.3334x1 = 0.3334 g

So, the percentagens of the compounds are:

C = 2.5416/3.54 = 0.718 = 71.8%

H = 0.3334/3.54 = 0.094 = 9.4%

For N, based on the other experiment:

N = 0.281/2.35 = 0.1196 = 11.96%

So, the percentage of oxygen must be 100 - (71.8 + 9.4 + 11.96) = 6.84%

In 1 mol of lidocaine, there is 234 g, so the masses and the number of moles of the compounds are:

C: m = 0.718x234 = 168 g

n = 168/12 = 14 mol

H: m = 0.094x234 = 22 g

n = 22/1 = 22 mol

N: m = 0.1196x234 = 28 g

n = 28/14 = 2 mol

O: m = 0.0684x234 = 16g

n = 16/16 = 1 mol

So, the molecular formula is C₁₄H₂₂N₂O.