Question

An extract from a toboacco leaves is completely miscible in water at temperatures below 60 C. What is the molality of a solution of this extract in water which freezes at -0.450 C. If this solution was obtained by dissolving 1.921 g of the extract in 48.92g of water, What is the molar mass of the extract? Combustion analysis of the extract yields 74.03% C, 8.70% H, and 17.27 % N by mass. What is the molecular formula of the extract?

Answer 1

1)

delta Tf = Kf x m

0.450 = 1.86 x molality

molality = 0.242 m

molality = moles / mass of solvent

0.242 = moles / 48.92 x 10^-3

moles = 0.01184

moles = mass / molar mass

0.0118 = 1.921 / molar mass

molar mass = 162.3 g/mol

74.03% C, 8.70% H, and 17.27 % N by mass

mass % of C = 74.03

moles of C = 74.03 / 12 = 6.169

moles of H = 8.70 / 1 = 8.70

moles of N = 17.27 / 14 = 1.234

ratio :    

           C             H         N

       6.169         8.70      1.234

          5             7            1

Empirical formula = C5H7N

mass of empirical formula = 81

n = 162 / 81 = 2

Molecular formula = n x empirical formula

                             = 2 x C5H7N

Molecualr formula = C₁₀H₁₄N₂