Question

A student failed to vaporize the entire sample prior to placing the Erlenmeyer flask in the ice bath. How did this error affect the calculated molar mass? Justify your answer using calculations.

Answer 1

Answer: The said error will lead to the erroneous measurement of the mass of condensed vapor, barometric pressure inside the flask and also the volume of the flask after cooling which are very critical values in calculating the molar mass of the unknown substance. To justify this, the following sets of data are taken to calculate the molar mass.

Set -1: (When the student failed to completely vaporize, the mass of condensed mater will relatively be higher and Volume and barometric pressure would relatively be lower. In such a case, the measured molar mass would be higher than the real value)

Mass of condensed vapor = 0.7 g

Volume at STP = V_obs x (P_obs/P_STP) x (T_STP/T_obs) = 130 mL x (750 torr / 760 torr) x (273 K / 373 K)

= 93.895 mL

Hence, The molar mass is

= (0.7 g / 0.093895 mL) x 22.4 L/mol = 166.995 g/mol

Set -2: (When the student completely vapourized, mass of condensed mater, volume and barrometric pressure would relatively be closer to the real values and hence could result in a molar mass value of great precision and accuracy.)

Mass of condensed vapor = 0.65 g

Volume at STP = V_obs x (P_obs/P_STP) x (T_STP/T_obs) = 135 mL x (758 torr / 760 torr) x (273 K / 373 K)

= 98.54695 mL

Hence, The molar mass is

= (0.65 g / 0.098547 mL) x 22.4 L/mol = 147.54 g/mol

If the accepted value of the molar mass of the unknown substance is 145.5, we can see that the second one is in closer agreement where the student vaporized the substance completely.