Data Sheet Chemicl A molar mass unknown identification code third first second m

Question

Data Sheet Chemicl A molar mass unknown identification code third first second mass of flask, boiling stone, foil cap, and unknown after cooling. g mass of flask, boiling stone, and foil cap, mass of unknown, g 2 water bath temperature at complete vaporization, "C water bath temperature, K barometric pressure, in. Hg 373 barometric pressure, atm volume of flask, ml volume of flask, L density of vaporized unknown, gL- molar mass of unknown, g mol average molar mass of unknown, g mol-1 accepted molar mass of unknown, g mol percent error, % 41

Explanation / Answer

First Trial:

The mass of unknown compound (m) = 81 - 79.3 = 1.7 g

The vapour density of unknown compound (d) = 1.7 g / 0.01 L = 170 g/L

(Note: Density = mass(m) / volume (V))

The molar mass of unknown compound can be calculated as shown below.

According to the ideal gas equation, PV = nRT

i.e. PV = (m/M) * RT

i.e. M = (m/V) * RT/P

i.e. The molar mass of unknown, M = dRT/P

Here, P = 1 atm, d = 170 g/L, R = 0.0821 L atm mol-1 K-1 and T = 373 K

i.e. M = 170 g/L * 0.0821 L atm mol-1 K-1 * 373 K / 1 atm

i.e. M ~ 5206 g/mol

Trial 2:

The mass of unknown compound (m) = 80.2 - 79.3 = 0.9 g

The vapour density of unknown compound (d) = 0.9 g / 0.01 L = 90 g/L

The molar mass of unknown, M = dRT/P

Here, P = 1 atm, d = 90 g/L, R = 0.0821 L atm mol-1 K-1 and T = 373 K

i.e. M = 90 g/L * 0.0821 L atm mol-1 K-1 * 373 K / 1 atm

i.e. M ~ 2756 g/mol

Trail 3:

The mass of unknown compound (m) = 80.1 - 79.3 = 0.8 g

The vapour density of unknown compound (d) = 0.9 g / 0.01 L = 80 g/L

The molar mass of unknown, M = dRT/P

Here, P = 1 atm, d = 80 g/L, R = 0.0821 L atm mol-1 K-1 and T = 373 K

i.e. M = 80 g/L * 0.0821 L atm mol-1 K-1 * 373 K / 1 atm

i.e. M ~ 2450 g/mol

The average molar mass of unknown sample = (5206 + 2756 + 2450) / 3 ~ 3471 g/mol

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