Examples of questions related to the Dumas method:
How many grams of ethanol would occupy 750 mL at 790.5 torr pressure at 100EC? (molar
mass of ethanol is 46.0 g mol-1)
1) PV = nRT
Ideal Gas Equation
2)
Definition of molar mass
R = 0.8206 L atm mol-1 K-1
Value of R
Parse
P = 750 torr, V = 0.750 L,
T = 100EC, M = 46.0 g mol-1
Multiply P by unit factor (1 atm/760 torr)
Evaluate right side
P = 0.987 atm
Use Definition of EC
3) T/K = t/EC + 273.15
Substitute into 3)
T/K =100EC/EC + 273.15
Evaluate right side
T/K = 373.15
Multiply both sides by unit K
T = 373.15 K
Substitute into 1)
(0.987 atm)(0.750 L) =
n (0.8206 L atm mol-1 K-1)(373 K)
Evaluate right and left side
0.740 atm L = n(30.6 L atm mol-1)
Divide both sides by (30.6 L atm mol-1)
Evaluate
0.0242 mol = n
Substitute into 2)
Multiply both sides by 0.0242 mol
(0.0242 mol)(46.0 g mol-1) = m
Evaluating
1.11 g = m
Answer
Dumas method:
Calculate the molar mass of a compound in the Dumas method for which a volume of the
experimental container was 452 mL and the pressure was 745.1 torr. The difference in mass
between the empty container and the final measurement was 1.129 g. (The method implies using
a boiling water bath, i.e. T = 100EC.)
1) PV = nRT
Ideal Gas Equation
2)
Definition of molar mass
R = 0.8206 L atm mol-1 K-1
Value of R
P = 745.1 torr, V = 452 mL
T = 100EC, m = 1.129 g
Parse
P = 745.1 torr, V = 452 mL,
T = 100EC, m = 1.129 g
Multiply P by unit factor (1 atm/760 torr)
Evaluate right side
P = 0.9804 atm
Use Definition of EC
3) T/K = t/EC + 273.15
Substitute into 3)
T/K =100EC/EC + 273.15
Evaluate right side
T/K = 373
Multiply both sides by unit K
T = 373 K
Multiply V by unit factor (1 × 10-3 L / 1 mL)
Evaluate
V = 0.453 L
Substitute into 1)
(0.9804 atm)(0.453 L) =
n (0.8206 L atm mol-1 K-1)(373 K)
Evaluate right and left side
0.444 atm L = n(30.6 L atm mol-1)
Divide both sides by (30.6 L atm mol-1)
Evaluate
0.01.45 mol = n
Substitute into 2)
Evaluate right side
M = 46.7 g mol-1
Answer