C HEMICAL E
NGINEERING P
RINCIPLES
1
,lution:
Average molecular weight = 0.2 * 44 + 0.3 * 28 + 0.4 * 16.04 + 0.1 * 2.02 = 23.8 kg/kg mol
Example 18
A liquefied mixture has the following composition: n-C 4 H 10 50% (MW=58), n-C 5 H 12 30% (MW=72), and n-C 6 H
14 20%
(MW=86). For this mixture, calculate: (a) mole fraction of each component. (b) Average molecular weight of the
mixture.
Solution
Example 19
In a ternary alloy such as Nd 4.5 Fe 77 B 18.5 the average grain size is about 30 nm. By replacing 0.2 atoms of Fe with
atoms
of Cu, the grain size can be reduced (improved) to 17 nm.
(a)What is the molecular formula of the alloy after adding the Cu to replace the Fe?
(b)What is the mass fraction of each atomic species in the improved alloy?
Solution
Basis: 100 g mol (or atoms) of Nd 4.5 Fe 77 B 18.5
(a) The final alloy is Nd 4.5 Fe 76.8 B 18.5 Cu 0.2.
(b) Use a table to calculate the respective mass fractions.
22
,Chemical Engineering principles - First Year
Example 20
A medium-grade bituminous coal analyzes as follows:
The residuum is C and H, and the mole ratio in the residuum is H/C = 9. Calculate the weight (mass) fraction
composition of the coal with the ash and the moisture omitted (ash - and moisture - free).
Solution
Take as a basis 100 kg of coal because then percent = kilograms.
Basis: 100 kg of coal
The sum of the S + N + O + ash + water is 2 + 1 + 6 + 11 + 3 = 23 kg
We need to determine the individual kg of C and of H in the 77 kg total residuum.
To determine the kilograms of C and H, you have to select a new basis.
Basis: 100 kg mol (Because the H/C ratio is given in terms of moles, not weight)
, H: (77kg) (0.43) = 33.15 kg
C: (77kg) (0.57) = 43.85 kg
Finally, we can prepare a table summarizing the results on the basis of 1 kg of the coalash-free and water-free.
23
NGINEERING P
RINCIPLES
1
,lution:
Average molecular weight = 0.2 * 44 + 0.3 * 28 + 0.4 * 16.04 + 0.1 * 2.02 = 23.8 kg/kg mol
Example 18
A liquefied mixture has the following composition: n-C 4 H 10 50% (MW=58), n-C 5 H 12 30% (MW=72), and n-C 6 H
14 20%
(MW=86). For this mixture, calculate: (a) mole fraction of each component. (b) Average molecular weight of the
mixture.
Solution
Example 19
In a ternary alloy such as Nd 4.5 Fe 77 B 18.5 the average grain size is about 30 nm. By replacing 0.2 atoms of Fe with
atoms
of Cu, the grain size can be reduced (improved) to 17 nm.
(a)What is the molecular formula of the alloy after adding the Cu to replace the Fe?
(b)What is the mass fraction of each atomic species in the improved alloy?
Solution
Basis: 100 g mol (or atoms) of Nd 4.5 Fe 77 B 18.5
(a) The final alloy is Nd 4.5 Fe 76.8 B 18.5 Cu 0.2.
(b) Use a table to calculate the respective mass fractions.
22
,Chemical Engineering principles - First Year
Example 20
A medium-grade bituminous coal analyzes as follows:
The residuum is C and H, and the mole ratio in the residuum is H/C = 9. Calculate the weight (mass) fraction
composition of the coal with the ash and the moisture omitted (ash - and moisture - free).
Solution
Take as a basis 100 kg of coal because then percent = kilograms.
Basis: 100 kg of coal
The sum of the S + N + O + ash + water is 2 + 1 + 6 + 11 + 3 = 23 kg
We need to determine the individual kg of C and of H in the 77 kg total residuum.
To determine the kilograms of C and H, you have to select a new basis.
Basis: 100 kg mol (Because the H/C ratio is given in terms of moles, not weight)
, H: (77kg) (0.43) = 33.15 kg
C: (77kg) (0.57) = 43.85 kg
Finally, we can prepare a table summarizing the results on the basis of 1 kg of the coalash-free and water-free.
23
