Process streams occasionally contain one substance; more often they consist of mixtures of liquids or gases, or solutions of one or more solutes in a liquid solvent. The following terms may be used to define the composition of a mixture of substances, including a species A:
Mass fraction mass of A total mass
: x =A (1.13)
Mole fraction moles of A totalnumber of moles
: y =A (1.14)
Mass fractions can be converted to mole fractions or vice versa by assuming a basis of calculation. Remember mass and mole fractions are unitless.
Example 1.10 Mole Fractions Problem
A mixture of methanol (CH3OH, Mw = 32.04) and ethanol (C2H5OH, Mw = 46.07) is flowing through a circular pipe at a flow rate of 3.0 m/s.
The mixture contains 30.0 wt% methanol and 70.0 wt% ethanol. The spe-cific gravity of the mixture is 0.80. If the inside diameter of the pipe is 0.10 m, what is the flow rate of the mixture in kg/s, kmol/s?
Solution
Known quantities: Mixer velocity, specific gravity, and inner pipe diam-eter are known.
Find: The molar flow rate of the mixture.
Analysis: The mass flow rate is obtained by multiplying its density by the volumetric flow rate:
m = A=⎛ ×
Since we know the total mass flow rate, we can compute the molar flow rate of methanol and ethanol:
nCH OH3
Ethanol molar flow rate
nC H OH2 5
The total molar flow rate then is
Example 1.11 Mass Fractions Problem
A liquid stream containing only benzene, toluene, and p-xylene flows through a conduit that has a square cross section. The total flow rate of the stream is 10.0 mol/s. If the mole fraction of benzene (Mw = 78) is 0.300, that of toluene (Mw = 92) is 0.500, and that of p-xylene (Mw = 106) is 0.200, what is the mass fraction of each of the three components? If the density of the liquid is 0.87 g/cm3 and the conduit is 0.10 m on a side, what is the mass fraction and volumetric flow rate of the liquid inside the conduit?
Solution
Known quantities: Mixture mole fraction of benzene, toluene, and p-xylene, density of mixture and conduit diameter, and molar flow rate are known.
Find: The mass fraction of each component.
Analysis: The mass flow rate for the three components is equal to their respective molar flow rate times their molecular weight:
m y ni= i× tot×Mwi
where
mi is the component mass flow rate yi is the component mole fraction ntot is the total molar flow rate
Mwi is the component molecular weight Mass flow rate of benzene:
mB mol
Mass flow rate of toluene:
mT mol
Mass flow rate of xylene:
mX mol
The total mass flow rate is the sum of these three:
mTotal m m mB T X g
= + + =234 460 212 906+ + = s
The mass fractions are the individual mass flow rates divided by the total mass flow rate.
Mass fraction of benzene: xB g s
The volumetric flow rate obtained by dividing the mass flow rate by the density is
Concentrations can be expressed in many ways: weight/weight fraction (w/w), weight/volume fraction (w/v), molar concentration (M), and mole
fraction. The weight/weight concentration is the weight of the solute divided by the total weight of the solution, and this is the fractional form of the per-centage composition by weight. The weight/volume concentration is the weight of solute divided by the total volume of the solution. The molar con-centration is the number of moles of the solute, expressed in moles, divided by the volume of the solution. The mole fraction is the ratio of the number of moles of the solute to the total number of moles of all species present in the solution [7].
Example 1.12 Air Compositions Problem
If 100 g of air consists of 77% by weight of nitrogen (N2, molecular weight = 28), and 23% by weight of oxygen (O2, molecular weight = 32), calculate (a) the mean molecular weight of air, (b) mole fraction of oxy-gen, (c) concentration of oxygen in mol/m3 and kg/m3, if the total pres-sure is 1.5 atm and the temperature is 25°C.
Solution
Known quantities: Mass of air and its composition in weight fraction are known.
Find: The air composition.
Analysis: The number of moles is obtained by dividing the mass by molecular weight.
a. The mass of N2 and O2 that 100 g of air contains:
Mass of N2 = 0.77 × 100 g = 77 g Mass of O2 = 0.23 × 100 g = 23 g Number of moles of N2 = 77g
28 gmol
mol
=2 75.
Number of moles of O2 = 23g 32 gmol
0.72 mol
=
Total number of moles = 2.75 + 0.72 = 3.47 mol
The mean molecular weight of air is the total mass/total moles.
Therefore, the mean molecular weight of air = 100 g/3.47 mol = 28.8 g/mol.
b. Mole fraction of oxygen (O2) = 0.72/3.47 = 0.21, and this is also the volume fraction.
c. In the ideal gas equation, n is the number of moles present, P is the pressure (in atm, in this instance), and T is the temperature (in K, in this case). The corresponding value of R is 0.08206 m3 atm/(mol K).
PV = nRT and molar concentration = n/V
O2 molar concentration
The O2 mass concentration is the molar concentration multiplied by the molecular weight:
Example 1.13 Concentration of Salt in Water Problem
Salty water is prepared by mixing salt (NaCl, Mw = 58.5) in water (Mw = 18). A solution is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg/m3. Calculate the concentra-tion of salt in this soluconcentra-tion as a (a) weight/weight fracconcentra-tion, (b) weight/
volume in kg/L, and (c) mole fraction.
Solution
Known quantities: Mass of salt and mass of water, mixture density.
Find: (a) Weight/weight fraction, (b) weight/volume in kg/L, and (c) mole fraction.
Analysis: Weight fraction is obtained by dividing the mass of each com-ponent by the total mass.
(a) Weight fraction of salt = 2 kg
2 kg 1 kg0 167
0 + 00 0
( )
= .(b) Salt concentration is the weight of salt/total volume. The total volume is the mass of the mixture divided by mixture density. Weight of salt/volume = 2 kg kg
m3 0 0910. 220
( )
=Weight/volume in kg/L = 220 kg
m m
1000L= 0.22 kg
3 L
3
(c) Moles of water and salts are obtained by dividing the mass of each component by its respective molecular weight:
Moles of water = 1 kg = 18 kg/kmol 00 5 56. Moles of salt = 2 kg salt =
58 5 kg/kmol .0 0.34
Mole fraction is obtained by dividing moles of each component by the total number of moles:
Mole fraction of salt =
(
5 565 56+. 34)
=. 0. 0 942.