Explanation:
Concentration tells us how much of one component (in a mixture) we have
relative the mixture (or another component in the mixture). That is, for a
mixture of sugar and water, a concentration can specify how much water we
have relative to the solution (water and sugar), or how much water we have
relative to the other component (sugar). These concentrations can be
expressed using units of amount (mols, molecules, ...), mass (kg, lb,
...), and volume (m^{3}, L, ...). The following are some ways to
express concentration, where the numerator signifies the solute and the
denominator signifies the solution.
Mass Concentration: g / cm^{3}, lb_{m} /
ft^{3}, kg / in^{3}, ...
Molar Concentration: kmol / m^{3}, lbmol /
ft^{3}, gmols / L
The last molar concentration listed, gmol/L is the Molarity of the
solute in the mixture. Thus, the definition of molarity follows:
Molarity of component A,
M_{A} = gmol_{A} / L_{total}
Mass and Mole Fractions
Mass and mole fractions will be used frequently in material and energy
balance problems. The notation used in the course and thus the notation
used here differs from the notation introduced in chapter 3 (here, x
designates a liquid or solid mass or mole fraction, and y designates mass
or mole fraction for a gas). Here are the definitions for mass and mole
fractions:
Equation 1:
x_{A} = m_{A} / m_{total}
Equation 2:
x_{A} = n_{A} / n_{total}
Equation 3:
y_{A} = m_{A} / m_{total}
Equation 4:
y_{A} = n_{A} / n_{total}
x_{A}, y_{A}  mass or mole fraction of component A.
m_{A}  the mass of component A
n_{A}  the mole
of component A
m_{total}  the mass of the total mixture
n_{total}  the moles of the total mixture
"x" is used for liquids and solids, thus the first two equations are
the mass and mole fractions for liquids and solids. Accordingly, the last
two equations refer to gases. As you can see, "x_{A}" can donote
several different things. For example, x_{H2O} can
refer to either the mass or mole fraction of water or ice in a mixture. It
will be up to the problem solver to keep track of how "x_{A}" is
defined in working a problem.
Related to the mass and mole fractions are the definitions for parts
per million (ppm) and the parts per billion (ppb). Here, we simply
multiply the mass or mole fraction by 10^{6} for ppm, and by
10^{9} for ppb.
parts per million
ppb=x, y(mole fraction)*10^{6}
parts per billion
ppm=x, y(mole fraction)*10^{9}
note: The mole and mass fractions must always be unitless, so
the mole/mass of the component (numerator) and the total mole/mass
(denominator) must both have the same units.
Example 1
A mixture of gases has the following composition by mass:
O _{2} 16%
CO 4.0%
CO_{2} 17%
N_{2} 63%To reach mass fractions from percent
composition, we would move the decimal place over two places, so, for
example, the mixture is 16% O_{2}, or we could say, the mass
fraction of O_{2} in the mixture is .16. Determine the molar
composition of this mixture. (hint: you first need to find the total moles
of the mixture)
Goto  Check
Answer  See
Solution
Example 2
Calculate the average molecular weight of air. Air is 79% N_{2}
and 21% O_{2}, and us the following formula to calculate the
average molecular weight:
MW_{avg} = y_{N2}MW_{N2} +
y_{O2}MW_{O2}
Goto  Check
Answer  See
Solution
Example 3
A material balance problem (chapter 4) gives us the following
quantities of compounds going into a condenser. Determine the mole
fractions of each stream.
Goto  Check
Answer  See
Solution
