Category: Chemical Engineering Math
"Published in Vacaville, California, USA"
Percent composition data for the chlorofluorocarbon Freon-12 by mass are: 9.933% carbon, 58.63% chlorine, and 31.44% fluorine. Based on these data, what is the empirical formula of Freon-12?
Solution:
Basis: 100 grams of sample
Weight of each components:
Weight of carbon is
Weight of chlorine is
Weight of fluorine is
Moles of each components:
Moles of carbon is
Moles of chlorine is
Moles of fluorine is
From the number of moles of each component, we need to divide all of them by their least number of moles which is carbon in order to get the number of atoms in a sample.
Number of carbon in a sample is
Number of chlorine in a sample is
Number of fluorine in a sample is
Therefore, the empirical formula for Freon-12 is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
Two gases, HBr and CH4, have molecular weights 81 and 16, respectively. The HBr effuses through a certain small opening at the rate of 4 mL/sec. At what rate will the CH4 effuse through the same opening?
Solution:
Since the rate of flow of the gases as well as their molecular weight are involved in the problem, then we have to us the Graham's Law of Effusion and Diffusion. What is the difference between effusion and diffusion? Effusion is the process in which a gas escapes from one chamber of a vessel by passing through a very small opening or orifice while diffusion is mixing of molecules of different gases by random motion and collision until the mixture becomes homogeneous.
The rate of effusion or diffusion of a gas is inversely proportional to the square root of its density or molecular weight. The equation of Graham's Law of Effusion and Diffusion can be written as follows
or
From the given problem, therefore, the rate of effusion of CH4 through the small opening is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
What pressure is required to compress 5 liters of gas at 1 atm pressure to 1 liter at a constant temperature?
Solution:
At the given problem, if the temperature is constant during the compression of a gas, then it is a principle of Boyle's Law. The volume of a gas is inversely proportional to its pressure. The equation for Boyle's Law can be written as follows
or
Therefore, the pressure of a gas after it is compressed to 1 liter is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
The vapor pressures of pure benzene and toluene at 60°C are 385 and 135 Torr, respectively. Calculate (a) the partial pressures of benzene and toluene, (b) the total vapor pressure of the solution, and (c) the mole fraction of toluene in the vapor above a solution with 0.60 mole fraction toluene.
Solution:
In the given problem, if mole fraction of toluene in a solution is
then mole fraction of benzene in a solution is
The sum of the mole fractions of the components in a solution must be equal to 1. In this case, benzene and toluene are the components in a solution.
The vapor pressures of pure benzene and toluene at 60°C are
(a) By Raoult's Law, we can calculate the partial pressures of benzene and toluene in the vapor.
Partial Pressure of Benzene:
Partial Pressure of Toluene:
(b) The total vapor pressure of the solution is
(c) Since we know already the partial pressures of benzene and toluene as well as the total vapor pressure of the solution, therefore, the mole fraction of toluene in the vapor is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
Four liters of octane gasoline weigh 3.19 kg. Calculate the volume of air required for its complete combustion at STP.
Solution:
The first thing that we have to do is to write the chemical equation for the combustion of octane gasoline as follows
Make sure that the above chemical equation is balanced. Next, we will calculate the molecular weight of octane gasoline as follows
Moles of octane gasoline:
From the chemical equation above, moles of oxygen is
At STP (Standard Temperature and Pressure), 1 mole of any gas is equal to 22.4 liters. Hence, the volume of oxygen is
Since air contains 21% oxygen and 79% nitrogen, therefore, the volume of air required for the complete combustion of octane gasoline is
