Category: Chemical Engineering Math
"Published in Vacaville, California, USA"
A compound containing only carbon and hydrogen is subjected to combustion analysis. From a sample of the compound, 67.5 g of CO2 and 13.8 g of H2O are obtained. What is the empirical formula of the compound?
Solution:
Since the products from the combustion of a hydrocarbon are given, then we can identify the unknown hydrocarbon by getting the empirical formula.
Moles of each components:
Moles of carbon is
Moles of hydrogen is
From
the number of moles of each component, we need to divide all of them by
their least number of moles which is hydrogen in order to get the number
of atoms in a sample.
Number of carbon in a sample is
Number of hydrogen in a sample is
Since the empirical formula of the hydrocarbon which is CH is not considered as a molecule, then we need to multiply all the atoms by 2 so that the empirical formula is considered as a molecule. There's an excess of 3 valence electrons for carbon and there's no way that carbon can make 8 valence electrons. If we multiply all the atoms by 2, then carbon atoms can make 8 valence electrons by covalent bonding. There's a triple bond between the two carbons and one bond between carbon and hydrogen. Each carbons have now 8 valence electrons. Therefore, the empirical formula for the hydrocarbon which is acetylene is
Category: Chemical Engineering Math
"Published in Vacaville, California, USA"
The tip of the head of a strike anywhere match contains a phosphorus-sulfur compound that readily ignites when drawn over a rough surface. What is the empirical formula of this ignitable compound given that a sample of the compound contains 0.5629 g of phosphorus and 0.4371 g of sulfur?
(Molecular Weights: P = 31, and S = 32)
Solution:
Moles of each components:
Moles of phosphorus is
Moles of sulfur is
From
the number of moles of each component, we need to divide all of them by
their least number of moles which is sulfur in order to get the number
of atoms in a sample.
Number of phosphorus in a sample is
Number of sulfur in a sample is
Since
the number of atoms for phosphorus is a fraction which is 4/3, then we need
to multiply all the atoms by 3 so that the number of atoms for
empirical formula are all whole numbers. Therefore, the empirical
formula for a strike anywhere match is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
To determine the Ka of a weak acid, a student took 20 mL of a solution of the acid and titrated it with a base. He found out that it took 32 mL of base to neutralize the acid. He then took a new 20 mL of acid and added exactly 10 mL of the same base to it. He measured the pH of this solution and found it to be 4.20. What is the value of Ka for the weak acid?
Solution:
In the given problem, the concentrations of a weak acid and a base are not given but their volumes are given. We can still solve for the value of Ka for the weak acid. Let's assume that the unknown weak acid is a monoprotic acid and the unknown base is a monoprotic base so that it will be easy for us to solve for the value of Ka.
After the titration of 20 mL of a weak acid with 32 mL of a base, the amount of weak acid at the start of the reaction is
or
where CBase is the concentration or molarity of a base in millimoles of solute per milliliter of solution. Let's assign HA for the weak acid and BOH for the base.
The equilibrium reaction for the ionization of a weak acid is
and the ionization constant is
At equilibrium,
where x is the mmoles of HA that partially ionized. Hence, the above equation becomes
If 10 mL of the same base is added to a new 20 mL of a weak acid, then the amount of weak acid that is converted into salt BA is
From the reaction of a weak acid with the base, salt BA is formed
By using mole to mole relationship of the reactants and products, if
then it follows that
As mentioned earlier that 10 mL of the base is added to a new 20 mL of a weak base. At equilibrium, the amount of weak acid will decrease and the amount of product which is the A ion will increase. The ionization constant for the weak acid becomes
Let's assume that the value of Ka is less than 1 x 10-3 so that we can neglect x at the numerator and denominator as follows
If the pH of the solution is 4.20, then the value of x which is the concentration of [H+] is
Take the inverse logarithm on both sides of the equation, we have
or
Therefore, the value of Ka for the weak acid is
