Quantitative Analysis
The content that follows is the substance of lecture 16. In this lecture we cover Titration and Combustion Analysis processes.
Titration
A titration is a process by which a known volume but unknown concentration of an acid, base or other solution is reacted with a volume of titrant of known concentration to reach an endpoint. The endpoint is where the moles of unknown and the moles of titrant are equal.
The theory is that if we can determine the moles of titrant that exactly matches the moles of unknown we can determine the concentration of that unknown. In order to do this process, we need a few peices of information: 1) A balanced chemical reaction between the unknown and the titrant compounds; 2) the mass or volume of the unknown; and 3) the concentration of the titrant.
Titration Problems:
The process of a titration problem follows the same process of any other stoichiometric problem.
Example: A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl?
Frist write the balanced chemical equation for the reaction between the unknown and the titrant.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Second convert the volume of titrant to the number of moles using the concentration.
# of moles NaOH = (0.5 M)(.025 L) = 0.0125 mol NaOH
Third use the balanced equation to calculate the moles of the unknown.
0.0125 mol NaOH x 1 mol HCl/ 1 mol NaOH = 0.0125 mol HCl
Finally convert the moles of unknown to concentration (M) using the volume of unknown that is given.
Molarity = # of moles/volume
Molarity of HCl = (0.0125 mol)/(0.050 L) = 0.25 M
Here are some practice problems for you to try:
Combustion Analysis
Another form of Quantitative Analysis is called Combustion Analysis and it involves the burning of substances, predominantly hydrocarbons (molecules made up of carbon and hydrogen) in oxygen.
A typical Combustion reaction:
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)
Pure Hydrocarbons burning in oxygen always produce the same two products: carbon dioxide and water. The quantitative analysis of a hydrocarbon by combustion analysis involves the burning of a hydrocarbon of unknown composition. Represented as CxHy the production of carbon dioxide and water is measured and then used to determine the molar ratio of carbon to hydrogen in the unknown. In a process similar to the one we used when determining the empirical formula of compounds by the mass percent, the empirical formula of the unknown hydrocarbon is determined.
The primary concept to understand is that all of the carbon in the unknown will be found in the CO2 and all of the hydrogen will be found in the H2O. Therefore if you know the mass of these two products you can use mass percentages and the same basic process as before to determine the empirical formula.
Here is a step by step example:
Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the empirical formula for menthol?
As before, the first step is to convert the masses given to moles. But unlike before we then need to convert the moles of CO2 and water to moles of Carbon and Hydrogen, respectively:
| 0.2829 g of CO2 x | 1 mol CO2 | =0.006430mol CO2 x | 1 mol C | =0.006430 mol C |
| 44.0g CO2 | 1 mol CO2 | |||
| 0.1159 g H2O x | 1 mol H2O | =0.006439mol H2O x | 2 mol H | = 0.01288 mol H |
| 18.0g H2O | 1 mol H2O |
Now we need to make an extra step here because we need to figure out the moles of oxygen too. In order to do that we need to figure out how much of the mass we were given for the overall sample was oxygen. Since the mass of carbon + mass of hydrogen + mass of oxygen = total mass. Wehave to convert the moles of hydrogen and carbon to grams to calculate the mass of oxygen. Then we will be right back on track to calculating the empirical formula as we have done it previously.
So convert the moles to grams of the carbon and hydrogen:
| 0.006430 mol C x | 12.0g C | =0.07716 g C |
| 1 mol C | ||
| 0.01288 mol H x | 1.0g H | =0.01288 g H |
| 1 mol H |
Find the mass of Oxygen by subtracting the C and H from the total mass of the sample
Total= mass C + mass H + mass O
0.1005g= 0.07716 g C + 0.01288 g H + mass O
mass O= 0.01046g O
Convert to moles of O
| 0.01046g O x | 1 mol O | =0.0006538 mol O |
| 16.0 g O |
| Finally find the mole ratio by dividing by the smallest quantity | Empirical Formula C10H20O |
| 0.006430 mol C/ 0.0006538 =9.83≈ 10 | |
| 0.01288 mol H/ 0.0006538 = 19.70 ≈20 | |
| 0.0006538 mol O/ 0.0006538 = 1 | |
If you would like another example explained as you go, here is a cool step by step set of videos. The first one goes over a simple process with just carbon and hydrogen but if you keep watching it will then take you to a more advanced problem.
Here are some problems for you to try:
1) A hydrocarbon fuel is fully combusted with 18.214 g of oxygen to yield 23.118 g of carbon dioxide and 4.729 g of water. Find the empirical formula for the hydrocarbon.
2) After combustion with excess oxygen, a 12.501 g of a petroleum compound produced 38.196 g of carbon dioxide and 18.752 of water. A previous analysis determined that the compound does not contain oxygen. Establish the empirical formula of the compound.
3) In the course of the combustion analysis of an unknown compound containing only carbon, hydrogen, and nitrogen, 12.923 g of carbon dioxide and 6.608 g of water were measured. Treatment of the nitrogen with H2 gas resulted in 2.501 g NH3. The complete combustion of 11.014 g of the compound needed 10.573 g of oxygen. What the compound’s empirical formula?
4) 12.915 g of a biochemical substance containing only carbon, hydrogen, and oxygen was burned in an atmosphere of excess oxygen. Subsequent analysis of the gaseous result yielded 18.942 g carbon dioxide and 7.749 g of water. Determine the empirical formula of the substance.
5) 33.658 g of oxygen was used to completely react with a sample of a hydrocarbon in a combustion reaction. The reaction products were 33.057 g of carbon dioxide and 10.816 g of water. Ascertain the empirical formula of the compound.