During this laboratory experiment, a gas chromatograph was used to analyze the ethanol and butanol content of solutions of known mass ratios, and then construct a calibration curve using calculated area ratios from the graphs produced by the gas chromatograph. A solution of unknown mass ratio was then analyzed with the gas chromatograph in order to determine its area ratio. This area ratio was compared to the calibration curve in order to find the mass ratio of the unknown. Seven solutions of ethanol and butanol were prepared using different mass ratios of the two components. The solutions were then one by one injected into the gas chromatograph using a syringe. As each solution went through the gas chromatograph, the apparatus heated the solution. Ethanol and butanol have different chemical compositions and boiling points, so they reached a detector at different times. As the components reached the detector, the voltage given off by the components was read and recorded as a graph on a computer of time versus voltage.
Specifically, the gas chromatograph functions using an inert gas, in the case of this experiment, nitrogen, to function as a carrier gas for the analyte. The analyte is injected through a septum using a microsyringe. The carrier gas aids the analyte flow through the column. As the analyte moves through the column, the different components are separated and reach the detector at different rates, as the components have different chemical properties. In the case of this experiment, once the analyte reached the detector, the thermal conductivity of the components were measured as voltage. The voltage readings were graphed as a function of time on a computer.
|Desired Ratio of Masses of Ethanol to Butanol||Mass of Ethanol (g)||Mass of Butanol (g)||True Ratio of Masses of Ethanol to Butanol||Ratio of Areas of Ethanol to Butanol|
*The graph of the original unknown was unreadable to determine the ratio of the areas, so
Anwar’s group’s graph of unknown was used to find a ratio of areas.
|Equation of Linear Regression Line||y = 1.134x – 0.0184|
|Mass Ratio of Unknown (Ethanol vs. Butanol)||1.750437|
|Substance||Weight of Cut-out Piece of Paper for 4:4 Composition (g)||Ratio of Areas|
The calibration curve came out fairly well. It follows almost exactly a straight line except for the value where the ratio of masses was 5:3. The ratio of the areas is much lower than it should be. There is a decrease in the ratio of the areas from the 4:4 composition (1.100032), to the 5:3 composition (0.846482), and then the area ratio goes back up in the 6:2 composition (2.95128). The value for the ratio of the areas should have been somewhere between 1.100032 and 2.95128. Other than that one value, all the other values are very linear. It should have been expected that the values would linear, considering that the ratio of the masses was not increased exponentially; the ratios were only decreased and increased by a constant value. This also means the slope of the area ratio versus the mass ratio should be close to being in unity (a value of one), because of the constant increase and decrease in the mass ratio.
The ratio of the areas for the original unknown was unreadable on the graph, so another group’s unknown graph was used to find a ratio of areas. That group did not record their original masses used for their unknown sample, so it is not possible to find the percent error. However, using the equation for the linear regression line, the expected ratio of masses could be found by subbing the ratio of the areas in for “x”. The mass ratio found for their unknown was 1.750437, which is close to the mass ratio for the 5:3 composition (1.677444), so it can be estimated that their unknown was composed of about 5 grams of ethanol and 3 grams of butanol. A percent error can be determined by comparing one of the points on the calibration curve to the actual equation of the line. For example, using the 6:2 mass compisition, the actual mass ratio was 3.083036. However, using the calculated area, according to the calibration curve the theoretical mass ratio is 3.31701. This yields a percent error of 7.59%. This is a fairly low percent error, but it should be, as the linear regression line is calculated to make it as close to the points on the graph as possible.
Finally, using the alternative method of finding the ratio of areas by physically weighing the two curves, the value was extremely close to the calculated value. The calculated ratio for the 4:4 mass ratio was 1.100032, while the ratio from weighing the curves was 1.094. This shows that weighing all the curves would have been an acceptable way of determining the area ratios, rather than using the trapezoidal rule to calculate the area ratios. The trapezoidal rule has error to it, so weighing the curves may even be more accurate than using the trapezoidal rule. If it was not a requirement to use the trapezoidal rule, it would have been preferred to simply weigh all the curves, as this would have been a much quicker way of determining the area ratios.