This experiment was conducted to determine the molecular weight of acetone and an unknown volatile liquid by using the Dumas Method. Three trials are conducted for each volatile liquid. The molecular weight of the acetone and the unknown volatile liquid was determined by first taking 2mL of acetone and the unknown volatile liquid in a Erlenmeyer flask and heating it to the point where all the of the liquid is completely vaporized. The temperature is measured using a temperature probe from the MeasureNet system after being calibrated. The next step in determining the molecular weight of the two volatile liquids is to find the mass of acetone and unknown after the flask is heated by finding the difference in the mass of the flask before and after
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On the other hand, water would not give good results because it is not very volatile. Water also has a higher boiling point than pentane. Therefore the experiment would take longer to conclude if water was used. If there were a non-volatile impurity present in the unknown, it would result in the calculated molecular weight to be higher. This is due to the fact that the non-volatile impurity will not evaporate. This will cause the impurity to remain in the unknown liquid increasing the mass. The calculated molecular weight would be higher, if the actual pressure of the vapour was less than the surroundings. This will occur if the actual pressure was entered in the formula resulting in the molecular weight to increase greatly. The pressure allows the substance to be less …show more content…
Using the Dumas Method, the molecular weight for acetone and the unknown were determined through experimental trials. The molecular weight of acetone from the three trials was 51.31g and the unknown liquids molecular weight was 12.13g. When the molecular weight for the unknown volatile liquid was determined from the experiment and calculated, methanol was found to the identity of the unknown volatile liquid. The calculated molecular weight values for the two liquids were compared to the actual molecular mass to see if the experiment was successful or not. The percent error calculations showed that for acetone the experiment was fairly successful because the percent error was only 11.66%. The percent error was 62.14% for the unknown volatile liquid making the experiment unsuccessful for
On the other hand, water would not give good results because it is not very volatile. Water also has a higher boiling point than pentane. Therefore the experiment would take longer to conclude if water was used. If there were a non-volatile impurity present in the unknown, it would result in the calculated molecular weight to be higher. This is due to the fact that the non-volatile impurity will not evaporate. This will cause the impurity to remain in the unknown liquid increasing the mass. The calculated molecular weight would be higher, if the actual pressure of the vapour was less than the surroundings. This will occur if the actual pressure was entered in the formula resulting in the molecular weight to increase greatly. The pressure allows the substance to be less …show more content…
Using the Dumas Method, the molecular weight for acetone and the unknown were determined through experimental trials. The molecular weight of acetone from the three trials was 51.31g and the unknown liquids molecular weight was 12.13g. When the molecular weight for the unknown volatile liquid was determined from the experiment and calculated, methanol was found to the identity of the unknown volatile liquid. The calculated molecular weight values for the two liquids were compared to the actual molecular mass to see if the experiment was successful or not. The percent error calculations showed that for acetone the experiment was fairly successful because the percent error was only 11.66%. The percent error was 62.14% for the unknown volatile liquid making the experiment unsuccessful for