Dumas Method To Identify The Unknown Molar Mass Of
Using a Dumas method, you are asked to identify the unknown pure liquid by comparing the molar masses of the given possible substances (ethanol, methanol, isopropanol, acetone).
Introduction
Dumas Method allows us to measure the molar mass of the substance, using the Ideal Gas Law. The Ideal Gas Law, PV=nRT, explains the behaviour of the gases that are near 100kPa and in the room temperature. This equation shows the product of the pressure(P) and volume(V) equals to the product of the number of moles(n), ideal gas constant (R, 8.31451J/molK), and the temperature in Kelvin(T). In order to calculate ‘n’, several assumptions must be made. As the substance goes through the boiling water bath, the beaker is assumed to only contain the vaporized from of the …show more content…
• The atmospheric pressure in the lab is 101.7kPa.
• The volume of the flask is 159mL.
TRIAL #3
Qualitative
• The solvent is clear liquid that has a water-like consistency.
• The unknown solvent is the sample #3
• After the heating, the organic solvent is still clear.
• As the heating continues, the liquid portion of the solvent gradually starts to disappear. Noticeable precipitation or significantly visible chemical reactions were not observed.
• Flask is empty and the liquid is no longer present in the flask.
• When the cold water was ran over the flask to speed up the cooling, the liquid formed again inside the flask.
Quantitative
• Empty Erlenmeyer flask + 2.5” x 2.5” aluminum foil + rubber band = 85.59g (before heating)
• 2mL of the unknown Solvent is dispensed into an Erlenmeyer flask.
• Temperature probe reads 100.2 degrees for the temperature of the water bath that was heating up the flask.
• Solvent + flask + 2.5”x2.5” foil + rubber band = 85.77g (after heating)
• The mass difference between the after heating and before heating is 0.18g.
• The atmospheric pressure in the lab is 101.7kPa.
• The volume of the flask is
Introduction
Dumas Method allows us to measure the molar mass of the substance, using the Ideal Gas Law. The Ideal Gas Law, PV=nRT, explains the behaviour of the gases that are near 100kPa and in the room temperature. This equation shows the product of the pressure(P) and volume(V) equals to the product of the number of moles(n), ideal gas constant (R, 8.31451J/molK), and the temperature in Kelvin(T). In order to calculate ‘n’, several assumptions must be made. As the substance goes through the boiling water bath, the beaker is assumed to only contain the vaporized from of the …show more content…
• The atmospheric pressure in the lab is 101.7kPa.
• The volume of the flask is 159mL.
TRIAL #3
Qualitative
• The solvent is clear liquid that has a water-like consistency.
• The unknown solvent is the sample #3
• After the heating, the organic solvent is still clear.
• As the heating continues, the liquid portion of the solvent gradually starts to disappear. Noticeable precipitation or significantly visible chemical reactions were not observed.
• Flask is empty and the liquid is no longer present in the flask.
• When the cold water was ran over the flask to speed up the cooling, the liquid formed again inside the flask.
Quantitative
• Empty Erlenmeyer flask + 2.5” x 2.5” aluminum foil + rubber band = 85.59g (before heating)
• 2mL of the unknown Solvent is dispensed into an Erlenmeyer flask.
• Temperature probe reads 100.2 degrees for the temperature of the water bath that was heating up the flask.
• Solvent + flask + 2.5”x2.5” foil + rubber band = 85.77g (after heating)
• The mass difference between the after heating and before heating is 0.18g.
• The atmospheric pressure in the lab is 101.7kPa.
• The volume of the flask is