Molar Mass Lab

713 Words2 Pages

Introduction and Background :
The molar mass of a gas relates to the number of grams in one mole of that particular gas and all molar masses can be found on the periodic table of the elements. The objective of this lab is to compare the theoretical value of butane gas’s, C4H10 (g), molar mass with an experimental value where the gas from a lighter was released and measured by testing if the water displacement method, where a graduated cylinder is inverted in water and gas is released beneath it and trapped in the cylinder, is reliable for collecting data due to it being combined with water vapour as well. After the volume and mass of the gas released are measured, the number of moles can be found by subbing all known variables, including temperature …show more content…

R is the universal constant in this equation with a value of 8.314 L·kPa/K·mol.
By finding the number of moles, the molar mass of butane can be calculated using the formula :

Where, again, n is moles, M is the molar mass of butane in g/mol, and m is the mass in grams, found by subtracting the resulting mass of the lighter after drying from the initial mass.

The theoretical and experimental values of the molar mass of butane gas will be very similar, proving the accuracy of the periodic table of elements. The prediction for the molar mass of butane gas can be found by adding together the molar masses of carbon and hydrogen in accordance to butane’s formula of C4H10:
Molar mass of butane = (periodic table molar mass of carbon x number of carbon atoms in …show more content…

This reading is the same for the total pressure occupying the graduated cylinder due to all measurements being taken when the level in the cylinder was equal with that of the bin water, equalizing the pressures on the outside and the inside.
The partial pressure of water vapour in the room at 21°C is 2.49 kPa (see table one in References). With both the pressure of the atmosphere leveled with the water in the cylinder and the partial pressure of the water vapour combined with the butane gas, Dalton’s law of partial pressures can be applied to determine the partial pressure of just the butane gas.

The partial pressure of the butane gas is 93.2 kilopascals. The temperatures of both the water used in the bin and of the room were measured and came to be 18°C for the room and 21°C for the bin water, which remained 21°C even at the end of the lab. In kelvins, this comes out to (273.15 + 21) 294.15, 294 K going to three significant digits. The volume at the end of the lab of the gas in the graduated cylinder was 0.400 L when the level of water in the cylinder was level with that of the water in the bin, so equal pressures were acting upon them

Open Document