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Aim: To find the molar mass of butane, by finding the number of moles of gas in the container and comparing it to the mass of butane in the container

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Investigation on finding the Molar Mass of Butane Aim: To find the molar mass of butane, by finding the number of moles of gas in the container and comparing it to the mass of butane in the container Theory Butane (C4H10), also called n-butane, is the unbranched alkane with four carbon atoms, CH3CH2CH2CH3. Its only other isomer is methylpropane: CH(CH3)3. It is an organic compound which belongs to the alkane group or organic compounds. It is a highly flammable, colourless and odorless gas at r.t.p. this, along with the fact that is an easily liquedified gas, is why it is used in lighters as a fuel. Its Relative Molecular Mass is 58.12g, and it is barely soluble in water like most organic compounds: 0.0061 g/100 cm3, at 20 �C. In the experiment we shall find the mass of butane by calculating the change in mass of the lighter before and after the experiment. We shall find the number of moles in the container by finding the volume, pressure and temperature of butane inside the container, and then use the formula PV = nRT (where P: Absolute Pressure measured in millibars, V: Volume of gas measured in dm3,T: absolute temperature in Kelvin, and R is the universal gas constant, which equals to 83.14472 dm3�mbar�K-1�mol-1). ...read more.


Safe Test * We removed the metal piece of the lighter, to ensure the butane could not catch fire. * We were careful while dipping the lighter in ethanol, not to release butane as ethanol is flammable. Results Table 1: Raw Data Trial No. Mass of Lighter/ g [� 0.01g] The volume marking the Water reached/cm3 [� 0.05 cm3] Temperature /�C [� 0.01�C] Atm. Pressure /mmHg [� 0.5mmHg] Initial Final Change Initial Final Change 1 11.11 11.03 0.08 0.0 45.0 45.0 28.2 748.5 2 12.06 12.00 0.06 0.0 29.6 29.6 28.2 748.5 3 11.01 10.95 0.06 0.0 36.7 36.7 28.3 748.5 4 11.58 11.53 0.03 0.0 24.0 24.0 28.1 748.5 Table 2: Finding Pressure of Butane Trial No. Temp. / �C [� 0.01�C] Pressure/ mmHg [� 0.5mmHg] Atmospheric Pressure Pressure Inside Burette Pressure of water vapour Pressure of Butane 1 28.2 748.5 748.5 28.3 720.2 2 28.2 748.5 748.5 28.3 720.2 3 28.3 748.5 748.5 28.3 720.2 4 28.1 748.5 748.5 28.3 720.2 Calculations * Atmospheric Pressure = Pressure Inside Burette * Found pressure of water vapour at 28 �C to be 28.3 using a table, which is mentioned in apparatus * Pressure of Butane = Pressure Inside Burette - Pressure of water vapour Table 3: Finding the Number of Moles of Butane Trial No. ...read more.


Regarding the precision of error, the margin of error has been calculated as 60.2% and is far to high, resulting in a poor precision. However the experiment was more precise than this value of 60.2% shows it to be, because from the graph one can see that all the points lie close to the line of best fit. In the experiment, various errors could have occured, such as: Systematic Errors: * Error in readings of pressure due to high uncertainty. Random Errors: * The pressure due to the water column has no been taken into consideration. * We are assuming that the temperature inside the burette is equal to atmospheric pressure, which may not be true. * Water droplets may have still stuck to the lighter, causing an error in mass. To reduce the margin of error we could have: * Calculated the pressure exerted do to the water column, and subtracted it from the atmospheric pressure to find the pressure inside the container. * Taken a larger container than a burrette, so that we could release more butane and cause a larger change in mass, so that the uncertainty of 0.01g would have a smaller affect on the margin of error. * Used a seperate lighter for each trial, eliminating the error caused due to water droplets clinging to the sides of the lighter. ?? ?? ?? ?? Gaurang Poddar Yr.11, B3 Chem HL Page 1 of 7 ...read more.

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