How to Determine Molecular Weight by Ideal Gas Law
The ideal gas law illustrates the simplest expression of the parameters determining gas behavior, namely pressure, volume and temperature. It is called ideal because it doesn't fit any real gas perfectly, but if atoms were infinitesimally small, taking up no volume themselves, and they had no interaction with each other, they should obey the ideal gas law exactly.
Even with the "imperfections" of real gases, the law yields fair-agreement results for the inert gases such as helium, and at least qualitative results for many other gases. As a bonus, other parameters can be calculated using the law, such as molecular weight.
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Instructions
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Derivation and Example
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Write the ordinary expression of the ideal gas law: PV = nRT, where "P" equals pressure, "V" equals volume of the container, "n" equals the number of moles, or mass units, involved, "T" equals temperature of the system, and "R" is the ideal gas constant.
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Convert the number of moles into a different form that includes molecular weight. One such form is the weight of the gas, "w," divided by its molecular weight, "MW": PV = (w/MW)RT.
Rearrange that equation so that its solution produces the molecular weight. Using simple algebra, transfer the molecular weight to the left-hand side of the equation and the pressure and volume to the right-hand side of the equation.
The result is MW = wRT/PV.
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3
Using the appropriate apparatus in the physical chemistry laboratory, introduce exactly 1 liter of helium gas into the container, under exactly 1 atmosphere of pressure. The temperature is 20 degrees Celsius. Calculate the weight of the container minus the weight of air it initially contained; the weight that is attributable to the helium gas is 0.167 grams. Since it is known that R equals 0.08206 liter-atmospheres/mole-degree Kelvin, and 20º C equals 293º K, the calculation is straightforward.
MW = (0.167 gm)(0.08206 lit-atm/mole-º K)(293º K)/(1.00 atm x 1.00 lit)
Therefore, MW equals 4.02.
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Notice that this determination is in close agreement with the actual molecular weight of helium gas, which is 4.00--well within experimental error. Attempting a molecular weight determination of a number of other gases would work fairly well; however, reactive gases such as chlorine gas would not work as well. Also, gases whose atoms take up a lot more space would introduce additional error. This is because atomic volume would need to be figured into the ideal gas equation.
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