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Question Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.538 g/L at 25ºC and 721 mmHg? Answer The density is of course key to solving this problem. It is possible to backtrack from the density to determine the molar percentage using the ideal gas law. The equation that we will be dealing with in this case is derived as follows: PV = nRT n/V = P/RT m/(molar mass)V = P/RT d = P(molar mass)/RT Now plugging in the values into the above equation will give you the average molar mass of the two components of the gas. Let us get that out of the way first after rearranging to get the molar mass: molar mass = dRT/P molar mass = (0.538 g/ molar mass = 13.8828 g/mol Now, this is the average molar mass. From here it is possible to calculate the number of mols of each component by knowing their molar mass. To do this simple algebra is necessary. (molar mass)ave = x(molar mass)He + (1-x)(molar mass)O2 To arrive at the above equation you have to realize that you are really calculating the percentage of each in the gas mixture. Your total should be 100% which means it should add up to 1. Knowing this one of your components is equal to 'x' (a variable) and your other component is equal to 1-x meaning that your first component is going to be less than the total amount of your mixture and your second component is just going to be the difference. Now, to plug and chug. (13.8828 g/mol)ave = x(4.003 g/mol)He + (1-x)(32.00 g/mol)O2 13.8828 = 4.003x - 32.00x + 32.00 -18.1172 = -27.997x .6471 = x = % He % O2 = (1 - % He) = 0.3528 So this is a roughly 65/35 mixture of helium to air. I tried to do some sleuthing to determine if this was close to the standard mixture. One such commercially available mixture is Heliox, it is approximately 80/20 helium to oxygen although from what I have found the ratio can vary widely depending on the specific application, though 80/20 appears to be the standard for medical use. So I was unable to confirm if this is indeed the correct answer with any great certainty since the method by which I achieved the answer is somewhat convoluted.
Reference Chang, Raymond. Chemistry. 6th ed. USA, McGraw-Hill. 1998 |
