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A 0.04358 g sample of gas occupies 10.0-ml at 292.0 k and 1.10 atm. upon further analysis, the compound is found to be 25.305% c and 74.695% cl. what is the molecular formula of the compound?

Answer

Answer:



A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?

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Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different.



25.305% C/12 = 2.108



74.695% Cl/35.5 = 2.104



So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.)



0.044 grams/10 ml = x/22.4 liters



0.044g/0.010 liters = x/22.4 liters



22.4 liters/0.010 liters = 2240 (ratio)



2240 x .044 = 98.56 (actual atomic weight)



CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole.



This is sufficiient to distinguish C2CL2, (dichloroacetylene)



from C6CL6 (hexachlorobenzene) which would



mass 3 times as much.
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