When 3.56 g of an unknownhydrocarbon was dissolved in 125 g of benzene, the free

Question

When 3.56 g of an unknownhydrocarbon was dissolved in 125 g of benzene, the freezing pointof the solution was found to be 4.41oC. Analysis of thehydrocarbon showed that it contained 93.71% carbon (by mass) and6.29% hydrogen. Determine the molecular formula of the unknownhydrocarbon. The freezing point of benzene is 5.50 °C andKfp = -5.12 °C/m. When 3.56 g of an unknownhydrocarbon was dissolved in 125 g of benzene, the freezing pointof the solution was found to be 4.41oC. Analysis of thehydrocarbon showed that it contained 93.71% carbon (by mass) and6.29% hydrogen. Determine the molecular formula of the unknownhydrocarbon. The freezing point of benzene is 5.50 °C andKfp = -5.12 °C/m.

Explanation / Answer

Multistep problems like this one are fun. It requires two largely separate steps; the first one I'llshow will be how to calculate the empirical formula of thecompound; the empirical formula being some unknown multiple of theactual molecular formula... if that doesnt make sense yet, try tofollow the steps below: First, ignore everything about the benzene and its freezingpoint. Just assume we have a 100g sample of the unknownhydrocarbon. Since we have exactly 100g, the percentages given in thequestion tell us that there will be 93.71g of Carbon in thatsample, and 6.29g of Hydrogen. The next thing we do is convert these two mass values intomoles: 93.71g x 1molC = 7.81molC               12.0g 6.29g x 1molH = 6.29molH             1.0g Now that we know the number of moles of each element in our100g sample, we divide the larger number of moles by the smallernumber, to find the proportion of each of the elements to eachother: 7.81mol   =1.24       Now, 1.24 is very close tothe fraction 5/4, so its safe to assume that this is withinexperimental error, 6.29mol                     and that the proportion of carbon to hydrogen in the compound is5:4. Therefore, the empirical formula of the compound isC5H4 Next, we move onto the given data with benzene, in order tocalculate the molar mass of the hydrocarbon. From thefreezing point depression formula, T = Kfmi (where m is the molality and i is a constant expressing thenumber of moles of dissociating ions in the dissolvingcompound), we can calculate the molality of the hydrocarbon. (i in theequation will just be 1 in this case and can be ignored, sincehydrocarbons do not dissociate into ions in solution) T = 4.41 - 5.50 = -1.09 degrees C Rearranging the formula, m = T   =   -1.09 = 0.213molal                                               Kf        -5.12 Next, we know that molality is defined as moles of solute perkilogram solvent, so using the amount of benzene given in thequestion, we can calculate the precise number of moles of thehydrocarbon: 125g x 0.213molHydrocarbon = 0.0266molHydrocarbon                 1000g Divide the given amount of mass of the hydrocarbonby this calculated number of moles to find its molecularmass:     3.56g = 133.8g/mol 0.0266mol Now that we know the molar mass, we add up the masses of theelements in the empical formula we calculated before: C5H4 --->   5 carbons + 4hydrogens = 5*12.0 + 4*1.0 = 64.0g   ---» thistells us the weight of the compound if its actual formula were thesame as the empirical formula. Finally, we just find the number by which you need to multiply64.0 to get our calculated molar mass of 133.8. Its not exact, but 2 times 64 will give 128, which should beclose enough for the purposes of a question at this level. What this means is that since the actual compound is twice asheavy as the empirical compound, the actual molecular formula istwice that of the empirical formula: Molecular formula = C5H4 x 2 =C10H8 (final answer) 0.0266mol Now that we know the molar mass, we add up the masses of theelements in the empical formula we calculated before: C5H4 --->   5 carbons + 4hydrogens = 5*12.0 + 4*1.0 = 64.0g   ---» thistells us the weight of the compound if its actual formula were thesame as the empirical formula. Finally, we just find the number by which you need to multiply64.0 to get our calculated molar mass of 133.8. Its not exact, but 2 times 64 will give 128, which should beclose enough for the purposes of a question at this level. What this means is that since the actual compound is twice asheavy as the empirical compound, the actual molecular formula istwice that of the empirical formula: Molecular formula = C5H4 x 2 =C10H8 (final answer)

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