Elemental ananalysis of a compound results in the following data: 65.45 %C, 5.49 %H, 29.06 %O :?

Question

5.51 g of the compound fills a 179.0 ml bottle with a pressure of 5183.3 mm Hg at a temperature of 24.4oC. What is the molecular formula of this compound ?

Answer 1

I love these kinds of problems...lol...

First you need to find the empirical formula of the compound. To do this, pretend that you have 100 grams of the compound. From your elemental analysis, that means that in that 100 grams, you have 65.45 grams of C, 5.49 grams of H and 29.05 grams of O. Convert each of those amounts into moles by dividing by each elements atomic weight. That gives you 5.45 moles of C, 5.49 moles of H and 1.80 moles of O. Now divide all of those by the smallest number. (What you're trying to find is the ratio of atoms of each element in the formula.). WHen you do that you find the empirical formula of the compound is C3H3O. From that empirical formula, the molar mass is 55.

Now, go to the gas law part of the problem. You need to somehow get to the actual molecular weight of the compound. Using PV=nRT, you can calculate the number of moles of the compound you have. (Remember to use the right units for each of the numbers you're given.) I got something like 0.05 moles.

Now, to get the actual molar mass, divide the mass you have in the flask by the number of moles you just calculated. This gives a molar mass of 110 g/mol.

Since your empirical formula has a mass of 55, the compound has a formula of C6H6O2.

Hope this makes sense...

Answer 2

Use the equation PV = nRT to unravel for n that's the style of moles (placed the quantity, tension, and temp into the the superb option contraptions for the value of R which you employ). Then, calculate the gmw of this compound with the help of dividing the burden (a million.80 g) with the help of the style of moles produced above. Then, multiply the gmw with the help of the p.c. carbon (0.4 hundred). this supplies you a weight which ought to be on the brink of a few diverse of the gaw of carbon (12,24,36, etc.). no remember if this is 36, then you definately could be conscious of that the formulation consists of 3 atoms of carbon. proceed to try this for H and O and you will locate the finished molecular formulation.