Molecular Formula?

Question

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

4.13 g of the compound fills a 184.0 ml bottle with a pressure of 2501.7 mm Hg at a temperature of 22.1oC. What is the molecular formula of this compound ?

Answer 1

Take a 100 grams sample of the compound, then you would have:

65.45 g C / 12.00 g/mole = 5.454 moles

5.49 g H / 1.00 g/mole = 5.49 moles

29.06 g O / 16.00 g/mole = 1.816 moles

Now divide each by the smallest to get the mole combining ratio:

Carbon = 5.454 moles / 1.816 moles = 3.00

Hydrogen = 5.49 moles 1.816 moles = 3.02

Oxygen = 1.816 moles / 1.816 moles = 1

So your mole ratio is 3 : 3 : 1, and the empirical formula is C3H3O.

The formula mass of the empirical formula is 3(12) + 3(1) + 16 = 55 grams / mole

From the ideal gas law:

P = gRT/mm V

mm = gRT / PV = (4.13 grams)(62.4 l torr/mole K)(295.1 K) / (2501.7 torr)(0.184 l)

mm = 165 grams / mole

So the molecular mass is 3 times the empirical formula mass, which means your molecular formula must be C9H9O3.

Answer: C9H9O3

Answer 2

a. Use PV=nrT to determine the number of moles ( n) of the compound in gas. Divide 4.13 by (n) to find the mole weight.

(b) Divide the elemental analysis percentages by 12, 1 and 16 respectively. This will give you a "pseudo-formula" of CxHyOz. For example, y=5.49.

(c) Divide by the mole wt by 100. Take that factor and multiply x, y and z by it. That will give you the formula.

Answer 3

65.45 divided by the rmm of C wgich is 12 = 5.454
5.49 divided by rmm of H which is = 5.49
29.06 divided by rmm of O which is 16 = 1.81

divide by each 1 by the smallest value so O = 1
and C and H are within a tenth of 3
so the emperical formula is C3H30

and now im not sure how to work the molecular formula but its to do with the mass of one mole of the molecule and the the RMM of them