Chem HELP!!! :) thanks in advance!?

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 187.0 ml bottle with a pressure of 2455.8 mm Hg at a temperature of 21.4oC. What is the molecular formula of this compound ?

Answer 1

To do these, first determine the empirical molecular formula:

Divide each weight % by the atomic wt of it's respective element to give relative numbers in terms of atoms:

65.45/12.01 = 5.45
5.49/1.001= 5.48
29.06/16.0 = 1.82

Now get them into the lowest whole numbers. First ,we'll try dividing everything by the smallest value 1.82 (oxygen value) and see what happens:

we get: C2.99H3.01O1

(this is close enough to count as whole numbers) so the empirical formula is: C3H3O

Remember, the empirical formula is not necessarily the actual formula! It will have the proper ratio of atoms but might not be the proper molecular weight. Actual wt might be 2 x or 3 x this value etc.

Now we need to use the other data to get the molecular weight of the material, first, use the gas equation, solving for moles:

Get everything into the proper units:

2455.8mmHg/760mmHg/atm = 3.231atm
187ml/1000ml/L = 0.187L
21.4 C + 273.1 = 294.5K

n = PV/RT =

3.231atm x 0.187L/0.08206Latm/molK x 294.5K =

0.0250moles

Now calculate molecular weight:

4.13g/0.0250mol = 165.2g/mole

Now, we combine the 2 pieces of data to get the true formula:

The empirical formula: C3H3O is 55.03g/mol; we need to see how many empirical formulas fit into the actual weight per mole (we just determined) to get the actual molecular weight [it should come out to a whole number (or nearly so) if there are no mistakes].

165.2/55.03 = 3.002 so the formula is 3 x the empirical

formula or C9H9O3

Answer 2

Sb: atomic mass = 121.seventy six S: atomic mass = 32.0.5 Molar Mass, Sb2S3 2 Sb x 121.seventy six = 243.fifty two 3 S x 32.0.5 = ninety six.18 total................339.70 % Sb = (243.fifty two/339.70) x one hundred =~seventy two% Use your calculator to get top discern.