Chemistry help please~!?!?

Question

So ive been working on this problem and i dont know if i got it right so please help me!! and show me the work so if i did get the wrong answer i can see where i went wrong.

The percentages by mass of C, H, N, and O, in an unknown compound are found to be 23.3%, 4.85%, 40.78% and the rest is oxygen respectively.

A) determine the emperical formula for this compound


B) If the molar mass of the compound is 206g mol(-)1 what is the molecular formula?

thanks a million!

Answer 1

Assume 100g sample for ease of calculation.

23.3 g C = 1.9399 moles C atoms (23.3/12.011)
4.85g H = 4.8115 moles H atoms (4.85/1.008)
40.78 g N = 2.9112 moles N atoms (40.78/14.008)
31.07 g O = 1.9419 moles O atoms (31.07/16)

divide all by the smallest
C = 1.00 (1.9399/1.9399)
H = 2.48 (4.8115/1.9399)
N = 1.50 (2.9112/1.9399)
O = 1.00 (1.9419/1.9399)

multiply all by 2 to get whole numbers
C2H5N3O2; molar mass = 103 empirical formula

must be twice the empirical formula
C4H10N6O4; molar mass = 206 molecular formula

Answer 2

1. State each percent in grams
2. Find the molar mass of each from the periodic table... like there are 12.01 grams in 1 mole of Carbon, etc.

23.3 g C * 1 mol C / 12.01 g C = 1.94 mol C

4.85 g H * 1 mol H / 1.01 g H = 4.80 mol H

40.78 g N * 1 mol N / 14 g N = 2.91 mol N

31.07 g O * 1 mol O / 16 g O = 1.94 mol O

So, the empirical formula is C2H5N3O2.

Molecular formula is twice that, C4H10N6O4

Answer 3

23.3% + 4.85% + 40.78% = 68/93%

100% - 67.93% = 32.07% O

23.2/12 = 1.93

4.85/1.00 = 4.85

40.78/14 = 2.91

32.07/16 = 2.00

1.93/1.93 = 1.00 C

4.85/1.93 = 2.51 H

2.91/1.93 = 1.51 N

2.00/1.93 = 1.04 O

Multiply by 2: C2H5N3O2

C2=24 H5=5 N3=42 O2=32 = 103

So the molecular weight is C4H10N6O4

Answer 4

I might suggest a google search to find a chemistry help site. I know there is one but I am not sure the acutal page it comes from Prudue University...I am in the bussiness degree progam, and chem is not a class I am taking!!