Topic: Chemistry Help

Yes, I know you're going to say "figure it out yourself", but I need help with this problem and I want you to check my work please. Here's the problem and my work:

10: Analysis shows that a compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Further analysis shows finds the molar mass to be 720g/mol. Determine the empirical & molecular formulas.

EF: 100% = 100g. 40g, 6.7g, 53.3g. (I got the 680 g/mol from my molar mass of problem #4)

40g X 1mol over 680. 40 X 1mol = 40mol / 680 = 0.05
6.7g X 1mol over 680. 6.7 X 1mol = 6.7mol / 680 = 0.009.
53.3g X1mol over 680. 53.3 X 1mol = 53.3mol / 680 = 0m07

0.05/0.05 = 1
0.009/0.05 = .18
0.07/0.05 = 1.4

CH18O

MF:
C = 12
H = 18 (Hydrogen's atomic mass * 18)
0 = 16.

12 + 18 + 16 = 46.

46/720 = 15.65. I rounded it to be 16.

I times 16 by my empirical formula subscripts:

1 * 16 = 16
18 * 16 = 288
1 * 16 = 16

C16H288O16

Carbon's atomic mass (12) * 16 = 192
Hydrogen's atomic mass (1) * 288 = 288
Oxygen's atomic mass (16) * 16 = 256

192 + 288 + 256 = 736.

Where did I messed up? Thank you.

Um er uh 2+2=24

Nailed it

If you don't round 720/46 to 16 you will get the right answer.

Thank you. So How would the MF subscripts look?

0m07 should me 0.07. My apologies.

I got it thank you @GregDavisBlows.

Well I haven't taken chemistry since I was a sophomore in high school which was 5 years ago so I'm not sure what it supposed to look like. I would just leave everything as mixed numerals but I have no idea if that's the right thing to do.

That looks like a long way of calculating empirical formula. What I do is that I find the atomic mass of the elements involved and use it to divide their given proportions respectively:
C. H. O.
40. 6.7 53.3 where atomic masses of C, H and O are 12, 1 and 16 respectively. Dividing the proportions by their respective atm. masses, we have;

C. H. O.
3.33 6.7 3.33 Then we divide all through by the smallest value here (in this case, 3.33) and approximate, so that we have;

C. H. O.
1 2 1 This is the no. of moles of each element in the empirical formula;
ÇH20 = Empirical formula.
I would have continued but my phone battery is low. Later!

Solved it now thanks to @GregDavisBlows.

NERDS

(looks down at full body Mario Costume... cries )

@baba_944 As a student taking a chem college course, I can say that you never want to round anything until the end. Leave it all in your calculator. And apply significant figures afterwards.

Wow, this wasn't something I was expecting to see here at Nintendolife! Chemistry students unite!

I hate chemistry.

Hey! Don't twist my words, young man!

Equola is spreading.