Question

You perform combustion analysis on 255 mg of a molecule that contains only C, H, and O, and you find that 561 mg CO2 is produced, along with 306 mg H2O. If the molecule contains only C, H, and O, what is the empirical formula?

Answer 1

@nincompoop @Jesstho.-. @abb0t @hartnn

Answer 2

first calculate how much carbon is produced from CO2, then calculate how much hydrogen is produced from watter and add those two numbers up. that number you get substract from begining mass of 255 mg and you will get oxygen, from there should be easy...

Answer 3

Is this right?
\[0.561gCO2 \times \frac{ 12gC }{ 44gCO2} = 1.53 \times 10^{-1} gC = 153mg C\]\
\[0.306gH2O \times \frac{ 1gH }{ 18gH2O } \times 2mol H = 3.4 \times 10^{-2} gH= 34mgH\]
153mgC + 34mgH = 187mg
255-187=68mgO
Now do I change everything to grams to convert to moles?

Answer 4

really? how to change to grams?! ok...
how many mg is there in 1 g?

Answer 5

1mg=0.001g. lol i wasn't asking how. i was asking if that's what i am supposed to do.

Answer 6

ok convert that to moles and arange them in growing order i.e.
n(O):n(H):n(C) = ? : ? : ?
and then you take least number and divide all numbers with that number untill you get whole or approx. whole numbers

Answer 7

ex.
n(O) = 1
n(H) = 0,5
n(C) = 2

n(H) : n(O) : n(C) = 0,5 : 1 : 2

then

0,5/0,5 = 1
1/0,5 = 2
2/0,5 = 4

and you get C4O2H1

Answer 8

:( how'd you do that? This is what I did:
C: 0.153gC/12g = 0.0127molC / 0.00425 = 3
H: 0.034gH/1g = 0.034molH/ 0.00445 = 8
O: 0.068gO / 16g = 0.00425molO / 0.00425 = 1

Answer 9

C3H8O

Answer 10

what am i doing wrong?

Answer 11

are hydrogen and oxigen gaseous molecules or atomized gas? i.e.
is it O2 and H2 or O and H?

Answer 12

I don't know..it says: the molecule contains only C, H, and O and you find that 561 mg CO2 is produced, along with 306 mg H2O

Answer 13

i am asking you for your calculated masses that you got from that data, is that mass m(O2) or is it m(O)? go back trough each step and check... i know its teadious but thats how you learn...

Answer 14

oh i did O not O2

Answer 15

I did H2

Answer 16

now correct entire calculation...

Answer 17

how?

Answer 18

make entire calculation for n(O), n(H), n(C) get it?

Answer 19

But what exactly did I do wrong? What should I fix?

Answer 20

somewhere in your calculation you took molar mass for oxygen molecule and not oxygen atom or vice versa, and check the same for hydrogen... dont believe you made misstake with carbon but you can check it too

Answer 21

so should i have done this:
\[.306g H20 \times \frac{ g1H }{ 18g H2O } \times 1mol H \] ?

Answer 22

dont know, try like that and see what you get

Answer 23

that would be 17mg of H

Answer 24

\[0.561gCO2 \times \frac{ 12gC }{44gCO2 } = 153mgC\]
\[0.306 \times \frac{ 1gH }{ 18gH2O} = 0.017 g H = 17mgH\]

Answer 25

153-17 = 136
255-136 = 119mgO

Answer 26

\[C: \frac{ 0.153gC }{ 12gC} = \frac{ 0.01275molC }{ 0.0074375 } = 1.714\]
\[H: \frac{ 0.017gH }{ 1gH } = \frac{ 0.017molH }{ 0.0074375 } = 2.29\]
\[O: \frac{ 0.119gO }{ 16g } = 0.0074375molO/00.0074375 = 1\]

Answer 27

C2H2O?

Answer 28

you didnt get quite whole numbers... try higher multiplier... ex.

N(C) : N(H) : N(O) = 1,714 :2,29 : 1 / * 2
= 3,428 :4,58 : 2 .... NOT A GOOD FIT now multiply upper numbers with 3 and see if you get better fit to a whole number

Answer 29

1.714 * 3 = 5.142
2.29 * 3 = 6.87
1*3 = 3

Answer 30

now round those numbers to one digit and tell me do you think its a proper fit or do you need to multiply higher? is it? C5H7O3 or not?

Answer 31

yeah C5H7O3. That's the empirical formula?

Answer 32

well if you calculated all correct it should be...

Answer 33

Hmm. Ok. Thanks.

Answer 34

and heres a bit of practice material now that i tortured you for long enough...
http://depts.washington.edu/chemcrs/bulkdisk/chem142A_aut07/notes_C142A%20Chapter%203%20Part%203&4.pdf

Answer 35

wow thanks so much!

Answer 36

now take a look at page 4 of that document i gave you... :D

Answer 37

I am sorry, but I think that she was right with the C3H8O1.

Answer 38

i did not check calculation, i just pointed and wanted to see if she/he would calculate it and use their head... now in that document are solved examples that should make all things clear...

Answer 39

what you did was confuse her

Answer 40

:o so the work i did for my first answer is correct?

Answer 41

I think so, but @Kryten as a different opinion.
I dont know why he said about the molecular gases O2, H2.

Then you get messed up with the 1gH/18gH2O that doesn't make sense. It is 2gH/18gH2O

Answer 42

yep i checked it first calculation should be right but i was meerely pointing out that you need to be carefull... sorry for confusion, and further steps are also useful to know if whole numbers dont come right away... thats the procedure of calculation and again im sorry for confusion and again caution with molecular forms (ex. O2 and O) lots of misstakes are made there

Answer 43

Okay. Thank you to both of you for your time. :)

Answer 44

no problem!

Answer 45

not sure who to give best response to...