Question

A 0.4481-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8326 g of CO2 and 0.1705 g of H2O. What is the empirical formula of the compound?

A) CHO
B) C2H2O
C) C3H3O2
D) C6H3O2
E) C3H6O2

i just need a GENERAL IDEA on what to do

Answer 1

@sami-21 help

Answer 2

Last time i saw my chemistry book was 4 years ago :P

Answer 3

of what?

Answer 4

is there any easy way than this one?
http://answers.yahoo.com/question/index?qid=20080927180546AAWEvyP

Answer 5

would be if that would be explained...

Answer 6

Find the stoichiometry of \[\text{C}_x\text H _y \text O_z+\text O_2\rightarrow\text {CO}_2+\text H_2\text O\]

Answer 7

by that you mean?

Answer 8

determine the number of moles of the products

Answer 9

so i have to balance first @UnkleRhaukus ?

Answer 10

i have no idea what you meant @UnkleRhaukus ....honestly....

Answer 11

0.8326 g of CO_2

how many moles is this?

Answer 12

i got 0.07

Answer 13

oh wait...

Answer 14

0.019

Answer 15

ok good, 0.02 moles of CO_2

how many moles of H_2O
0.1705 g

Answer 16

weird... i got 0.0095

Answer 17

ok so we have
\[\frac{0.448}{x\times12.0107+y\times1.0079+z\times15.9994}[\text C_x\text H_y \text O_z ]\]\[\rightarrow\]\[\frac{ 0.8326}{12.0107+2\times15.9994}[\text {CO}_2]+\frac{0.1705}{2\times1.0079+15.9994}[\text H_2\text O]\]

solve for x,y,z

Answer 18

whops 0.448 should be 0.4481

Answer 19

uhmm explain?

Answer 20

oh nevermind

Answer 21

i went backwards one step because the accuracy looks like it is important

Answer 22

so what happens next? cross multiplication?

Answer 23

you could do that ,

im gonna try simply those fractions again to high accuracy

Answer 24

\[\small \frac{0.448}{x\times12.0107+y\times1.0079+z\times15.9994}[\text C_x\text H_y \text O_z ]\rightarrow 0.01892[\text {CO}_2]+ 0.009464 [\text H_2\text O]\]

Answer 25

i forgot how imprecise theses calculations are

comparing 0.01892
& 0.009464
we see the moles of CO_2 is approximately twice the moles of H_2O

Answer 26

i think this is starting to get more complicated than it's supposed to be o.O

Answer 27

multiply by one hundred

Answer 28

\[ \frac{44.8}{x\times12.0107+y\times1.0079+z\times15.9994}[\text C_x\text H_y \text O_z ]\rightarrow 2[\text {CO}_2]+ 1 [\text H_2\text O]\]

Answer 29

determine the stoichiometry by balancing the atomic species,

for some reason i left oxygen off the LHS,

Answer 30

\[\text{C}_x\text H _y \text O_z+\text O_2\rightarrow2\text {CO}_2+1\text H_2\text O\]

balance the carbons first

Answer 31

on the RHS

\[2\text C : 2\text H : 5\text O\]

Answer 32

\[2\text C_x\text H_x \text O_{z}+\text O_2\rightarrow2\text {CO}_2+1\text H_2\text O\]\[4\text C_x\text H_x \text O_{z}+\text O_2\rightarrow4\text {CO}_2+2\text H_2\text O\]

Answer 33

\[4\text C\text H \text O+\text O_2\rightarrow4\text {CO}_2+2\text H_2\text O\]