[kaeyeon]

A 0.100g sample of a compound containing C, H, and O is burned oxygen producing 0.1783g of CO2 and 0.0734g of H2O. Determine the empirical formula of the compound.

[opti384]

Hint: the combustion of the compound only produces CO2 and H2O.

[kaeyeon]

I got H2CO3 but my friend says C3H6O2.  
Gave you a snack for the quick response! 

[kaeyeon]

Can anyone help? I'm still in a huge debate between H2CO3 or C3H6O2. D:

[yoyoils]

how exactly do you figure H2CO3 or C3H6O2? I'm curious how you worked it out (keeping things simple) given those masses

[Vidya]

okay follow this
calculate moles of  C in CO2 and  H in H2O
I give you hint for getting it
mass of CO2 ---> moles of CO2 -----> moles of C
mass of H2O ----> moles of H2O ----> moles of H
take the simplest ratio of C:H  to get the empirical formula

[yoyoils]

okay follow this
calculate moles of  C in CO2 and  H in H2O
I give you hint for getting it
mass of CO2 ---> moles of CO2 -----> moles of C
mass of H2O ----> moles of H2O ----> moles of H
take the simplest ratio of C:H  to get the empirical formula

When I take the simplest ratio of C:H, what about Oxygen?
Plus the question states "A .1g compound sample containing C,H, and O is burned oxygen producing...etc.." Did you mean to say burned *IN* oxygen? because you said it produced .1783g of CO2 and .0734g of H2O and when combined are totaled as .2517g mass. I'm just not sure if the original question is correctly worded

and I only came to .004074319mol H2 with .004051398mol Carbon if that's right?

[Vidya]

yes you need to take out O also
now when I calculated
mass of C = 0.04862727
mass of H = 0.00815555
now total of these two is less than total mass of the substance mentioned there
you need to subtract total of H and C from given mass and that is going to be of the O
so nothing is wrong the question
try it again
get the moles of O,H  and C and take their simplest whole number ratio

[yoyoils]

A 0.100g sample of a compound containing C, H, and O is burned oxygen producing 0.1783g of CO2 and 0.0734g of H2O. Determine the empirical formula of the compound.

What I meant was the starting sample weighed .1grams and ended up .2517grams of CO2+H2O....
Should I just ignore that fact? How does it amount to something more unless there was extra oxygen present? lol that's why I'm extremely picky about the question

[Vidya]

your calculations are not correct
you are adding masses of CO2 and H2O which you are not supposed to do so.
actually in making of CO2 and H2O oxygen of the atmosphere is also used  .
please do the calculations as I have mentioned  in my earlier post

[opti384]

Let's clarify things. As I said above,

the whole reaction will be something like this

C_xH_xO_x + O₂  xCO₂ + xH₂O

We can see that all the carbon from the unknown compound goes to form carbon dioxide. Therefore, the mass of carbon in CO2 will equal to that in the compound. This will be same for hydrogen in water as well. For oxygen, you can find its mass by subtracting the mass of carbon and hydrogen from .1g

[AWK]

I got H2CO3 but my friend says C3H6O2.  
Gave you a snack for the quick response! 

Give congratulations to your friend

[maivu]

A 0.100g sample of a compound containing C, H, and O is burned oxygen producing 0.1783g of CO2 and 0.0734g of H2O. Determine the empirical formula of the compound.

Empirical Formula shows only the ratio of number of atoms of each element in a compound.

Let empirical formula be CxHyOz. In this question, you are given that Oxygen is presented, it will be easier for you to set up the empirical formula.

Mole of CO2 = 0.00405   --> Mole of C = 0.00405 mol
Mole of H2O = 0.00408   --> Mole of H = 0.008156  mol

Find the mass of oxygen contained in the compound
mass of Oxygen = mass of compound - mass of C - mass of H
                       = 0.1  -  12*0.00505   -   1*0.008156
                       = 0.043244

Find the number of mole of oxygen: Mole of oxygen = 0.043244/16 = 0.0027 (mol)

Setting up the ratio. ( In order to set up the ratio, you need to know number of mole of each element in the compound)

C : H : O = 0.00405             : 0.008156            : 0.0027
             =  0.00405/0.0027  : 0.008156/0.0027 : 0.0027/0.0027
              =  1.5                  :     3                    : 1
               = 3  :  6  : 2

Conclusion: the required Empirical formula is C3H6O2


Hello kaeyeon, this is my step-by-step calculation for your question. Hope that it helped. If you find any mistake or any question in my solution, dont hesitate to ask me.