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francesca

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Can someone help me with this problem?

 

A 8.48 g sample of an unknown organic compound containing only carbon, hydrogen, and oxygen was burned completely to produce 12.42 g of carbon dioxide and 5.08 g of water. Find the percentage of carbon, hydrogen, and oxygen in the compound.

 

I just need to know how to get started, and what to do, my teacher confused herself on this one and no one in the class can do it.

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This article explains it nicely: Wikipedia: Combustion. Scroll down to the heading "Chemical Equation".

 

Now your compound isn't a hydrocarbon, because it also contains oxygen. So you will start with the following equation:

 

[math]C_xH_yO_z + aO_2 \longrightarrow bCO_2+cH_2O[/math]

 

You will need to determine [imath]a,b,c[/imath] in terms of [imath]x,y,z[/imath]. Once you have that it should be easy, because you know the mass of the organic compound, so you should be able to figure out how many moles of [imath]C,H,O[/imath] you are starting with.

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Can you explain what you don't understand about what we said? This is basically a problem in stoichiometry. For starters, since you have 1 mole of [imath]C_xH_yO_z[/imath] on the left side and you have [imath]a[/imath] moles of [imath]CO_2[/imath] on the right side, it is clearly the case that [imath]x=a[/imath]. You need to do a similar analysis to find [imath]b[/imath] and [imath]c[/imath].

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First thing, i don't know what stoichiometry is, my class doesn't have a real science teacher. We're on our own, and she has no idea what's going on, so she can't teach us. And she relies on us to correct her mistakes and if we don't know we're going to sit there and go through trial and error until we get something.

 

Okay enough with my ranting.

 

So how do I get the percent from the formula?

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First thing' date=' i don't know what stoichiometry is,

[/quote']

 

I'm a bit confused by this remark. You posted this in the Organic Chemistry subforum. I was under the assumption that you are taking organic chemistry, in which case you would have already taken at least one course in basic chemistry in which you would have learned stoichiometry.

 

Am I to take it from your remark above that this assumption is wrong?

 

my class doesn't have a real science teacher. We're on our own, and she has no idea what's going on, so she can't teach us. And she relies on us to correct her mistakes and if we don't know we're going to sit there and go through trial and error until we get something.

 

Okay enough with my ranting.

 

So how do I get the percent from the formula?

 

Well you originally asked how to get the problem started, and you have been given plenty of clues as to how to do that much. Honestly if your teacher isn't teaching you then you can't rely on an internet forum to do the job. People post here for fun, not to do your teacher's job. You and your classmates should go at once to your school's administration and explain the situation. If it is as bad as you say then that is unacceptable.

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If what you are saying is the case, then I would strongly complain at the school director or whatever appropriate place.

 

Without basic understanding of the concept of stoichiometry this kind of exercises cannot be solved.

The idea is that at both sides of the arrow (see post of Tom Mattson), the number of atoms for each involved element must be the same.

 

For carbon, this means that x = b

For oxygen this means that z + 2a = 2b + c

For hydrogen this means that y = 2c

 

Another constraint is that the total weight at both sides of the arrow must be the same. Using atomic masses for the elements C,H,O, this gives another set of equations in x,y,z and a,b,c. Solving this set of equations will give you the answer.

 

EDIT: @Tom Mattson: While I was typing, you just added your reply. But, indeed, I totally agree with you :).

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The techer figured it out last night at her house. But anyway I come to find out she is really a math techer, they couldn't find a science teacher so they said science has alot of math it so hire a math teacher. So everything worked out.

 

Thanks everybody who helped!

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