Chemical Formulas from percents/ratios

[QuarkCharmer]

Homework Statement

1.)
A compound containing only C, H, and O, is subjected to elemental analysis. Upon completion of the combustion, a .4266g sample of the compound produces .5518g of CO_2, and .2259g of H_2O. What is the empirical formula of the compound?

2.)
Maleic Acid contains 41.4% Carbon, 3.47% Hydrogen, and 55.1% Oxygen by mass. A 0.050 mol sample of this compound weighs 5.80g. What is the molecular formula of maleic acid?

Homework Equations

The Attempt at a Solution

Once I get the empirical formula, I know how to get the molecular without any problem. I have done questions like this where there is one product, and it's easy to calculate, but these are worded strangely, have products involving elements from more than one of the reactants, and I am not sure how to begin.

Can you please point me in the right direction? I would have work to show, but I can't figure out how to start!

 

[QuarkCharmer]

My initial thought for #1, is that converting the grams of CO_2 expressed as moles would also be the moles of Carbon right? I could do the same thing to find the moles of Oxygen if I am correct. Finding Hydrogen, I could take the H_2O, express it in moles, and then multiply it by 2? I am not sure if this is a realistic way to approach this.Number 2:
I am thinking that I can treat the percents like grams, and then calculate the mole ratios of all three elements. That should get me the empirical? From there I can express the 5.80g/.050mol unit factor in terms of 1 mole, and divide the masses to get the multiplier that goes through the empirical formula?

 

[iRaid]

The first one you make them all into moles and you can find out how many grams of each element there is then compare ratios.
Also do you have the answer, if you do I would help more, but I don't want to give you the wrong advice :p

 

[Borek]

For the first - you are right about carbon and hydrogen, but oxygen is what is left after you subtract carbon and hydrogen from the sample mass.

In the second 5.80g/.050mol gives you molar mass - compare it with molar mass of empiricial formula to find out what the true formula is.

 

[QuarkCharmer]

I figured out the second one, but that first one has me stumped. No matter how I go about it, I keep thinking that because the mass of the reactant(s) is less than that of the products, the hydrocarbon must be the limiting reactant and have an infinite supply of oxygen. Under those circumstances, is it even possible to solve it?

 

[Borek]

that first one has me stumped. No matter how I go about it, I keep thinking that because the mass of the reactant(s) is less than that of the products, the hydrocarbon must be the limiting reactant and have an infinite supply of oxygen. Under those circumstances, is it even possible to solve it?

Yes, please reread what I wrote in my previous post. Your compound is made of carbon, hydrogen and oxygen. You can calculate mass of of carbon and hydrogen, and you know mass of the sample. Let's assume you have 3g of a compound and you calculated (from the data given) it contains 1g of C and 1g of H. 3-1-1=1, that's mass of the oxygen in the compound. Products contain this 1 g and whatever excess oxygen reacted with the sample.