Calculating the Empirical Formula of a Material | 1.235g of C, O, and H

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In summary, the material being discussed weighs 1.235 grams and is composed of carbon, oxygen, and hydrogen. When burned in oxygen, it produces 3.425 grams of CO_2 and 0.971 grams of H_2O. The empirical formula for the material is still unknown and is being solved for using coefficients and basic algebra-arithmetic skills. Based on the number of moles of CO_2 and H_2O produced, the coefficients for carbon and hydrogen can be determined, but determining the coefficient for oxygen may require an alternative approach due to oxygen being present in both the initial material and the oxygen gas supplied.
  • #1
transgalactic
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a matirial which weighs 1.235gram
made of carbon oxigen and hydrogen
and this material burned in oxigen
which gives 3.425 gram of CO_2
and 0.971 of H_2O ??
what is the ampiric formula of the material
??
 
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  • #2
1258 posts and you still don't know that you have to try by yourself before asking for help?
 
  • #3
transgalactic said:
a matirial which weighs 1.235gram
made of carbon oxigen and hydrogen
and this material burned in oxigen
which gives 3.425 gram of CO_2
and 0.971 of H_2O ??
what is the ampiric formula of the material
??

Hi transgalactic! :smile:

What equations do you think are involved, and what constants do you need to know? :wink:
 
  • #4
This one was reasonably difficult, but I found an answer. The trick is to account for both moles formula balance AND mass balance. (Or I should say, atoms balance AND mass balance). Initially, you have variables for the subscripts in the unknown C-H-O compound, and the way I proceeded, a coefficient variable for the reactant oxygen molecule and a coefficient 1 for the unknown compound. Like I say, I came up with what I believe is a very reasonable empirical formula result. I'd like to plainly state the result here, now, but doing so might not be acceptable at this state of progress on the forum.
 
  • #5
i have 0.077824106 moles of CO_2
and 0.05389869 moles of H_2O

what to do next??

it not a whole numbers
and i don't know how to build balancing equations for this
??
 
  • #6
Well, so you have been thinking about and studying this exercise some more. Good.

transgalactic said:
i have 0.077824106 moles of CO_2
and 0.05389869 moles of H_2O

One of those values looks familar; the other I do not remember. I solved the problem on my own several days ago.

what to do next??

See my last post on this topic. At first, you do not yet know the actual coefficients for the reactant compounds.

it not a whole numbers
and i don't know how to build balancing equations for this
??
The coefficients are unknow to start with, just as the subscripts for the atoms in the unknown compound. You need basic algebra-arithmetic skills. The decision on which final WHOLE numbers to find will come as one of your last steps. I initially found nonwhole values for one or more subscripts in the atoms of the compound also. The process was finished by finding a whole number factor which would keep all subscripts as extremely close to whole OR whole; and just good judgement.
 
  • #7
ok
i have to find coefficient for C O H
C=x
O=y
H=z
x+0=0.077824106
y+2=2*0.077824106+0.05389869
z+0=2*0.05389869

how to get whole cofficients from this decimal numbers
??
 
  • #8
hint:

can you calculate the number of moles of CO2 you obtain?! If the products are CO2 and H2O only, then all the Carbon from the initial material have gone into CO2.

so, if you have x mol of CO2 formed, then you should have had x mol of Carbon in the initial substance.

Using this same logic you can find out for Hydrogen.

but for Oxygen it is going to be a little bit tricky since the oxygen in the product comes from both the initial material and the oxygen gas supplied. you have find an alternative path!
 

FAQ: Calculating the Empirical Formula of a Material | 1.235g of C, O, and H

What is the empirical formula of a material?

The empirical formula of a material is the simplest whole number ratio of elements present in a compound. It represents the relative number of atoms of each element in a compound.

How do you calculate the empirical formula of a material?

To calculate the empirical formula of a material, you need to determine the mass of each element present in the compound and then convert it into moles. Next, divide each mole value by the smallest mole value, and then round the resulting numbers to the nearest whole number. These whole numbers represent the subscripts in the empirical formula.

What information do I need to calculate the empirical formula of a material?

You need to know the mass of each element present in the compound and the molar mass of each element. This information can be obtained from the periodic table.

Can the empirical formula of a material be the same as its molecular formula?

In some cases, the empirical formula of a material may be the same as its molecular formula. However, in most cases, the molecular formula will be a multiple of the empirical formula. This means that the molecular formula contains more atoms than the empirical formula.

What does it mean if the empirical formula and molecular formula are the same?

If the empirical formula and molecular formula of a material are the same, it means that the compound is made up of simple whole number ratios of elements. This is also an indication that the compound is a simple molecule or a simple ionic compound.

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