Question

Caffeine consist of carbon, hydrogen oxygen and Nitrogen. When 0, 1920g of caffeine combost in axcess of Oz, forms 0,3482g CO can someone explain all steps on how to solve this thx
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Answer #1

We must do an analysis:
CO2 and H2O are obtained from a combustion reaction, from caffeine and oxygen in excess, therefore, to form the products the limiting reagent was the amount of carbon and hydrogen present in the caffeine, we can calculate the moles of carbon and of hydrogen involved to know the existing quantity in the compound.

Then, knowing the % of nitrogen in the sample we can calculate the mass of nitrogen, knowing these 3 species (carbon, hydrogen and nitrogen) we can calculate the amount of oxygen.
Having the mass quantity of all the elements, we can calculate the% of each to find the empirical formula of the compound, and then the molecular one.

We start with:

0.3482g CO2 * 1mol 44.019 7.912. 10-3 molco

lmolc 7.912:r 10-3 molCO2 * 1mol CO2 7.912.-10-3molc

7.912. 10-3 molC* 12.011g 1mol 0.0950gC

Then:

0.0891g H20* Imol 18.029 4.945x 10-3 mol H2O

4.945x10-3 mol H2O * 2mol H 1mol H2O 9.89.r 10-3 mol H

9.89.c10-3molt 1.008g Imol - 9.969.r 10-3 gH

If:

0.19209caffeine 28.849N 100gcaffeine = = 0.05549N

For oxygen:

go Scaffeine - gH - OC - ON

gO = 0.1920g - 0.09509C - 9.969.r 10-39H - 0.05549 N

gO = 0.03169

Taking the masses of all the components, we calculate the% of each in the caffeine:

C= 0.0950g 0.19209 100 = 49.48

9 H 9.969x 10- 0.1920g * 100 = 5.19

N= 0.05549 0.19209 * 100 = 28.84

0 = 0.0316g 0.1920g 16.47

We divide the % by the atomic weights:

C= 49.48 12.011 = 4.12

H 5.19 1.008 = 5.149

N= 28.84 14.01 = 2.06

O= 16.47 15.999 1.029

We take the smallest value, in this case that of oxygen, and divide the others by it to have the ratio in the empirical formula:

C= 4.12 1.029 = 4

Η 5.149 1.029 =5

N = 2.06 1.029 2

0= 1.029 1.029

So the empirical formula is: C4H5N2O1 (97.103g/mol)

Knowing that the molecular weight ranges from 190 to 200, multiplying the empirical formula by 2:

C8H10N4O2 (194.206 g/mol).


answered by: ANURANJAN SARSAM
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