# What is the Empirical and Molecular Formula of Caffeine?

1. Mar 13, 2010

### haengbon

1. The problem statement, all variables and given/known data

What is the Empirical and Molecular Formula of caffeine, given these information:

0.376g caffeine would yield to 0.682 CO2 , 0.174g H2O and 0.110g N. The molecular weight if caffeine is 194 g/mole.

2. Relevant equations

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3. The attempt at a solution

0.682 CO2 $$\rightarrow$$ gC = 0.186 gC
0.174g H2O$$\rightarrow$$ gH = 9.735 x 10-03 gH
0.110gN $$\rightarrow$$0.110 gN

0.376g sample = gC + gH + gN + gO
0.376g sample - 0.186 gC - 9.735 x 10-03 gH - 0.110gN = gO
gO = 0.070

C= $$0.186/12.01$$ = $$0.15/.004375$$ = 3.42 $$\approx$$ 3

H= $$.009735/1.008$$ = $$.009735/.004375$$ = 2

N= $$0.110/14.01$$ = $$.00785/.004375$$ = 1.7 $$\approx$$ 2

O= $$0.070/16.00$$ = $$.004375/.004375$$ = 1

EF: C3H2N2O

n= $$\frac{MW}{EFW}$$

n= $$\frac{194}{82.066}$$

n=2

MF= n(EF)
MF= 2(C3H2N2O)

MF= C6H4N4O2

Is this correct? :/ I think It's not but I hope it is! :)

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