Question

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): c 74.03%, h 8.70%, n 17.27%
b. caffeine (found in coffee beans): c 49.48%, h 5.19%, n 28.85%, o 16.48%

Answer 1

a.
elements C H N
percentage composition 74.03 8.70 17.27
Molecular mass 12 1 14

# of mole 6.17 8.70 1.23

÷smallest mole 5.0 7.0 1.0

mole ratio 5 : 7 ; 1

THE EMPERICAL FORMUKLA FOR A. IS C5H7O

NOTE: #of mole = percentage composition ÷ Mr
and the ÷ smallest mole is used to find the ratio...for the above question a. it is 1.23


and b. should be done using the same procedure

Answer 2

A.
`C_5H_7N`

B.
`C_4H_5N_2O`

Step-by-step explanation:

Hello,

a. In this case, based on the given percentages, one computes the moles of C, H and N as follows:

`n_C=0.7403gC*(1molC)/(12gC)=0.062molC\\n_H=0.087gH*(1molH)/(1gH)=0.087molH\\n_N=0.1727gN*(1molN)/(14gN)=0.0123molN\\`

Now, we divide each element's moles by the smallest amount of moles, that is nitrogen's moles:

`C=(0.062)/(0.0123)=5;H=(0.087)/(0.0123)=7;N=(0.0123)/(0.0123)=1`

So the empirical formula is:

`C_5H_7N`

b. In this case, based on the given percentages, one computes the moles of C, H, N and O as follows:

`n_C=0.4948gC*(1molC)/(12gC)=0.0396molC\\n_H=0.0519gH*(1molH)/(1gH)=0.0519molH\\n_N=0.2885gN*(1molN)/(14gN)=0.021molN\\n_O=0.1648gO*(1molO)/(16gO)=0.0103molO\\`

Now, we divide each element's moles by the smallest amount of moles, that is oxugen's moles:

`C=(0.0396)/(0.0103)=4;H=(0.0519)/(0.0103)=5;N=(0.021)/(0.0103)=2;O=(0.0103)/(0.0103)=1`

So the empirical formula is:

`C_4H_5N_2O`

Best regards.