Question

Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19 g/mol and a mass percentage composition of 49.48% C, 5.19% H, 28.85% N, and 16.48% O. What is the molecular formula of caffeine?

Answer 1

Answer : The molecular formula of a caffeine is,
`C_8H_(10)N_4O_2`

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (49.48g)/(12g/mole)=4.12moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (5.19g)/(1g/mole)=5.19moles`

Moles of N =
`\frac{\text{ given mass of N}}{\text{ molar mass of N}}= (28.85g)/(14g/mole)=2.06moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (16.48g)/(16g/mole)=1.03moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
`(4.12)/(1.03)=4`

For H =
`(5.19)/(1.03)=5.03\approx 5`

For N =
`(2.06)/(1.03)=2`

For O =
`(1.03)/(1.03)=1`

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =
`C_4H_5N_2O_1=C_4H_5N_2O`

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

`n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}`

`n=(194.19)/(97)=2`

Molecular formula =
`(C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_(10)N_4O_2`

Therefore, the molecular of the caffeine is,
`C_8H_(10)N_4O_2`