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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles$

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

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

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

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

For C = $\frac{4.12}{1.03}=4$

For H = $\frac{5.19}{1.03}=5.03\approx 5$

For N = $\frac{2.06}{1.03}=2$

For O = $\frac{1.03}{1.03}=1$

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 :

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

$n=\frac{194.19}{97}=2$

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, $C_8H_{10}N_4O_2$