Question

Caffeine is a compound found in some natural coffees and teas and in some colas. a. Determine the empirical formula for caffeine, using the following composition of a 100.00-g sample. 49.47 grams of carbon, 28.85 grams of nitrogen, 16.48 grams of oxygen, and 5.20 grams of hydrogen b. If the molar mass of caffeine is 194.19 g/mol, calculate its molecular formula.

Answer 1

Answer: The molecular formula will be $C_8N_4H_{10}O_2$

Explanation:

Mass of C= 49.47 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Mass of H = 5.20 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.47g}{12g/mole}=4.12moles$

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

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

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.20g}{1g/mole}=5.20moles$

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

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

For H = $\frac{5.20}{1.03}=5$

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

Hence the empirical formula is $C_4N_2OH_5$

The empirical weight of $C_4N_2OH_5$ = 4(12)+2(14)+1(16)+5(1)= 97 g.

The molecular weight = 194.19 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{194.19}{97}=2$

The molecular formula will be = $2\times C_4N_2H_5O=C_8N_4H_{10}O_2$