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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.

User Jalmarez
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Answer: The molecular formula will be
C_8N_4H_(10)O_2

Step-by-step 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}}= (49.47g)/(12g/mole)=4.12moles

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

Moles of H =
\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (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 =
(4.12)/(1.03)=4

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

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

For H =
(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}}=(194.19)/(97)=2

The molecular formula will be =
2* C_4N_2H_5O=C_8N_4H_(10)O_2

User SMP
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