Answer
Empirical Formula: $C_{4}$$H_{5}$$N_{2}$O
Molecular formula: $C_{8}$$H_{10}$$N_{4}$$O_{2}$
Work Step by Step
Step 1: Determine the moles of each element.
Given percentages by mass:
Carbon (C): 49.5%
Hydrogen (H): 5.15%
Nitrogen (N): 28.9%
Oxygen (O): 16.5%
Assume you have 100 grams of the compound, which makes the calculations easier.
Moles of C: $\frac{49.5 g}{12.01g/mol}$
Moles of H:$\frac{5.15 g}{1.01g/mol}$
Moles of N:$\frac{28.9 g}{14.01g/mol}$
Moles of O:$\frac{16.5 g}{16.0g/mol}$
Step 2: Determine the smallest whole number ratio of moles.
Divide each of the moles by the smallest number of moles you calculated.
Step 3: Write the empirical formula.
Write the ratio obtained in Step 2 as subscripts in a chemical formula.
Step 4: Determine the molecular formula.
Given molar mass: 195 g/mol
Calculate the molar mass of the empirical formula. Then, find the ratio between the molar mass of the molecular formula and the molar mass of the empirical formula. Finally, multiply the subscripts in the empirical formula by this ratio.
Let's perform the calculations:
Step 1:
Moles of C: $\frac{49.5 g}{12.01g/mol}$= 4.12
Moles of H:$\frac{5.15 g}{1.01g/mol}$ =5.10
Moles of N:$\frac{28.9 g}{14.01g/mol}$ =2.07
Moles of O:$\frac{16.5 g}{16.0g/mol}$ = 1.03
Step 2:
Divide each by the smallest number of moles (1.03 in this case):
C: 4.12 / 1.03 ≈ 4
H: 5.10 / 1.03 ≈ 5
N: 2.07 / 1.03 ≈ 2
O: 1.03 / 1.03 = 1
Step 3:
Empirical Formula: $C_{4}$$H_{5}$$N_{2}$O
Step 4:
Molar mass of empirical formula: 4(12.01)+5(1.01)+2(14.01)+1(16.00)=97.15 g/mol4(12.01)+5(1.01)+2(14.01)+1(16.00)=97.15g/mol
Calculate the ratio: $\frac{195}{97.15}$ $\approx$ 2
Molecular Formula: Multiply the subscripts in the empirical formula by 2: $C_{8}$$H_{10}$$N_{4}$$O_{2}$
So, the empirical formula of caffeine is $C_{4}$$H_{5}$$N_{2}$O and the molecular formula is $C_{8}$$H_{10}$$N_{4}$$O_{2}$
