Question

1. vanillin, the active component of the flavoring vanilla, has the composition 63.15% c, 5.30% h, and 31.55% o.

a. what is its empirical formula?
b. if the molar mass of vanillin is 152.14 g/mol, what is its molecular formula?

Explanation:

Step1: Calculate the moles of each element

Assume we have a 100 - g sample of vanillin. Then the mass of \(C = 63.15\space g\), \(H=5.30\space g\), \(O = 31.55\space g\).
The molar mass of \(C\) is \(M_{C}=12.01\space g/mol\), the molar mass of \(H\) is \(M_{H}=1.008\space g/mol\), and the molar mass of \(O\) is \(M_{O}=16.00\space g/mol\).
The number of moles of \(C\): \(n_{C}=\frac{63.15\space g}{12.01\space g/mol}\approx5.26\space mol\)
The number of moles of \(H\): \(n_{H}=\frac{5.30\space g}{1.008\space g/mol}\approx5.26\space mol\)
The number of moles of \(O\): \(n_{O}=\frac{31.55\space g}{16.00\space g/mol}\approx1.97\space mol\)

Step2: Find the mole ratio

Divide each number of moles by the smallest number of moles (\(n_{O}\approx1.97\space mol\))
For \(C\): \(\frac{5.26}{1.97}\approx2.67\)
For \(H\): \(\frac{5.26}{1.97}\approx2.67\)
For \(O\): \(\frac{1.97}{1.97} = 1\)
Multiply by \(3\) to get whole - number ratios.
\(C:2.67\times3 = 8\), \(H:2.67\times3=8\), \(O:1\times3 = 3\)

Step1: Calculate the empirical formula molar mass

The empirical formula is \(C_{8}H_{8}O_{3}\).
The molar mass of the empirical formula \(M_{empirical}=(8\times12.01 + 8\times1.008+3\times16.00)\space g/mol=(96.08+8.064 + 48.00)\space g/mol=152.144\space g/mol\)

Step2: Determine the molecular formula

The molar mass of vanillin \(M = 152.14\space g/mol\)
Since \(\frac{M}{M_{empirical}}=\frac{152.14\space g/mol}{152.144\space g/mol}\approx1\)

Answer:

a. The empirical formula of vanillin is \(C_{8}H_{8}O_{3}\)