Question

nicotine contains 74.0% carbon, 8.72% hydrogen, and 17.3% nitrogen. what is the empirical formula of nicotine? o c₁₀h₁₄n₂ o c₂h₄n₂ o c₃h₄n₂ o c₄h₁₀n₂ o c₅h₇n

Explanation:

Step1: Assume 100g of nicotine

Since the percentages are given, assuming 100g of the compound makes the calculations easier. So, we have 74.0g of C, 8.72g of H, and 17.3g of N.

Step2: Calculate the number of moles of each element

The molar - mass of C is 12.01g/mol, H is 1.01g/mol, and N is 14.01g/mol.
The number of moles of C, $n_C=\frac{74.0g}{12.01g/mol}\approx6.16mol$
The number of moles of H, $n_H=\frac{8.72g}{1.01g/mol}\approx8.63mol$
The number of moles of N, $n_N=\frac{17.3g}{14.01g/mol}\approx1.23mol$

Step3: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles (1.23mol in this case).
For C: $\frac{6.16mol}{1.23mol}\approx5$
For H: $\frac{8.63mol}{1.23mol}\approx7$
For N: $\frac{1.23mol}{1.23mol}=1$

Answer:

C₅H₇N