Question

Nicotine has a molar mass of \(160 \mathrm{~g} / \mathrm{mol}\). If the percent composition is \(74.0 \% \mathrm{C}, 8.70 \% \mathrm{H},\) and \(17.3 \% \mathrm{~N},\) what is the molecular formula of nicotine?

Short answer

The molecular formula of nicotine is \(\text{C}_5\text{H}_7\text{N}\).

Step-by-step solution

Determine the Moles of Each Element

First, calculate the number of moles of each element using their percent compositions and the molar mass of nicotine.- Carbon: \(74.0\%\) of \(160\, \text{g/mol} = 0.74 \times 160 = 118.4\, \text{g} \) Moles of carbon: \(\frac{118.4}{12.01} \approx 9.86\, \text{mol}\)- Hydrogen: \(8.70\%\) of \(160\, \text{g/mol} = 0.087 \times 160 = 13.92\, \text{g} \) Moles of hydrogen: \(\frac{13.92}{1.008} \approx 13.81\, \text{mol}\)- Nitrogen: \(17.3\%\) of \(160\, \text{g/mol} = 0.173 \times 160 = 27.68\, \text{g} \) Moles of nitrogen: \(\frac{27.68}{14.01} \approx 1.98\, \text{mol}\)

Calculate the Simplest Whole Number Ratio

Divide the number of moles of each element by the smallest number of moles to find the simplest whole number ratio.- Carbon: \(\frac{9.86}{1.98} \approx 5\)- Hydrogen: \(\frac{13.81}{1.98} \approx 7\)- Nitrogen: \(\frac{1.98}{1.98} = 1\)

Write the Molecular Formula

Using the determined simplest ratio, write the molecular formula. The simplest whole number ratio of C:H:N is 5:7:1, leading to the molecular formula:\(\text{C}_5\text{H}_7\text{N}\)

Key concepts

Percent Composition

Percent composition is a way to express the proportion of each element within a compound. It tells us the percentage by mass of each element in the compound. To calculate it, take the mass of an individual element in one mole of the compound, divide it by the molar mass of the compound, and multiply by 100. For our example, nicotine has a percent composition of 74.0% Carbon, 8.70% Hydrogen, and 17.3% Nitrogen. These percentages tell us how each element contributes to the total weight of nicotine's molar mass, which is 160 g/mol.

• Carbon contributes 74.0% which is 118.4 g out of 160 g/mol.
• Hydrogen contributes 8.7% or 13.92 g.
• Nitrogen makes up 17.3% or 27.68 g.

Understanding percent composition is essential for determining the number of moles of each element in a compound.

Molar Mass

Molar mass refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a critical concept because it allows us to convert between moles and grams, which is often necessary in chemistry calculations. For nicotine, the molar mass is given as 160 g/mol. This value is essentially the weighted sum of each element's contribution to the compound based on its atomic mass and quantity.
Molar mass helps you determine how many grams of nicotine are needed to have one mole of the compound. By knowing the molar mass and the percent composition, you can determine the actual weight of each element in the compound, as we demonstrated with Carbon, Hydrogen, and Nitrogen in nicotine.

Moles Calculation

Calculating moles is a fundamental skill in chemistry. Moles represent the amount of a substance and are calculated by dividing the mass of the substance by its molar mass. In the nicotine example, we calculated the number of moles for each element based on their percent mass in the compound.

• For Carbon: Given 118.4 g and a molar mass of 12.01 g/mol, the calculation is: \ \(\frac{118.4}{12.01}=9.86\text{ moles}\)
• For Hydrogen: From 13.92 g, dividing by 1.008 g/mol gives: \ \(\frac{13.92}{1.008} = 13.81 \text{ moles}\)
• For Nitrogen: With 27.68 g and a molar mass of 14.01 g/mol: \ \(\frac{27.68}{14.01} = 1.98 \text{ moles}\)

These calculations allow us to compare the amount of each element in the nicotine molecule on a like-for-like basis, which is particularly important when determining empirical formulas.

Empirical Formula

The empirical formula of a compound provides the simplest whole-number ratio of the elements in the compound. It doesn't give the actual number of atoms but rather their relative proportions. Using our moles calculations from nicotine, we determine the empirical formula by dividing each element’s mole value by the smallest number of moles calculated, which is 1.98 moles for Nitrogen in this case.

• Carbon becomes \( \frac{9.86}{1.98} \approx 5 \)
• Hydrogen becomes \( \frac{13.81}{1.98} \approx 7 \)
• Nitrogen is \( \frac{1.98}{1.98} = 1 \)

Thus, the empirical formula for nicotine is C\(_5\)H\(_7\)N. This formula indicates that for every one nitrogen atom, there are five carbon atoms and seven hydrogen atoms. Understanding empirical formulas is vital for deducing the actual molecular structure when combined with information on molar mass.