Question

The molecular weight of nicotine, a colorless oil, is \(162.1\) and it contains \(74.0 \%\) carbon, \(8.7 \%\) hydrogen, and \(17.3 \%\) nitrogen. Using three significant figures for the atomic weights, calculate the molecular formula of nicotine.

Short answer

The molecular formula of nicotine is C10H14N2.

Step-by-step solution

Calculate Moles of Each Element in 100 g of Nicotine

Assume we have 100 g of nicotine. This will simplify calculations while keeping the same percentage composition. Calculate the mass of each element and then find their moles by dividing the mass of the element by its respective molecular weight: Carbon: \(74.0 \% \) of 100 g = 74 g, Moles of carbon = \(\frac{74}{12.01}\) = 6.16 moles (rounded to three significant figures and using the atomic weight of C as 12.01) Hydrogen: \(8.7\%\) of 100 g = 8.7 g, Moles of hydrogen = \(\frac{8.7}{1.008}\) = 8.63 moles (rounded to three significant figures and using the atomic weight of H as 1.008) Nitrogen: \(17.3\%\) of 100 g = 17.3 g, Moles of nitrogen = \(\frac{17.3}{14.01}\) = 1.23 moles (rounded to three significant figures and using the atomic weight of N as 14.01)

Find the Empirical Formula of Nicotine

To find the empirical formula, determine the ratio of moles of each element relative to the lowest number of moles. Divide the moles of each element by the lowest moles: Ratio of moles of C: \(\frac{6.16}{1.23}\) = 5 Ratio of moles of H: \(\frac{8.63}{1.23}\) = 7 Ratio of moles of N: \(\frac{1.23}{1.23}\) = 1 The empirical formula is C5H7N.

Calculate the Molecular Weight of the Empirical Formula

Now calculate the molecular weight of the empirical formula C5H7N: Molecular Weight = (5 × 12.01) + (7 × 1.008) + (1 × 14.01) = 81.1

Determine the Molecular Formula from the Empirical Formula and Given Molecular Weight

Now find the ratio between the molecular weight of nicotine and the molecular weight of the empirical formula: Ratio = \(\frac{162.1}{81.1}\) = 2 Multiply each subscript in the empirical formula by the ratio to obtain the molecular formula: C5H7N × 2 = C10H14N2 So, the molecular formula of nicotine is C10H14N2.

Key concepts

Empirical Formula

Understanding the empirical formula is crucial for studying compounds in chemistry. It represents the simplest whole number ratio of the atoms of each element in a compound. Consider it as a reduced-to-the-basics version of the actual molecular makeup. For example, if we determine that a compound contains 6 moles of carbon, 14 moles of hydrogen, and 2 moles of oxygen, we can simplify this to the ratio of 3:7:1, resulting in the empirical formula C₃H₇O.

In our example with nicotine, we used the percentage composition of carbon, hydrogen, and nitrogen to calculate the number of moles of each element in a 100 gram sample and then simplified the ratios to whole numbers to find nicotine's empirical formula, which turned out to be C₅H₇N. The process involves a straightforward conversion of mass percentages into moles and then finding the lowest common denominator to express these moles as the smallest possible whole number ratio.

Atomic Weight

The atomic weight, or atomic mass, is a crucial value used to calculate the mass of an element in a given sample. It reflects the average mass of atoms of an element, measured in atomic mass units (amu), and it takes into account the isotopes of that element and their abundance. For instance, the atomic weight of carbon is approximately 12.01 amu.

When solving for the molecular formula, as we did with nicotine, the atomic weights of carbon (12.01 amu), hydrogen (1.008 amu), and nitrogen (14.01 amu) were used to convert the grams of each element to moles. This helps in determining the ratios of atoms in the empirical formula. Remember to always use accurate atomic weights and round to the appropriate number of significant figures for precision.

Percent Composition

The percent composition is the percentage by mass of each element in a compound. This information is pivotal in determining the empirical formula and, ultimately, the molecular formula of a substance. Knowing that nicotine contains 74.0% carbon, 8.7% hydrogen, and 17.3% nitrogen allowed us to calculate the amount of each element present in a hypothetical 100 gram sample. From there, the amount in grams could easily be converted to moles.

Grasping the concept of percent composition is important because it connects the mass of a compound to its formula. A solid understanding of percentage calculations and conversions to moles is necessary for this concept. In educational materials, representing these calculations clearly helps students to follow along with the process and apply it to other exercises.

Mole Concept

The mole concept is a fundamental cornerstone of chemistry that provides a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. One mole corresponds to Avogadro's number (approximately 6.022 × 10²³) of particles, whether they are atoms, molecules, ions, or electrons.

For the calculation of the molecular formula of nicotine, we applied the mole concept to determine the number of moles of each element. This allowed us to find the empirical formula and compare it with nicotine's molecular weight to deduce the molecular formula. Since all measurements in chemistry are ultimately compared to this standard unit, proficiency in converting grams to moles and moles to atoms or molecules is essential for any student studying chemistry.