Question

Strychnine (rat poison) has the composition \(75.42 \%\) C, \(6.63 \% \mathrm{H}, 8.38 \% \mathrm{~N}\), and \(9.57 \% \mathrm{O}\). What is the empirical formula of strychnine?

Short answer

The empirical formula of strychnine is \(C_{10}H_{11}NO\).

Step-by-step solution

Convert percentages to masses

We're assuming a 100g sample, so the percentage mass is also the actual mass of each element in the sample. Therefore C: 75.42g, H: 6.63g, N: 8.38g, and O: 9.57g.

Convert masses to moles

Using the atomic weights for each element (C: 12.01 g/mol, H: 1.008 g/mol, N: 14.01 g/mol, and O: 16.00 g/mol), which can be found on the periodic table, one converts each element mass to moles, by dividing the mass by the atomic weight: For carbon (C), \(\frac{75.42g}{12.01g/mol} = 6.28~mol; \) For hydrogen (H), \(\frac{6.63g}{1.008g/mol} = 6.58~mol; \) For nitrogen (N), \(\frac{8.38g}{14.01g/mol} = 0.60~mol; \) For oxygen (O), \(\frac{9.57g}{16.00g/mol} = 0.60~mol; \)

Calculate Ratio

Determine the ratio of the number of moles of each element to the smallest number of moles calculated (which in this case is 0.60 mol). For carbon, \(\frac{6.28}{0.60} = 10.47 \approx 10; \) For hydrogen, \(\frac{6.58}{0.60} = 10.97 \approx 11; \) For nitrogen and oxygen, \(\frac{0.60}{0.60} = 1.00 \approx 1; \)

Empirical Formula

The empirical formula can now be written using the whole number ratios: \(C_{10}H_{11}N_1O_1\), which simplifies to \(C_{10}H_{11}NO\).

Key concepts

Percent Composition

The concept of percent composition is essential when determining the empirical formula of a compound. It refers to the percentage by mass of each element in a compound. To illustrate, let's consider strychnine, with the composition of 75.42% carbon (C), 6.63% hydrogen (H), 8.38% nitrogen (N), and 9.57% oxygen (O). These percentages represent how much of each element's mass contributes to the total mass of the compound.

To make calculations easier, it is customary to consider a 100 gram sample of the compound. This simplifies the conversion, as the percent composition directly translates to mass in grams. For instance, in a 100 gram sample of strychnine, 75.42 grams would be carbon, 6.63 grams hydrogen, and so forth. Understanding this concept lays the groundwork for further calculations, such as determining the moles of each element present.

Moles Calculation

Once you have the mass of each element from the percent composition, the next step is to convert these masses into moles. The mole is a fundamental unit in chemistry that provides a bridge between the atomic scale and macroscopic measurements. This involves using the atomic weight of each element, typically found on the periodic table.

For example, to find the number of moles of carbon in strychnine, you divide its mass by its atomic weight, which is 12.01 g/mol. Thus, the calculation is \( \frac{75.42\,\text{g}}{12.01\,\text{g/mol}} = 6.28\,\text{mol} \). This calculation is repeated for each element, which ultimately gives you the moles: 6.28 mol for carbon, 6.58 mol for hydrogen, 0.60 mol for nitrogen, and 0.60 mol for oxygen. These quantities are crucial for determining the ratio needed for the empirical formula.

Atomic Weight

Understanding atomic weight is crucial for converting mass into moles. The atomic weight of an element represents the average mass of atoms of an element, measured in atomic mass units (amu), considering the various isotopes of the element and their abundances. For convenience in empirical formula calculations, this average mass is also expressed in grams per mole (g/mol), which facilitates the conversion from grams to moles.

For example, carbon's atomic weight is 12.01 g/mol, hydrogen's is 1.008 g/mol, nitrogen's is 14.01 g/mol, and oxygen's is 16.00 g/mol. These values are used to convert the mass of each element in the given sample into moles, a core step in determining the empirical formula. Mastering this concept will empower you in many areas of chemistry, enabling you to engage in meaningful and accurate chemical calculations.