Problem 33 - General Chemistry: Principles and Modern Applications

Question

Determine the empirical formula of (a) the rodenticide (rat killer) warfarin, which consists of \(74.01 \% \mathrm{C}\) \(5.23 \% \mathrm{H},\) and \(20.76 \%\) O, by mass; (b) the antibacterial agent sulfamethizole, which consists of \(39.98 \%\) C. \(3.73 \% \mathrm{H}, 20.73 \% \mathrm{N}, 11.84 \% \mathrm{O},\) and \(23.72 \% \mathrm{S},\) by mass.

Step-by-Step Solution

Answer

The empirical formula for Warfarin is C9H4O while for Sulfamethizole it's C9H10N4O2S2.

Step 1: Determine the empirical formula of Warfarin

First, convert the given percentages to grams (since it is a percentage, it would be percentage out of 100 g). This gives us \(74.01 \mathrm{g}\) C, \(5.23 \mathrm{g}\) H, and \(20.76 \mathrm{g}\) O. Next, convert these amounts into moles using atomic masses of the elements. For C, \( \frac{74.01 g}{12.01 g/mol} = 6.16 \mathrm{mol} \). For H, \( \frac{5.23 g}{1.008 g/mol} = 5.18 \mathrm{mol} \). For O, \( \frac{20.76 g}{16.00 g/mol} = 1.30 \mathrm{mol} \). Now we divide by the smallest number of moles to get the empirical formula.This leads to: C \((\frac{6.16}{1.30} = 4.74)\), H \((\frac{5.18}{1.30} = 3.98)\), and O \((\frac{1.30}{1.30} = 1)\). Since these are all close to whole numbers, the empirical formula is thus C9H4O.

Step 2: Determine the empirical formula of Sulfamethizole

Now, repeat the same process for Sulfamethizole. The given percentages convert to \(39.98 \mathrm{g}\) C, \(3.73 \mathrm{g}\) H, \(20.73 \mathrm{g}\) N, \(11.84 \mathrm{g}\) O, and \(23.72 \mathrm{g}\) S. Next, convert these amounts into moles using atomic masses of the elements. For C, \( \frac{39.98 g}{12.01 g/mol} = 3.33 \mathrm{mol} \). For H, \( \frac{3.73 g}{1.008 g/mol} = 3.70 \mathrm{mol} \). For N, \( \frac{20.73 g}{14.007 g/mol} = 1.48 \mathrm{mol} \). For O, \( \frac{11.84 g}{16.00 g/mol} = 0.74 \mathrm{mol} \). For S, \( \frac{23.72 g}{32.06 g/mol} = 0.74 \mathrm{mol} \). Now divide all the moles by the smallest number of moles (0.74), leading to C \((\frac{3.33}{0.74} = 4.5)\), H \((\frac{3.70}{0.74} = 5)\), N \((\frac{1.48}{0.74} = 2)\), O \((\frac{0.74}{0.74} = 1)\), and S \((\frac{0.74}{0.74} = 1)\). Since the ratio for C is not a whole number, double all of them to get the empirical formula C9H10N4O2S2.

Key Concepts

Chemical Composition

Chemical composition refers to the arrangement, types, and ratios of atoms in molecules of substances. In chemistry, understanding the chemical composition of a compound helps in determining the formula representing that compound.
For example, consider the compound warfarin. The chemical composition is given as 74.01% carbon (C), 5.23% hydrogen (H), and 20.76% oxygen (O). These percentages indicate the proportion of each element by mass in the compound. In a substance, these figures can be used to determine the empirical formula which is the simplest ratio of moles of each element present in the compound.
This foundational principle aids in translating bulk experimental data (mass percentages) into a clear representation of the compound’s basic building blocks, that is, its empirical formula.

Mole Calculations

Mole calculations are crucial in converting the mass of substances to number of moles, utilizing atomic or molecular masses. This method bridges mass and molecular quantities, simplifying chemical equations and reactions.
An example would be the transformation of mass percentages into moles in our exercise. For instance, converting 74.01 g of carbon into moles is achieved by dividing by the molar mass of carbon, yielding 6.16 moles. This is calculated using the formula:

• For Carbon: \( \frac{74.01 \text{ g}}{12.01 \text{ g/mol}} = 6.16 \text{ mol} \)
• For Hydrogen: \( \frac{5.23 \text{ g}}{1.008 \text{ g/mol}} = 5.18 \text{ mol} \)
• For Oxygen: \( \frac{20.76 \text{ g}}{16.00 \text{ g/mol}} = 1.30 \text{ mol} \)

By determining the moles for each element, we gain insights into the compound's empirical formula. Understanding this conversion is essential for chemists as it provides a quantitative understanding of how much of each element is present and allows for the scaling of chemical reactions to a macroscopic level.

Percent Composition

Percent composition provides the relative amount, in percentage by mass, of each element in a compound. It is a critical concept as it lays the groundwork for determining a compound's empirical formula.
Consider sulfamethizole as an example. It is composed of 39.98% carbon, 3.73% hydrogen, 20.73% nitrogen, 11.84% oxygen, and 23.72% sulfur. These percentages allow us to understand what part each element contributes to the overall mass of the compound. When expressed in such clear numerical terms, it simplifies the process of deducing the simplest whole number ratio of atoms in a molecule.
Percent composition is essential not just for empirical formula determination but also for validating assumptions about a compound's identity and purity, ensuring the accurate production and application of chemical substances.