Question

The amino acid cysteine plays an important role in the three-dimensional structure of proteins by forming "disulfide bridges." The percent composition of cysteine is 29.74 percent \(\mathrm{C}, 5.82\) percent \(\mathrm{H}, 26.41\) percent \(\mathrm{O}\) 11.56 percent \(\mathrm{N},\) and 26.47 percent \(\mathrm{S}\). What is the molecular formula if its molar mass is approximately \(121 \mathrm{~g}\) ?

Short answer

The molecular formula of cysteine is \(\mathrm{C}_3\mathrm{H}_7\mathrm{O}_2\mathrm{N}\mathrm{S}\).

Step-by-step solution

Identify the Given Percent Composition

We are given the percent composition of each element in cysteine: 29.74% \(\mathrm{C}\), 5.82% \(\mathrm{H}\), 26.41% \(\mathrm{O}\), 11.56% \(\mathrm{N}\), and 26.47% \(\mathrm{S}\). Each percentage represents the proportion of each element in cysteine.

Determine the Mass of Each Element in 100g Sample

Assume we have a 100g sample of cysteine, allowing us to directly convert percentages to grams. This gives us: \(29.74\,\mathrm{g\,C}\), \(5.82\,\mathrm{g\,H}\), \(26.41\,\mathrm{g\,O}\), \(11.56\,\mathrm{g\,N}\), and \(26.47\,\mathrm{g\,S}\).

Convert Masses to Moles

Use the molar masses of \(\mathrm{C}\), \(\mathrm{H}\), \(\mathrm{O}\), \(\mathrm{N}\), and \(\mathrm{S}\) to convert grams to moles:\[\text{C: } \frac{29.74}{12.01} \approx 2.48\,\text{moles}\]\[\text{H: } \frac{5.82}{1.008} \approx 5.77\,\text{moles}\]\[\text{O: } \frac{26.41}{16.00} \approx 1.65\,\text{moles}\]\[\text{N: } \frac{11.56}{14.01} \approx 0.825\,\text{moles}\]\[\text{S: } \frac{26.47}{32.07} \approx 0.826\,\text{moles}\]

Calculate the Simplest Whole Number Ratio

Divide each of the mole amounts by the smallest number of moles calculated:\[\text{C: } \frac{2.48}{0.825} \approx 3.006\]\[\text{H: } \frac{5.77}{0.825} \approx 6.997\]\[\text{O: } \frac{1.65}{0.825} \approx 2.000\]\[\text{N: } \frac{0.825}{0.825} \approx 1.000\]\[\text{S: } \frac{0.826}{0.825} \approx 1.001\]

Determine the Empirical Formula

Using the simplest whole number ratio, the empirical formula is \(\mathrm{C}_3\mathrm{H}_7\mathrm{O}_2\mathrm{N}\mathrm{S}\).

Calculate Molar Mass of Empirical Formula

Calculate the molar mass of \(\mathrm{C}_3\mathrm{H}_7\mathrm{O}_2\mathrm{N}\mathrm{S}\): \[(3 \times 12.01) + (7 \times 1.008) + (2 \times 16.00) + (14.01) + (32.07) = 121.18\,\text{g/mol}\]

Confirm the Molecular Formula

The empirical formula mass (121.18 g/mol) is nearly identical to the known molar mass of cysteine (121 g/mol), indicating that the empirical formula is also the molecular formula.

Key concepts

Disulfide Bridges in Protein Structure

Disulfide bridges are crucial in determining the three-dimensional shape of proteins. These bonds form when two cysteine amino acids, each containing a sulfur atom in the form of a thiol group (-SH), come close to one another. The sulfur atoms bond together to form a disulfide linkage (-S-S-). This bond is covalent and is much stronger than other interactions that maintain protein structure, like hydrogen bonds or ionic interactions.

Disulfide bridges add stability to the protein's 3D shape, helping maintain its structure under various conditions. This is essential for proteins to function properly, as the biological activity of proteins depends heavily on their shape.

In short, the presence of disulfide bridges, primarily involving cysteine, allows proteins to withstand different environments, making them more robust in their biological roles.

Understanding Percent Composition

Percent composition tells us how much of each element is present in a compound relative to its total mass. In the case of cysteine, the percent composition was given for each element. For example, if you have 100 grams of cysteine, 29.74 grams will be carbon, 5.82 grams hydrogen, and so forth.

This information allows chemists and researchers to understand how elements are distributed in a compound. By calculating percent composition, we can make an educated guess about the possible structure and formula of an unknown compound.

Knowing percent composition is fundamental. It serves as a stepping stone for determining molecular or empirical formulas, especially when working with an unknown substance.

Defining Empirical Formula

An empirical formula gives the simplest whole-number ratio of elements in a compound. It doesn't necessarily show the exact number of atoms per molecule, like a molecular formula, but provides the basic ratio of the elements involved.

To determine the empirical formula, we convert the mass of each element (from percent composition) into moles. Then, we find the simplest ratio of these moles by dividing each by the smallest number of moles calculated. For cysteine, the process led us to the empirical formula \[ \mathrm{C}_3\mathrm{H}_7\mathrm{O}_2\mathrm{N}\mathrm{S} \].

Once we have the empirical formula, we can further analyze it against the known molar mass to decide if the empirical formula is the same as the molecular formula, or if additional steps are required to determine the exact formula of the compound.

Elemental Analysis Explained

Elemental analysis involves determining the amount of each element in a chemical compound. This analysis is done using various methods, such as mass spectrometry or combustion analysis, to identify the percentage of different elements in a substance.

In our exercise, elemental analysis is reflected in the percent composition data. This data gives a clear depiction of each element's role within cysteine. By having precise details, researchers can then calculate moles and deduce molecular structure, leading to discovering empirical or molecular formulas.

Elemental analysis is the backbone of analytical chemistry, allowing us to uncover detailed information about substances. It forms the basis for several significant discoveries and analyses in chemistry, giving insight into both simple and complex compounds.