Question

A 1.00 -mg sample of a pure protein yielded on hydrolysis \(0.0165 \mathrm{mg}\) of leucine and \(0.0248 \mathrm{mg}\) of isoleucine. What is the minimum possible molar mass of the protein? \((\mathrm{MM}\) leucine \(=\mathrm{MM}\) isoleucine \(=131 \mathrm{~g} / \mathrm{mol})\)

Short answer

Answer: The minimum possible molar mass of the protein is 131.1 g/mol.

Step-by-step solution

Identifying given values

The problem provides us with the following information: - Total mass of protein: 1.00 mg - Mass of leucine: 0.0165 mg - Mass of isoleucine: 0.0248 mg - Molar mass of leucine and isoleucine: 131 g/mol

Convert masses to moles

We need to determine the number of moles for each amino acid present in the protein sample. We will use the equation: Moles = \(\frac{\text{mass}}{\text{molar mass}}\) For leucine: Moles of leucine = \(\frac{0.0165 \mathrm{mg}}{131 \mathrm{~g} / \mathrm{mol}} \cdot \frac{1 \mathrm{~g}}{1000 \mathrm{~mg}}\) = \(1.26 \times 10^{-4} \mathrm{~mol}\) For isoleucine: Moles of isoleucine = \(\frac{0.0248 \mathrm{mg}}{131 \mathrm{~g} / \mathrm{mol}} \cdot \frac{1 \mathrm{~g}}{1000 \mathrm{~mg}}\) = \(1.89 \times 10^{-4} \mathrm{~mol}\)

Calculate the ratio of amino acids in the protein

Since we are trying to find the minimum molar mass of the protein, we need to assume that the protein contains only one leucine and one isoleucine residue per molecule. Therefore, the protein contains the same number of moles of leucine and isoleucine in it.

Find the mass of the amino acids in the protein

Now that we have the moles of leucine and isoleucine in the protein, we can find their respective masses. Mass of leucine = moles of leucine × molar mass of leucine Mass of leucine = \(1.26 \times 10^{-4} \mathrm{~mol} \cdot 131 \mathrm{~g} / \mathrm{mol}\) = 0.0165 g Mass of isoleucine = moles of isoleucine × molar mass of isoleucine Mass of isoleucine = \(1.89 \times 10^{-4} \mathrm{~mol} \cdot 131 \mathrm{~g} / \mathrm{mol}\) = 0.0248 g Mass of both amino acids in the protein = Mass of leucine + Mass of isoleucine = 0.0165 g + 0.0248 g = 0.0413 g

Calculate the minimum possible molar mass of the protein

To find the minimum molar mass of the protein, we can now divide the mass of both amino acids in the protein by the total moles of amino acids. Moles of amino acids in the protein = Moles of leucine + Moles of isoleucine = \(1.26 \times 10^{-4} \mathrm{~mol}\) + \(1.89 \times 10^{-4} \mathrm{~mol}\) = \(3.15 \times 10^{-4} \mathrm{~mol}\) Minimum molar mass of the protein = \(\frac{\text{mass of both amino acids in the protein}}{\text{moles of amino acids in the protein}}\) Minimum molar mass of the protein = \(\frac{0.0413 \mathrm{~g}}{3.15 \times 10^{-4} \mathrm{~mol}}\) = 131.1 g/mol The minimum possible molar mass of the protein is 131.1 g/mol.