Problem 19 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

If isotopic distribution of \(\mathrm{C}^{12}\) and \(\mathrm{C}^{14}\) is \(98.0 \%\) and \(2.0 \%\), respectively, then the number of \(\mathrm{C}^{14}\) atoms in \(12 \mathrm{~g}\) of carbon is (a) \(1.032 \times 10^{22}\) (b) \(1.20 \times 10^{22}\) (c) \(5.88 \times 10^{23}\) (d) \(6.02 \times 10^{23}\)

Step-by-Step Solution

Answer

The number of \(\mathrm{C}^{14}\) atoms in 12 g of carbon is \(1.20 × 10^{22}\) atoms.

Step 1: Determine the mole quantity of carbon

Calculate the number of moles of carbon in 12 grams by using the molar mass of carbon (approximately 12 g/mol for \(\mathrm{C}^{12}\)). The number of moles \(n\) is equal to the mass \(m\) divided by the molar mass \(M\): \[\frac{m}{M} = \frac{12 \: \text{g}}{12 \: \text{g/mol}} = 1 \text{mol}.\]

Step 2: Calculate the total number of carbon atoms

Use Avogadro's number to find the total number of atoms in 1 mole of carbon, which is \(\frac{6.022 × 10^{23} \text{ atoms}}{1 \text{ mol}} \). Multiplying by the number of moles gives you the total number of carbon atoms: \[1 \text{ mol} × 6.022 × 10^{23} \text{ atoms/mol} = 6.022 × 10^{23} \text{ atoms.}\]

Step 3: Find the proportion of \(\text{C}^{14}\) atoms

Since \(\text{C}^{14}\) makes up 2.0% of the carbon atoms, multiply the total number of carbon atoms by 0.02 (2.0%) to find the number of \(\text{C}^{14}\) atoms: \[6.022 × 10^{23} \text{ atoms} × 0.02 = 1.2044 × 10^{22} \text{ atoms.}\]

Step 4: Round the answer to the correct number of significant figures

The percentage of \(\text{C}^{14}\) was given to 2 significant figures, so the final answer should be rounded accordingly to 2 significant figures. Thus you have: \[1.20 × 10^{22} \text{ atoms of } \text{C}^{14}.\]

Key Concepts

Molar Mass Calculation

Understanding how to calculate molar mass is a fundamental chemistry skill. It refers to the mass (in grams) of one mole of a substance. The molar mass of an element is determined by its atomic weight listed on the periodic table and is expressed in grams per mole (g/mol).

For isotopes, the atomic weight is an average, taking into account the relative abundance of each isotope. In practical terms, you can calculate the number of moles of an element by dividing the provided mass of the element by its molar mass. For instance, the molar mass of carbon-12 is approximately 12 g/mol, so 12 grams of carbon-12 equals exactly one mole.

By finding the moles, one can then proceed to use this information to relate mass to the number of atoms or molecules, bridging the gap between the macroscopic world we can measure and the microscopic world of atoms and molecules.

Avogadro's Number

Avogadro's number is a key constant in chemistry, representing the number of constituent particles (typically atoms or molecules) found in one mole of a substance. The value is 6.022 x 10^23 particles per mole.

It's used to convert between the number of atoms or molecules and moles, allowing chemists to work with amounts of a substance that can be physically measured. When we calculate the number of atoms in a sample, we multiply the number of moles by Avogadro's number. As shown in the exercise solution, calculating the total number of carbon atoms involves applying Avogadro's number to find how many atoms are in a mole of carbon.

Significant Figures

The concept of significant figures in measurements reflects the precision of those measurements. They are the digits in a number that carry meaningful information about its precision. When performing calculations, the number of significant figures in your final answer should correspond to the number in the least precise measurement you have.

In the example provided, given that the isotopic distribution is known to 2 significant figures, the final answer should be expressed with 2 significant figures as well. This is why the number of C^14 atoms is rounded to 1.20 x 10^22, maintaining the clarity and correctness of our calculation's precision according to the rules of significant figures.