Problem 96 - Chemistry: The Central Science

Question

Gallium (Ga) consists of two naturally occurring isotopes with masses of \(68.926\) and \(70.925\) amu. (a) How many protons and neutrons are in the nucleus of each isotope? Write the complete atomic symbol for each, showing the atomic number and mass number. (b) The average atomic mass of Ga is \(69.72\) amu. Calculate the abundance of each isotope.

Step-by-Step Solution

Answer

Gallium has 31 protons. Its two isotopes have approximately 38 and 40 neutrons, with atomic symbols \(_{31}^{68}\text{Ga}\) and \(_{31}^{70}\text{Ga}\). The abundances of these isotopes are approximately 60% and 40%, respectively.

Step 1: Find the Atomic Number of Gallium

To find protons, we need to know the atomic number (Z) of Gallium, which can be found in the periodic table. The atomic number of Gallium is 31, which means Gallium has 31 protons.

Step 2: Calculate the Number of Neutrons in Each Isotope

Now that we know the atomic number of Gallium, we can calculate the number of neutrons in each isotope. The mass number (A) is the sum of protons and neutrons. For the first isotope with a mass of 68.926 amu: \(A_1 = \text{Protons} + \text{Neutrons}\) \(\text{Neutrons} = A_1 - \text{Protons}\) \(\text{Neutrons} = 68.926 - 31\) \(\text{Neutrons} \approx 38\) For the second isotope with a mass of 70.925 amu: \(A_2 = \text{Protons} + \text{Neutrons}\) \(\text{Neutrons} = A_2 - \text{Protons}\) \(\text{Neutrons} = 70.925 - 31\) \(\text{Neutrons} \approx 40\)

Step 3: Write the Complete Atomic Symbol for Each Isotope

The atomic symbol of Gallium is Ga. We will write the complete atomic symbol for each isotope, showing the atomic number (Z) and mass number (A). For the first isotope with 31 protons and 38 neutrons: \(_{31}^{68}\text{Ga}\) For the second isotope with 31 protons and 40 neutrons: \(_{31}^{70}\text{Ga}\)

Step 4: Set up the Equation for Average Atomic Mass

We are given the average atomic mass of Gallium, the mass numbers of the two isotopes of Gallium, and their respective unknown abundances. Let's say the abundance of the first isotope with a mass of 68.926 amu is x, and the second isotope with a mass of 70.925 amu is y. Since there are only two isotopes, the sum of their abundances must be 1. The equation for the average atomic mass will be: \(69.72 = 68.926x + 70.925y\) With the constraint that x + y = 1.

Step 5: Solve the Equation for the Abundances

We can rewrite the first equation, subtracting y along the way to make it a single variable equation. \(69.72 = 68.926x + 70.925(1-x)\) Now, we will solve for x (the abundance of isotope with mass 68.926 amu): \(69.72 - 70.925 = -1.999x\) \(x \approx 0.60\) Now we can solve for y (the abundance of isotope with mass 70.925 amu) using the constraint equation: \(y = 1 - x\) \(y \approx 0.40\) The abundances of the two isotopes are approximately 60% for the first isotope with mass 68.926 amu (\(_{31}^{68}\text{Ga}\)) and 40% for the second isotope with mass 70.925 amu (\(_{31}^{70}\text{Ga}\)).