Antimony has many uses, for example, in infrared devices and as part of an alloy in lead storage batteries. The element has two naturally occurring isotopes, one with mass 120.904 amu and the other with mass 122.904 amu. (a) Write the \({ }_{Z}^{A} \mathrm{X}\) notation for each isotope. (b) Use the atomic mass of antimony from the periodic table to calculate the natural abundance of each isotope.

Isotopes are variations of elements that have the same number of protons but different numbers of neutrons. The isotope notation helps us to identify these variations clearly. It is generally written in the form \({ }_{Z}^{A} \mathrm{X}\). Here,
Z represents the atomic number,
A is the mass number (protons + neutrons),
and X stands for the chemical symbol of the element.
For example, antimony (Sb) has an atomic number (Z) of 51. Two isotopes of antimony can be written using the isotope notation:
For the isotope with a mass of 120.904 amu: \({ }_{51}^{121} \text{Sb}\)
For the isotope with a mass of 122.904 amu: \({ }_{51}^{123} \text{Sb}\).
Using this notation, we can easily identify specific isotopes of any element.

The atomic number is fundamental to understanding elements and their isotopes. The atomic number (Z) tells us the number of protons in the nucleus of an atom. It is this number that defines the element.
For example, antimony (Sb) has an atomic number of 51, meaning each antimony atom contains 51 protons. The atomic number is always represented as the subscript in the isotope notation \({ }_{Z}^{A} \mathrm{X}\).
It's important to note that while isotopes of an element have the same atomic number (same protons), they have different mass numbers due to the varying number of neutrons.

Natural abundance refers to the percentage of a particular isotope naturally found on Earth.
Since most elements exist as a mixture of isotopes, their atomic mass is usually a weighted average of the masses of these isotopes.
Let's consider antimony (Sb), which has two naturally occurring isotopes.
The atomic mass of antimony given in the periodic table is a weighted average, considering the natural abundances and masses of its isotopes.
If we let the natural abundance of the first isotope, \({ }_{51}^{121} \text{Sb}\), be 'x' and the second isotope, \({ }_{51}^{123} \text{Sb}\), be '1 - x', we can set up an equation to find these abundances.

Weighted average calculation is crucial for determining the atomic mass of elements with multiple isotopes. Here's a step-by-step approach:
1. Let x be the natural abundance of the first isotope.
2. Therefore, the natural abundance of the second isotope will be (1 - x).
3. Use the atomic mass from the periodic table as the weighted average of the isotopes.
For example, with antimony: \[ 120.904x + 122.904(1 - x) = 121.760 \]
4. Distribute and combine like terms: \[ 120.904x + 122.904 - 122.904x = 121.760 \]
5. Combine and simplify: \[ -2x + 122.904 = 121.760 \]
6. Solve for x: \[ x = \frac{1.144}{2} = 0.572 \]
So, the first isotope has a natural abundance of 57.2%, and the second isotope has an abundance of \[ 1 - 0.572 = 0.428 \] or 42.8%.
Using weighted averages, we can find the natural abundances and understand the contribution of each isotope to the element's atomic mass.