As we saw in the previous problem, lithium has two naturally occurring isotopes: Li-6 (natural abundance 7.5\%; mass 6.0151 amu) and Li-7 (natural abundance \(92.5 \% ;\) mass 7.0160 amu). Without doing any calculations, determine which mass is closest to the atomic mass of Li. a. 6.00 amu b. 6.50 amu c. 7.00 amu

Isotopes are variants of a particular chemical element that share the same number of protons but have different numbers of neutrons in their atomic nuclei. This means that while they have the same atomic number, they have different atomic masses. For example, lithium, the element in our exercise, has isotopes such as Li-6 and Li-7. Li-6 has 3 protons and 3 neutrons resulting in an atomic mass of 6.0151 atomic mass units (amu), while Li-7 has 3 protons and 4 neutrons, leading to an atomic mass of 7.0160 amu.

Understanding isotopes is crucial because they explain why elements may not have an atomic mass that's a whole number. Isotopes also have different properties, which can lead to their varied applications in fields such as medicine, archaeology, and energy production.

The weighted average is an essential mathematical concept that's applied when calculating the atomic mass of an element. Rather than a simple average that treats all values equally, a weighted average multiplies each value by a weight factor, which represents its relative importance or frequency. The result of the weighted average calculation gives us the atomic mass, which is reflective of the different isotopes and their abundances in nature.

In our exercise, even without calculations, understanding the nature of a weighted average can guide us to conclude that because Li-7 has a much higher abundance than Li-6, its mass will have a much greater effect on the final atomic mass of lithium. Essentially, when applied to atomic masses, the weighted average provides an adjusted mean that accounts for the distribution of each isotope in nature.

Natural abundance refers to the occurrence of isotopes of a chemical element as they are found on Earth. It is often expressed as a percentage and represents the ratio of a particular isotope in relation to all isotopes of that element present in a natural sample. In our exercise involving lithium isotopes, natural abundance is a critical factor because it determines the weight each isotope has in the weighted average calculation for the element's atomic mass.

Since Li-7 has a natural abundance of 92.5%, it is much more prevalent than Li-6, which has a natural abundance of 7.5%. Consequently, in the context of lithium's atomic mass, Li-7 will have a much bigger influence on the weighted average than Li-6. Thus, when we look at which mass is closest to the atomic mass of Li, it's logical to choose the mass of the more abundant isotope, Li-7.