Question

Calculate the atomic mass for lithium given the following data for its natural isotopes: $$ \begin{array}{ccc} { }^{6} \mathrm{Li} & 6.015 \mathrm{amu} & 7.42 \% \\ { }^{7} \mathrm{Li} & 7.016 \mathrm{amu} & 92.58 \% \end{array} $$

Short answer

The atomic mass of lithium is approximately 6.94 amu.

Step-by-step solution

Understand the Problem

You are given the masses and natural abundances of two isotopes of lithium: \( ^6 \mathrm{Li} \) and \( ^7 \mathrm{Li} \). The goal is to calculate the atomic mass of lithium using this data.

Convert Percentages to Decimal Form

To calculate the atomic mass, you need the natural abundance in decimal form. Convert the percentages by dividing by 100:- \( ^6 \mathrm{Li}: 7.42\% \rightarrow 0.0742 \)- \( ^7 \mathrm{Li}: 92.58\% \rightarrow 0.9258 \)

Calculate Contribution of Each Isotope

Calculate the contribution of each isotope to the atomic mass by multiplying its mass by its abundance:- \( ^6 \mathrm{Li}: 6.015 \times 0.0742 = 0.446331 \)- \( ^7 \mathrm{Li}: 7.016 \times 0.9258 = 6.491093 \)

Sum the Contributions

Add the contributions of each isotope to find the atomic mass of lithium:\[ 0.446331 + 6.491093 = 6.937424 \]

Round to Appropriate Significant Figures

Since the data provided has four significant figures, round the atomic mass to four significant figures:\[ 6.9374 \approx 6.94 \] amu (due to rounding conventions).

Key concepts

Isotopes

Isotopes are different forms of the same element that have equal numbers of protons but different numbers of neutrons. This slight change in neutron count results in different masses for the isotopes of an element. For example, lithium has two naturally occurring isotopes:

• \( ^6 \mathrm{Li} \) with a mass of 6.015 amu
• \( ^7 \mathrm{Li} \) with a mass of 7.016 amu

Both isotopes are chemically identical in reactions since the number of protons (which determines chemical identity) remains unchanged. However, the differing number of neutrons gives them different physical properties, such as atomic mass.
Understanding isotopes is crucial for calculating atomic mass, as the average atomic mass of an element depends on the mass of its isotopes and their respective abundances.

Natural Abundance

Natural abundance refers to the relative percentage of a particular isotope present in a naturally occurring sample of an element. When we talk about natural abundance, we are considering how much of each isotope exists compared to the total amount of that element. In the case of lithium:

• \( ^6 \mathrm{Li} \) has a natural abundance of 7.42%
• \( ^7 \mathrm{Li} \) has a natural abundance of 92.58%

These percentages show how much each isotope contributes to the overall sample of lithium found in nature.
To calculate the average atomic mass of an element, it is essential to weigh each isotope's mass by its natural abundance. This involves converting these percentages into decimal form by dividing by 100. Therefore,

• 7.42% becomes 0.0742
• 92.58% becomes 0.9258

Including natural abundance allows a more accurate representation of the element's true atomic weight as found in nature.

Significant Figures

Significant figures are crucial in scientific calculations, as they communicate the precision of a measured or calculated value. They include all the certain digits in a number, plus one final uncertain digit. When calculating the atomic mass of lithium, the figures used in the data have a certain level of precision:

• The mass of isotopes was given to four significant figures (e.g., 6.015 and 7.016 amu)
• The natural abundances also provided precision with two decimal places (e.g., 0.0742 and 0.9258 after conversion)

In calculating the atomic mass of an element from its isotopes, the number of significant figures should be consistent with the data's precision. After computing contributions of the isotopes, adding them gives a result which, in the case of lithium, was initially 6.937424 amu.
Rounding this number to four significant figures gives us 6.937 amu, reflecting the precision of the initial data. Proper use of significant figures in reporting final calculations helps ensure the credibility and accuracy of scientific communication.