What is the average atomic mass calculator

The Correct Answer and Explanation is:

To calculate the average atomic mass of an element, you need to know the masses of its isotopes and their relative abundances. The formula for average atomic mass is:

[
\text{Average Atomic Mass} = \sum (\text{Isotope Mass} \times \text{Relative Abundance})
]

Here’s how you calculate it step by step:

Step 1: Identify the isotopes and their masses

An element may have several isotopes, each with a different mass number (mass). For example, carbon has two main isotopes: carbon-12 and carbon-14.

Carbon-12 has an isotope mass of 12.000 amu (atomic mass unit).
Carbon-14 has an isotope mass of 14.000 amu.

Step 2: Find the relative abundance of each isotope

Relative abundance refers to the percentage of each isotope in nature. For instance, if carbon-12 makes up 98.93% of natural carbon, and carbon-14 makes up 1.07%, these would be the relative abundances.

Step 3: Convert the relative abundances to decimals

Relative abundance should be in decimal form for the calculation. For example:

Carbon-12: 98.93% becomes 0.9893.
Carbon-14: 1.07% becomes 0.0107.

Step 4: Multiply the mass of each isotope by its relative abundance

For each isotope, multiply the mass by the relative abundance:

Carbon-12: ( 12.000 \times 0.9893 = 11.8716 )
Carbon-14: ( 14.000 \times 0.0107 = 0.1498 )

Step 5: Add the results together

Add the results of each isotope’s contribution:
[
\text{Average Atomic Mass} = 11.8716 + 0.1498 = 12.0214 \, \text{amu}
]

Final Answer:

The average atomic mass of carbon is approximately 12.0214 amu.

Explanation:

The average atomic mass reflects the weighted average mass of an element’s isotopes, accounting for their natural abundance. This value is critical in chemistry for determining the mass of an atom or molecule when performing calculations, such as in stoichiometry. The more abundant isotopes have a larger impact on the average mass, which is why the average atomic mass of carbon is closer to 12 amu rather than 14 amu, even though both isotopes are present.