By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Atomic Mass is the average weight of all the different versions (isotopes) of an element. Think of it like a big box of cereal with different flavors (isotopes) - the average weight of all the cereal pieces (atoms) is the atomic mass.
This matters in real life because knowing the atomic mass helps us understand how elements react with each other, which is crucial in fields like medicine, energy, and materials science. For example, without understanding atomic mass, we wouldn't be able to develop medicines that target specific cells or create new materials with unique properties.
Example:
Step 1: Find the atomic masses of each isotope.12.01 amu (carbon-12)13.01 amu (carbon-13)
Step 2: Find the natural abundance of each isotope.98.9% (carbon-12)1.1% (carbon-13)
Step 3: Multiply the atomic mass of each isotope by its natural abundance.12.01 amu x 0.989 = 11.89 amu (carbon-12)13.01 amu x 0.011 = 0.143 amu (carbon-13)
Step 4: Add up the results of step 3 for each isotope.11.89 amu + 0.143 amu = 12.033 amu
Step 5: Round the result to the correct number of significant figures.12.03 amu
The average atomic mass of carbon is 12.03 amu.
Calculate the average atomic mass of chlorine (Cl) using the following data:
Step 1: Find the atomic masses of each isotope.34.97 amu (Cl-35)36.97 amu (Cl-37)
Step 2: Find the natural abundance of each isotope.75.8% (Cl-35)24.2% (Cl-37)
Step 3: Multiply the atomic mass of each isotope by its natural abundance.34.97 amu x 0.758 = 26.49 amu (Cl-35)36.97 amu x 0.242 = 8.94 amu (Cl-37)
Step 4: Add up the results of step 3 for each isotope.26.49 amu + 8.94 amu = 35.43 amu
Step 5: Round the result to the correct number of significant figures.35.4 amu
The average atomic mass of chlorine is 35.4 amu.
Calculate the average atomic mass of oxygen (O) using the following data:
Step 1: Find the atomic masses of each isotope.15.99 amu (O-16)16.99 amu (O-17)17.99 amu (O-18)
Step 2: Find the natural abundance of each isotope.99.8% (O-16)0.2% (O-17)0.2% (O-18)
Step 3: Multiply the atomic mass of each isotope by its natural abundance.15.99 amu x 0.998 = 15.96 amu (O-16)16.99 amu x 0.002 = 0.04 amu (O-17)17.99 amu x 0.002 = 0.04 amu (O-18)
Step 4: Add up the results of step 3 for each isotope.15.96 amu + 0.04 amu + 0.04 amu = 16.04 amu
Step 5: Round the result to the correct number of significant figures.16.0 amu
The average atomic mass of oxygen is 16.0 amu.
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