Average Atomic Mass Calculator
Calculate the average atomic mass (atomic weight) of elements from isotope masses and their percent abundances. Also known as relative atomic mass, this weighted average accounts for all naturally occurring isotopes. Essential for chemistry calculations involving chlorine, magnesium, silicon, rubidium, and other elements with multiple isotopes.
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Average Atomic Mass Formulas
Formula for Relative Atomic Mass
1. Average Atomic Mass Formula (Weighted Average):
\( \text{Average Atomic Mass} = \sum_{i=1}^{n} \left(\text{Mass}_i \times \frac{\text{Abundance}_i}{100}\right) \)
Where Mass\(_i\) is the atomic mass of isotope \(i\) in amu and Abundance\(_i\) is the percent abundance. Sum over all isotopes. For chlorine: (34.969 × 75.77/100) + (36.966 × 24.23/100) = 35.45 amu.
2. Two Isotope Formula:
\( \text{Average Mass} = (M_1 \times A_1) + (M_2 \times A_2) \)
Simplified for two isotopes where \(M\) is mass, \(A\) is decimal abundance (percent/100). If \(A_1 = 0.7577\) and \(A_2 = 0.2423\), ensure \(A_1 + A_2 = 1\).
3. Percent Abundance Formula:
\( \text{Percent Abundance} = \frac{\text{Isotope Abundance}}{\text{Total Abundance}} \times 100\% \)
Total abundance should equal 100%. If given fractions, convert to percentages. For two isotopes: if one is 75.77%, other is 100% - 75.77% = 24.23%.
4. Atomic Weight Formula (Same as Average Atomic Mass):
\( \text{Atomic Weight} = \frac{\sum (\text{Isotope Mass} \times \text{Natural Abundance})}{\sum \text{Natural Abundance}} \)
Atomic weight and relative atomic mass are synonymous terms. Both represent weighted average of all naturally occurring isotopes. Used on periodic table (e.g., Cl = 35.45, Si = 28.09, Rb = 85.47, Mg = 24.31).
Average Atomic Mass Practice Problems with Answers
Problem 1: Calculate the Atomic Mass of Chlorine
Question: Chlorine has two isotopes: Cl-35 (34.969 amu, 75.77% abundance) and Cl-37 (36.966 amu, 24.23% abundance). Calculate the average atomic mass.
Solution:
Average atomic mass = (34.969 × 75.77/100) + (36.966 × 24.23/100)
= (34.969 × 0.7577) + (36.966 × 0.2423)
= 26.496 + 8.957 = 35.453 amu
Answer: 35.45 amu (matches periodic table value)
Problem 2: Calculate the Atomic Mass of Silicon
Question: Silicon has three isotopes: Si-28 (27.977 amu, 92.23%), Si-29 (28.976 amu, 4.67%), Si-30 (29.974 amu, 3.10%). Calculate average atomic mass.
Solution:
Average = (27.977 × 0.9223) + (28.976 × 0.0467) + (29.974 × 0.0310)
= 25.803 + 1.353 + 0.929 = 28.085 amu
Answer: 28.09 amu (Si-28 dominates due to 92% abundance)
Problem 3: Calculate the Atomic Weight of Rubidium
Question: Rubidium has Rb-85 (84.912 amu, 72.17%) and Rb-87 (86.909 amu, 27.83%). Calculate atomic weight.
Solution:
Atomic weight = (84.912 × 0.7217) + (86.909 × 0.2783)
= 61.274 + 24.191 = 85.465 amu
Answer: 85.47 amu (closer to Rb-85 due to higher abundance)
Problem 4: Calculate Relative Atomic Mass of Magnesium
Question: Magnesium: Mg-24 (23.985 amu, 78.99%), Mg-25 (24.986 amu, 10.00%), Mg-26 (25.983 amu, 11.01%). Find relative atomic mass.
Solution:
Relative atomic mass = (23.985 × 0.7899) + (24.986 × 0.1000) + (25.983 × 0.1101)
= 18.948 + 2.499 + 2.861 = 24.308 amu
Answer: 24.31 amu (Mg-24 is most abundant)
Frequently Asked Questions
What is average atomic mass?
Average atomic mass (also called atomic weight or relative atomic mass) is the weighted average of the masses of all naturally occurring isotopes of an element. It accounts for both the mass of each isotope and its natural abundance. For example, chlorine's average atomic mass is 35.45 amu because nature contains 75.77% Cl-35 (34.969 amu) and 24.23% Cl-37 (36.966 amu). This weighted average appears on the periodic table and is essential for stoichiometry calculations in chemistry.
How do you calculate relative atomic mass?
To calculate relative atomic mass: (1) List all isotopes with their masses and percent abundances. (2) Convert percentages to decimals by dividing by 100. (3) Multiply each isotope's mass by its decimal abundance. (4) Sum all products. Formula: Σ(mass × abundance/100). For chlorine: (34.969 × 75.77/100) + (36.966 × 24.23/100) = 26.50 + 8.96 = 35.45 amu. Verify abundances sum to 100%. The result should match the periodic table value within rounding error. This method works for any element with 2+ isotopes.
What is the atomic mass formula?
The atomic mass formula is: Average Atomic Mass = Σ(Isotope Mass × Fractional Abundance). For each isotope, multiply its mass in amu by its fractional abundance (percent/100), then sum for all isotopes. Alternative notation: M = (M₁A₁) + (M₂A₂) + (M₃A₃)... where M is mass and A is abundance. The abundances must sum to 1.0 (or 100%). Example for two isotopes: M = (M₁ × %₁/100) + (M₂ × %₂/100). This weighted average reflects the natural distribution of isotopes and gives the atomic weight listed on periodic tables.
How do you calculate percent abundance of isotopes?
To calculate percent abundance when given average atomic mass: (1) Set up equation with unknown abundances (let x = abundance of isotope 1, then 1-x = abundance of isotope 2 for two isotopes). (2) Write: Average Mass = (Mass₁ × x) + (Mass₂ × (1-x)). (3) Solve for x. (4) Convert to percent. Example: If Cl average is 35.45, Cl-35 mass is 34.969, Cl-37 mass is 36.966: 35.45 = 34.969x + 36.966(1-x). Solving: 35.45 = 34.969x + 36.966 - 36.966x. 35.45 = 36.966 - 1.997x. x = 0.7577 = 75.77% for Cl-35.
How to calculate relative atomic mass from mass spectrum?
In mass spectrometry, peak heights (or areas) represent relative abundances. Steps: (1) Identify each peak's mass-to-charge ratio (m/z) - this is the isotope mass for singly charged ions. (2) Measure peak heights or areas. (3) Calculate relative abundance: (peak height ÷ sum of all peak heights) × 100%. (4) Apply weighted average formula. Example: If spectrum shows peaks at m/z 35 (height 152) and m/z 37 (height 49): Total = 152+49 = 201. Abundance: 35 → 152/201 = 75.6%, 37 → 49/201 = 24.4%. Average mass = (35 × 0.756) + (37 × 0.244) = 35.49 amu.
Why is atomic mass not a whole number?
Atomic mass on the periodic table is not a whole number because it's a weighted average of multiple isotopes. While individual isotopes have nearly whole-number masses (due to protons and neutrons), elements exist as mixtures of isotopes in nature. Chlorine is 35.45 amu (not 35 or 37) because it's 75.77% Cl-35 and 24.23% Cl-37. The decimal reflects this natural mixture. Only elements with one naturally occurring isotope (like fluorine-19 or sodium-23) show atomic masses close to whole numbers on the periodic table. The weighted average accounts for isotopic distribution found in Earth's crust.