Relative atomic mass: weight each isotope by its abundance
| English | 中文 | Pinyin |
|---|---|---|
| relative atomic mass/ˈrelətɪv əˈtɒmɪk mæs/ | 相对原子质量 | xiāng duì yuán zi zhì liàng |
| isotope abundance/ˈaɪsətəʊp əˈbʌndəns/ | 同位素丰度 | tóng wèi sù fēng dù |
What would explain this observation?
- A periodic-table mass need not be a whole number. It describes an average over the natural mixture of isotopes rather than a nucleus containing a fraction of a neutron.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Relative atomic mass 相对原子质量, Ar, is an average that accounts for isotope abundance 同位素丰度. For GCSE calculations using supplied isotope mass numbers, multiply each mass by its percentage abundance, add the products and divide by 100. More abundant isotopes contribute more to the mean. If frequencies rather than percentages are supplied, divide the weighted sum by total frequency.
- relative atomic mass: The abundance-weighted average atomic mass relative to one twelfth of carbon-12; isotope abundance: The proportion of atoms belonging to a particular isotope in a sample.
Why is the weighted mean closer to the more abundant isotope’s mass?
For a model chlorine sample containing 75% chlorine-35 and 25% chlorine-37, Ar=(35×75+37×25)/100=35.5. The mean is closer to 35 because that isotope is more abundant. An ordinary unweighted mean would give 36 and incorrectly treat both abundances as equal. Ar is a relative quantity without a gram unit; it is not the mass number of an individual atom.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- For a model chlorine sample containing 75% chlorine-35 and 25% chlorine-37, Ar=(35×75+37×25)/100=35.5. The mean is closer to 35 because that isotope is more abundant. An ordinary unweighted mean would give 36 and incorrectly treat both abundances as equal. Ar is a relative quantity without a gram unit; it is not the mass number of an individual atom.
- Check that percentage abundances total 100 before using the percentage formula. A missing second abundance can be found as 100 minus the first only when the sample has exactly two stated isotopes. Estimate where the answer should lie between the isotope masses, then calculate without premature rounding. A bead model with labelled masses illustrates weighted counting without implying all natural samples have the chosen fictional proportions.
Which two habits make the investigation or model in this case more defensible?
Check that percentage abundances total 100 before using the percentage formula. A missing second abundance can be found as 100 minus the first only when the sample has exactly two stated isotopes. Estimate where the answer should lie between the isotope masses, then calculate without premature rounding. A bead model with labelled masses illustrates weighted counting without implying all natural samples have the chosen fictional proportions.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: a fictional magnesium sample contains 80% mass-24, 10% mass-25 and 10% mass-26. Weighted sum=24×80+25×10+26×10=2,430. Divide by 100 to obtain Ar=24.3. It lies between 24 and 26 and close to the dominant mass-24 isotope. No atom in this model has mass number 24.3.
A fictional sample has 60% mass-10 and 40% mass-11. Calculate its relative atomic mass. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A fictional sample has 60% mass-10 and 40% mass-11. Calculate its relative atomic mass.
The result is 10.4 . Known: a fictional magnesium sample contains 80% mass-24, 10% mass-25 and 10% mass-26. Weighted sum=24×80+25×10+26×10=2,430. Divide by 100 to obtain Ar=24.3. It lies between 24 and 26 and close to the dominant mass-24 isotope. No atom in this model has mass number 24.3.
Check the conclusion and its limits
- A percentage is divided by 100 once, not twice. Isotope abundance is not the proportion of neutrons inside a nucleus. A changing mean in different samples can reflect a different isotope mixture; it does not show that a single atom acquired a fractional neutron.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Every atom of an element has mass number equal to the decimal value on its periodic-table entry. This claim is false: A percentage is divided by 100 once, not twice. Isotope abundance is not the proportion of neutrons inside a nucleus. A changing mean in different samples can reflect a different isotope mixture; it does not show that a single atom acquired a fractional neutron.
Relative atomic mass: weight each isotope by its abundance: For a model chlorine sample containing 75% chlorine-35 and 25% chlorine-37, Ar=(35×75+37×25)/100=35.5. The mean is closer to 35 because that isotope is more abundant. An ordinary unweighted mean would give 36 and incorrectly treat both abundances as equal. Ar is a relative quantity without a gram unit; it is not the mass number of an individual atom.
Every atom of an element has mass number equal to the decimal value on its periodic-table entry.
A percentage is divided by 100 once, not twice. Isotope abundance is not the proportion of neutrons inside a nucleus. A changing mean in different samples can reflect a different isotope mixture; it does not show that a single atom acquired a fractional neutron.
The abundance-weighted average atomic mass relative to one twelfth of carbon-12: write the technical term.
relative atomic mass means The abundance-weighted average atomic mass relative to one twelfth of carbon-12.