AQA GCSE Combined Science (8464), Higher tier · Chemistry › Atomic structure and the periodic table › A simple model of the atom, symbols, relative atomic mass,
Practise Relative atomic mass. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Relative atomic mass (Ar) is an average value for the atoms of an element that takes account of the abundance of each of its isotopes. You must be able to calculate it from percentage abundances: a common 2 to 3 mark calculation.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Explain what relative atomic mass meansAn average value for the atoms of an element that takes account of the abundance of each isotope.
5
Calculate Ar for two isotopesMultiply each mass number by its percentage abundance, add the results, then divide by 100.
5
Explain why Ar is rarely wholeIt is an average of isotopes with different mass numbers, weighted by how common each one is.
6
Decide which isotope is more abundantThe Ar is closer to the mass number of the more abundant isotope.
7
Calculate Ar for three isotopesSame method with three terms; check that the abundances add up to 100%.
8
Find an unknown abundance from ArCall one abundance x and the other (100 − x), then solve the equation.
Notes
What relative atomic mass means
Many elements exist as a mixture of isotopes, which have different masses.
The relative atomic mass (symbol Ar) of an element is an average value that takes account of the abundance (the percentage of atoms) of each of its isotopes.
Ar is the number shown for each element in the periodic table on the data sheet, e.g. chlorine 35.5. It has no units.
Ar is often not a whole number, because it is an average. It is closer to the mass number of the more abundant isotope.
Calculating Ar
Ar = (mass number × % abundance of the first isotope + mass number × % abundance of the second isotope + …) ÷ 100
Worked example: chlorine is 75% chlorine-35 and 25% chlorine-37. (35 × 75) + (37 × 25) = 2625 + 925 = 3550 3550 ÷ 100 = 35.5
Sense check: the answer must lie between the smallest and largest mass numbers, and nearer to the mass number of the most abundant isotope.
If you are given the number of atoms of each isotope instead of percentages, divide by the total number of atoms instead of by 100.
Finding an unknown abundance grade 8+
If you know Ar and there are two isotopes, call the abundance of one isotope x and the other (100 − x).
Worked example: boron (Ar = 10.8) is made of boron-10 and boron-11. 10x + 11(100 − x) = 10.8 × 100 1100 − x = 1080, so x = 20 Boron is 20% boron-10 and 80% boron-11.
Cheatsheet
Ar = average value for an element's atoms, taking account of the abundance of its isotopes
Ar = sum of (mass number × % abundance) ÷ 100
Ar has no units
Ar is nearer to the mass number of the most abundant isotope
Chlorine: 75% chlorine-35 + 25% chlorine-37 gives Ar = 35.5
Unknown abundance: use x and (100 − x) and solve grade 8+
How to answer each type of question
Calculate relative atomic mass from two isotopes
2 marks5
Multiply each mass number by its percentage abundance.
Add the results.
Divide by 100 and round as the question asks.
Example. Copper has two isotopes: copper-63 (69%) and copper-65 (31%). Calculate the relative atomic mass of copper. Give your answer to 1 decimal place.
Calculate relative atomic mass from three isotopes
2 to 3 marks7
Check the abundances add up to 100%.
Multiply each mass number by its abundance and add all three results.
Divide by 100.
Example. Magnesium is 79% magnesium-24, 10% magnesium-25 and 11% magnesium-26. Calculate the relative atomic mass of magnesium. Give your answer to 1 decimal place.
Use the words 'average', 'isotopes' and 'abundance'.
To decide which isotope is more common, see which mass number the Ar is closer to.
Example. Lithium has two isotopes, lithium-6 and lithium-7. The relative atomic mass of lithium is 6.9. (a) Explain why the relative atomic mass of lithium is not a whole number. (b) Which isotope is more abundant? Give a reason.
Show the model answer
(a) It is an average of the masses of the isotopes, taking account of how abundant each isotope is (1). (b) Lithium-7, because 6.9 is much closer to 7 than to 6 (1).
Calculate an unknown percentage abundance
3 marks8
Let the abundance of one isotope be x; the other is (100 − x).
Write the equation: mass × x + mass × (100 − x) = Ar × 100.
Expand the brackets, collect terms and solve for x.
Example. Gallium has two isotopes, gallium-69 and gallium-71. The relative atomic mass of gallium is 69.8. Calculate the percentage abundance of gallium-69.
Show the model answer
69x + 71(100 − x) = 69.8 × 100 (1) 69x + 7100 − 71x = 6980, so 2x = 120 (1) x = 60, so gallium-69 is 60% (1)
Shortcuts and memory tricks
Sense check: Ar must lie between the lightest and heaviest mass numbers.
If two isotopes are 50% each, Ar is exactly halfway between their mass numbers.
Type it in one go with brackets to avoid rounding errors: (35 × 75 + 37 × 25) ÷ 100.
For a real element, compare your answer with the Ar in the periodic table: it should be very close.
Where marks are lost
Dividing by the number of isotopes (e.g. by 2) instead of by 100.
Adding the mass numbers and halving, which ignores the abundances.
Rounding part-way through the calculation, or giving the wrong number of decimal places.
Giving Ar units such as grams.
Exam technique
Write out the whole expression before you calculate, so you earn the method mark even if you slip on the arithmetic.
Give the answer to the number of decimal places the question asks for (often 1 decimal place).
In 'explain' questions about Ar, use the key words: average, isotopes, abundance.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Explain why the relative atomic mass of an element is usually not a whole number.
It is a weighted mean of the masses of isotopes with different mass numbers.
Bromine has two isotopes, bromine-79 and bromine-81. The two isotopes have almost equal abundances. The atomic number of bromine is 35. Suggest the approximate relative atomic mass of bromine.
80
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy2 marks
Relative atomic mass takes into account the different isotopes of an element.
(a) What is the relative atomic mass of an element? Tick (✓) one box.[1]
The mass of the most common isotope only
An average mass of the isotopes, taking into account their abundance
The total mass of all the isotopes added together
The mass of the heaviest isotope only
(b) An element has two isotopes, with mass numbers 63 and 65. The two isotopes have equal abundances. What is the relative atomic mass of the element? Tick (✓) one box.[1]
63
64
65
128
Show the answer and mark scheme
(a)Answer: An average mass of the isotopes, taking into account their abundance
(b)Answer: 64
Question 2Medium2 marks
Chlorine has two isotopes, chlorine-35 and chlorine-37. The relative atomic mass of chlorine is 35.5.
A student says: 'Because chlorine has isotopes with mass numbers 35 and 37, its relative atomic mass must be exactly 36.' Evaluate this claim.[2]
Show the answer and mark scheme
Answer: The student is wrong: relative atomic mass is a weighted average that depends on the abundance of each isotope, not simply the midpoint; since the actual value (35.5) is closer to 35, chlorine-35 must be the more abundant isotope.
relative atomic mass is a weighted average that depends on the (percentage) abundance of each isotope, not just the midpoint of the mass numbers
the actual value (35.5) is closer to 35, so chlorine-35 must be more abundant than chlorine-37
Question 3Hard7 marks
Boron has two isotopes, 10B and 11B. The relative atomic mass of boron is 10.8.
(a) Calculate the percentage abundance of 10B.[3]
(b) A student said: 'The relative atomic mass of boron should be 10.5 because boron has two isotopes.' Explain why the student is wrong.[2]
(c) Boron trifluoride, BF3, contains boron. Fluorine has only one isotope, 19F. Explain how many different relative formula masses BF3 molecules can have. Give the values.[2]
Show the answer and mark scheme
(a)Answer: 20%
abundance of 10B = x: 10x + 11(100 − x) = 1080
1100 − x = 1080
x = 20 (%)
(b)Answer: 10.5 would only be correct if the two isotopes were equally abundant; there is more 11B (80%), so the average is closer to 11.
10.5 would only be correct if both isotopes had equal abundance / 50% each
there is more 11B / 11B is 80%, so the mean is closer to 11
(c)Answer: Two, because boron has two isotopes: 67 and 68.
two, because boron has two isotopes (and fluorine has one)