Isotopes Worked ExamplesIGCSE Chemistry 0620

In a nuclide symbol the top number is the nucleon (mass) number, the total of protons and neutrons, and the bottom number is the proton (atomic) number.

(a) Isotopes of one element have the same proton number. A, C and D all have proton number $28$ (nickel), so A, C and D are isotopes.

(b) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Here A, C and D each have $28$ protons but have different nucleon numbers, so different numbers of neutrons.

(c) Number of neutrons $=$ nucleon number $-$ proton number. Work it out for every atom:

C and E both have $32$ neutrons. (d) An element is defined by its proton number. B has $27$ protons (cobalt) and C has $28$ protons (nickel), so they are different elements even though their nucleon numbers are the same.

Protons $=$ proton number (bottom number). Neutrons $=$ nucleon number $-$ proton number. For a neutral atom, electrons $=$ protons.

$^{78}_{34}\text{Se}$ is a neutral atom: $34$ protons, $78 - 34 = 44$ neutrons, $34$ electrons.

$^{80}_{34}\text{Se}^{2-}$: $34$ protons and $80 - 34 = 46$ neutrons. A $2-$ charge means the atom has gained two electrons, so it has $34 + 2 = 36$ electrons.

$^{65}_{29}\text{Cu}^{2+}$: $29$ protons and $65 - 29 = 36$ neutrons. A $2+$ charge means two electrons have been lost, so it has $29 - 2 = 27$ electrons. The charge never changes the number of protons.

Chemical reactions involve the electrons, especially those in the outer shell. Both atoms have $12$ protons and therefore $12$ electrons, arranged $2,8,2$. The same electronic configuration gives the same chemical properties. Answer: C.

A and B are false: $^{24}\text{Mg}$ has $12$ neutrons and $^{26}\text{Mg}$ has $14$, so their neutron numbers and nucleon numbers are different — that is what makes them isotopes.

D is false: the two atoms have different masses. Mass affects physical properties such as density, not the way the atoms react.

(a) Relative atomic mass is the average mass of the isotopes, weighted by their abundance. Multiply each isotope's mass by its percentage and add.

The percentages add up to $100$, so divide by $100$.

(b) $^{69}\text{Ga}$ is the more abundant isotope ($60.1\%$ compared with $39.9\%$), so it contributes more to the weighted average.

(a) Treat the ratio as $2$ atoms of $^{6}\text{Li}$ for every $23$ atoms of $^{7}\text{Li}$. Find the total mass of this group of atoms.

The group contains $2 + 23 = 25$ atoms, so divide by $25$ — not by $100$.

(b) $2$ atoms out of every $25$ are $^{6}\text{Li}$.

(a) From the chart: $^{20}\text{Ne}$ $90.5\%$, $^{21}\text{Ne}$ $0.3\%$, $^{22}\text{Ne}$ $9.2\%$. Check they add up to $100\%$: $90.5 + 0.3 + 9.2 = 100.0$.

Multiply each mass by its abundance and add.

Divide by the total abundance, $100$.

(b) The least abundant isotope is $^{21}\text{Ne}$ ($0.3\%$). Neutrons $= 21 - 10 = 11$.

(a) Relative atomic mass compares the average mass of an element's atoms with $\frac{1}{12}$ of the mass of an atom of carbon-12, $^{12}\text{C}$.

(b) If every atom has the same nucleon number, there is only one isotope of fluorine in nature. The average of identical masses is that mass, which is why $A_r = 19.0$.

(c) Chlorine is a mixture of isotopes with different masses (chlorine-35 and chlorine-37). $A_r$ is their abundance-weighted average, which lies between the two masses and is not a whole number.

(d) $A_r$ of sodium is $23.0$, a whole number, which suggests sodium has only one isotope, $^{23}\text{Na}$. Its atoms all contain $23 - 11 = 12$ neutrons, so the student is correct.

(a) Credit goes to things you can see: bubbles (fizzing) rise from the calcium; the calcium gets smaller and eventually disappears; the universal indicator turns from green to blue/purple (an alkaline solution forms); the water level in the test-tube falls as gas collects.

(b) The observations are the same. $^{40}\text{Ca}$ and $^{44}\text{Ca}$ are isotopes: both have $20$ protons and $20$ electrons, so they have the same electronic configuration $(2,8,8,2)$ and therefore the same chemical properties.

(c) The gas is hydrogen. Hold a lighted splint at the mouth of the test-tube: it burns with a squeaky pop.

(d) Calcium forms calcium hydroxide and hydrogen. Check the atoms: left Ca $1$, H $4$, O $2$; right Ca $1$, O $2$, H $2 + 2 = 4$.

(e) An atom of $^{44}\text{Ca}$ has $44 - 20 = 24$ neutrons; an atom of $^{40}\text{Ca}$ has $40 - 20 = 20$. It has four more neutrons (the numbers of protons and electrons are the same).

The two masses are $2$ units apart. $121.8$ is $0.8$ above $121$ and $1.2$ below $123$, so the average lies closer to $121$: more than half the atoms must be $^{121}\text{Sb}$. That rules out A and B.

The fraction of the heavier isotope equals how far $A_r$ has moved from the lighter mass, divided by the gap: $\dfrac{0.8}{2} = 0.40$, so $40\%$ is $^{123}\text{Sb}$ and $60\%$ is $^{121}\text{Sb}$. Answer: C.

Check by working forwards:

B ($40\%$) is the abundance of $^{123}\text{Sb}$ — the tempting answer if you use $0.8$ as the fraction of the lighter isotope. A and D do not give $121.8$ when checked.

(a) Let the percentage of $^{191}\text{Ir}$ be $x$. The percentage of $^{193}\text{Ir}$ is then $100 - x$. Write the $A_r$ expression.

Multiply both sides by $100$ and expand.

Collect terms.

So $^{191}\text{Ir}$ is $40\%$ and $^{193}\text{Ir}$ is $60\%$. Sense check: $192.2$ is nearer $193$, so the heavier isotope should be more abundant — it is. (Quick route: $192.2 - 191 = 1.2$ out of a gap of $2$, so $\frac{1.2}{2} = 60\%$ is the heavier isotope.)

(b) Neutrons $=$ nucleon number $-$ proton number.

Identify each particle from its proton number, then compare electrons with protons to find its charge. V: $11$ protons (sodium), $10$ electrons, charge $1+$. W: oxygen, neutral. X: neon, neutral. Y: oxygen, neutral. Z: fluorine, $10$ electrons, charge $1-$.

(a) W and Y both have $8$ protons but different numbers of neutrons ($10$ and $8$): they are isotopes of oxygen.

(b) $2,8$ is $10$ electrons: V, X and Z. (c) $10$ neutrons: W, X and Z.

(d) Nucleon number $=$ protons $+$ neutrons. V: $11 + 12 = 23$ with charge $+$. Z: $9 + 10 = 19$ with charge $-$.

(e) W. W and Y are isotopes of oxygen; both have $8$ electrons arranged $2,6$, and the same electronic configuration gives the same chemical properties.

(a) Let the unknown mass be $m$. Write the $A_r$ expression using the abundances from the chart.

Multiply by $100$ and substitute $107 \times 52.0 = 5564$.

Solve for $m$.

(b) Neutrons $= 109 - 47 = 62$.

(c) $A_r$ is the weighted average of the masses of both isotopes. Because silver is a mixture of $^{107}\text{Ag}$ and $^{109}\text{Ag}$ in similar amounts, the average ($108$ to 3 s.f.) lies between the two nucleon numbers.