Sulfur, S, and polonium, Po, are both elements in Group VI of the Periodic Table. Sulfur has three isotopes. (a) Explain the meaning of the term isotope . (b) A sample of sulfur has the following isotopic composition by mass. isotope mass 32 33 34 --- --- --- --- % by mass 95.00 0.77 4.23 Calculate the relative atomic mass, $A_r$, of sulfur to two decimal places. $A_r$ = ............... (c) Isotopes of polonium, proton number 84, are produced by the radioactive decay of…

Chemistry 9701/23 — October/November 2011

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Topics Atoms, Molecules and Stoichiometry · Introduction to Organic Chemistry · Atomic Structure · Carboxylic Acids and Derivatives · Hydroxy Compounds · Chemical Periodicity · +2 more

Q1Atomic StructureAtoms, Molecules and StoichiometryFree sample

Sulfur, S, and polonium, Po, are both elements in Group VI of the Periodic Table.

Sulfur has three isotopes.

(a)

Explain the meaning of the term isotope.

(b)

A sample of sulfur has the following isotopic composition by mass.

isotope mass	32	33	34
% by mass	95.00	0.77	4.23

Calculate the relative atomic mass, $A_r$ , of sulfur to two decimal places.

$A_r$ = ...............

(c)

Isotopes of polonium, proton number 84, are produced by the radioactive decay of several elements including thorium, Th, proton number 90.

The isotope $^{213}\text{Po}$ is produced from the thorium isotope $^{232}\text{Th}$ .

Complete the table below to show the atomic structures of the isotopes $^{213}\text{Po}$ and $^{232}\text{Th}$ .

isotope	protons	neutrons	electrons
$^{213}\text{Po}$
$^{232}\text{Th}$

(d)

Radiochemical reactions, such as nuclear fission and radioactive decay of isotopes, can be represented by equations in which the nucleon (mass) numbers must balance and the proton numbers must also balance.

For example, the nuclear fission of uranium-235, ${}^{235}_{92}\text{U}$ , by collision with a neutron, ${}^{1}_{0}\text{n}$ , produces strontium-90, xenon-143 and three neutrons.

{}^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \rightarrow {}^{90}_{38}\text{Sr} + {}^{143}_{54}\text{Xe} + 3 {}^{1}_{0}\text{n}

In this equation, the nucleon (mass) numbers balance because: $235 + 1 = 90 + 143 + (3 \times 1)$ .

The proton numbers also balance because: $92 + 0 = 38 + 54 + (3 \times 0)$ .

In the first stage of the radioactive decay of ${}^{232}_{90}\text{Th}$ , the products are an isotope of element $E$ and an alpha-particle, ${}^{4}_{2}\text{He}$ .

(i)

By considering nucleon and proton numbers only, construct a balanced equation for the formation of the isotope of $E$ in this reaction.

{}^{232}_{90}\text{Th} \rightarrow \text{.....................} + {}^{4}_{2}\text{He}

Show clearly the nucleon number and proton number of the isotope of $E$ .

nucleon number of the isotope of $E$ ...........

proton number of the isotope of $E$ .............

(ii)

Hence state the symbol of the element $E$ .

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