Question

Study the following table.

| Compound (mol. Mass)| Compound mass of the (in gm) taken
---|---|---
I.| $C{O_2}(44)$ | $4.4$
II.| $N{O_2}(46)$ | $2.3$
III.| ${H_2}{O_2}(34)$ | $6.8$
IV.| $S{O_2}(64)$ | $1.6$

Which two compounds have the least mass of oxygen?
(Molecular masses of compounds are given in brackets)
A.II and IV
B.I and III
C.I and II
D.III and IV

Answer 1

Chemists generally use the mole as the unit for the number of atoms or molecules of a material. One mole is equal to $6.022 \times {10^{23}}$ atoms/molecules/particles. This $6.022 \times {10^{23}}$ is also known as Avogadro's number.

Complete step by step answer:
As we all know, the molar mass of an oxygen atom is $16$ $gm/mol$.
So the molar mass of oxygen gas $({O_2})$ is $32$ $gm/mol$ .
We have to calculate mass of oxygen so firstly we will calculate the number of moles of oxygen in a molecule and then convert these moles of oxygen into mass and afterwards we will compare those masses of oxygen of different molecules.
We will calculate the mass of oxygen present $C{O_2}$.
Number of moles of $C{O_2}$ = $\dfrac{{4.4}}{{44}} = 0.1$
Number of moles of oxygen in $C{O_2}$ = $0.1 \times 2 = 0.2$
And the weight of one mole of oxygen is $16gm$
Therefore,
Mass of oxygen = $0.2 \times 16 = 3.2gm$ -----------(1)
So the mass of oxygen present in $C{O_2}$ is $3.2gm$ .
Now we will calculate the mass of oxygen present $N{O_2}$.
So,
Number of moles of $N{O_2}$ = $\dfrac{{2.3}}{{46}} = 0.05$
Number of moles of oxygen in $N{O_2}$ = $2 \times 0.05 = 0.1$
And the weight of one mole of oxygen is $16gm$
Therefore,
Mass of oxygen= $0.1 \times 16 = 1.6gm$ ------------------(2)
So the mass of oxygen present in $N{O_2}$ is $1.6gm$ .
Now we will calculate the mass of oxygen present ${H_2}{O_2}$.
So,
Number of moles of ${H_2}{O_2}$ = $\dfrac{{6.8}}{{34}} = 0.2$
Number of moles of oxygen in ${H_2}{O_2}$ = $2 \times 0.2 = 0.4$
And the weight of one mole of oxygen is $16gm$
Therefore,
Mass of oxygen= $0.4 \times 16 = 6.4gm$ ------------------(3)
So the mass of oxygen present in ${H_2}{O_2}$ is $6.4gm$ .
Now we will calculate the mass of oxygen present $S{O_2}$.
So,
Number of moles of $S{O_2}$ = $\dfrac{{1.6}}{{64}} = 0.025$
Number of moles of oxygen in $S{O_2}$ = $0.025 \times 2 = 0.05$
And the weight of one mole of oxygen is $16gm$
Therefore,
Mass of oxygen= $0.05 \times 16 = 0.8gm$ -------------------(4)
So the mass of oxygen present in $S{O_2}$ is $0.8gm$ .
Hence we can see that the mass of oxygen is least in $N{O_2}$ and $S{O_2}$.
So our correct option is (A), II and IV.

Note:
Oxygen’s atomic weight is 16 amu. One mole of oxygen contains $6.022 \times {10^{23}}$ number of atoms. The molar mass of an atom is always equal to atomic weight of an atom. The unit of molar mass is $gm/mole$.