Question

What will be the weight of $CO$ having the same number of oxygen atoms as present in $22g$ of $C{O_2}$ ?
A.$28g$
B.$22g$
C.$44g$
D.$72g$

Answer 1

To answer the question, first find the mass of the $C{O_2}$ , then find the number of oxygen atoms that are present in the $C{O_2}$ molecules and the moles of oxygen. This approach will help you to reach the answer.

Complete answer:
Let us start by the mass of $C{O_2}$= $44g$
In a molecule of $C{O_2}$ the number of oxygen present is = $2$
This means that there are $2mol$ of oxygen present in $44g$ of $C{O_2}$
Here the given quantity is $22g$ of $C{O_2}$
therefore in $22g$ of $C{O_2}$ number of $moles$ of oxygen is = $\dfrac{2}{{44}} \times 22 = 1mol$
$1mol$ $CO$= $28g$
Therefore$28g$ of $CO$ contains $1mol$oxygen atom
And hence the correct answer is option A.

Additional information: One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly $12g$ of the $^{12}C$ isotopes. It is similar to the quantities as we denote one dozen or a gross for $12$ or $144$ items. The S.I. unit is $mole$ and the symbol for mole is $mol$. The mole of a substance always contains the same number of entities, no matter what the substance may be. This is represented by which is equal to $6.022 \times {10^{23}}atoms/mol$
Also the molar mass is the mass of one mole of a substance in $grams$. The molar mass in $grams$ is numerically equal to atomic or molecular formula mass in $u$
The number of moles is given by the formula-
$No.{\text{ }}of{\text{ }}moles = \dfrac{{Given{\text{ mass}}}}{{Molar{\text{ mass}}}}$

Note:
Atoms and molecules are extremely small in size and their numbers in even a small amount of any substance is really very large. To handle such large numbers, a unit of similar magnitude is required. Therefore the mole concept was introduced.