Question

How many moles of O atoms are present in 88 gm${\text{C}}{{\text{O}}_{\text{2}}}$?

Answer 1

Find the molecular weight of ${\text{C}}{{\text{O}}_{\text{2}}}$ first, which is 44 gm. Therefore 44 gm ${\text{C}}{{\text{O}}_{\text{2}}}$ = 1 mol of ${\text{C}}{{\text{O}}_{\text{2}}}$. Now, 1 mol $C0_2$ contains 2 moles of oxygen atoms. Therefore 88 gm of $C0_2$ must contain 4 moles of oxygen atoms. Note that one mole of any substance contains $6.022 \times {10^{23}}$ atoms. Hence find the number of atoms present in 4 moles.

Formula used: 1 mol = $6.022 \times {10^{23}}$

Complete step by step answer:
1 mol of ${\text{C}}{{\text{O}}_{\text{2}}}$ has a molecular weight of = 12 + 32 = 44 gm
Number of oxygen atoms present in 1 mol of ${\text{C}}{{\text{O}}_{\text{2}}}$ is 2
Therefore, 44 gm of ${\text{C}}{{\text{O}}_{\text{2}}}$ contains 2 moles of oxygen atoms.
88 gm of ${\text{C}}{{\text{O}}_{\text{2}}}$ contains 2×2=4 moles of oxygen atoms.
Now, 1 mol atoms =$6.022 \times {10^{23}}$ atoms
Therefore, 4 moles of atoms = $4 \times 6.022 \times {10^{23}}$=$24.088 \times {10^{23}}$ atoms

So, 88 gm of $C0_2$ contains $24.088 \times {10^{23}}$ oxygen atoms.

Note: Note that the gram atomic weight of Carbon and Oxygen are 12 gm and 16 gm respectively.
Also, the Avogadro number is $6.022 \times {10^{23}}$. Any substance of 1mol contains $6.022 \times {10^{23}}$ atoms in it.