Question

Calculate the number of atoms $ C,H $ and $ O $ in $ 72.5g $ of iso-propanol ( $ CH,OH $ ) having molar mass $ 60g $ mol ?

Answer 1

Hint: To solve the given question, we should have information regarding mole concept, atomicity and avagadro’s number. Mole is the amount of substance which contains present in $ 12gm $ of $ C-12 $ isotope of carbon.
Atomicity is the total number of elements present in a molecule or composed of a molecule. Avagadro’s number is the number of ions, atoms or molecules present in a mole of a substance. It has a constant value of $ 6.022 \times 10^{22} $ .
The formula for number of moles is :
Total-mass $= \dfrac{\text{Given mass}}{\text{Molar mass}} $
The other formula for the same is the number of moles and the number of molecules by Avagadro’s number denoted by $ N_A $ .

Complete answer:
Step-1 :
We have given $ 72.5g $ of iso-propanol ($ C_3H_7OH $) and molar mass of iso-propanol is given to be $ 60g $ mol. So, using the given formula, moles can be calculated as :
$ moles = \dfrac{\text{Given mass}}{\text{Molar mass}} $
$ moles = \dfrac{72.5}{60} = 1.2 $
So, we have $ 1.2 $ moles of iso-propanol.

Step-2 :
We also have another formula for number of moles :
$ moles = \dfrac{\text{Total molecules}}{N_A} $
To calculate the number of atoms, multiplicity the number of molecules by atomicity.
For $ C $ atoms we have, number of atoms :
$ 3 \times N_A \times 1.2 = 2.128 \times 10^{ 24 } $
For $ H $ atoms we have, number of atoms :
$ 8 \times N_A \times 1.2 = 5.82 \times 10^{ 24 } $
For $ O $ atoms we have, number of atoms :
$ 1 \times N_A \times 1.2 = 7.2 \times 10^{ 23 } $

Note:
We should know that a molecule is made up of a group of atoms considered as elements or compounds. To calculate the number of atoms, the number of molecules is multiplied by atomicity, i.e. atoms present in a molecule.