Question

What is the total number of atoms in 4.65 g of aniline (C6H5NH2)?

Answer 1

Molar mass of C6H5NH2,
MM = 6*MM(C) + 7*MM(H) + 1*MM(N)
= 6*12.01 + 7*1.008 + 1*14.01
= 93.126 g/mol


mass(C6H5NH2)= 4.65 g

use:
number of mol of C6H5NH2,
n = mass of C6H5NH2/molar mass of C6H5NH2
=(4.65 g)/(93.126 g/mol)
= 4.993*10^-2 mol
use:
number of molecules = number of mol * Avogadro’s number
number of molecules = 4.993*10^-2 * 6.022*10^23 molecules
number of molecules = 3.007*10^22 molecules

Count of each element in compound is:C=6, H=7, N=1
Total number of atoms in 1 molecule = 14
use:
number of atoms = number of molecules * Total number of atoms in 1 molecule
number of atoms = 3.007*10^22 * 14
number of atoms = 4.21*10^23
Answer: 4.21*10^23