Question

How many sodium atoms are there in 6.0 g of Na₃N?

Answer 1

I am doing this problem by two methods:

1)

Molecular weight for Na3N = 83 g/mol
6.0g /( 83g/mol) = 0.0723 mol Na3N

0.0723 mol Na3N * 3mol Na/1mol Na3N = 0.217 mol Na

one mole contain= 6.022*10²³ atoms (avagadro number)
0.217 mol * 6.022*10^23 atoms/mol = 1.306*10^23 sodium atoms in 6 grams of Na3N

2)

%Na in Na3N = (23x3)gmol^-1 / [(23X3) + 14]gmol^-1 = 83.1325301%

in 6g --> mass of nitrogen = (83.1325301/100) x 6g = 4.987951807g

n=m/M
n=moles
m=mass(g)
M=molar mass(gmol^-1)

n(N) = 4.987951807g / 23gmol^-1 = 0.21686747 moles

1 mole of nitrogen atoms = 6.022x10^23 nitrogen atoms
0.21686747 moles = 0.21686747 moles x 6.022x10^23
= 1.30598 x 10^23 nitrogen atoms

thus, answer = c. 1.3 × 10^23 atoms