How many atoms are in 5.9g of lithium carbonate, Li2(CO3)?

The number of atoms = 4.8 x 10²² atoms

Given

5.9 g of Lithium carbonate

Required

The number of atoms

Solution

mol Lithium carbonate(MW=73.9 g/mol) :

= mass : MW

= 5.9 g : 73.9 g/mol

= 0.08

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

So for 0.08 mol :

= 0.08 x 6.02.10²³

= 4.8 x 10²² atoms