Question

calculate the number of lithium ions, sulfate ions, S atoms, and O atoms in 73.2g of lithium sulfate. (a) lithium ions (b) sulfate ions (c) S atoms (d) O atoms

Answer 1

you need the following data for this excersise :

Li = 7; S = 32; O = 16

The compound is Li₂SO₄

M (Li₂SO4) = 7.2 + 32.1 + 16.4

M = 110 g which means there is 1 mol of Li₂SO₄ in 110 g of Lithium sulfate. You can use cross multiplication to solve this:

1 mol Li₂SO₄ ---------------110 g

X -------------------------------73.2 g

X = (1 x 73.2) / 110

X = 0.66 mol of Li₂SO4

a) The number of lithium ions is 1.32 mol or 7.94 x 10²³ since there are 2 mol of Li+ ions for every mol of Lithium sulfate (multiply 0.66 by 2)

b) The number of sulfate ions is 0.66 mol or 3.97 x 10²³, since for every mol of Li₂SO₄ there is also a mol of sulfate ion (SO₄^2-)

c) Same answer as (b), the number of S atoms equals the number of sulfate ions

d) Simply multiply 4 by 0.66 ------ 2.64 mol or 1.59 x 10²⁴