story Booikmarks Window Help ork 6 022518 0 This is a Chemic

story Booikmarks Window Help ork 6 02/25/18 0 This is a Chemical Equation question / It is worth 1 point / You have unlimited attempts / There is a 1% attempt penalty Question (1 point a See page 292 1st attempt l See Periodic Table See Hint 18.79% Li, 16.25% C. and 64.96% Q Spodumene is processed chemically to convert it to a material having the percent What is the empirical formula of this lithium compound? 07:20 O OF 20 QUESTIONS COMPLETED

Solution

consider 100 g of the compound then

mass of Li = 18.79 g

mass of C = 16.25 g

mass of O = 64.96 g

now

moles of Li = 18.79 g / 6.941 (g/mol) = 2.7071 mol

moles of C = 16.25 g / 12 (g/mol) = 1.3541 mol

moles of O = 64.96 g / 16 (g/mol) = 4.06 mol

now

consider the ratio of moles

Li : C : O = 2.7071 : 1.3541 : 4.06

Li : C : O = 2 : 1 : 3

so

the empirical formula is Li₂CO₃