How many grams of cl are in a 38.0-g sample of cfcl3

38.0 g CF2Cl2) / (120.9135 g CF2Cl2/mol) x (2 mol Cl / 1 mol CF2Cl2) x (35.4532 g Cl/mol) =

22.3 g Cl in CF2Cl2

(38.0 g CFCl3) / (137.3681 g CFCl3/mol) x (3 mol Cl / 1 mol CFCl3) x (35.4532 g Cl/mol) =

29.4 g Cl in CFCl3

(38.0 g C2F3Cl3) / (187.3756 g C2F3Cl3/mol) x (3 mol Cl / 1 mol C2F3Cl3) x (35.4532 g Cl/mol) =

21.6 g Cl in C2F3Cl3

(38.0 g CF3Cl) / (104.4589 g CF3Cl/mol) x (1 mol Cl / 1 mol CF3Cl) x (35.4532 g Cl/mol) =

12.9 g Cl in CF3Cl

29.4 g Cl in CFCl3

Explanation:(38.0 g CFCl3) / (137.3681 g CFCl3/mol) x (3 mol Cl / 1 mol CFCl3) x (35.4532 g Cl/mol) =

29.4 g Cl in CFCl3

1. they vary with a higher number because boiling is way hotter than melting