How many grams of cl are in 38.0 g of each sample of chlorofluorocarbons (cfcs)?

If there are 38 g of CFC, then there will be 120.9135 g of CFCl2 per mol. You will multiply this times the number of moles of Cl (2) for every mole of CFCl2, and then by the number of grams of Cl per mole, which is 35.4532: (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 Next, if there are 38 g of CFC, there will be 137.3861 g of CFCl3 per mole. You will multiply this times the number of moles of Cl (3 this time) for every mole of CFCl3. You will then multiply this by 35.4532 again: (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 Continue following these steps until you are able to multiply 1 mole of Cl per 1 mol CF3Cl by 35.4532: (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

Ithink the answer is a or d.