Question

How many grams of Cl are in 38.0 g of each sample of chlorofluorocarbons (CFCs)?

Answer 1

Without the specific chemical formula of the chlorofluorocarbon (CFC), we cannot determine the precise mass of chlorine in a 38.0 g sample. If we had the formula, we would use the molar mass and stoichiometry to convert the given mass of CFC to grams of Cl.

Step-by-step explanation:

To determine how many grams of chlorine (Cl) are in a 38.0 g sample of a chlorofluorocarbon (CFC), we first need to know the chemical formula of the CFC to find the molar mass and the amount of Cl it contains. However, without the specific formula, we cannot provide an exact value. If we had the formula, we would calculate the number of moles of CFC using its molar mass, then use the mole ratio from the balanced chemical equation to find the moles of Cl present. Finally, using the molar mass of Cl (approximately 35.45 g/mol for 35Cl, 37.0 g/mol for 37Cl), we would convert the moles of Cl to grams.

For instance, if the CFC was CCl2F2, we would calculate the moles of CCl2F2 in the 38.0 g sample and then calculate the moles and mass of Cl within that sample.

Answer 2

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