Final answer:
The amount of carbon in an 83.5 g sample of carbon tetrachloride is approximately 6.55 g, computed using the molar masses of carbon and carbon tetrachloride; hence none of the provided options are correct.
Step-by-step explanation:
To determine how many grams of carbon are present in a 83.5 g sample of carbon tetrachloride (CCl4), we first need to know the molar mass of carbon tetrachloride and the molar mass of carbon.
The periodic table lists the atomic mass of carbon as approximately 12.011 g/mol, and the atomic mass of chlorine as approximately 35.453 g/mol. Carbon tetrachloride consists of one carbon and four chlorine atoms, so its molar mass is:
- (1 × 12.011 g/mol) + (4 × 35.453 g/mol) = 12.011 g/mol + 141.812 g/mol = 153.823 g/mol
To find out how much carbon is in the 83.5 g sample, we use the ratio of the mass of carbon in one mole of carbon tetrachloride to the molar mass of carbon tetrachloride:
(12.011 g C / 153.823 g CCl4) × 83.5 g = 6.553 g of carbon (rounded to three significant figures).
Therefore, the amount of carbon in the 83.5 g sample of carbon tetrachloride is closer to 6.55 g, so none of the given options (A, B, C, D) are correct.