Question

1.) How many grams of B are there in a sample of B that contains 6.55×1023 atoms?

2.) A 19.35 gram sample of chromium is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 33.49 g. Determine the empirical formula of the metal fluoride. (Enter the elements in the order Cr , F)

3.) A 25.34 gram sample of chromium is heated in the presence of excess iodine. A metal iodide is formed with a mass of 210.9 g. Determine the empirical formula of the metal iodide. (Enter the elements in the order Cr , I)

Answer 1

1. molecular mass of B (boron) = 10g. Thus, 1 mole boron = 10g. Therefore, 6.022*10²³ atoms = 10g. Therefore, mass of 6.55*10²³ atoms = (6.55/6.022) * 10 = 10.87 grams.
2. molecular mass of chromium = 52g. So, 19.35g = 0.37 moles. mass of fluorine = mass_fluoride - mass_cr = 33.49-19.35 = 14.14g. molecular mass of fluorine = 19g Therefore, moles = 0.744g. Therefore, ratio of fluorine to chromuim = 2.011 = 2 (approx.). Therefore, empirical formula of compound = CrF₂.
3. molecular mass of chromium = 52g. So, 25.34g = 0.487. moles. mass of iodine = mass_iodide - mass_cr = 210.9-25.34 = 185.56g. molecular mass of iodine = 127g Therefore, moles = 1.45. Therefore, ratio of iodine to chromuim = 2.9 = 3 (approx.). Therefore, empirical formula of compound = CrI₃.