Final answer:
To determine the atomic mass of the metal in the metal chloride, we calculate the molar mass from the vapor density and subtract the mass of chlorine, based on its isotopes, from it. We verify this using the percentage of the metal in its oxide form.
Step-by-step explanation:
To find the atomic mass of the metal in a metal chloride with a vapor density of 66 and wherein the oxide contains 53% metal, we use the concept of vapor density and the percentage composition of the metal oxide. The vapor density (VD) is half the molar mass, so the molar mass (MM) of the metal chloride is VD × 2, this gives us MM = 66 × 2 = 132 g/mol. Next, we must consider the composition of chlorine in the compound.
From the periodic table, we have two main isotopes of chlorine, 35Cl and 37Cl, with atomic masses 34.969 amu and 36.966 amu, and percent abundances 75.77% and 24.23%, respectively. The average atomic mass of chlorine, considering its isotopes, can be calculated as follows:
(34.969 amu × 0.7577) + (36.966 amu × 0.2423) = 35.45 amu approximately
Since metal chloride would generally have the formula MCl2, where M is the metal, the mass of chlorine in the compound is 2 × 35.45 g = 70.90 g. The remaining mass would then be that of the metal (MM of MCl2 - MM of 2Cl = MM of Metal).
Therefore, 132 g/mol (MCl2) - 70.90 g/mol (2× Cl) = 61.10 g/mol, which is the atomic mass of the metal. However, to ensure this aligns with the oxide composition, we verify using the given oxide contains 53% metal by mass. The 61.10 g/mol should be 53% of the metal oxide's molar mass, confirming our calculation.