Final answer:
The percent by mass of chloride ion in the original alkali metal chloride is found using the mass of AgCl precipitate. The identity of the alkali metal chloride is determined by comparing the calculated percent by mass with theoretical values for known alkali metal chlorides.
Step-by-step explanation:
To determine the percent by mass of chloride ion in the original alkali metal chloride compound, we need to analyze the precipitate formed when the compound reacted with silver nitrate (AgNO3). The balanced equation for the reaction is:
AgNO3 (aq) + MCl (aq) → AgCl (s) + MNO3 (aq)
Where M represents the alkali metal. When an alkali metal chloride reacts with AgNO3, a precipitate of AgCl forms. The mass of the AgCl precipitate gives us information about the amount of chloride in the original sample because the stoichiometry of the reaction is 1:1.
The mass of AgCl formed is 3.03707 g from a 2.5624-g sample of alkali metal chloride. To calculate the percent by mass of chloride ion (Cl-), we need the formula:
Percent by Mass = (Mass of Cl- in AgCl / Mass of original sample) × 100%
The formula weight of AgCl is 143.4 g/mol, which comprises 35.45 g/mol of Cl- (from its atomic weight). Using the ratio of the mass of Cl- in AgCl to the molar mass of AgCl, we find:
Mass of Cl- in AgCl = (3.03707 g AgCl) × (35.45 g/mol Cl- / 143.4 g/mol AgCl)
This gives the mass of Cl- present in the precipitate. Plugging this value back into the percent by mass formula and comparing with the original sample mass of 2.5624 g, we can solve for the percentage of Cl-.
To determine the identity of the alkali metal chloride, we compare the percent by mass of Cl- with theoretical values for different possible alkali metal chlorides and choose the one that matches our calculated value.