Final answer:
To calculate the percentage of bromine in the molar mass of an ionic compound based on the mass of a precipitate formed, you must find the moles of the bromide ions in the precipitate and use this to calculate the mass of bromine, which is then divided by the initial mass of the compound and multiplied by 100 to give the percentage.
Step-by-step explanation:
The question is asking to calculate the percentage of bromine in the ionic compound's molar mass based on the mass of the precipitate AgBr (silver bromide) formed when the compound is treated with AgNO₃ (silver nitrate). To do this, we can use the mass of the precipitate formed and the stoichiometry of the reaction between the bromide ions and AgNO₃. Since one mole of bromide ions reacts with one mole of AgNO₃ to form one mole of AgBr, the molar mass of AgBr can be used to find the moles of bromide in the precipitate.
Firstly, we calculate the moles of AgBr:
Moles of AgBr =
Mass of AgBr / Molar mass of AgBr
= 0.6964 g / 187.77 g/mol (molar mass of AgBr)
Next, we can find the moles of bromide ions, which will be the same since the reaction ratio is 1:1, and then find the mass of bromide ions by multiplying with the atomic mass of bromine (79.904 g/mol).
Mass of bromide ions
= Moles of bromide ions x Atomic mass of Br
Finally, to compute the percentage of bromine by mass in the original compound, we divide the mass of bromine calculated by the initial mass of the ionic compound and multiply by 100:
Percentage of bromine = (Mass of bromide ions / Mass of ionic compound) x 100