A sample of 0, 3220g of an ionic compound containing bromide ions (Br-) is dissolved in water and treated with excess of AgNO3. If the mass of the precipitate
of AgBr formed is 0, 6964g, answer: a) what percentage of bromine in molar mass? b) Give the mass of bromine in the compound formed. c) Which the percentage by mass of bromine in the original compound?
To find the answers to these questions, we'll need to use stoichiometry and some basic calculations.
a) To find the percentage of bromine in molar mass, we first need to determine the molar mass of AgBr. We know that Ag has a molar mass of 107.87 g/mol and Br has a molar mass of 79.90 g/mol.
The molar mass of AgBr is:
Molar mass (AgBr) = molar mass (Ag) + molar mass (Br)
Molar mass (AgBr) = 107.87 g/mol + 79.90 g/mol
Molar mass (AgBr) = 187.77 g/mol
Next, we can calculate the percentage of bromine in the molar mass using the formula:
Percentage of bromine in molar mass = (molar mass of Br / molar mass of AgBr) x 100
Percentage of bromine in molar mass = (79.90 g/mol / 187.77 g/mol) x 100
Percentage of bromine in molar mass ≈ 42.58%
b) To find the mass of bromine in the compound formed (AgBr), we can use the mass of the precipitate formed (0.6964g) and the molar mass of AgBr.
Number of moles of AgBr = mass of AgBr / molar mass of AgBr
Number of moles of AgBr = 0.6964g / 187.77 g/mol
Now, since AgBr has a 1:1 stoichiometric ratio with Br, the molar mass of bromine is also its atomic weight, so we can find the mass of bromine using the number of moles obtained earlier.
Mass of bromine in the compound = number of moles of AgBr x molar mass of Br
Mass of bromine in the compound = (0.6964g / 187.77 g/mol) x 79.90 g/mol
c) Finally, to find the percentage by mass of bromine in the original compound, we can use the mass of bromine in the compound (from part b) and the mass of the original compound (3220g).
Percentage by mass of bromine in the original compound = (mass of bromine in the compound / mass of the original compound) x 100
Percentage by mass of bromine in the original compound = [(0.6964g / 187.77 g/mol) x 79.90 g/mol] / 3220 g x 100
Please plug in the values and calculate the final answer.
To answer these questions, we need to consider the chemical reactions taking place and the stoichiometry involved. Here's how to find the answers step by step:
Step 1: Determine the balanced chemical equation for the reaction between bromide ions (Br-) and silver nitrate (AgNO3) to form silver bromide (AgBr) precipitate:
2AgNO3 + Br- → 2AgBr + 2NO3-
This equation shows that two moles of silver nitrate react with one mole of bromide ions to give two moles of silver bromide.
Step 2: Calculate molar mass:
The molar mass of silver bromide (AgBr) can be calculated by adding the atomic masses of silver (Ag) and bromine (Br):
Ag: 107.87 g/mol
Br: 79.90 g/mol
AgBr: 107.87 g/mol + 79.90 g/mol = 187.77 g/mol
Step 3: Calculate the number of moles of AgBr formed:
Given the mass of precipitate AgBr as 0.6964 g, we can use the molar mass calculated in Step 2 to determine the moles of AgBr formed:
moles of AgBr = mass of AgBr / molar mass of AgBr
moles of AgBr = 0.6964 g / 187.77 g/mol
Step 4: Calculate the moles of bromine (Br) in the compound formed:
From the balanced equation, we know that one mole of AgBr contains one mole of bromide ions (Br-). Therefore, the moles of bromine in the compound formed is equal to the moles of AgBr.
moles of Br = moles of AgBr
Step 5: Calculate the mass of bromine (Br) in the compound formed:
mass of Br = moles of Br x molar mass of Br
mass of Br = moles of AgBr x molar mass of Br
Step 6: Determine the percentage by mass of bromine in the compound formed:
The percentage by mass is given by:
% by mass of Br = (mass of Br / mass of compound formed) x 100
Now, let's calculate each part of the question using the given values.
a) What percentage of bromine in molar mass?
To find the percentage of bromine in molar mass, we need to calculate the molar mass of bromine in the compound formed:
molar mass of Br = 79.90 g/mol
Therefore, the percentage of bromine in molar mass is 79.90%.
b) Mass of bromine in the compound formed:
This is calculated using the moles of AgBr determined:
mass of Br = moles of Br x molar mass of Br
c) Percentage by mass of bromine in the original compound:
To calculate the percentage by mass of bromine in the original compound, we need to compare the mass of bromine to the mass of the original compound:
% by mass of Br = (mass of Br / mass of original compound) x 100
Unfortunately, the mass of the original compound is not provided in the given information, so we cannot calculate this value without additional data.
I hope this helps! Let me know if you have any further questions.
a. I have no idea what you're talking about.
b. AgBr is compound formed.
mass Br (not Br2) in the AgBr is
0.6964 x (atomic mass Br/molar mass AgBr) = ?
c.
%Br (not Br2) in compound is
%Br = [(answer b)/mass sample)]*100 = ?