1. A 12.5843g sample of zrbr4 was dissolved and after several steps all of the combined bromine was precipitated as ArBr. The silver content of AgBr was found to be 13.12160g Assume the atomic masses of silver and bromine to be 107.868 and 79.904.what value was obtained for the atomic masses of Zr from the experiment?

This is almost the same as the MO2/M2S3 problem but I would convert 12.13160 g Ag to the mass of AgBr. Post your work if you get stuck.

To find the atomic mass of Zr from the experiment, we need to use stoichiometry and the atomic mass of bromine.

Step 1: Calculate the amount of bromine in the form of AgBr.
The molar mass of AgBr can be calculated as follows:
Atomic mass of Ag = 107.868 g/mol
Atomic mass of Br = 79.904 g/mol
Therefore, the molar mass of AgBr = (107.868 + 79.904) g/mol = 187.772 g/mol

Using the molar mass of AgBr, we can calculate the number of moles of bromine present in 13.12160 g of AgBr:
Moles of bromine = Mass of AgBr / Molar mass of AgBr
Moles of bromine = 13.12160 g / 187.772 g/mol

Step 2: Calculate the number of moles of Zr in the sample.
Since the bromine in ZrBr4 is precipitated as AgBr, the number of moles of bromine in AgBr is equal to the number of moles of bromine in ZrBr4.

Therefore, the number of moles of Zr = Moles of bromine

Step 3: Calculate the atomic mass of Zr.
To calculate the atomic mass of Zr, we need the mass of Zr and the number of moles of Zr.

The mass of Zr can be calculated using the mass of the sample and the mass of bromine:
Mass of Zr = Mass of sample - Mass of bromine

Now we can calculate the atomic mass of Zr:
Atomic mass of Zr = Mass of Zr / Moles of Zr

Using the given data, we can now perform the calculations to find the atomic mass of Zr.