A 12.5843 g sample of ZrBr4 was dissolved and, after several chemical steps, all of the combined Bromine was precipitated as AgBr. The silver content of the AgBr was found to be 13.2160 g. Assume the atomic weights of silver and bromine to be 107.870 and 79.909 respectively. What value was obtained for the atomic weight of Zr for this experiment?

To find the atomic weight of Zr, we need to determine the moles of AgBr formed and the moles of bromine.

Step 1: Calculate the moles of AgBr
Molar mass of AgBr = atomic weight of Ag + atomic weight of Br
= 107.870 g/mol + 79.909 g/mol
= 187.779 g/mol

Moles of AgBr = mass of AgBr / molar mass of AgBr
= 13.2160 g / 187.779 g/mol
= 0.070271 mol

Step 2: Calculate the moles of bromine
Moles of Br = Moles of AgBr (since 1 mole of AgBr contains 1 mole of Br)
= 0.070271 mol

Step 3: Calculate the moles of Zr
Molar ratio ZrBr4 : Br = 1 : 4 (from the formula ZrBr4)
Moles of Zr = Moles of Br / 4
= 0.070271 mol / 4
= 0.017568 mol

Step 4: Calculate the atomic weight of Zr
Atomic weight of Zr = mass of Zr / moles of Zr
= 12.5843 g / 0.017568 mol
= 715.6857 g/mol

Therefore, the value obtained for the atomic weight of Zr for this experiment is approximately 715.6857 g/mol.

To determine the atomic weight of Zr (Zirconium) in this experiment, we can use the concept of stoichiometry and the law of conservation of mass.

First, we need to determine the moles of AgBr precipitated. We can do this by dividing the mass of AgBr by its molar mass. The molar mass of AgBr can be calculated by summing the atomic weights of silver (Ag) and bromine (Br):

Molar mass of AgBr = atomic weight of Ag + atomic weight of Br
= 107.870 g/mol + 79.909 g/mol
= 187.779 g/mol

Moles of AgBr = mass of AgBr / molar mass of AgBr
= 13.2160 g / 187.779 g/mol
= 0.0703 mol

Since ZrBr₄ is a 1:4 molar ratio with AgBr, we can infer that the moles of ZrBr₄ dissolved are four times the moles of AgBr precipitated:

Moles of ZrBr₄ = 4 * moles of AgBr
= 4 * 0.0703 mol
= 0.2812 mol

To calculate the atomic weight of Zr, we can divide the mass of ZrBr₄ dissolved by its moles:

Atomic weight of Zr = mass of ZrBr₄ / moles of ZrBr₄
= 12.5843 g / 0.2812 mol
= 44.7267 g/mol

Therefore, the value obtained for the atomic weight of Zr in this experiment is approximately 44.7267 g/mol.

107.870 + 79.909 = 187.779

%Ag = [(107.870 / 187.779](100) = 57.4452%
%Br = [(79.909/187.779)](100) = 42.5548%
(13.2160 g AgBr)(0.42.5548) = 5.624g Br in AgBr

[(5.624g Br)/(12.5843g ZrBr4)](100) = 44.691% Br in ZrBr4
100.000 - 44.691 = 55.309% Zr in ZrBr4
Let the atomic mass of Zr = x
The formula mass of ZrBr4 = x + (4)(79.909), or
ZrBr4 = x + 319.636

Set up the equation:
x / (x + 319.636) = 0.44691
Solve for x to get the atomic mass of Zr based on the information given here.