Question

A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide was precipitated as silver bromide, AgBr. The mass of the AgBr(s) was found to be 23.0052 grams. What is the identity of the unknown metal? Justify your answer with calculations.

Answer 1

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

$Moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}$

$Moles\ of\ AgBr= 0.1225\ mol$

The reaction taking place is:

$MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4$

From the reaction,

4 moles of AgBr is produced when 1 mole of $MBr_4$ undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of $MBr_4$ undergoes reaction

0.1225 mole of AgBr is produced when $\frac {1}{4}\times 0.1225$ mole of $MBr_4$ undergoes reaction

Moles of $MBr_4$ got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of $MBr_4$ = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

$0.030625 = \frac{12.5843}{319.616 + x}$

Solve for x,

we get, x = 91.2999 g/mol

The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.