Question

A mixture of CrBr3 and inert material is analyzed to determine the Cr content. First the mixture is dissolved in water. Then all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. CrBr3(aq) + 3AgNO3(aq) 3AgBr(s) + Cr(NO3)3(aq) In one experiment, a 0.8409 g sample of the mixture resulted in 1.0638 g of AgBr. Determine the percent (by mass) of Cr in the mixture.

Answer 1

Answer:

Mass% Cr = 85.5%

Explanation:

Given:

Mass of CrBr3 sample = 0.8409 g

Mass of the AgBr precipitate = 1.0638 g

To determine:

The mass percent of Cr in the sample

Calculation:

The reaction of CrBr3 with silver nitrate results in the precipitation of the bromide ion as silver chloride (AgBr) and Cr as soluble Cr(NO3)2

CrBr3(aq) + 3AgNO3(aq)→ 3AgBr(s) + Cr(NO3)3(aq)

Molecular weight of AgBr =187.77 g/mol

Moles of AgBr precipitated is:

$Moles(AgBr)=\frac{Mass(AgBr)}{Mol.wt(AgBr)}=\frac{0.8409g}{187.77g/mol}=0.004478moles$

Since 1 mole of AgBr contains 1 mole of Cl, therefore:

# moles of Cl = 0.004478 moles

At wt of Cl = 35.45 g/mol

$Mass(Chloride)=moles*at.wt = .004478moles*34.45g/mol=0.1543$

$Mass%(chloride)=\frac{mass(chloride)}{mass(sample)}*100=\frac{0.1543}{1.0638}*100 = 14.50%$

Mass%(Cr) = 100 - 14.50=85.5%