Question

To determine the percentage of bromide in a mixture containing a soluble bromide salt, the procedure from Experiment 10 is used except the bromide is precipitated with aqueous lead (II) nitrate, Pb(NO3)2, forming PbBr2(s). Initially, a 0.5014 g sample of an unknown bromide salt is dissolved in water. Excess Pb(NO3)2 is added to the solution to quantitatively precipitate the bromide from the solution. The resulting mixture is filtered through a piece of filter paper weighing 1.3258 g. After several days of drying, the filter paper and precipitate weighed 1.7895 g. Calculate the percent bromide in the original sample.

Answer 1

40.27% Bromide in the original sample

Step-by-step explanation:

The reaction of bromide ion, Br- with Pb(NO3)2 is:

2 Br- + Pb(NO3)2 → PbBr2(s)

Where the precipitate produced is PbBr2

To solve this question we must find the moles of PbBr2 filtered using its mass. Twice these moles we can find the moles of Br- and its mass. Percent bromide is:

Mass Bromide / Mass Sample (0.5014g) *100

Mass PbBr2:

1.7895g - 1.3258g = 0.4637g PbBr2

Moles -Molar mass: 367.01g/mol-

0.4637g PbBr2 * (1mol / 367.01g) =

0.001263 moles PbBr2 * (2mol Br / 1mol PbBr2) = 0.002527 moles Br

Mass Br -Molar mass: 79.904g/mol-

0.002527 moles Br * (79.904g / mol) =

0.2019g Br

Percent Bromide:

0.2019g Br / 0.5014g * 100 =

40.27% Bromide in the original sample