gravimetric analysis

1)Calculate the titer in mg/ mL of As2O3 of a 0.100N solution of I2: HAsO2 + I2 + 2H2O = H3AsO4 + 2I- + 2H+ (convert muna natin molarity, nd ako sanay normality eh *wink*)

N = M*f

M = 0.100N I/2 = .050M I

To get the titer 0.050mol/L I x [(1mol HAsO2/1mol I x 1mol As2o3/ 2mol HAsO2 x 197.841g/1mol As2O3 x 1000mg/ 1g)]

= 4946.025mg As2O3/L I x (1L / 1000 mL)

Titer 4.946 mg As2O3/ mL I

2)A student analyzed a limestone sample but forgot to dry it. He found 31.20% CaO, whereas the correct value was 31.60. If the error was caused by the failure to dry the sample, what % moisture did the sample contain?

Soln: assume 1g of the sample

.3120 * 1.000g = .3120 g CaO

Determine the corrected weight of the supposed to be dried sample (x)

.3120g / x = .3160

X = .3120/.3160 = 0.9873 g dried sample

Then get the % moisture in the sample

(1.0-0.9873)/1 = 1.27% moisture

3)A 0.5000g sample that contains KCl and KI gives the precipitate of AgCl and AgI that weighs 0.2818g. A 0.8000g sample is titrated with 0.1053M AgNO3 , requiring 26.74 mL. Calc. the percentages of KCl and KI in the sample. (AgNO3 = 169.873; KCl 74.551; KI = 166.003; AgCl = 143.321; AgI = 234.773)

Soln

0.2818g = wt AgCl + wt AgI

(0.1053 * 0.02674) *107.8682 = wt silver = 0.3037 g Ag

Ratio and proportion

A .500 g sample would require 16.7125 mL of 0.1053 M AgNO3 = 0.001760 mol Ag

.001760 mol Ag * 107.8682g/mol Ag = .1898 g Ag

Let X = AgCl and Y = AgI

0.1898g Ag = .753X + .459Y

X = (0.1898 - 459Y)/ .753 = 0.2520 – 0.610 Y

Substitute X

0.2818 = X + Y = 0.2520 – 0.610 Y +Y

0.2818 – 0.2520 = Y – 0.610Y

0.0298 = 0.39 Y

AgI = Y = .07641 g

AgCl = .2818-.07641 = 0.2071 g AgCl

.2071g AgCl x 74.551 gmol^-1 KCl/143.321gmol^-1 AgCl = 0.1077g KCl

0.1077 KCl/0.500 g = 21.54%

.07641g Agl x 166.003 gmol^-1 Kl/234.773gmol^-1 Agl = 0.05402g Kl

0.5402 Agl/0.500 g = 10.81%

4)A fish meal sample has the following composition: protein 20%, moisture 12%, additive 66%. The sample is placed in a drying oven and subsequent analysis gave a value of 22% protein. Calculate the moisture content of the dried sample.

Soln. Assume that the sample is 1g

Initially protein constitute 0.20g of the sample and 0.12g of the moisture

After drying the protein constitute 22% of the total weight.

0.20g/ wt of the dried sample (x) = 22% protein

X = 0.20g/0.22 = .9091g sample

Wt received sample – wt dried = wt of the water removed

1.000g – 0.9091g = 0.0909g moisture

% moisture in the dried sample is (.14g- 0.0909g)/0.9091g = 5.40 %moisture

5)1.000g of soil, as received, gave a moisture content of 14%. The oven- dried sample, completely moisture free showed 18.00% K. Find the percentage of K in the sample as received.

Soln: First get the weight of the dried sample:

Wt sample –(% moisture* wt sample) = wt sample

1.000g – (1.000g*0.14) = 0.8600g moisture

Then get the weight of K in the dried sample

.1800 = wt of K / wt dried sample

X = .18*.86= .1548g K

Then get the weight in received sample

.1548/1.000g * 100 = 15.48% K