Analytical Gravimetric

Determination

Suh Kwon

Purpose• To measure the number of a To measure the number of a

given substance in a solution given substance in a solution by precipitation, filtration, by precipitation, filtration, drying, and weighingdrying, and weighing– Approximating the amount of Approximating the amount of SOSO44

2-2- ions in a sample of alum, ions in a sample of alum, KAl(SOKAl(SO44))22· · 12H12H2200

Background Information• Gravimetric analysis Gravimetric analysis is used to is used to

determine the amount of a determine the amount of a substance by finding its mass, substance by finding its mass, and then using the mass to and then using the mass to find the quantity of the find the quantity of the substancesubstance– Example: to find the measurement of Example: to find the measurement of

solids suspended in a water sample, a solids suspended in a water sample, a known volume of water is filtered and known volume of water is filtered and the collected solids are weighedthe collected solids are weighed

For this experiment…• The precipitate that will be formed in The precipitate that will be formed in

this experiment is barium sulfate, this experiment is barium sulfate, BaSOBaSO44

– Forms very fine crystalsForms very fine crystals

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• Beforehand, Beforehand, calculate how much calculate how much 0.200 M Ba(NO0.200 M Ba(NO33))22 you you would need to totally would need to totally precipitate all of the precipitate all of the sulfate ion present in sulfate ion present in solutionsolution

!!!• Make sure you add twice the Make sure you add twice the

calculated amount of calculated amount of Ba(NOBa(NO33))2 2 because for every because for every 11 mole of alum, mole of alum, there are there are 22 moles of moles of Ba(NOBa(NO33))22

KAl(SOKAl(SO44))22·· 12H12H22O + 2Ba(NOO + 2Ba(NO33))2 2

KNO KNO33 + Al(NO + Al(NO33)) 3 3 + 2BaSO + 2BaSO44 + 12H + 12H22OO

Materials• Buchner Funnel Buchner Funnel • A piece of Whatman No. 42 filter paperA piece of Whatman No. 42 filter paper• Filter flaskFilter flask• 1.059 g of alum compounds1.059 g of alum compounds• 50.0 mL of distilled water50.0 mL of distilled water• 22.3 mL (calculated) of 0.200 M 22.3 mL (calculated) of 0.200 M

Ba(NOBa(NO33))22

• Heating PlateHeating Plate• Analytical BalanceAnalytical Balance• Drying oven set at 50°CDrying oven set at 50°C

Filter Flask

Filter Paper

Analytical Balance

Buchner Funnel

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An analytical balance is an instrument that's used to measure mass at a very high degree of precision.

Procedures1.1. Dissolve the 1.059 g of alum in the Dissolve the 1.059 g of alum in the

50.0 mL of distilled water50.0 mL of distilled water

2.2. Add twice the amount of 0.200 M Add twice the amount of 0.200 M Ba(NOBa(NO33))2 2 to alum solution, stirring to alum solution, stirring constantlyconstantly

3.3. Heat this solution for 15 minutes and Heat this solution for 15 minutes and allow the solution to stand overnightallow the solution to stand overnight

Next Day: You will discover that fine Next Day: You will discover that fine crystals had appeared in the solutioncrystals had appeared in the solution

Procedures (2)4. Weigh the filter paper separately Weigh the filter paper separately

and determine its mass on the and determine its mass on the analytical balanceanalytical balance

5. Filter the solution through the 5. Filter the solution through the Buchner funnel, containing the Buchner funnel, containing the filter paper to collect the crystalsfilter paper to collect the crystals

6. Remove the filter paper and allow 6. Remove the filter paper and allow it to dry in the 50°C drying ovenit to dry in the 50°C drying oven

7. Weigh the paper and crystals 7. Weigh the paper and crystals together and determine their total together and determine their total massmass

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DataINITIALINITIAL• Mass of alum: 1.059 gMass of alum: 1.059 g• Volume of distilled water: 50.0 mLVolume of distilled water: 50.0 mL

• Amount of 0.200 M Ba(NOAmount of 0.200 M Ba(NO33))2 2 : 22.3 mL: 22.3 mL

COLLECTEDCOLLECTED• Mass of filter paper (alone): 1.675 gMass of filter paper (alone): 1.675 g• Mass of filter paper + filtered Mass of filter paper + filtered

crystals: 2.715 gcrystals: 2.715 g

Equations• Balanced Chemical Formula:Balanced Chemical Formula:

*** KAl(SO*** KAl(SO44))22· 12H· 12H22O + 2Ba(NOO + 2Ba(NO33))2 2

KNO KNO33 + Al(NO + Al(NO33)) 3 3 + 2BaSO + 2BaSO44 + 12H + 12H22OO

*** Ba*** Ba2+2+ + SO + SO442-2- BaSO BaSO4 (s)4 (s)

• % Sulfate = (Mass of Sulfate/Mass of Sample) * 100%% Sulfate = (Mass of Sulfate/Mass of Sample) * 100%

• Percent Error = (Theoretical Value – Actual Value) * Percent Error = (Theoretical Value – Actual Value) * 100%100%

Theoretical Value Theoretical Value

Calculations• How much 0.200 M Ba(NOHow much 0.200 M Ba(NO33))2 2 would be would be

needed to totally precipitate all of the needed to totally precipitate all of the sulfate ion present in the alum solution:sulfate ion present in the alum solution:

1.059 1.059 g alum g alum · 1 · 1 mol of alum mol of alum · 2 · 2 mol Ba(NOmol Ba(NO33))2 2 · 1 · 1 L of Ba(NOL of Ba(NO33))2 2 · 1000 · 1000 mL = 22.3 mLmL = 22.3 mL

474.46 474.46 g alum g alum 1 1 mol of alum mol of alum 0.200 0.200 mol Ba(NOmol Ba(NO33))22 11 L L

• The percent of sulfate ion in the alum The percent of sulfate ion in the alum based upon the experiment:based upon the experiment:

2.715 g filter paper + barium sulfate2.715 g filter paper + barium sulfate- 1.675 g paper1.675 g paper

1.040 g BaSO1.040 g BaSO44

1.040 1.040 g BaSOg BaSO44 · 1 · 1 mol BaSOmol BaSO44· 1 · 1 mol SOmol SO442-2-· 96.04 g SO· 96.04 g SO44

2- 2- = = 0.4281 g SO0.4281 g SO44

2-2-

233.37 233.37 g BaSOg BaSO4 4 1 1 mol BaSOmol BaSO44 1 1 mol SOmol SO442-2-

% sulfate = 0.4281 g SO% sulfate = 0.4281 g SO442-2- · 100% = 40.43% · 100% = 40.43%

1.059 g alum1.059 g alum

1

2

3

• The theoretical % of sulfate ion in the The theoretical % of sulfate ion in the alum:alum:

192.08 g192.08 g SO SO442- 2- · 100% = 40.48 %· 100% = 40.48 %

474.46 g alum474.46 g alum

• The percent error in this investigation:The percent error in this investigation:

% error = 40.43 – 40.48 · 100% = -% error = 40.43 – 40.48 · 100% = -0.15%0.15%

40.4840.48

Conclusion• With the help of this experiment, With the help of this experiment,

you were able to find the amount of you were able to find the amount of sulfate ions in a sample of alum sulfate ions in a sample of alum compound, comparing that value to compound, comparing that value to the theoretical amount necessarythe theoretical amount necessary

• A negative percent error signifies A negative percent error signifies that there was a production of a that there was a production of a solid throughout the reactionsolid throughout the reaction