Name: Date Checked: Determination of Calcum as Calcum icium as Calcium Oxalate R

Question

Name: Date Checked: Determination of Calcum as Calcum icium as Calcium Oxalate Read the experiment, and then answer the following questions. Show all calculations; attach additional pages as needed For reference, see Chapter 7: "Gravimetric and Combustion Analysis". Exploring Chemical Analysis by Daniel C. Harris, Fifth Edition or look up gravimetry in any analytical chemistry textbook. 1. Wha t is a gravimetric analysis? Is it a qualitative or quantitative technique? 2. What role does urea play in this experiment? 3. Why is oxalate ion used in this experiment? experiment. If the mass of CaC,0 H:O obtained was 0.6894 g. calculate the molarity of calcium in the unknown solution. 4. A 25.00-mL aliquot of unknown calcium chloride was analyzed by the method given in this 40

Explanation / Answer

#1.

Gravimetric analysis is an analytical technique in chemistry which uses sets of methods for determination of the analyte based on the mass of solid formed during the experimeriment. It is a quantitative technique.

#2

Calcium can be precipitated as calcium oxalate by treating with ammonium oxalate but the precipitate is soluble in acidic conditions.In order to have pure, large crystals of calcium oxalate, precipitation is carried out slowly in acidic conditions which gives soluble precipitate & then the pH of solution is raised by adding urea & heating the reaction mixture

Upon heating urea decomposes as (NH2)2CO + 3H2O + Heat ------> CO2 + 2NH4^+ + 2 OH-

presence of OH- makes the solution basic & facilitates precipitation.

#3

Oxalate ion is used in this experiment to precipitate out Calcium ions from the given analyte.

The reaction is between oxalate ion & calcium ion is given as follows.

Ca^2+ (aq) + C2O4^2- (aq) + H2O ---------> CaC2O4.H2O (s)

#4

We have got 0.6894 g CaC2O4.H2O. let us calculate number of moles for it.

Molar mass of CaC2O4.H2O is 146.1

Moles of CaC2O4.H2O = 0.6894/146.1 = 0.00472 mol

Acording to the above reaction, 1 mol Ca^2+ ion react with oxalte ion to form 1 mol CaC2O4.H2O

So 0.00472 mol Ca^2+ would form 0.00472 mol CaC2O4.H2O

So, moles of Ca^2+ are 0.00472

We used 25 mL of calcium solution which is 0.025 L

Molarity, M = moles /L = 0.00472/0.025 = 0.1888 M

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