Procedure Act 2 31254w

PROCEDURE Mix thoroughly in a mortar and pestle 0.5 gm of sugar and 1.0 gm of cupric oxide. Transfer the mixture to a big, dry hard glass test tube. Clamp to an iron stand in an almost horizontal position. Cover the test tube with a stopper to which is attached a bent glass tube, its end dipping into a small test tube containing 10 mL of clear limewater. HAVE YOUR SET-UP CHECKED BY YOUR INSTRUCTOR. Heat the mixture gently at first then strongly. Observe what forms in the limewater. Then remove first the delivery tub from the limewater before removing the burner. Result with limewater:

RESULTS EXPLANATION A. TEST FOR CARBON, HYDROGEN AND OXYGEN

(Set-up for the detection of Carbon, Hydrogen and Oxygen)

Carbon and hydrogen are detected by heating the organic compound with cupric oxide (CuO) strongly, where carbon is oxidized to carbon dioxide and hydrogen to water. Carbon dioxide is tested by lime water test, whereas water is tested by anhydrous copper sulphate test.

By conducting this experiment, we were able to detect different gases that are involve in the reaction of organic compounds. Moreover, we gained knowledge on the products created by the different reactions that are involve in this experiment.

A milky white precipitate was produced during the reaction.

The white precipitate is due to the presence of carbon gas in the reaction. Lime water turns milky as the Calcium hydroxide reacts with carbon dioxide to form Calcium Carbonate which is insoluble in water and thus forms a milky white precipitate.

What gas causes this?

(Milky white precipitate)

CONCLUSION

Observe what collects on the cooler end near the mouth of the test tube.

The collected substance is blue due to the formation of CuSO4.5H2O (by water vapor) then the compound contains hydrogen.

What is it?

(A blue substance was collected) Write a balanced equation for the reaction between: a) Cane oxide

sugar

and

cupric

C12H22O11 + 24CuO 12CO2 + 11H2O + 24Cu Ca(OH)2 + CO2

b) Limewater + produced in (a)

the

gas

CaCO3 + H2O

The products of the reaction are ed over (white) anhydrous copper sulfate and then bubbled through lime water. If copper sulfate turns blue due to the formation of CuSO4.5H2O (by water vapor) then the compound contains hydrogen.

The reactants in Reaction A are C12H22O11 and CuO while the products are CO2, H2O and Cu. When substances undergo reactions, there are always products. The reactions involve are decomposition (Reaction A) and The reactants in Reactant B are double displacement (Reaction B), Ca(OH)2 and CO2 while the respectively. products are CaCO3 and H2O.

PROCEDURE

RESULTS

EXPLANATION B. TEST FOR NITROGEN

CONCLUSION

Mix in a mortar and pestle 0.5 gm of urea and 1 gm of soda lime. Put the mixture in a dry test tube and heat. While heating, place a piece of moistened red litmus paper at the mouth of the tube. (mixing urea and soda lime)

The reaction produced a horrid smell. It stinks because of the presence of ammonia. When the moist red litmus paper was put inside, it turned into blue.

The presence of ammonia indicates the presence of nitrogen. Ammonia is a weak base. Hence, the red litmus paper turned into blue because of the presence of ammonia in the substance tested.

The gas produced is ammonia.

Through the soda lime test, we were able to detect nitrogen gas by the presence of ammonia which has a horrid smell and a basic substance.

(Heating the mixture of urea and soda lime) : : left (A red litmus paper is placed inside the test tube) : : right

Equation:

CH4N2O + 2NaOH Na2CO3

2NH3 +

PROCEDURE

RESULTS

EXPLANATION C. TESTS FOR HALOGENS

1. BEILSTEIN TEST Heat a copper wire in the nonluminous flame until no further color is imparted to the flame. Cool the wire and dip in a few mL of chloroform. Heat the moistened wire in the flame. Repeat the test using iodoform and carbon tetrachloride.

The copper wire is being heated at the non-luminous flame.

(heating copper wire in the nonluminous flame)

CONCLUSION The copper wire was introduced to fire to form Copper (II) oxide or cupric oxide. Then it was dipped to the given sample and heated again to test if the sample is a halogen. A positive test is indicated by the presence of green flame because of the formation of copper halide.

A. CHLOROFORM After heating the copper wire, it Since the flame produced turned was dipped into the chloroform. into green, the sample is halogen Then, the dipped copper wire was specifically Chlorine (Cl). introduced to heat and it produced a green flame.

B. IODOFORM The same test was also done but The present halogen in the sample instead of chloroform, we used is Iodine (I). iodoform. It also produced a green flame.

C. CARBON TETRACHLORIDE The same is also conducted but the compound used is carbon tetrachloride. Like the others, it also produced a green flame.

Carbon tetrachloride is an organic compound which is a halogen because it became positive in testing the presence of halogen.

When the silver nitrate was added to the prepared solution, a white precipitate was formed and the color of the solution is white.

A white precipitate was formed in the solution because of the presence of chloride ion in the solution. The silver nitrate test indicates the presence of a certain halogen in a given solution.

2. SILVER NITRATE TEST Place 0.2 gm of soda lime in a dry test tube and heat to redness. Cool a little and add two drops of chloroform. Heat again, cool a little and add once more two drops of chloroform. Allow to cool and add 5 mL of dilute nitric acid. Let it stand for a few minutes. Pipet off the clear the liquid and add three drops of silver nitrate solution.

(Silver Nitrate Test)

PROCEDURE 1. Weigh 0.2 gm of egg albumin, 05 gm of potassium nitrate and 0.5 gm of sodium carbonate. Mix the three solids thoroughly in a mortar and pestle. Put the mixture in a crucible and heat strongly until a gray or white ash is formed. Allow to cool and dissolve in 20 mL of water. Acidify the solution with nitric acid. Filter. Divide the filtrate into two portions.

RESULTS EXPLANATION D. TEST FOR SULFUR AND PHOSPHORUS

(Mixture of egg albumin, potassium nitrate and sodium carbonate)

(Heated mixture of egg albumin, potassium nitrate and sodium carbonate)

CONCLUSION

(White ash with dilute nitric acid)

(The filtrate is divided into two portions) 2. To one portion, test for the sulfate ion by adding 1 mL of barium chloride solution.

(with Barium chloride) 3. To the other portion, test for the phosphate ion by adding 5 drops of ammonium of molybdate

If barium chloride solution is added to a sample of water containing A white precipitate was formed in sulfate ions, barium sulfate is the solution when Barium chloride formed. Barium sulfate is insoluble was added in the solution. in water, and will be seen as a white precipitate. By adding barium chloride, we can test if the solution contains sulfate ions. A yellow precipitate is formed in When a solution containing the reaction. phosphate ions is heated with a solution of ammonium molybdate [(NH4)2MoO4] and dilute nitric acid

solution and 5 drops dilute nitric acid. Warm gently and then cool. (with Ammonium molybdate and dilute nitric acid)

a bright yellow precipitate of ammonium phosphomolybdate [(NH4)3PO4 • 12MoO4] is formed. The yellow precipitate is extremely insoluble in nitric acid.