Lesson Notes By Weeks and Term - Senior Secondary School 1

LAWS OF CHEMICAL COMBINATION

SUBJECT: CHEMISTRY

CLASS:  SS 1

DATE:

TERM: 2nd TERM

REFERENCE BOOKS

  • New Chemistry for Senior Secondary School by Osei Yaw Ababio; U.T.M.E Past Questions and Answers.
  • Practical Chemistry for Senior Secondary Schools by Godwin Ojokuku
  • Outline Chemistry for Schools & Colleges by Ojiodu C.C.
  • Chemistry Pass Questions for S.S.C.E and UTME.

 

 
WEEK FIVE                                        DATE------------------

TOPIC: LAWS OF CHEMICAL COMBINATION

CONTENT

  • Law of conservation of mass
  • Law of definite proportion or constant composition
  • Law of multiple proportion

 

LAW OF CONSERVATION MASS

This law states that during chemical reactions, matter can neither be created nor destroyed but changes from one form to another.

 

EXPERIMENT TO VERIFY THE LAW

AIM: To verify the law of conservation of mass

THEORY: The equation of the reaction chosen for study is as follows:

        HCl(aq) + AgNO3(aq) → AgCl(s) + HNO3(aq)

                    White ppt

 

APPARATUS: Weighing balance, conical flask, small test tube, string cork stopper.

REAGENTS NEEDED: Solutions of HCl and AgNO3 stored in two different reagent bottles.

METHOD: The dilute HCl is poured into a conical flask. The small test tube is filled with     AgNO3solution and by means of a string tied around the neck of the test tube, it is  suspended inside the conical flask containing the acid in such a way that the two solutions  do not mix together. The conical flask and its content are weighed using a weighing   balance and the result recordeD. The two solutions are mixed together by swirling the  conical flask and the weight of the conical flask and its content is taken again.

 

DIAGRAM











RESULT: After mixing the two solutions, a white precipitate of AgCl was formed  indicating that a chemical reaction has taken place.

DISCUSSION: The masses of the conical flask and its content before and after the  reaction remained the same indicating that the mass of the reactants equal that of the   products.

 

CONCLUSION:Since the two masses obtained are equal, it confirms that matter was not created nor destroyed during the chemical reaction.

 

LAW OF DEFINITE PROPORTION OR CONSTANT COMPOSITION

The law states that all pure samples of a particular chemical compound contain the same    elements combined in the same proportion by mass.

 

EXPERIMENT TO VERIFY THE LAW

 

AIM: To verify the law of definite proportion

 

APPARATUS: Crucible, test tube, combustion boats, combustion tube, weighing balance,     Bunsen burner, U-tube and two retort stands with clamps.

 

REAGENTS NEEDED: CuCO3 crystals, Na2CO3 solution, Cu(NO3)2 solution, dry hydrogen     gas and CaCl2 crystals.

 

METHOD: Two samples of black CuO are prepared using different methods. Sample A is   prepared by placing the CuCO3 crystals in a crucible and heating it strongly until it   decomposes into black CuO. The equation for the reaction is:

        CuCO3(s) → CuO(s) + CO2(g)

 

Sample B is prepared by reacting a solution of Na2CO3 in a test tube with a solution of  Cu(NO3)2. A green precipitate of CuCO3 is formeD. This is filtered off and then heated  strongly in a crucible to obtain black CuO. The equation for the reaction is:

                Na2CO3(aq) + Cu(NO3)2 → CuCO3(s) + 2NaNO3(aq)

            CuCO3(s) → CuO(s) + CO2(g)

The two samples of black CuO are placed in two dried and weighed combustion boats  labelled A and B and weighed again. These boats are then placed in a combustion tube and  heateD. A stream of dry hydrogen is passed through the combustion tube to reduce the  CuO to metallic Cu. After heating for sometimes, a reddish-brown residue shows that all  the CuO has been reduced to metallic copper. The flame is removed but the passing in  hydrogen gas continues to prevent the re-oxidation of the hot copper residues by  atmospheric oxygen. Any water formed during the reaction is absorbed by the fused  CaCl2 in the adjacent U-tube. When the boat is cool, the weight of it is taken. From the  results, the percentage of Cu in each sample is calculated.



DIAGRAM:







   










RESULT: Assuming the following result was obtained:

Sample                     A            B

Mass of boat                    3.16g            3.31g

Mass of boat + CuO                 5.15g            5.29g

Mass of boat + Cu                4.76g            4.90g

Mass of CuO = (ii) – (i)            1.99g            1.98g

Mass of Cu = (iii) –(i)            1.60g            1.59g

% of Cu in CuO                1.60 x 100        1.59 x 100

                            1.99      1        1.98      1

                            80.40%        80.30%

Therefore, % of Cu in CuO            80%            80%

                  % of O2 in CuO            20%            20% 

 

DISCUSSION: The % of Cu residue in the two samples is approximately 80% irrespective of the method of preparation of the CuO samples.

 

CONCLUSION: In pure CuO, Cu and O are always present in a definite proportion by mass of approximately 4:1.

 

LAW OF MULTIPLE PROPORTIONS

This states that if two elements combine to form more than one compound, the masses of one of the elements which separately combine with fixed mass of the other element are in simple ratio

 

EXPERIMENT TO VERIFY THE LAW

AIM: To verify the law of multiple proportions

 

APPARATUS: Combustion boats, combustion tube, weighing balance, Bunsen burner, U-tube and retort stand with clamp

 

REAGENTS NEEDED: Cu2O crystals, CuO crystals, dry hydrogen gas and calcium chloride crystals

 

METHOD: The two boats are dried and weighed. Cu2O is placed in one and labelled A and CuO is placed in the other and labelled B. The two boats are weighed again and placed in a combustion tube to reduce the oxides to copper by passing hydrogen gas into the combustion tube. When the samples are cooled, the residues obtained are weighed.

 

RESULT: Assuming the following result was obtained:

Sample                    Cu2O            CuO

Mass of sample (oxide)             3.04g            1.91g

Mass of Cu residue                2.55g            1.35g

Mass of oxygen removed from oxide    0.49g            0.53g

 

CALCULATION: Calculating the various masses of copper which combine separately with fixed mass (say 1g of oxygen)

For Cu2O,

0.49g of O2 combines with 2.55g of Cu

1.0g of O2 will combine with Xg of Cu

Xg of Cu = 2.55g x 1.0g

                      0.49g

              = 5.20g

For CuO,

0.53g of O2 combines with 1.38g of Cu

1.0g of O2 will combine with Xg of Cu

Xg of Cu = 1.38g x 1.0g

                      0.53g

              = 2.60g

Oxides of copper            Cu2O        CuO

Mass of copper            5.20g        2.60g

Ratio of copper              2    :      1

 

CONCLUSION: The masses of copper which combines with a fixed mass of oxygen in Cu2O and CuO are in simple ratio of 2:1.



GENERAL EVALUATION/REVISION

  1. State the law of (a) definite proportion (b) multiple proportion
  2. Balance the following chemical equation
  3. Ca(HCO3)2(s) → CaCO3(s) + H2O(l) + CO2(g)
  4. SO2(g) + H2O(l) + 02(g) → H2SO4(aq)
  5. Determine the oxidation number of: (a) Cu in CuCO3 (b) P in H3PO4 and name the compound

 

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O.Y.Ababio, Pg 34-37

 

WEEKEND ASSIGNMENT

  1. All pure samples of chemical compound contain the same element in the same proportion    by mass is a the law of------ A. definite proportion B. reciprocal proportion C. multiple    proportion 
  1. conservation of matter
  1. What is used to measure the mass of atom and molecules? A. Beam balance B. Spring balance   
  1. Chemical balance   D. Mass spectrometer
  1. What is the ratio by mass of oxygen and hydrogen in 1 mole of water?A. 3:1 B. 2:1 C.1:2 D. 2:4
  2. In two separate experiments 0.18g and 0.36g of chlorine combine with a metal M, to give A and B respectively. An analysis showed that A and B contain 0.10g and 0.20g of M     respectively. Which law is illustrated by the data? A A. Law of multiple proportions.  B.Law of conservation of mass.
  1. Law of constant composition.  D. Law of simple proportion
  1. An element E forms the following compounds with bromine: EBr2, EBr3, and EBr4. This observation illustrates the A. Law of conservation of mass.B. Law of definite    proportion.C. Law of multiple proportion.D. Law of chemical combination

  

THEORY

  1. 8.50g of CuO when heated in a current of dry hydrogen gas gave 6.58g of copper and 2.16g of water. Calculate the proportion of oxygen to hydrogen by mass in water.
  2. Balance the following equations:

    C4H10 + O2  → CO2 + H2O

    H2SO4 + NaOH → Na2SO4 + H2O





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