Laws of Chemical Combination

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In today’s class, we will be talking about the laws of chemical combination. Enjoy the class!

LAWS OF CHEMICAL COMBINATION

CONTENT

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

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 the study is as follows:

HCl_(aq) + AgNO_3(aq) → AgCl_(s) + HNO_3(aq)

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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.

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 equals 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:

CuCO_3(s) → CuO_(s) + CO_2(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 CuCO₃is formed. This is filtered off and then heated strongly in a crucible to obtain black CuO. The equation for the reaction is:

Na₂CO_3(aq) + Cu(NO₃)₂ → CuCO_3(s) + 2NaNO_3(aq)

CuCO3_(s) → CuO_(s) + CO_2(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.

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 = (1.60/1.99) x 100 = 80.40% (1.59/1.98) x 100 = 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 a fixed mass of the other element are in a 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

The ratio of copper 2 : 1

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

GENERAL EVALUATION/REVISION

State the law of (a) definite proportion (b) multiple proportions
Balance the following chemical equation
Ca(HCO₃)_2(s) → CaCO_3(s) + H₂O_(l) + CO_2(g)
SO_2(g) + H₂O_(l) + 0_2(g) → H₂SO_4(aq)
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

All pure samples of chemical compound contain the same element in the same proportion by mass is the law of _____ A. definite proportion B. reciprocal proportion C. multiple proportions D. conservation of matter
What is used to measure the mass of atom and molecules? A. Beam balance B. Spring balance C. Chemical balance D. Mass spectrometer
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
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. Law of multiple proportions. B. Law of conservation of mass. C. Law of constant composition. D. Law of simple proportion
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 proportions. D. Law of chemical combination

THEORY

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.
Balance the following equations:

C₄H₁₀ + O₂ → CO₂+ H₂O

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

In our next class, we will be talking about Identification and Types of Alloys. We hope you enjoyed the class.

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Laws of Chemical Combination