CHEMISTRY

LECTUER BY ER. JASBIR SINGH BHANGU E-MAIL ADDRESS:- bhangujs@hotmail.com LAND LINE NUMBER:183-5006326 MOBILE NUMBER:- 9878041411 LAWS OF CHEMICAL COMBINATIONS It is interesting to note from the followings that chemical reactions obey certain rules and these rules are generally called the laws of chemical combinations. There are five laws of chemical combinations in which different elements enter to form compounds. The formation of the compounds have been classified under these five kinds of aspects/rules. These aspects have been called the laws of chemical combination. These laws are:- 1.Law of conservation of mass. [Lavoisier-1774-French] Lavoisier, Antoine (1743-1794) 2.Law of constant composition OR Law of definite Proportions [ Proposed in 1799] [Joseph Proust -French] Joseph Proust was one of the forefront scientists in the development of the atomic theory. He was a French chemist born at Angers in northwest France, in September 26, 1754. He discovered grape sugar and established the law of definite proportions, also known as Proust's law, which states that in any compound the elements are present in a fixed proportion by weight. He died some time last in the year 1826. 3.Law of multiple proportions. [Proposed in 1803] [Dalton -English school teacher] Born 6 September 1766(1766-09-06)Eaglesfield, Cumberland, England Died 27 July 1844 (aged 77)Manchester, England Notable students James Prescott Joule Known for Atomic Theory, Law of Multiple Proportions, Dalton's Law of Partial Pressures, Daltonism 4.Gay Lussac’s law of CombiningVolumes. [ proposed in 1808] Joseph Louis Gay-Lussac Born 6 December 1778 (1778-12-06)Saint-Léonard-de-Noblat Died 9 May 1850 (1850-05-10)Saint-Léonard-de-Noblat Nationality French Fields Chemistry Known for Gay-Lussac's law The volume 5.Law of reciprocal proportions [Proposed in 1792] Jeremias Richter-Germany German chemist Born March 10, 1762(1762-03-10)london, Jelenia Gora,Germany (now Poland) Died April 14, 1807 (aged 45)Berlin,Germany The term "stoichiometry" was devised by German chemist Jeremias Richter in 1792 to describe the measurement of the combining ratios of chemical elements by mass. The term has since been expanded to include the combining ratios of substances in any chemical reaction. Richter studied mathematics with philosopher Immanuel Kant and wrote a thesis on the use of mathematics in chemistry. He was convinced that all chemical changes could be described in terms of simple whole-number ratios. He put forward the Law of Reciprocal Proportions, stating that if two chemical elements unite separately with a third element, the proportion in which they unite with the third element will be the same or a multiple of the proportion in which they unite with each other. This law has disappeared from most chemistry textbooks, but a companion law, the Law of Multiple Proportions, has survived. (1)LAW OF CONSERVATION OF MASS: [ Lavoisier- French Scientist] IT STATES AS UNDER: DEFINITION:- During any physical or chemical change, the total mass of the products is equal to the total mass of the reactants.That is there is no change in mass. No destruction of mass. Mass of reactants is equal to the mass of products. Still more, mass can neither be created nor be destroyed. Chemical equations are based on this law. However, this law does not hold good for nuclear reactions. In nuclear fission reaction, mass is destroyed and converted in to heat energy. We balance the chemical equations on account of this law. A balanced chemical equation will have to conserve the mass,atoms & electrons . but balance equation is not to follow the law of combining the volumes. SOLVE THE FOLLOWING PROBLEM WITH THE HELP OF LAVOISIER LAW THAT STATES THAT MASS IS CONSERVED IN PHYSICAL OR CHEMICAL CHANGES. THAT IS THERE IS NO LOSS OF MASS: PROBLEM:- WHAT WEIGHT OF SODIUM CHLORIDE WOULD BE DECOMPOSED BY 4.0 g OF SULPHURIC ACID IF 6.0g OF SODIUM BISULPHATE.( NaHSO4) AND1.825 g OF HYDROGEN CHLORIDE WERE PRODUCED IN THE REACTION. SOLUTION:- Chemical reaction taking place is:- NaCl+H2SO4 NaHSO4 +HCl Mass of NaCl is not known. Assume it to be X g. Now apply law of conservation of mass on the above equation that say that mass do not changes in physical or chemical changes. That is mass of reactants and products remains same and there is no loss or generation of mass after the chemical change.Chemical equations are based on this law. (X g) NaCl +(4.0g) H2SO4 = (6.0g) NaHSO4 +(1.825g) HCl Equalty sign represents the Law of conservation of mass. X+4.0 = 6.0+1.825 X=6.0+1.825-4.0 =2+1.825=3.825g SO with the help of this law of conservation of mass in a chemical change helped us to calculate the mass (x=3.825g) of NaCl required to complete the reaction PROBLEM:- AMMONIA CASE If 0.17 g of ammonia on decomposition gives 0.14g of nitrogen and 336 ml hydrogen at NTP we find:- AMMONIA HEAT NITROGEN+HYDROGEN NH3 (0.17g) = 0.14g NITROGEN+336ml of hydrogen 22400 ml of H2 at NTP weigh=(H 2 )2g 336 ml of hydrogen =2/22400x336=0.03g So, NH3(0.17g)=0.14g(NITROGEN)+0.03g(HYDROGEN) Thus in this experiment we saw that there is no change of mass in the reaction . That is mass remain conserved . Hence the reaction took its name as the law of conservation as there is neither destruction nor creation of mass. NTP:- NORMAL TEMPERATURE & PRESSURE NTP - Normal Temperature and Pressure - is defined as air at 20oC (293.15 K, 68oF) and 1 atm ( 101.325 kN/m2, 101.325 kPa, 14.7 psia, 0 psig, 29.92 in Hg, 760 torr). Density 1.204 kg/m3 (0.075 pounds per cubic foot) TO PROVE THROUGH - NACL & AgNO3- CASE Law of conservation of mass can be verified by also by the following experiment:- Landolt’s experiment to illustrate the law of conservation of mass: It means that in chemical reactions there is no loss of mass or increase in mass. In other words we say it that matter can not be created nor destroyed in chemical combinations or chemical reactions. Landolt put the two solutions in a H type tube. The tubes were sealed and weighed. Sodium chloride solution was in the one tube and silver nitrate solution in the second tube . After weight both solutions were mixed thoroughly by shaking the tube. Reaction took place with the formation of a white precipitate of silver chloride as: AgNO3+NaCl AgCl (White ppt.) +NaNO3 After reaction weight of tube was taken and found that there was no change in the weight. Hence the law of conservation of mass was confirmed. PROBLEM WHAT ARE THE DEFECTS IN THE FOLLOWING CHEMICAL REACTION EQUATIONS? NAME THE LAW THAT INDI CATE THE DEFECT IN EACH EQUATION? (1) Mg + H Cl MgCl2 +H2 (2) KClO3 K Cl +3O (3) Mg+CO2 Mg O+C (4) C6H6 +O2 CO2 +H2 (5) Fe3O4 +H2 Fe+H2O (6) FeS2+O2 Fe2O3+SO2 (7) As2O3+SnCl2+HCl SnCl4+2As+3H2O (8) H2S+O2 H2O+S (9) PbO2+HCl PbCl2+Cl2+H2 (10) Si +NaOH+H2O Na2Si3+H2O PROBLEM:- 4.90g of KClO3 when heated produced 1.92g of oxygen and the residue(KCl)left behind weighs 2.96g. Show that these results illustrate the law of conservation of mass. SOLUTION:- Heated KClO3 KCl+O2 (4.90) = (2.96+1.92)g = 4.88g As 4.88 is very close to 4.90 g . Minor difference is on account of experimental error. So there is no loss of mass. Hence figures shows the law of mass conservations. PROBLEM:- To find the un known weight of reactant with the help of law of conservation of mass. What weight of silver nitrate will react with 5.85g of sodium chloride to produce 14.35g of silver chloride and 8.5g of sodium nitrate, if the law of conservation of mass is true . To find the un known weight of one of the reactants with the help of law of conservation of mass SOLUTION:- AgNO3 (W g) + Na Cl ( 5.85 g) Ag Cl(14.35g)+NaNO3(8.5g) Law of conservation of mass states that during any physical or chemical change, the total mass of the products is equal to the total mass of the reactants. So in above chemical change the mass on both sides of equation should remain same. So, AgNO3 (W g) = (14.35+8.5)g - 5.85g = (22.85) g -5.85 = 17.0g PROBLEM:- Do the data is as per the law of of mass of conservation? When 4.2g of NaHCO3 is added to a solution of acetic acid (CH3COOH) weighing 10.0g, it is observed that 2.2g of CO2is released in to the atmosphere .The residue left behind is found to weigh 12g.Show that these observations are in the agreement with the law of conservation of mass. SOLUTION:- NaHCO3 (4.2 g) +CH3 COOH (10.0g) CO2 (2.2g) + Residue (12g) Let the reaction is complete and all reactants have been converted in to products. As per the law of conservation of mass:- L.H.S = R.H.S 4.2 g + 10.0 g = 2.2 g +12 g 14.2g = 14.2 So, mass of reactants turns out to be equal to the mass of products. Hence the law of conservation of mass is true. PROBLEM:- HOW THE LAW IS HELPFUL TO DETERMINE THE MASS OF UNKNOWN PRODUCT? If 6.3 g of NaHCO3 is added to 15.0g of CH3COOH solution, the residue is found to weigh 18.0g . What is the mass of CO2 released in the reaction? SOLUTION:- NaHCO3 [6.3g] +CH3COOH [15.0g] CO2 [W g]+Residue[1.8.0g] As per law of mass conservation, the mass in a chemical change is conserved. So mass of reactants should be equal to the mass reactants. That is mass on the left hand side [L.H.S] of the reaction equation should be equal to the mass on the right hand side[R.H.S] of the equation. 6.3 g + 15.0g = W g + 18.0 g 6.3g + 15g - 18 .0 g = W g 21.3 g - 18.0 g = W g 3.3 g = W g So CO2 produced is 3.3 g PROBLEM WHICH IS TRUE? IF LAW OF CONSERVATION OF MASS WAS TO HOLD TRUE, THEN 20.8g OF BaCl2 ON REACTION WITH 9.8g OF H2SO4 WILL PRODUCE 7.3g OF HCl AND BaSO4 EQUAL TO:- A) 11.65g B) 23.3g C) 25.5g D) 30.6g SOLUTION:- BaCl2 + H2SO4 BaSO4 + 2HCl (20.80 + (9.8) = BaSO4 (W) + 7.3 30.6 = BaSO4(W) +7.3 30.6 – 7.3 = BaSO4 (W) 23.3 = BaSO4 (W) There fore BaSO4 is equal to 23.3g . Hence answer (B) is correct PROVE THAT DATA REPRESENTS THE LAW OF CONSERVATION OF MASS:- 1.5g. of hydrocarbon [HXCY] on combustion in excess of oxygen produces 4.4 g of CO2 and 2.7g of H2o. PLANING FOR SOLUTION: (A) HOW MUCH CARBON IS THERE IN PRODUCED 4.4 g. OF CO2 ? FROM WHERE THIS CARBON CAME? (B) HOW MUCH HYDROGEN IS THERE IN 2.7g. OF H2O. FROM WHERE THIS HYDROGEN CAME? SOLUTION:- HOW MUCH CARBON ?:- (12/44)x 4.4= (12/44)x(44/10)=1.2 g of carbon in 4.4 g of CO2 HOW MUCH HYDROGEN ? (2/18)x2.7=(2/18) x(27/10)= (2/2) x(3/10)=0.3g of HYDROGEN in 2.7g. Of H2O SO 1.2g. OF CARBON+0.3g.OF HYDROGEN CAME FROM [ HXCY] . ITS SUM IS 1.2+0.3=1.5g. BUT GIVEN WEIGHT OF HYDROCARBON [HXCY]IS ALSO= 1.5g. So mass ( for carbon & hydrogen) in the reaction remained conserved. That is there is no change in the mass in the chemical reaction. Hence data is in accordance to the law of mass conservation. (2)LAW OF CONSTANT COMPOSITION OR DEFINITE PROPORTION:- [Proust- French Scientist] Definition:- IT STATES THAT A PURE CHEMICAL COMPOUND IS MADE UP OF THE SAME ELEMENTS COMBINED TOGETHER IN THE SAME FIXED PROPORTION BY WEIGHT IRRESPECTIVE OF THE PROCESS OF FORMATION OF THAT COMPOUND. ELEMENTS ALWAYS COMBINE IN FIXED RATIO OF THEIR WEIGHTS. FOR EXAMPLE PURE WATER (H2 O) ALWAYS CONTAINS 1:8 RATIO OF HYDROGEN AND OXYGEN BY WEIGHT. TESTING [ CO2 - CASE] That the pure compound might have taken its formation through different processes but ultimately the formed compound has its elements {From the reactants} and constant proportions by weight in the formed compound. Let CO2 be the pure compound formed in different processes. In each process the formed Co2[compound] has its elements carbon & oxygen by weight ( A constant proportion) 12 and 32 or ratio 3:8, parts by weight- that is fixed proportion of weight OR ratio, by weight. PRODUCTION OF CO2 COMPOUND (1) FIRST PROCESS :- BY BURNING OF CARBON IN THE AIR. C+O2 co2-----------(12:32)=3:8- carbon 3 and oxygen 8 by weights (2) SECOND PROCESS:- BY HEATING CALCIUM CARBONATE HEAT CaCO3 CaO+CO2----------(12:32)=3:8 (3) THIRD PROCESS:- BY HEATING CALCIUM BICARBONATE HEAT Ca(HCO3) CaCO 3+H2 O+CO 2-----------(12:32)=3:8 In each of the above three processes CO 2 ( compound) is produced and it has same ratio for carbon element and oxygen element in its CO 2 compound. That is 3:8 parts by weights---------(three parts by weight of carbon and eight parts by weight of oxygen.) WHAT IS THE PERCENTAGE OF OXYGEN IN NaOH ? (Na)=23 (O)= 16 (H)= 1 Therefore: The ratio is 23:16:1[parts by weight] In case the weight of the samle of NaOH is 40g, It contains 16gram of sodium. If weight ofNaoh is 100gram , the weight of oxygen in it according to its ratio is :-16g DIVIDED BY 40 X 100g=16/40X100=40g So, percentage of oxygen is 40 % in the compound of NaoH. It means if the compound of NaoH has the weight 40gram , the weight of oxygen is 16grams. But if the compound carries more quantity that is 100gram the oxygen contained in it will be 40 g in other words 40 % [As weight of NaOH was taken to be 100g] PROBLEM:- If 6.537g. of zinc reacts with exactly 7.0906g of chlorine to form the only compound of chlorine &zinc, how much zinc will react with(a)14.18g of chlorine?(b)with 28.36g of chlorine? (c)with 100.0g of chlorine? SOLUTION: ZnCl2=65+71=136 As each time the compound formed is zinc chloride. So compound is having constant composition of zinc & chlorine or in other words we can say that compound is of fixed proportions of zinc & chloride. It is understood that reaction is complete and the whole quantity of Zn & Cl has completely converted in to the product of ZnCl2 This proportion is given to us and is :- 7.0906 (Cl) ______ 6.537 (Zn) As the fixed proportion (ratio) is to remain the same as per the laws of constant composition:- (a) So of chlorine used has doubled (14.18g), the zinc used will also be required to double. As ratio of the elements in the compound is to remain the same . In case the amount of zinc is less the reaction will not complete and correspondingly will use less chlorine. So for complete reaction, the amount of zinc will be doubled and thus the fixed ratio remained same (b)In this case chlorine has tripled, so zinc will also be tripled as ratio of both elements is constant. This constant ratio is achieved if and only if we triple the amount of zinc as well. (c) Now let us see how many times chlorine has increased. The same figure times will have to increase the zinc. 100 divided by7.0906= (100/7.0906) times. Now increase the amount of zinc to same times: 6.537 g x (100/ 7.0906) times =92.19g of zinc 7.0906 14.18 21.2718 100 Cl -------- = ----- = --------- = ------- =1.085= ------- 6.537 13.074 19.611 92.19 Zn So we have seen that in each case the compound of zinc chloride has the same ratio for its elements zinc and chlorine to participate in the compound IN OTHER WAY WE CAN ALSO WRITE AS UNDER: 7.0906 (Cl) Chlorine /zinc ratio = ______ = 1.0846871 6.537 (Zn) (a) 2x7.0906 (Cl) ------------------- = 1.0846871 Zn 2x7.0906 (Cl) ------------------- = Zn 1.0846871 14.1812 ------------ = Zn 1.0846871 14.1812 Zn = ------------ = 13.074 1.0846871 (b) In this case chlorine has tripled:- 3x7.0906 (Cl) 7.0906 (Cl) -------------------- = Chlorine /zinc ratio = ----------------- (Zn) 6.537 (Zn) =1.0846871 3x7.0906 ------------- =1.0846871 Zn 3x7.0906 ------------------ = Zn 1.0846871 21.2718 -------------------- = Zn 1.0846871 19.611 = ZN (C) 100(Cl) 7.0906 (Cl) -------------- =Chlorine /zinc ratio = ----------------- = 1.0846871 Zn 6.537 (Zn) 100 Zn== -------------- =92.192485 1.0846871 NOW LET US CHECK THAT WHEATHER THE RATIO OF PARTICIPATING AMOUNTS OF ZINC & CHLORINE IN EACH CASE REMAINED THE SAME TO THE FACT OF THE LAW . 7.0906 14.18 21.2718 100 -------- = ----- = --------- = ------- =1.085 6.537 13.074 19.611 92.19 TESTING : [ CU & CUO- CASE] A CASE SHOWING THAT CUO CONSITS OF FIXED PROPORTION:- DO THE COPPER OXIDE[CUO ] COMPOUND HAS FIX RATIO FOR COPPER AND OXYGEN? TO EXAMINE THE PARTS BY WEIGHT IN COPPER OXIDE:-[Testing by two ways:- (A) By formation of CUO (B)By reducing CUO] In case the results of two cases are compared and found to be the same, it means the law is true and validated for the truth that a pure chemical compound is made up of the same elements combined in a fixed proportion by weight irrespective of the process of formation of that compound or the elements always combine in fixed ratio of their weights (A) Heat Cu+HNO3 CUO+H2O+2NO heat 2.16 g of Copper+Nitric Acid CUO (2.70g) FROM ABOVE, CHECK THE:- 1. PERCENTAGE OF COPPER IN CUO ? 2. PERCENTAGE OF OXYGEN IN CUO ? PERCENTAGE OF COPPER:- 2.16[Cu] X 100[CUO] ------------ ------------ = 80g[Cu] 2.70[CUO] [ That is 80 grams of copper from 100grams of CUO] PERCENTAGE OF OXYGEN:100g of CUO-80 g of CU=20g[O] It means compound CUO contains 80percent parts by weight of copper and 20percent parts by weight of oxygen. As CUO is 100 g then it contains 80g of copper and 20g of oxygen. THE RATIO IS 80:20 OR 4:1 (B)Now let us see if the same relation of 80:20 [Copper and Oxygen] exists for some other case in which we reduce 1.15 gms of copper oxide. By reducing copper oxide (1.15 gms), we got Copper 0.92gms So let us check the percentage of each [copper & oxygen) in CUO: Weight of copper oxide = 1.15g Weight of copper left after reduction =0.92g 0.92[CU] x 100[CUO] ________________ 1.15 [CUO] Percentage of copper in copper oxide = 80gms [CU] SO, it means 100g of copper oxide gives 80g of copper and remaining (100-80=20) 20gms out of 100g is oxygen So under these two different processes for copper and oxygen we checked that copper oxide compound contains the fixed or constant composition of copper and oxygen, by weight that is 80 by weight copper and 20by weight of oxygen. THAT IS CU:OXYGEN {CUO}=80:20=4:1 IT MEANS WHENEVER FORMATION OF CUO WILL TAKE PLACE THE RATIO OF CU AND OXYGEN (BY WEIGHT) WILL BE FIX RATIO (4:1) PROBLEM HOW THE LAW IS HELPFUL TO US :BY APPLYING THIS LAW OF CONSTANT COMPOSITION OR DEFINITE PROPORTION WE CAN SOLVE DIFFERENT KINDS OF PROBLEMS:- One kind of problems is discussed here as under:- HOW MUCH WEIGHTS ARE PRESENT IN A SAMPLE? What weights of calcium, carbon and oxygen are present in 1.15 g of calcium carbonate. Given that a sample of calcium carbonate from another sample contains the following percentage composition. Ca = 40.0 %, C = 12.0 % ; O = 48.0 % As we know the law of constant composition is true that calcium, carbon and oxygen elements have a fixed ratio by weight in calcium carbonate. The above ratio 40:12:48[ 10:3:12] in an other sample is also the true ratio for the sample of 1.15 g of calcium carbonate. So with the help of this data we can find calcium, carbon and oxygen in the sample of 1.15 grams of calcium carbonate. 100 grams of CaCO3 contains = 40 grams of Ca 40 1gram of Ca CO 3 will contain =----- 100 1.5 grams of CaCO3 will contain = 40/100x1.5 = 0.6g of Ca Similarly, as per the pattern of above calculation we can calculate that weight of carbon is 0.18 g and weight of oxygen is 0.72 g in CaCO3 of weight 1.5 g 0.6+0.18+0.72= 1.5 g CHECK UP RATIO= (0.6) : (0.18) : (0.72)=60:18:72=10:3:12 THE RATIOS IN BOTH CASE ARE SAME THAT VALIDATE THE TRUTHNESS OF LAW So by knowing the ratio from one sample, we can apply that ratio to the other sample of the same compound to know that how much weight of each element in that sample is present. NAME THE DIFFERENT COMPOUNDS WHICH ARE FORMED FROM ITS TWO ELEMENTS? H2O, H2O2 HYDROGEN PEROXIDE NITROUS OXIDE N2O NITRIC OXIDE NO OR 2NO NITROGEN TRIOXIDE (N2O3) NITROGEN PENTOXIDE N2O5 NH3 and N2H4 (DIAMINE) SnCl2, SnCl4 CO,CO2 FeO, Fe2O3, Fe3O4 FeCl2 ferrous chloride, FeCl3 ferric chloride SO2, SO3 H2O, H2O2 HYDROGEN PEROXIDE DO THERE ANY RELATION BETWEEN THE TWO ELEMENTS WHILE FORMING THEIR VARIOUS COMPOUNDS? (3) LAW OF MULTIPLE PROPORTIONS:- Scientist DALTON-[ENGLAND] Definition:- The law of multiple proportions is one of the basic laws in chemistry, and is a major tool of chemical measurement (stoichiometry). This law states that when elements combine they do so in a ratio of small whole numbers. For example, carbon and oxygen react to form CO or CO2, but not CO1.3. Furthermore, it states that if two elements form more than one compound between them then the ratios of the masses of the second element combined with a fixed mass of the first element will also be in ratios of small whole numbers. The English chemist John Dalton first expressed this observation in 1803 and it is sometimes called Dalton's Law (although this term usually refers to his law of partial pressures). In short, the law of multiple proportions states: "When two elements form more than one compound, the different masses of one element that are combined with the same mass of the other element are in a ratio of small whole numbers." It states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers. SIMPLE WHOLE NUMBER RATIO? this sounds confusing, but an example clarifies this statement. Consider the carbon oxides, and let carbon be element B and oxygen be element A. Take a fixed given mass of carbon (element B), say 1 gram. The mass of oxygen which combines with 1 gram of carbon to form the first oxide is 1.33 grams. The mass of oxygen which combines with 1 gram of carbon to form the second oxide is 2.66. These masses are in ratio 2.66:1.33=2:1, a simple whole number ratio According to the law of multiple proportions, when two elements combine to form two or more than two compounds, the weights of one of the elements[Say B] which combines with the fixed weight of the other[Say A] , bear a simple ratio to one another. Say there are two elements. Element “A”and “B”. Let these elements form a variety of compounds by chemical combining with each others. Let the compounds formed be:- A(W)B(W1) First kind of compound with A&B elements A(W)B(W2) Second kind of compound with A&B elements A(W)B(W3) Third kind of compound with A&B elements A(W)B(W4) Fourth kind of compound with A&B elements A(W)B(W5) Fifth kind of compound with A&B elements The fix weight (Say-W) of element “A” will remain the same for each compound made with B. Where W1,W2,W3,W4 & W5 are the weights of element (B) used in the respective compounds :-A(W)B(W1), A(W)B(W2) , A(W)B(W3) , A(W)B(W4) & A(W)B(W5) respectively. It means in each above five compounds weight used of element “A” IS “W” where as for B element it varied from W1 TO W5. According to this multiple law, the ratio of weights used by element “B” to form different compounds by combining with fixed(W) weight of ”A” from compound A(W)B(W1) TO A(W)B(W5) has a simple ratio or multiple among its ( various weights W1,W2,W3,W4,W5 of B )weights used in the formation of compounds by combining with element “A” .That is ratio of weights W1:W2:W3:W4:W5 is a simple ratio or a multiple of the simple ratio among its own range. W1,W2,W3,W4 & W5 are also the simple multiple [number of folds] of the fixed weight (W) of element “A” . This law is applicable to the compounds where various compounds are formed from the same two elements Law of multiple proportions problems? CARBON OXYGEN COMPOUND 12 16 CO 12 32 CO2 RATIO 16:32 1:2 MULTIPLE 16 HYDROGEN OXYGEN COMPOUND 2 16 H2O 2 32 H2O2 RATIO 16:32 1:2 MULTIPLE 16 SULPHUR OXYGEN COMPOUND 32 32 SO2 32 48 SO3 RATIO 32:48 2:3 [ MULTIPLE IS 16] Sn[Tin] CHLORINE COMPOUND 119 2X35.5=71 Sncl2 119 4X35.5=142 Sncl4 RATIO 71;142 1:2 [ MULTIPLE IS 71] Let us see the case of :NH3 and N2H4(DIAMINE) 28g 6g 28g 4g Multiple :2 Ratio 6:4 3:2 OXIDES OF NITROGEN-CASE In order to understand the law of multiple proportional, let us consider two elements nitrogen (N) and oxygen (O), they can form as many as five oxides. Let us keep the weight of nitrogen in these oxides to be fixed i. e 28 parts.The weights of oxygen which combine with a fixed weight of nitrogen in these oxides are in the ratio 16:32:48:64:80 i. e 1:2:3:4:5.since the ratio is a simple whole number ratio, the law of multiple proportions stands verified. NAME OF OXIDE WEIGHT OF NITROGEN WEIGHT OF OXYGEN NITROUS OXIDE N2O 28 16X1=16 NITRIC OXIDE NO OR 2NO 28 16X2=32 NITROGEN TRIOXIDE N2O3 28 16X3=48 NITROGEN PEROXIDENO2 OR N2O4 28 16X4=64 NITROGEN PENTOXIDE N2O5 28 16X5=80 From above we see that nitrogen by taking its weight (28) each time has reacted with oxygen to form above five kinds of compounds [ N2O, NO OR 2NO , N2O3 , N2O4 ,N2O5] . In each compound, oxygen has used its different weight to combine with the fixed weight (28) of nitrogen. When we check this ratio of the weights of oxygen used in above five compounds it comes out to be 16:32:48:64:80 or its simple ratio is 1:2:3:4:5 and16 is its multiple. PROBLEM IS DATA CONSISTENCE WITH THE LAW OF MULTIPLE PROPORTIONS? TWO DIFFERENT COMPOUNDS ARE FORMED BYTHE ELEMENTS CARBON AND OXYGEN.THE FIRST COMPOUND CONTAINS 42.9 % BY MASS CARBON AND 57.1 % BY MASS OXYGEN.THE SECOND COMPOUND CONTAINS 27.3% BY MASS CARBON AND 72.7% BY MASS OXYGEN. SHOW THAT THE DATA ARE CONSISTENT WITH THE LAW OF MULTIPLE PROPORTIONS T-1 ELEMENT ELEMENT COMPOUND RATIO COMPOUND FOUND C O 12 16 CO 0.75 42.9 57.1 CX OY 0.75 CO T-2 ELEMENT ELEMENT COMPOUND RATIO COMPOUND FOUND C O 12 32 CO2 0.375 27.3 72.1 CX OY 0.375 CO2 T-3 ELEMENT ELEMENT COMPOUND RATIO C O 12 16 CO 0.75 12 32 CO2 0.375 Ratio:1:2 As the ratios (0.75) & (0.375) in Table NO-1 & TABLE NO-2 are in accordance to the ratios of Table not-3. But ratios in table no 3 shows the law of multiple proportions. Hence the given data is in consistence with the law of multiple proportions. HOW THE DATA CONFIRMS THE LAW OF MULTIPLE PROPORTION? Consider 100.0 g of two different compounds, consisting only of carbon and oxygen. One Compound contains 27.2 g of carbon and the other has 42.9 g of carbon. How can these data support the law of multiple proportions if 42.9 is not a multiple of 27.2? FIRST SAMPLE OF 100 g. OF COMPOUND OF CARBON & OXYGEN:-O = 100 - 27.2=72.8 g Carbon attached with 72.8 g of Oxygen=27.2g. 27.2g. of carbon needs oxygen to form a compound= 72.8g 1g. of carbon would need oxygen= 72.8 / 27.2=2.67647g. SECOND SAMPLE OF 100 g. OF COMPOUND OF CARBON & OXYGEN:-Mass of “O” = 100 - 42.9=57.1 g of Oxygen 42.9g.of carbon needs oxygen to form the carbon-oxygen compound=57.1g 1g. of carbon would need oxygen : to form carbon-oxygen compound=57.1 / 42.9 =1.3310 Element Element Name of compound Actual compound C O Ca Ob C1OI.33=CO CO C O Ca Oc C1O2.6=CO2 CO2 Ratio = b : c RATIO (1.33): (2.6) 1:2 So data of two samples each of 100g , containing 27.2 g & 42.9g of carbon respectively is consistence with the law of multiple proportions. OXIDES OF PHOSPHOROUS -CASE ANOTHER EXAMPLE:- [ DO THE FIGURES SHOWS THE LAW OF MULTIPLE PROPORTIONS ?] PHOSPHOROUS IS FOUND TO FORM THREE OXIDES CONTAINING 43.668, 49.212, & 56.365 PERCENT OF PHOSPHOROUS RESPECTIVELY. SHOW THAT THESE FIGURES ILLUSTRATE THE LAW OF MULTIPLE PROPORTIONS ---- HOW MUCH OXYGEN IS COMBINED WITH ONE GRAM OF PHOSPHOROUS TO FORM THE OXIDE OF PHOSPHOROUS?- IN EACH CASE AND THEN MAKE COMPARIGON OF OXYGEN USED IN RESPECTIVE THREE CASES FIRST SAMPLE: (A) 100g OF PHOSPHOROUS OXIDE CONTAINS PHOSPHOROUS=43.668 g[P] 100g OF PHOSPHORUS OXIDE CONTAINS OXYGEN= 100-43.668=56.332g [OXYGEN] 43.668 GRAMS OF PHOSPHOROUS COMBINES WITH OXYGEN=56.332 GRAMS[OXYGEN] 1GRAMS OF PHOSPHORUS COMBINES WITH OXYGEN=56.32(OXYGEN)/43.66(PHOSPHOROUS)=1.290 GRAMS OXYGEN 1.290 g Of OXYGEN IN ONE GRAM OF PHOSPHOROUS SECOND SAMPLE (B) 100GRAMS OF PHOSPHOROUS OXIDE CONTAINS PHOSPHORUS= 49.212 GRAMS PHOSPHORUS 100GRAMS OF PHOSPHORUS OXIDE CONTAINS OXYGEN=100-49.212=50.788 GRAMS OF OXYGEN AS, FROM ABOVE 49.212 GRAMS OF PHOSPHORUS COMBINES WITH OXYGEN=50.788 GRAMS[OXYGEN] THEREFORE, 1 GRAM OF PHOSPHORUS COMBINES WITH OXYGEN= 50.788(OXYGEN)/49.212(PHOSPHOROUS)=1.032 GRAMS[OXYGEN] 1.032 g Of OXYGEN IN ONE GRAM OF PHOSPHOROUS THIRD SAMPLE (C) 100GRAMS OF PHOSPHORUS OXIDE CONTAINS PHOSPHORUS=56.365 GRAMS[PHOSPHOROUS] 100GRAMS OF PHOSPHORUS OXIDE CONTAINS OXYGEN=100-56.365=43.635GRAMS[OXYGEN] 1GRAM OF PHOSPHORUS COMBINES WITH OXYGEN=43.635[oxygen]/56.365[phosphorous]=0.774 0.774 g of OXYGEN IN ONE GRAM OFPHOSPHOROUS 1.032 g Of OXYGEN IN ONE GRAM OF PHOSPHOROUS 1.290 g Of OXYGEN IN ONE GRAM OF PHOSPHOROUS 0.774 g of OXYGEN IN ONE GRAM OFPHOSPHOROUS 1.290 1.032 0.774 OXYGEN USED IN THREE CASES SO FROM ABOVE THREE SAMPLES/CASES WE HAVE SEEN THAT PHOSPHOROUS HAS COMBINED WITH DIFFERENT WEIGHTS OF OXYGEN TO FORM THE FOLLOWING OXIDES: FIRST :- PHOSPHOROUS OXIDE 1:1.290 SECOND:PHOSPHOROUS OXIDE: 1:1.032 THIRD: PHOSPHOROUS OXIDE1:0.774 SO OXIDE HAS RATIO AMONG ITSELF AS: (1.290) : ((1.032)) : (0.774) DIVIDING BY 0.774 TO MERELY BRINGING IN TO SIMPLE RATIO 1.290 1.032 0.774 -------: -------- :--------- =1.67 : 1.33 :1= 5 :4 :3 0.774 0.774 0.774 SO simple whole ratio is 5:4:3 Thus samples illustrates the laws of multiple proportions. FOLLOWINGS ARE THE DIFFERENT OXIDES OF PHOSPHOROUS. PO33- ( PHOSPHITE) , PO43- ( PHOSPHATE) , P2O5 PENT OXIDE Kind of oxide of phosphorous Weight of phosphorous Weight of oxygen used as oxide Dividing by0.774, figure comes out to be By multiplying 3 figures comes out to be First sample of oxide of phosphorous One gram 1.290 grams 1.67 5 Second sample of another kind of oxide of phosphorous One gram 1.032 grams 1.33 4 Third sample of oxide of phosphorous different than the above two kinds of oxides One gram 0.774 grams 1 3 From above it is concluded that to form three kinds of oxides of phosphorous, phosphorous has used its weight one gram for each kind of its oxides. But oxygen has taken different weights to combine with one gram of phosphorous to form the corresponding oxides. Here, we see that for three kinds of oxides of phosphorous, oxygen has used 1.67grams, 1.33 grams & 1 gram of oxygen. When we see it in terms of ratio it comes out to be 5:4:3 A simple whole number ratio. So oxygen beares a simple ratio among its attached weight to the fixed weight of phosphorous for forming different kinds of oxides of phosphorous. Hence the figures 43.668,49.212 & 56.365 of the samples are corresponding to the 1gram of phosphorous of respective sample . Thus it gave us the above ratio of 5:4 :3. Hence the figures are corresponding to the law of multiple proportions ----------------------------------------------------------------------- Jseph Gay Lussac’s law Of CombiningVolumes. [1778-1850] DEFINITION:- Whenever gases take part in chemical combination ,they do so in volumes which bear simple ratio to one another and also the gaseous products, the measurements being made under, similar conditions of temperature and pressure. Avogadro's law has been useful in substantiating a number of important laws and concepts. It has helped in the following areas: In explaining Gay Lussac's law of gaseous volumes. In determining the atomicity of gases. In determining the molecular formula of a gas. In establishing the relationship between relative molecular mass and vapour density. COMBINATION BETWEEN NITROGEN AND HYDROGEN: Berthelot studied the composition between nitrogen and hydrogen and observed that 1volume of nitrogen combines with 3volumes of hydrogen to form two volumes of ammonia. NITROGEN+HYDROGEN AMMONIA N2 + 3H2 2NH3 1Volume 3Volume 2Volume The ratio of volumes of gases are 1:3:2 EXAMPLE: Combination between hydrogen & chlorine:- Two volumes of hydrogen combines with one volume of oxygen to form two volumes of steam when an electric spark is passed through the gaseous mixture. Electric HYDROGEN+OXYGEN WATER(STEAM) 2H2(g) O2(g) SPARK 2H2O(g) 2Volumes 1Volume 2 Volume RATIO: 2:1:2 EXAMPLE:- Combination between Hydrogen and chlorine:- One volume of hydrogen combines with one volume of chlorine in the presence of sun light to form two volumes of hydrochloric acid gas. SUN LIGHT HYDROGEN + CHLORINE HYDRO CHLORIC ACID 1 VOLUME 1VOLUME 2VOLUMES RATIO: 1:1:2 PROBLEM:-80cm3 of methane is mixed with 200cm3of pure oxygen at room temperature and pressure. the mixture is then ignited when it burns as illustrated by equation. Calculate the composition of resulting mixture if it is cooled to initial room temperature and pressure. EQUATION IS:- CH4 (g) +2O2 (g) CO2 (g)+ 2H2O (l) BY GAY LUSSAC’S LAW:- CH4 + 2O2 CO2 + 2H2O(l) 1VOL. 2VOL. 1VOL. 1VOL=80cm3 2VOL.=2X80=160 cm3 So products produced:- (1) Methane (CH4)= 80cm3 – 80cm3 =0 [That is Methane is totally consumed to form CO2+2H2O] (2) Carbon dioxide formed (1VOL.) = 80cm3 (3) Oxygen =Available (200)- Used, 2VOL.(2X80=160)=40cm3 (4) Water=Negligible PROBLEM:- 200cm3 of co2 is collected at S.T.P. when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at STP in original mixture. 2C2H2(g) +5O2 (g) 4CO2 +2H2O(l) 2VOL. 5VOL. 4VOL. BY APPLYING GAY LUSSAC’S LAW OF VOLUME:- CO2 MEASURED=200cm3 Therefore, 4volumes=200cm3 1VOL.=200 DIVIDED BY4=50cm3 VOL. OF Acetylene required =2VOL.=2(50cm3)=100cm3 VOL. OF Oxygen required= 5VOL.=5(50cm3)=250cm3 PROBLEM:- CALCULATE THE VOLUME OF HCL GAS FORMED AND CHLORINEGAS REQUIREDWHEN 40mlOF METHANE REACTS COMPLETELY WITH CHLORINEAS ILLUSTRATED IN THE REACTION EQUATION.ALL MEASUREMENTS ARE MADE AT SAME TEMPERATURE AND PRESSURE. CH4 (g) + 2Cl2 (g) C2H2Cl2 (g) + 2HCL (g) SOLUTION:- CH4 (g) + 2Cl2 (g) C2H2Cl2 (g) + 2HCL (g) 1vol. 2vol. 1vol 2vol. 40ml ml? ml? 1Vol. =40ml Therefore chlorine required: 2Vol.=2(40ml)=80ml HCl produced= 2(40ml.)=80ml PROBLEM:- 12dm3 ammonia gas formed when a mixture of nitrogen gas and hydrogen gas reacts catalytically. Calculate the volume of hydrogen and nitrogen in initial mixture. All measurements are made at same temperature and pressure. N2 (g) + H2 (g) NH3 (g) SOLUTION:- BALANCE THE EQUATION:- N2 (g) + 3H2 (g) 2NH3 (g) 1vol. 3vol. 2vol. By applying the Gay Lussac’s law that volumes of gases of reactants and products at same temperature bears a simple ratio. 2vol of NH3 =12dm3 1vol=12 divided by2=6dm3 So, 1 vol. of N2(g) = 6dm3 3vol. of H2 (g) = 3(6)=18dm3 TRY THE FOLLOWING QUESTION-OT OS COMING WRONG PROBLEM:-- What law enables us to answer, if the reaction of 6.54 g of Zinc and 3.2g of oxygen produces 9.74g of zinc oxide,the only compound of these elements. How much zinc oxide will be produced if 6.54g of zinc and 5.0g of oxygen were mixed and allowed to react.zn=65 Zn+o ZnO 81g of zinc oxide contains zinc=65g 1g of zinc oxide containsnzinc= 65/ 81 9.74g. of zinc oxide contains = 65/81x9.74=7.8160[Zn] 81g of zinc oxide contains=16g of oxygen 1g of zn oxide contains= 16/81 9.74 of zinc oxide contaons= 16/81x9.74 [oxygen]=1.9239g 7.8160+1.9239=9.7399=9.74g 65g Zn can produce, ZnO=81G OF ZnO 6.54g of Zn can produce,ZnO= 81/65X6.54=3.464g of ZnO 16g of oxygen 81g of ZnO REQUIRES OXYGEN=16g 3.46 g of ZnO requires OXYGEN=16/81X3.46=0.6834g of OXYGEN M SO 5g of oxygen is more than sufficient 0.6834g 31 PROCESS NO-1 CO2 RATIO-3:8 DEFINITE PROPORTION PROCESS NO-3 CO2 RATIO-3:8 DEFINITE PROPORTION PROCESS NO-2 CO2 RATIO-3:8 DEFINITE PROPORTION