stoichiometric mole concept, Cheat Sheet of Chemistry

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Mole Concept MEASUREMENT IN CHEMISTRY 11 1.2 Every measured physical quantity consists of two parts viz a numerical value and the unit. Physical quantity = numerical value » unit The numerical value of e physical quantity is determined experimentally. Every scientific measurement has same degree of uncertainty due to two reasons. @ Skill of the observer (ii) Limitation of measuring instrument There are two ways of expressing this uncertainty. @ One method of expressing il is to use the nolalion + 1 along wilh the doubtful digil 63.7 + O.1om. (i) Another method is expressing itis to use the concept of significant figures Rules for determining the number of significant figures: Rule (i) All nonzero digits are significant 6.324 has four significant figures. 7.92 has three significant figures. 1.2 has two significant figures. Rule (ji) Azeroappearing at a beginning of a number is not significant 0.32 has two significant figures 0.018 has two significant figures. 0.0004 has one significant figure 0.324 has three significant figures. Rule (ii) Azer appearing in the middle of a number or at the end of a number that included a decimal point is significant. 8.023 has four significant figures. 3.01 has three significant figures. 4.050 has four significant figures. 1.5000 has five significant figures. Rule (iv) Ifa number ends in zeros bul these zeros are not to the right side of a decimal point, then they may or may nat be significant. 6500 may have two or three or four significant figures if 6500 Is expressed as @ 65x 10% then it has two significant figures. (i) 6.50 x 103 then ithas three significant figures. (ji) 6.500 x 102 then it has four significant figures. (iv) In order to avoid the ambiguity in the rule (generally very large and very smalll numbers are expressed in exponential form or scientific notation. In this notation a numbers expressed as N x 10" where N = anumber with at least a single nonzero digit to the left of the decimal point. n = an integer. MOLE CONCEPT Ex.6 Ex.7 1 In Avogadro's number 6.023 x 10”. 6.023 is a number of four significant figures ‘.e.. in the exponential notation, the numerical portion represents the number of significant figures. 4.5 Calculations involving significant figures. Rule 1 : The result of addition or subtraction should be reported to the same number of decimal places as that of the term with least number of decimal places Addition example: 6.23 + 2.1 + 1.712 = 10.042 This value should be taken as 10.0 only because 2.1 has only one decimal place. Subtraction example: 16.3215 — 2.706 = 13.6155 This value should be taken as 13.615 because 2.706 have only three decimal places. Rule 2 : The result of multiplication or division should be reported to the same number of significant figures as possessed by the least precise term. : for multiplication: 6.102 x 2.1 = 12.8142. This value should be taken as 12 because 2.1 has two significant figures. : for division : §.2765/1.25 = 4.2212. This value should be taken as 4.22 because 1.25 have three significant figures. DIMENSIONAL ANALYSIS 2.4 The units of mass, length and time are independent units and as they are not derived from any other unit, these are called fundamental units. System International (S.|.) has seven basic units. Quantity Unit Symbol Length Meter m Mass. Kilogram Kg Time Second s Temperature Kelvin K Amountofsubstance mole mol. Electric Current ampere A Luminous intensity Candela Cd 2.2 Derived Units : (i) Area length « breadth == mxm m (i) Velocity displacement/lime = mis ms! (iii) Acceleration = velocity / time = ms‘/s =ms? (vy) Force = mass acceleration = mxa = Kg ms? ) Pressure = Force / area a Kg mt 52 = Nm2 (i) Volume length cube = mxm«m m3 (vil) Density mass / volume = Kgim* Kg m? (vii) Energy Work = Force « distance tavelled= Kgms2xm_ = Kg m?s2 Energy and work have same units. (x) Frequency = no, of cycles per second = cyclesisecond = s"' or Hz () Electric charge = current x time = ampere x second = Axs = coulomb. (xi) Power = Energy / time = Kg m2? 53 = Watt (xii) Potential difference = — Power/Ampere - Kgm@sS At = JAts+ MOLE CONCEPT Ex.11 Exi2 Ex.13 Ex.14 Ex.15, 20a - 0, + 2C20 2x40 1x32 2(40+16) = 80g = 32g =112g9 Total mass reactant =Total mass product = 112g, 5.2 The law of constant composition or definite proportion (i) This law was given by Proust. (ii) This law was verified by Stass & Richards. (iii) This law states that “All pure samples of the same chemical compound contain the same elements combinad in the same proportion by mass, irrespective of the method of preparation” Different samples of carbon di oxide contain carbon and oxygen in the ratio of 3:8 by mass. Similarly in water ratio of weight of hydrogen to oxygen is 1:8. 5.3 The Law of multiple proportion (i) This law was given by John Datton. (ii) The law was verified by Berzelius. (ill) This law states that “when two elements A and B combine together to form, more than one compound, then several, masses of A which separately combine with a fixed mass of B. are in a simple ratio” co and co, 12:16 12:32 ratio= 16:32 1:2 Hs and H,S, 2:32 264 ratio= 32:64 Ao 5.4 The Law of reciprocal proportions (i) This law was given by Ritcher. (ii) This law states thal “when two elements combines separately with third element and form differant types of molecules, their combining ratio is directly reciprocated if they combine directly.” C combines with © to form CO, and with H to form CH,. In CO, 12g of C reacts with 32g of O, whereas in CH, 12g of C reacts with 4g of H. Therefore when O combines with H, they should combine in the ratio of 32:4 (ie. 8:1) or in simple multiple of it. The same is found to be true in H,O molecules. The ratio of weight of Hand OinH,Ois 1:8 5.5 ‘The Law of Gaseous volume. (i) This law was given by Gaylussac (ii) This law states that ‘when gas combine, they do so in volume which bear a simple ratio to each other and also to the product formed provided all gases are measured under similar conditions.” Or in other words volume of reacting gasses and product gases have a simple numerical ratio to one another. Hag) + Cl, (g) > — 2HCl(g) 1 unil vol. 1 unil vol 2 unit vol ratio= 1:1:2 N+ 3H, > 2NH, unit vol. 3 unit vol 2 unit vol ratio= 1:3:2 5.6 The Avogadro Law (i) This law states that “equal volume of all gaseous under similar conditions of temperature and pressure contain equal number of molecules" MOLE CONCEPT 2H +O, > 2,0 2vol. 1 vol. 2val. ratio2:1:2 (ii) This law helped to remove anomaly between Dallon's alomic theory and Gay lussac’s law of volume by making a clear distinction between atoms and molecules (iii) Itreveals that common elements gases like hydrogen, nitrogen, oxygen etc. are diatomic. (iv) It provides a method to determine the atomic mass of gaseous elements. (¥) Il provides a relationship between vapour density and molecular mass of substances. 2% vapour density (VD) = molecular mass of gas. SOLVED PROBLEMS 1, 23.4 of NaCl on reacting with 68g of AgNO, formed 57.4g of AgC! and 34g of NaNO, Thisis in accordance with (1) The law of conservation of mass. (2) The law of constant composition. (3) The law of reciprocal proportion. (4) None of these Ans. (1) NaCl + AgNO, > = AgCl_— + NaNO, 23.49 68g S74g 3g mass of reactant = mass of product = 91.4g. Hence the law of conservation of mass is obeyed 2. 8g of CaCO, on heating gave 3.25g CO, gas. The mass of residue leftis - (1)49 (2)4.48g (3) 12g (4) 16g Ans. (2) CaCO,(s) > — CaO(s) + CO,(g) 8g x 3.529 According to law of conservation of mass mass of reactants = mass of products. 8 =x + 3.52 (where x is mass of residue) x=8-3.52= 4.489 as In one experiment 12g of Mg combine with 8g of O, to form 20g of MgO. In another experiment when 150g of Mg combine with 100g of ©, then 250g of MgO is formed. Above two experiment follows- (1) The law of conservation of mass (2) The law of constant composition (3) The law of definite proportion (4)Alllof the above massofMgreacted 12 _ 3 massofOpreacted 8 2 Ans. (4) | experiment mass of Mg reacted experiment Z&SS° “great mass of O2 reacted Hence both law of conservation of mass and constant composition is obeyed. 4. Hcombines with 0 to form two compounds water (H,O) and hydrogen peroxide (H,0,). If 2g of H reacts with © completely to form 18g of water and 34g of H,O, then what is ratio of mass of © combining with H. (1329 (2)4:16 (31:2 (4) None of these Ans. (3) The ratio of masses of oxygen which combine with 2g of hydragen to give H,O and H,O, are 16:32 = 1:2. That is, one mass of O is a multiple of the other mass of O combining with the same mass of H to form different compounds. 5. The vapour density of a gas is 11.2, then 11.2g of this gas at N.T.P. will occupy a volume- (1)4.12 litres. (2) 0.142 litres (3) 112 litres (4)11.2litres Ans. (4) Vapour density of any gas at N.T.P occupies a volume of 11.2 litres. MOLE Ex.18 Ex.19 NOTE: Ex.20 CONCEPT MASS, MOLE AND NUMBER CONVERSIO! 6.0210" Number of entites Weight of substance 6.0210 GAM or GMM ices Mole = Gram Molecular Vass (GMM) , fam Niecider Mase 22.4litres at STP ‘Gram Atemic Mass (GAR!) a (GMV) (in litre) at STP 22.4 Mole = ATOMIC MASS SCALE Carbon as standard : The modem standard reference for atomic mass is carbon isotope of mass number 12. £ Mass of talom of element Atomic mass of an elernent = 7p Mass of 1atom of c? IMPORTANT POINTS - Atomic mass is not a mass (weight) but a number on Atomic mass is not absolute but relative to the mass of the standard reference element C'2 MOLECULAR MASS Itis number of times a molecule is heavier than 1/12th of an alom of C'2, ut if lecul Molecular Mass -—Wass of tmoleoule 49 *Mass of one atam of cont] IMPORTANT POINTS . Molecular mass is not a mass (weight) but a number. all Molecular mass is relative and not absolute. _ Molecular mass expressed in grams and is called Gram Molecular Mass (GMM). ie Molecular mass is calculated by adding all the atomic mass of all the atoms in a molecule. CO, = 124 (2x 16) = 44, NH,CI = (14 x 1) + (1 4) + (1 * 35.5) = 53.5 MORE ABOUT ATOMIC MASS AND MOLECULAR MASS 11.1 Atomic Mass : The relative atamic mass (atomic weight) of an elementis the mass of one atom of the element compared with the mass of an atom of ,C'? (carbon -12 isotope) taken as 12.0000 units. Gram Atomic Mass (GAM) : The gram atomic mass of an element is the atomic mass of it, expressed ing. 1. gram atom of hydragen = 1.0089 ‘gram atom of carbon = 12g 1 gram atom of chlorine = 35.59 (i) Atoms of the same element which have different relative masses are called isotopes. (ii) In case of isotopes, atomic mass of the elements is average of relative masses of differentisotopes of the element. There are two isotopes of chlorine. cl and ch? relative mass 35 37 relative abundance 3 1 MOLE CONCEPT Ex.21 (At.mass of | isotope «relative abundance, + (At. mass of Isotope relative ebundancey|) At.mass of element ~ Total relative abundance At mass of Cle (85%3)+87x1) _ 105437 _ 45 5 344 4 11.1.1 Determination of Atomic mass : atomic mass can be determined by Dulog and Petit's method (1819). It states the “the product of atomic mass and specific heat of an element is 6.4 approximately”. Mathematically : - Atomic mass * specific heat = 6.4 The specific heat of metal is 1J/g/K. If equivalent mass of metal is 9. Calculate its exact atomic mass. 64 specific heat = 1Wigik = 54 oF 0.24 cal /g/K 64 : 64 alomic mass (app.)= Specific heat = gaa = 28-75 Atamic mass(app) _ 26.75 Equivalent mass ~~ 9 now valency = =2.9 ~3.(-» valency is integer) Exact atomic mass = Equivalent mass « valance =9x3=27 Atomic mass can also be determined from molecular mass and alomicily. Atomicity :- It may be defined as the number of atoms present in a molecule of an element. molecular mass Atomic mass = ~~ Siomicily 11.2.) MOLECULAR MASS (i) Molecular mass (the relative molecular mass) :- The relalive molecular mass (weight) of an element or compound is the mass of one molecule of the element or compound compared with the mass of atom of ,C'2 which is arbitrarily assigned as 12.0000. (ii) Gram molecular mass (GMM) :- The molecular mass expressed in grams is called gram molecular mass (GMM), or gram mole or mole. (iii) Molar volume or gram molecular volume (GMY) ; - The volume occupied by one gram mole or one mole of a gas at STP is called molar volume or gram molecular volume (GMV), 11.2.1 DETERMINATION OF MOLECULAR MASS (i) Regnault method (a) _ Bythis method vapour density (V.D) of the gas are determined by direct weighing weight of a certain vol. of gas or vapour under certain temp. and pressure ©) 'V.D.= ‘weight of the same vol. of Hunder same temp. and pressure (il) Diffusion method :- (a) Itis based on Graham’s Law of diffusion. (0) Graham's Lawstates that ‘the rate of diffusion of different gases, under similar conditions of tempera- ture and pressure, are inversely proportional to the square roots of their densities (molecular mass)’ MOLE CONCEPT 14. Ex.23 Ex.25 STOICHIOMETRY (i) Stoichiometry is a Greek word (sloicheio = element and metran = element) (i) Stoichiometry is a calculation of the quantities of reactant and product involved in a chemical reaction. (iii) Stoichiometry can be classified into two groups - (a) Gravimetric Analysis (b) Volumetric analysis 14.1 Stoichiometry and Problem Solving In problem solving we shall first discuss gravimetric analysis of chemical reaction. In gravimetric analysis we relate the weights of two substances or a weight of a substance with a volume of a gas or volumes of two or more gases. 14.2 Problem involving Mass-Mass relationship- What amount of MgO is formed when 12g of Mg reacts with oxygen completely. Following ara the staps to solve the above problem where mass of reactants given and mass of productis to be calculated. Step 1 Write balance equation Mg +O, + MgO (unbalanced reaction) 2Mg +0, + 2MgO (balanced reaction) Step 2. Write the moles below the formula 2Mg +0, + 2MgO moles 2 1 2 (this represents simplest molar ratio among reagents.) Step 3. Write the relative weights of the reactant and product 2Mg +O, +» 2MgO (2% 24) 2x (24418) = 48g = 80g Step 4 Apply unitary method 48g of Mg gives 80g of MgO 1280, 4° 12g of Mg gives of MgO = 20g of MgO 14.3 Problem involving Mass-Volume relationship. By heating 10g CaCO, 5.69 of CaO is formed. What volume of CO, obtained in this reaction at STP. Step 1 Write balance equation CaCo,(s) + — Ca0(s)+C0,(9) Step 2. Write the moles below the formula Caco,(s) > Ca0(s)+C0,(¢9) moles 1 * 1 Step 3. Write the relative weights of the reactant and volume of product CaCco,(s) + Cad(s)+C0,(9) 400g 22.4. at STP Step 4 Apply unitary method 100g of CaCO, gives 22.4L of CO, 10g CaCO, gives teazee =2.24L of CO, at STP 14.4 Problem invelying Volume-Volume relati hip- Hydrogen reacts with nitrogen to produce ammonia according to this equation. 3H,(9)+N,g) > ~—- 2NH,G) Determine how much ammonia would be produced if 200L of hydrogen react completely with nitrogen to form ammonia Page # 11 MOLE CONCEPT Sol. 13. Sol. Sol. Step 1 Write males below the balance equation 3H{g)+N(g) > 2NH,(9) 3 1 2 Step 2. Write relative volume of reactants and product 3H,g) + Ng) > 2NH.(9) 3x 22.4 1x 22.4 2% 22.4 = 67.2L = 22.4L = 44.8L Step 3. Apply unitary mathod 67.2L of H, gives 44.8L of NH, 200 x 44.8 67.2 Note ; Quantity of a substance consumed or produced can be determined only if we use a balance chemical equation. 200L of H, gives =133.3 L of NHg LIMITING REAGENT ()) Limiting Reagent (reactant) : The reactant which is completely consumed during the reaction. (li) Excess Reagent (reactant) : The reactant that is not completely consumed ina reaction. The moles of product formed are always determined by the initial moles of limiting reagent. Calculate the weight of iron which will be converted into its oxide by the action of 18g of steam. 3Fe + (4H,O + FeO, + 4H, 3Fe + 4H, +> FeO, + 4H, 3% 56 4«18 1% 232 = 168 =72 = 232 now 72g of steam (H,O) reacts with 168g of Fe 18 x 168 72 18g of steam will react with = 42gof Fe Calculate the weight of lime (CaO) obtained by heating 200kg af 90% pure limestone (CaCO,) 100kg impure sample has CaCO, = 90kg 200k: wapaneacy: < 228208 ennai, g impure sample has CaCO, = 755 = 180 kag now CaCO, > + Cad + CO, 100g 569 44g 100kg CaCO, gives 56kg of Cad 56 «180 180kg of CaCO, gives “4 = 100.8kg of Cad Oxygenis prapared by catalytic decomposition of potassium chlorate (KCIO,). Decomposition af KCIO, gives potassium chloride (KC!) and oxygen (0,). If 4.2 mole of oxygen is needed for an experiment, how many grams of KCIO, must be decomposed. step KCIOJs) > —-KClis) + OQ) Step 2 2KCIO,(s) > 2KC\K(s) + 20,9) 2mole 3mole Step3 (2x 122.5) 3 mole = 245g 3 mole Step 4 If 2.45g of KCIO, gives 3 mole of oO, then 4.2 mole of oO, will be obtained from 42245 =———— = 343g of KCIO,, 700 e 3 MOLE CONCEPT M2 M/5 M/10 M/100 M/L000 M SM. 10M Semi Penti Deci Centi Milli Penta Deca molat molar molar molar molar molar molac molar solution solution solution solution solution solution solution solution T2 mel) WS moll | WOmoliL | 1100 molt | 1/1000 mo | 1 mol/L Smoll 10 mol Ex.26 4g NaOH is present in 100mI of its aqueous solution. What is the molarity 12M (2)1M (3) 10M (4)0.1M Ans. (2) w 1900 4 1000 Sol, Molarity=—_,,__1000_ 4 ,, 1000 _ GMM ~ volume (ml) 40" 100 1M solution of NaOH Ex.27 The solution of H,SO, contains 80% by mass. Specific gravity (density) of solution is 1.71 g/cc. Find its Molarity, so, M=10x¢2percent ma19*1.71, 99 -13.05 cM 98 Ex.28 Toneutralizes 20mL NaOH, the volume of 1M HN (1)4mL (2)3mL (3)2mL (4)4mL Sol. NaOH HNO, MY, = MN, 02x20= 1xV, V,=4mL 17, MORE ABOUT EXPRESSION OF STRENGTH) CONCENTRATION OF SOLUTION “The amount of solute which is dissolved in unit volume of solution is called concentration of solution." amount of solute Concentration - 2mount of Solute volume of solution 17.1 Weight - weight pereent (w/W) : Weight of solute present in 100g of the solution weight of solute (g) wi Weight percent = ®) 5.100 W ere weight of solution (g) weightpercent 7 * 100 Ex.29 Whalis the weight percentage of NaCl solution in which 20g NaCl is dissolved in 60g of water. (1)10% (2)5% (3) 25% (4) 15% weight of NaCl weight percentage of Nac| = “SSM oT NAL". 199 Ans. (4) omrercemag: Weight of solution 20 fa a =~ «100=25% NaCl solution (w/W) 20+ 60 17.2 Volume - volume percent (v/V) : (In liquid - liquid solution) Volume of solute in ml. present in 100ml of the solution is called volume - volume percentage. Volumia= volume percentage = “UM Of eottte (mL) volume of solution (mL) volume percent = -%100 Ex.30 A solution is prepared by mixing of 10ml ethanol with 120ml of methanol, What is volume percentage of ethanol (1)10% (2)7.7% (3) 20% (4) 15% volume of ethanol