Chapter 04: Stoichiometry

Mole: Mole is defined in relation to Avogadro's number as follows. Amount of substance that contain 6.022 X 1023 no. of particles is known as mole. One can calculate the number of moles by dividing the number of particles by Avogadro's number Number of moles =_ No.of Particles Avogadro's Number

(a) Molar mass The mass of one mole of a substance expressed in grams is called molar mass. It is represented by M. Its unit is g/mol. The molar mass is the sum of masses of component atoms. (b) Molar Volume The volume of one mole of ideal gas at STP is called molar volume. Its value is equal to 22.414 dm which is rounded to 22.4 dm'. It is denoted by Vm. (c) Molar concentration Number of moles of a substance dissolved per volume of a solution in dm Its unit is mol/dm? It is denoted by C. C= V (dm ) ee 1

Molar Volume The volume of one mole of ideal gas at STP is called molar volume. Its value is equal to 22.414 dm? which is rounded to 22.4 dm. It is denoted by Vm. Relationship between Avogadro's Number and Molar Volume One mole of an ideal gas at 273.16 K and late pressure has a volume of 22.414 dm Since one mole of a gas has Avogadro's number of particles so 22.414 dm? of various ideal gases at STP will have Avogadro's 224dm of particles iP contains 023 gas at STP contains 1 mole and 6.6.02 × 10^23 particles. Example 22.4dm' of CO2 at STP = 1 mole = 6.6.02 × 10^23 molecules 22.4 dm of H2 at STP = 1 mole = 6.6.02 x 10^23 molecules

See Q3. of theory. NUMERICAL PROBLEMS (EXERCISE)

39g of K = 1 mole of K = 6.6.02 × 10^23 56g of Fe = 1 mole Fe = 6.6.02 × 10^23 atoms So, 39g of K and 56g of Fe has equal number of atoms in them, because they have equal number of moles.

See Q4. of theory solution to raise the volume of solution to 500 cm , what would be the concentration now? Ans. Data Mass of NaOH = 2g Initial Volume of NaOH 250 cm° - = 0.250 dm 1000 Final volume of NaOH 500cm = - = 0.500 dm 1000 Required MNaOH=? =? " NaOH Cinitial = ? Cfinal =? Solution (a) MNaoH =23 + 16 + 1 = 40g/mol Answer MNaOH (b) I NaOH M NaOH 2g - = 0.05 mol 40g mol Answer mole = (c) Initial concentration = C volume 0.0500 - = 0.200 mol/ dm Answer 0.250 (d) Final concentration = CF =? 1= moles CF = volume 0.0500 = - = 0.100 mol/ dm Answer 0.500

4g of He = 1 mol of He = 22.414 dm volume 17g of NH3 = 1 mol of NH3 = 22.414 dm volume 96g of SO4 = 1 mol of SO4? = 22.414dm' volume The above data indicates that all the three gases have different masses but occupy same volume because size and masses do not affect the volume, and the number of moles are same.

1 mole of H2 at 0°C and late = 22,414 dm = 6,6.02 × 10^23 molecules. 1 mole of NH3 at 0°C and 1atm = 22.414dm = 6.6.02 x 10^23 molecules. The above data shows that 1 mole of H2 and 1 mole of NH3 at 0°C and late will have Avogadro's number of particles because 1 mole of any substance have same no. of particles i.e. 6.6.02 × 10^23

Data →4 NO+ 6H2O 4 NH, +: 502 Mass of NH3 = 34g = 96g Mass of O2 Molar mass of NH3 (MNH,) = 17g/mol Molar mass of NO (MNo) = 14 + 16 = 30g/mol = 32g/mol Molar mass of (Mo,) Molar mass of H (ArH) = 1g Molar mass of N (Ar N) = 14g Molar mass of O (Ar O) = 16g Required (a) Limiting Reactant =? (b) Mass of NO=? (c) Mass of excess reagent left behind =? Solution MNH, (a) INH, M NH3 34 g = - = 2 moles 17g/ mol 96 g for = 32g/ mol Mor = 3 moles пог To determine limiting reactant, moles of NH3 and O2 are compared with moles of NO as product. NO NH3 4 4 2 2 Ammonia produce 2 moles of NO, now moles of Oz are compared with NO. NO 4 : 1 4/5 : 4/5 × 3 3 2.4moles Limiting reactant is NH3 as it produces less Answer moles of product. (b) Mass of NO To get the moles of NO, its moles can be calculated from balanced chemical equation by using moles of NH3 because it is limiting reactant. Stoichiometry Ratio : NO NH3 4 4 2 2 So 2 moles of NO are formed. MNO= MINOX MNO = 2 x 30g/mol Answer MNo= 60g (c) Mass of excess reactant remaining To get the excess mass of O2 as non-limiting or excess reactant. First, its number of moles left behind calculated by using moles of ammonia from balanced chemical equation. : NH3 02 5 4 5/4 2 5/4 × 2 2 2.5 So, 2.5 moles of 02 completely react with 2 moles of NH3. = 3 moles no, taken = 2.5 no, used left behind = 3 - 2.5 = 0.5 moles mo, left behind = no, x Mo, = 0.5 × 32 mo, left behind = 16g Answer

Definition: The branch of chemistry in which the quantitative relationship between reactants and products in a balanced chemical equation is studied. Assumptions of Stoichiometric calculations are: (1) All the reactants should be converted into products. (ii) No side reaction occurs.

Data Fe,O3+3CO- →2Fe+3CO2 Mass of Fe,O, = 1kg = 1000 g Mass of Fe (Actual yield) = 650 g Molar mass of Fe (Ar Fe) = 56 g/mol Molar mass of Fe203 (M, Fe203) = 2 × 56 + 3 × 16 = 160 g/mol Molar mass of O (Ar O) = 16 g/mol Required % yield of Fe =? Formula Actual yield - ×100 % yield = Theoretical yield Solution NFe, 03 MFejo, 1000g = - = 6.25 mol 160g mol To get the number of moles of Fe produce in a reaction theoretically, its theoretical yield is calculated from balance chemical equation in the following : Fe Fe203 2 1 6.25 × 2 6.25 12.5moles So, 12:5 moles Fe are formed theoretically. mass of Fe (Theoretical yield) = number of moles of Fe x molar mass of Fe = 12.5 moles x 56g/mol mass of Fe (Theoretical yield) = 700g So, the mass of Fe produced is 700g which is called theoretical yield. Now percentage yield is calculated using formula. % yield of Fe = 650g x 100 700 g = 92.86% Result % yield of Fe = 92.86%. Answer

Limiting Reactant The reactant which controls the amounts of products formed in a chemical reaction and is consumed earlier. In a chemical reaction, it controls the quantity of product formed because limiting reactant consumed first and once It is consumed the reaction cannot proceed further, thus limiting the amount of product. Differentiate theoretical and actual

Data: Mass of C2H6 = 1.5g Molar mass of C = 12g/mol Molar mass of H = 1g/mol Molar mass of O = 16g/mol Avogadro's number = 6.022x1023 Molar volume (Vm) = 22.4dm' Molar mass of C2H6 = 30.08g/mol Required (a) Number of molecules of solid CO2 =? (b) Number of molecules of 02=? (c) Number of CH bonds of CzH6 broken =? Solution 1.5g (a) Number of moles of C2H6 = 30.080 = 0.0499moles 2C,H, +70, 4CO, + 6H2O From balanced chemical equation: 2 moles of CHo produces moles of CO2 = 4 moles 0.0499 moles of CHo produces moles of CO2 =. =x 0.0499 = 0.0998moles Number of moles of CO2 = 0.0998moles From balanced chemical equation 2 moles of C2H6 react with moles of O2 = 7 moles 0.499 moles of C2H6 react with moles of O2 -x 0.0499 = 1.7465 moles = ? Number of CO2 molecules = 0.0998 × 6.6.02 x 10^23 Number of CO2 molecules = 6.008 × 1022 molecules Answer (b) Number of O2 molecules = 1.7465 x 6.022 × 1023 Number of 02 molecules = 1.051 × 1024 molecules Answer (c) Number of C2H6 molecules = 0.04998 × 6.022 × 1023 = 3.004 x 1022 molecules One molecule of C2H6 has C - H bonds = 6 bonds Number of C - H bonds broken = 3.002 × 1022x6 Number: of C-H bonds broken = x1023 bonds Answer

Actual Yield Theoretical Yield Maximum amount of Amount of product obtained in a product produced by amount of chemical reaction. given It is smaller than reactant according to balanced chemical theoretical yield. equation. It is greater than actual yield. Percentage Yield of Reaction A chemist is usually interested in the efficiency of a reaction. The efficiency of a reaction is expressed by comparing the actual and theoretical yield in the form of percentage (%) yield. Actual Yield -x100 % yield = - Theoretical

There are following reasons for that: • The processes like filtration, separation by distillation, washing etc. if not properly carried out, decrease the actual yield. • Some of the reactants might take part in a competing side reaction and reduce the amount of desired product. So, in most of the reactions the actual yield is less than the SLO BASED SHORT QUESTION ANSWERS Concept of Mole

The mole is the amount of a substance which contains as many elementary entities as there are atoms in 0.012 kg (12 g) of carbon-12, The elementary entities may be atoms, molecules, ions, electrons, and other particles. It is represented by n. The number of moles of a substance can be calculated by dividing mass in grams by molar mass. The formula for number of moles is: Given mass Number of moles = Molar mass m n= M

The number of entities present in one mole of a substance is a constant number, named Avogadro's Number, i.e. 6.6.02 × 10^23 It is represented by Na. This value is attributed to an Italian scientist Amedo Avogadro (1776 - 1856). Avogadro's number is a physical constant representing the molar number of entities. The exact value of it is 6.02214179X1023 mol. In calculations we use the rounded off value 6.6.02 × 10^23

Juglone, is a dye and is produced from the husks of black walnuts. The formula for Juglone is C6H1003. Juglone, is a natural herbicide (weed killer). It kills off competitive plants around the black walnut tree but does not affect grass ants around he and other noncompetitive plants. theoretical yield. • A reaction may be reversible. Therefore, the amount of product will be reduced by the backward reaction.

Isopentyl acetate CzH1402 is the compound responsible for the scent of bananas. Interestingly, bees release about 1 Mg (1X10° g) of this compound when they sting. The resulting scent attracts other bees to join the attack

Stoichiometric calculations obey law of conservation of mass and law of definite proportions. According to the law of conservation of mass, "matter (mass) can neither be created nor destroyed". It states in terms of stoichiometry that the total mass of reactants is equal to the total mass of products in a balanced equation. According to the law of definite proportions, a pure compound always contains the same element combined in the same ratio by mass is balanced chemical

The balanced chemical equation has the same number of atoms of each element on both sides of equation. It has definite ratios of reactants and products just as compounds have definite ratios of elements. Such ratios are used to calculate the mass or mole of other substances.

The following types of relationship can be studied with the help of a balanced chemical equation involving quantities of reactants) and products). i. Mole-Mole Relationship ii. Mass-Mass Relationship ill. Volume-Volume Relationship iv. Mole-Mass Relationship v. Mole-Volume Relationship vi. Mass-Volume Relationship

Approach to do Stoichiometric Calculations • Mass of known solid or volume of a known gas, or molar concentration of know solution. • Calculate number of moles from the mass of known solid or volume of a known gat or molar concentration of known solution using the relevant formula. • Find the ratio of the known and the unknown reactant or product from the balanced chemical equation. • Calculate the number of moles of the unknown reactant or product using the relevant formula. • Convert the number of moles of the unknown to mass, volume, or concentration of the substance. Molar volume

If the number of moles of a gas is known, one can calculate its volume by multiplying number of moles of the gas with molar volume. Volume of a gas-Number of moles x Molar volume V = nXVm Density of Gases give its

Density is defined as the mass per unit volume of a substance Density = Mass/Volume (d = m/V) As molar mass of all the gases occupies same volume at STP, therefore, density of a gas depends on its molar mass. A gas having higher molar mass will have higher density and vice verse. If the density of the gas at STP is known, its molar mass can be calculated. M=d ×Vm Molar Concentration

Molar concentration of the solution is the defined as the number of moles of substance dissolved per dm' of the solution. Units: Mole / dm? Limiting and Excess Reactant

amounts according to stoichiometry of reaction) and remain unreached at the end of the reaction are called "excess reagents" (or excess reactants). Consider the reaction between hydrogen and oxygen to form water. 2H 2(g) +02(g) 2H,0(0) When we take 2 moles of hydrogen (4g) and allow it to react with 2 moles of oxygen (64g), then we will get only 2 moles (36g) of water.

i. Calculate the number of moles from the given amounts of reactants. ii. Find out the number of moles of product with the help of a balanced chemical equation. wii. Identify the reactant which produces the least amount of product as limiting reactant and the other as an excess reactant.

A large amount of inexpensive reactant is supplied because of the following reasons: a. To ensure that whole of the mass of the reactant is completely expensive converted to the desired product. b. To produce maximum amount of product. c. To increase the rate of reaction

Efficiency of reaction is checked by measuring the percentage yield of the reaction. If the percentage yield of reaction is high than the efficiency of reaction is better. of the Give

field of medicine and is used: i. In the preparation of antibiotics, the stoichiometry ensures that each dose DESCRIPTIVE QUESTIONS (EXERCISE)