Some basic concepts of chemistry present

PPt on some basic concepts of chemistry

1. Grade 11 Chemistry-Introductory session

2. Some basic concepts of chemistry Unit-1

3. Learning objectives appreciate the contribution of India in the development of chemistry understand the role of chemistry in different spheres of life; explain the characteristics of three states of matter; classify different substances into elements, compounds and mixtures; use scientific notations and determine significant figures; differentiate between precision and accuracy; define SI base units and convert physical quantities from one system of units to another; explain various laws of chemical combination; appreciate significance of atomic mass, average atomic mass, molecular mass and formula mass; describe the terms – mole and molar mass; calculate the mass per cent of component elements constituting a compound; determine empirical formula and molecular formula for a compound from the given experimental data; and perform the stoichiometric calculations. .

4. Unit I: Some Basic Concepts of Chemistry General Introduction: Importance and scope of Chemistry. Nature of matter, laws of chemical combination, Dalton's atomic theory: concept of elements, atoms and molecules. Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.

5. Introduction Chemistry deals with the composition, structure and properties of matter . These aspects can be best described and understood in terms of basic constituents of matter: atoms and molecules. Importance of Chemistry: Chemistry is the branch of science that studies the composition, properties and interaction of matter. Chemists are interested in knowing how chemical transformations occur.

6. Chemistry plays an important role in meeting human needs for food, health care products and other materials aimed at improving the quality of life. This is exemplified by the large scale production of a variety of fertilizers, improved varieties of pesticides and insecticides. Similarly many life saving drugs such as cisplatin and taxol , are effective in cancer therapy and AZT ( Azidothymidine ) used for helping AIDS victims, have been isolated from plant and animal sources or prepared by synthetic methods.

7. Nature of Matter Everything around us, for example, book, pen, pencil, water, air, all living beings etc. are composed of matter. Matter is something that has mass and occupies space.

8. NOTE When bonds are formed, there will be release of energy.(Exothermic) When bonds are to be broken, you supply energy(endothermic) In solids, the atoms are tightly packed and have strong bonds between them and hence the energy is low In gases, the atoms are loosely packed and have weak bonds between them and hence the energy is high Fluidity is the tendency to flow. Diffusion is the movement of particles of particles from higher concentration to lower concentration.

9. At the macroscopic or bulk level, matter can be classified as mixtures or pure substances. These can be further sub-divided as follows:

10. Many of the substances present around you are mixtures. For example, sugar solution in water, air, tea etc., are all mixtures. A mixture is a material made up of two or more different substances which are physically combined A mixture may be homogeneous or heterogeneous. In a homogeneous mixture , the components completely mix with each other and its composition is uniform throughout . Sugar solution, and air are thus, the examples of homogeneous mixtures. In a heterogeneous mixtures , the composition is not uniform throughout and sometimes the different components can be observed. For example, the mixtures of salt and sugar, grains and pulses along with some dirt (often stone) pieces, are heterogeneous mixtures

11. Pure substances can be further classified into elements and compounds. Pure substance has definite & constant composition An element consists of only one type of particles . These particles may be atoms or molecules. Two or more atoms combine to give molecules of the element. When two or more atoms of different elements combine, the molecule of a compound is obtained.

12. Molecules

13. The atoms of different elements are present in a compound in a fixed and definite ratio and this ratio is characteristic of a particular compound. Also, the properties of a compound are different from those of its constituent elements. For example, hydrogen and oxygen are gases whereas the compound formed by their combination i.e., water is a liquid . It is interesting to note that hydrogen burns with a pop sound and oxygen is a supporter of combustion, but water is used as a fire extinguisher.

14. Recap Atom is the fundamental unit of matter Matter is something that has mass & occupies space Group of similar atoms combine to form elements/molecules Group of different elements/molecules combine for form compounds Mixtures is a material made up of 2 or more different substances. Homo-uniform composition Hetero-non uniform composition

15. 1.3-Properties of matter & their measurement Every substance has unique or characteristic properties. These properties can be classified into two categories – physical properties and chemical properties . Physical properties are those properties which can be measured or observed without changing the identity or the composition of the substance . Some examples of physical properties are colour , odour , melting point, boiling point, density etc. The measurement or observation of chemical properties require a chemical change to occur. The examples of chemical properties are characteristic reactions of different substances; these include acidity or basicity, combustibility etc

16. If the density of ice is more, then why does it float on water. Answer: Ice has a cage structure making it a hollow compound. Because of this space in between it floats on water.

17. Question The formation of ice layer on water in colder places. Will this phenomenon keep the aquatic animals warmer or colder or do they adapt? For liquid(water) to change into solid(ice), the bonds will get stronger. When the bonds get stronger, they release energy(exothermic) Due to this phenomenon, the aquatic animals feel warm. This phenomenon is also called as Anomalous expansion of water.( 4 degrees to zero degrees)

18. 1.3.1-The International system unit The SI system has seven base units. These units pertain to the seven fundamental scientific quantities. The other physical quantities such as speed, volume, density etc. can be derived from these quantities.

19. 1.3.2-Mass & Weight Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on an object. The mass of a substance is constant whereas its weight may vary from one place to another due to change in gravity.

20. Density Density of a substance is its amount of mass per unit volume. So SI units of density can be obtained as follows: This unit is quite large and a chemist often expresses density in g cm –3 , where mass is expressed in gram and volume is expressed in cm 3 .

21. Temperature It is the degree of hotness or coldness of a body. There are three common scales to measure temperature — °C (degree celsius ), °F (degree fahrenheit ) and K (kelvin).

22. Volume The amount of space that a substance or object occupies or that is enclosed within a container. Units: Cubic metres Litres Kilo litres Milli litres C Cm Gallon

23. Atomic no(Z) is the number of protons [OR] electrons in an atom. Atomic mass(A)-It is the number of nucleons present in an atom. Nucleons are protons+neutrons Molecular mass or molar mass:

24. Molecular mass It is the summation of the atomic masses(but not atomic numbers ) of all the elements present in a molecule. For example: Water(H 2 O) H(Z=1); O(Z=16) The molecular masses of H and O are 1 and 16 respectively. Therefore, H 2 O (2 x 1)+ (16)=18

25. Question Calculate the molar mass of the following : Sodium hydroxide( NaOH ) Carbon dioxide(CO 2 ) Methane(CH 4 ) NaOH-40 CO 2 -44 CH 4 -16

26. 1.4:Uncertainity in measurement Youtube link: https://www.youtube.com/watch?v=oKXzl0g9eOY In chemistry, most of the time, we come across both, theoretical as well as experimental calculations. There are many methods which can help in handling these numbers conveniently and with minimal uncertainty.

27. 1.4.1:Scientific notation Atoms  and molecules have extremely low masses, but they are present in large numbers. Chemists deal with the figures which are as large as  602,200,000,000,000,000,000,000 ,(6.022 x 10 23 ) which is the number of molecules of 2g of hydrogen molecule, i.e., 1 mole of H 2 . To help us in handling these numbers we use the following notation:  m × 10 n , which is,  m  times ten raised to the power of  n . In this  n  is an exponent having positive and negative values and  m  is that number which varies from 1.000… and 9.999…

28. Examples: The scientific notation 578.677 can be written as  5. 78677 × 10 2 . In this the decimal has to be moved left by two places and if it is moved three places left then the power of 10 will be 3. In the same way 0.000089 can also be written as 8.9 × 10 −5 . In this, the decimal is moved five places towards the right, (−5) is the exponent in the scientific notation. TIP-for any scientific notation, the decimal place has to be moved in such a way that, the point is right after the first digit.

29. i) 0.0048  (ii) 234,000  (iii) 8008 (iv) 500.0 (v) 6.0012 600.12

30. Questions Express the following in the scientific notation: (i) 0.0048 4.8 x 10 -3 (ii) 234,000  2.34 x 10 5 (iii) 8008  8.008 x 10 3 (iv) 500.0 5 x 10 2 (v) 6.0012 6.0012 600.126.0012 x 10 2

31. Multiplication & Division Examples (3.9 × 10 6 ) × (2.1 × 10 5 )  3.6 × 10 −5 /2.0 × 10 −4

32. Addition & Subtraction Examples 5.43×10 4 +0.345×10 5 5.43×10 4 -0.345×10 5

33. Definitions of Accuracy & Precision Precision refers to the closeness of various measurements for the same quantity.(Closeness between the values) Accuracy is the agreement of a particular value to the true value of the result.(Closeness with the accurate value) Accuracy is the average value being close to the true value Precision is the closeness between the different values Important 1 mark question

34. Recap Uncertainty in measurement(Scientific notation) m x 10 n Accuracy in measurement: Proper instrument Skill of the operator Accuracy & Precision Significant figures

35. Significant figures The total number of digits in a number including the last digit whose value is uncertain is called the number of significant figures. Are 5 and 5.0 the same? Listed below are the basics of the law: All non zero digits are significant. Zeroes between non zero digits are significant. A trailing zero or final zero in the decimal portion only are significant.

36. Rules for determining the number of Significant figures. All non-zero digits as well as the zeros between the non zero digit are significant. Examples: 576 has three significant figures 576- 3 5004 has four significant figures 0.48 has 2 significant figures 2.05 has three significant figures 0.05 has 1 significant figure 0.0200804 6 SF

37. Zeroes to the left of the first non zero digit in a number are not significant. Examples: 0.0 5 has one significant figure 0.00 403 has 3 0.00 45 has two significant figures 0.0 402 has three significant figures 0.502 has 3 0.000078 has 2 0.0700000802- 9

38. If the number ends with a zero but these zeroes are to the right of the decimal point then the zeroes are significant. Examples: 5.0 has 2 significant figures 2.500 has 4 significant figures 0.0200 has 3 significant figures 0.005800 has 4 significant figures 500.0 1 5.02 50.004 In short, if there are zeroes after the decimal point, they are considered to be significant

39. Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. Examples; 1000-1 1.000-4 0.00100-3 1.00- 3 0.01-1 Examples: The terminal zeros are not significant if there is no decimal point. For example, 100 has only one significant figure, but 100. has three significant figures and 100.0 has four significant figures. Such numbers are better represented in scientific notation. We can express the number 100 as 1×10 2 for one significant figure.

40. 1.4357 & 83.27= In additon & subtraction, the answer must be written in such a way that it has the significant figures equal to the number of least decimal places.

41. Dimensional analysis Often while calculating, there is a need to convert units from one system to other. The method used to accomplish this is called factor label method or unit factor method or dimensional analysis .

42. How many significant figures are present in the following? (i) 0.0025-2 (ii) 208-3 (iii)5005-4 (iv)126,000-3 (v) 500.0-4 (vi)2.0034 -5 Zeroes to the left of the non-zero digit are not considered as significant 0.050 500-1 500.0-4

43. 1.5 LAWS OF CHEMICAL COMBINATIONS

44. 1.5.1:Law of Conservation of Mass Law of conservation of mass states that the matter cannot be created nor be destroyed. This law was put forth by Antoine Lavoisier in 1789. He performed careful experimental studies for combustion reactions for reaching to the above conclusion.

45. 1.5.2:Law of multiple proportions(Dalton) According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. For example: - Hydrogen combines with oxygen to form 2 compounds, water and hydrogen peroxide. H 2 (2g) + (1/2)O 2  (16g)  H 2 O (18g) H 2 (2g) + O 2  (32g)  H 2 O 2  (34g) O in water: O in H 2 O 2 16:32 1:2

46. C(12g) +1/2 O 2 (16g)  CO C(12g) + O 2 (32g)CO 2 Therefore, the ratio of both the oxygen is 16:32,i.e., 1:2 This law states that when two elements(Carbon & oxygen) combine to form more than one compound( CO & CO 2 ), the masses of one element that combine with the fixed mass of other element(C=12g), will always be in whole number ratio.

47. 1.5.3:Law of definite proportions(Proust) According to this law, a given compound always contains exactly the same proportion of elements by weight. This law was given by French chemist, Joseph Proust in 1806. Water: H 2 O 2:16 1:8 Dihydrogen monoxide (Chemical name) CO 2 12:323:8

48. 1.5.4:GayLussac’s law Gay Lussac’s law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure. For example: - H(Hydrogen)(100mL)+O(Oxygen)(50mL)  Water(100mL) The volumes of hydrogen(H) and oxygen(O) which combine together and bear a simple ratio of 2:1. H2 + ½ O2 1 H2O

49. 1.5.5:Avogadro’s law In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules. He made distinction between atoms and molecules. Avogadro’s number: 6.022 x 10 23 N A

50. 1.6:Dalton’s atomic theory In 1808, Dalton published ‘A New System of Chemical Philosophy’ in which he proposed the following : Matter consists of indivisible atoms. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass. Compounds are formed when atoms of different elements combine in a fixed ratio. Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction. Dalton’s theory could explain the laws of chemical combination.

51. 1.7: Atomic & Molecular mass One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon - 12 atom. 1 AMU  1/12 mass of C-12 isotope(98.89%) Today, ‘ amu ’ has been replaced by ‘ u ’ which is known as unified mass Average atomic mass: Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence), the average atomic mass of that element can be computed.

52. Avg. Atomic mass of Chlorine Chlorine has two isotopes, Cl 35 and Cl 37 and their abundance in nature is 75% and 25% respectively. H 1 1 , H 1 2 , H 1 3 X Z A

53. Molecular mass Molecular mass is the sum of atomic masses of the elements present in a molecule. Example: CaCO 3 : 40+ 12+ (16 x 3)=100 u SiO 2 28+(16 x 2)= 60 u

54. Formula mass The  formula mass  of a molecule (also known as  formula weight)  is the sum of the Atomic weights of the atoms in the empirical formula of the compound. Empirical formula is defined as the simplest ratio of a molecule Examples: Glucose(C 6 H 12 O 6 ) E.F:(CH 2 O)12+2+1630 u Hydrogen peroxide(H 2 O 2 )-E.F-HO1+1617 u Benzene,C 6 H 6 E.FCH12+1=13

55. 1.8-Mole concept and molar masses

56. A mole is defined as the amount of substance comprising the same number of fundamental entities as the number of atoms present in a pure sample of carbon weighing exactly 12 g or A mole is defined as the amount of a substance that contains exactly 6.0221023 elementary entities of the given substance.

57. STP stands for Standard temperature and pressure and the conditions are as follows, Temperature is 273.15K/ 273 K/ 0 degrees Pressure is 1 atm  10 5 Pa 1 atm 1.013 bar 1 atm760 torr 1 atm760 mm of Hg Pressure is calculated using barometer 1 mole of a substance will have 6.022 x 10 23 1 mole of carbon atom 6.022 x 10 23 1 mole of Na + ion 6.022 x 10 23 ions

58. The mass of one mole of a substance in grams is called its molar mass. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12 C isotope. 1 mole= 6.022 x 10 23 Number of moles(n)=mass(m)/molar mass(M) In short, n=m/M Calculate the no. of moles of 2g of Calcium n=? m=2g:M=40g n=2/400.05 mol Calculate the mass of 0.5 mol NaOH

59. Round up the following upto three significant figures: (i) 34.216 34.2 (ii) 10.4107 10.4 (iii)0.04597 0.0460 (iv)2808 2810

60. 1.9: Percentage composition The percentage composition of an element in a compound is the mass percentage of the element present in the compound. It tells the mass percentage of each element present in a compound.

61. Example

62. C 2 H 5 OH C=12;H=1;O=16 Molar mass= 2 x12 + 6 x 1 + 16= 46 % of C= 24/46 x 100 % of H= 6/46 x 100 % of O= 16/46 x 100

63. Example-CaCO 3 Total mass= 40+12+(16 x 3)=100 Mass % of Ca =? C=? O=?

64. Magnesium Nitrate Mg(NO 3 ) 2 % of Mg=? 24 % of N=? 14 % of O=? 16 24+28+96= 148 g % of Mg= 24/148 x 100 % of N= 28/148 x 100 % of O=96/148 x 100

65. Recap of mole concept It is not possible to count the number of atoms or molecules involved in chemical reactions, since the molecules are so small, and so many are involved, even in a very small-scale reaction. Instead, it is necessary to measure amounts of molecules by using their mass. The relationship between sub-microscopic quantities like atoms and molecules, and macroscopic quantities like grams, is made using the mole concept. Using moles allows us to count particles by weighing them. Mole=amount of a substance

66. The mole

67. The molar mass of an element The molar mass (M or MM) of an element is the mass in grams of one mole of atoms of the element. It is numerically equal to the atomic mass of the element in amu’s : molar mass in g/ mol = atomic mass in amu

68. 1.9.1-Empirical formula & molecular formula An empirical formula(E.F) represents the simplest whole number ratio of various atoms present in a compound. C 6 H 12 O 6 -(CH 2 O) H 2 O 2 HO C 6 H 6 CH Molecular formula(M.F) shows the exact number of different types of atoms present in a molecule of a compound.(C 6 H 12 6 ) Formulae to remember: No. of moles= mass/molar mass(n=m/M) Molecular formula= (Empirical formula) x n M.F/E.F= n (CH 2 O) 6 C 6 H 12 O 6

69. Steps for solving numerical Step 1: Obtain the mass of each element present in grams Step 2: Determine the number of moles of each type of atom present Step 3: Divide the number of moles of each element by the smallest number of moles Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.

70. Question-1 An oxide of aluminum is formed by the reaction of 4.151g of aluminum and 3.692g of oxygen. Calculate the empirical formula for this compound. What do we know? The compound contains 4.151 g of aluminum and 3.692 g of oxygen We also know the atomic masses by looking them up on the periodic table. Aluminum (27g/ mol ) and oxygen (16.00 g/ mol ).

71. Let's go through the steps to solve this: Step 1: Determine the masses We have these: 4.151 g of Al and 3.692 g of O Step 2: Determine the number of moles by dividing the grams by the atomic mass So let's do that now: n=m/M (4.151/ 27g Al) = 0.1539 mol Al atoms n=m/M(3.692 / 16.00 g O) = 0.2398 mol O atoms Step 3: Divide the number of moles of each element by the smallest number of moles 0.1539 mol Al / 0.1539 = 1.000 mol Al atoms x 2= 2 0.2398 mol O / 0.1539 = 1.500 mol O atoms x 2= 3

72. Step 4: Convert numbers to whole numbers 1.000 Al * 2 = 2.000 Al atoms and 1.500 O atoms * 2 = 3.000 O atoms The compound contains 2 Al atoms for every 3 O atoms Empirical Formula = Al 2 O 3

73. Question-2 When a 0.3546 g sample of vanadium metal is heated in air, it reacts with oxygen to achieve a final mass of 0.6330 g. What is the empirical formula of this vanadium oxide? What do we know: Mass of vandium = 0.3546 g Total mass of V and O= 0.6330 g Mass of oxygen= Total mass- mass of V 0.6330-0.3546=0.2784 g Mass of oxygen=0.2784 g Mol. Mass of Vanadium= 51 g Mol. Mass of oxygen= 16

74. Answer-2 What do we know? The compound contains 0.3546 g of vanadium and a total mass of 0.6330 g We know the atomic masses of vanadium (50.94 g/ mol ) 51g and oxygen (16.00 g/ mol ) Let's go through the steps to solve: Step 1: Determine the masses We are given vanadium as 0.3546g and that must be present in the final mass. mass of the oxygen = final mass - vanadium mass 0.6330 g - 0.3546 g = 0.2784 g

75. Step 2: Determine the number of moles by dividing the grams by the atomic mass 0.3456 g V x (1 mol V / 50.94 g V) = 0.006961 mol V atoms 0.2784 g O x (1 mol O /1 6.00 g O) = 0.01740 mol O atoms Step 3: Divide the number of moles of each element by the smallest number of moles 0.006961 mol V / 0.006961 = 1.000 mol V atoms 0.01740 mol O / 0.006961 = 2.500 mol O atoms Step 4: Convert numbers to whole numbers 1.000 V * 2 = 2.000 V atoms 2.500 O atoms * 2 = 5.000 O atoms The compound contains 2 V atoms for every 5 O atoms Empirical Formula = V 2 O 5

76. Question-3 A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass/Molecular mass is 98.96 g. What are its empirical and molecular formulas? Empirical formula  CH 2 Cl49.5 g Molar mass/molecular mass98.96 g Molecular formula= E.F mass X n n=M.F/E.F98.96/49.5=2 Molecular formula (CH 2 Cl) 2 C 2 H 4 Cl 2

77. Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound? What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 grams? Answer: C 8 H 10 N 4 O 2

78. Given data:

79. The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O. What is the empirical formula for this gas? E.F= CO 2 S.No Element M.Mass n Simplest n ration Whole no ratio 1 C=27.29 g 12 27.29/12=2.27 2.27/2.27=1 one C 2 O=72.71 g 16 72.71/16=4.54 4.54/2.27=2 Two Oxygens

80. A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass/Molecular mass is 98.96 g. What are its empirical and molecular formulas? Molecular formula mass/empirical formula mass=n Fond the E.F Calculate the EFM

81. If the density of methanol is 0.793 kg L–1 , what is its volume needed for making 2.5 L of its 0.25 M solution? Molar mass of methanol(CH 3 OH) 32 g0.032 Kg Molarity= density/molar mass0.793/0,03224.78 mol /L M1V1=M2V2 M1= 24.78 V1? M2= 0.25 V2= 2.5 L

82. Determine the empirical formula of an oxide of iron which has 69.9% iron(A=55.8) and 30.1% oxygen(A=16) by mass. Determine the empirical and molecular formulas for a compound that has a molar mass of 62 g/ mol and is 38.7% C, 9.74% H, and 51.6% O by mass. Empirical formula- CH 3 O E.F mass- 31 Molecular mass-62 n=62/31=20 Molecular formula  (CH 3 O) 2 C 2 H 6 O 2

83. Question-4 Determine the empirical formula of an oxide of iron which has 69.9% iron(A=55.8) and 30.1% oxygen(A=16) by mass. Empirical formula: Fe 2 O 3

84. 1.10-Stoichiometry & stoichiometric calculations The word ‘stoichiometry’ is derived from two Greek words - stoicheion (meaning element) and metron (meaning measure). Stoichiometry deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a balanced chemical reaction. This is done using balance chemical equation.

85. Chemical reaction A chemical reaction takes place whenever a chemical change occurs. Reactants undergo a chemical change forming products. For example: - Magnesium (Mg) + Oxygen (O)  Magnesium oxide ( MgO ). Reactants  are the substances which undergo chemical change in the reaction. Products  are the new substances, formed during the reaction. For example: - Magnesium (Mg) + Oxygen (O)  Magnesium oxide ( MgO ). In this equation Magnesium and Oxygen are reactants and Magnesium oxide is the product formed.

86. Balanced Chemical equation According to law of conservation of mass; mass can neither be created nor be destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants. Number of atoms on each element remains the same, before and after a chemical reaction. For example:- (a) 2 Mg + O 2    2 MgO 2 x 24 + 32  2(24+16) 48+32 2 x 40 80=80 (b) Zn + H 2 SO 4    ZnSO 4  + H 2

87. Balanced chemical equation in Stoichiometry 1 CH  4 (g) + 2O 2  (g)  1 CO 2  (g) + 2H 2 O (g). The above reaction gives the information as follows:- One mole of CH 4  (g) reacts with two moles of O 2  (g) to give one mole of CO 2  (g) and two moles of H 2 O (g). One molecule of CH 4  (g) reacts with 2 molecules of O 2  (g) to give one molecule of CO 2  (g) and 2 molecules of H 2 O (g). 16 g of CH 4  (g) reacts with 2×32 g(64g) of O 2  (g) to give 44 g of CO 2  (g) and 2×18 g(36g) of H 2 O (g). Hints: Write in terms of moles Write in terms of molecules Write in terms of gram molecular weight(molar mass)

88. What do coefficients mean?

89. Problem Find the amount of water produced in the following eqn : 1 CH 4  (g) + 2O 2  (g)  CO 2  (g) + 2 H 2 O (g) From the above equation, 1 mol of CH 4  (g) gives 2 mol of H 2 O (g). 1 mol H 2 O  18 g H 2 O 2 mol H 2 O  ? = 2 × 18 g H 2 O = 36 g H 2 O

90. 1.10.1-Yield of chemical reactions & Limiting reagent In the examples we’ve seen, we have assumed that all of the reactions “go to completion” — that is, that all reactant molecules are converted into product molecules. In real life, some product is almost always lost due to: – too much heating – too little heating – small amounts of contamination in the glassware – impurities in the reactants – incomplete reactions – reactants evaporating into the air – side reactions that form other products

91. Percentage yield: The quantity of a product obtained from a reaction is expressed in terms of the yield of the reaction. The theoretical yield is the amount that would be obtained if the reaction goes to completion (i.e., the maximum amount that could be made). The actual yield of a reaction is the amount that is actually obtained. The percent yield (% yield) is the actual yield expressed as a percentage of the theoretical yield:

92. Numerical problem: Determine the theoretical yield of the formation of geranyl formate from 375 g of geraniol . A chemist making geranyl formate uses 375 g of starting material and collects 417g of purified product. Percentage yield is given as 94.1%. Given data: Actual yield  417 g % yield 94.1 % Theoretical yield? % yield= Actual yield/theoretical yield x 100 94.1= 417/ T.Y x 100 Theoretical yield= 443 g

93. Limiting reagent When we are given a reaction between two or more reactants, one may be completely consumed before the other(s). The reaction must stop at this point, leaving us with the remaining reactants in excess. The amount of this reactant, then, determines the maximum amount of the product(s) that can form, and is known as the limiting reactant. In short, The reactant that is consumed faster or the reactant that limits the formation of products is known as Limiting reagent. The reactant that is not consumed completely/ left over is known as Excess reagent.

94. Example

95. Example of L.R

96. Example-1 Find the limiting reagent 1 mole of silver nitrate is reacting with 1 mole of HCl to give 1 mole of AgCl and 1 mole of HNO 3 1 mol of AgNO 3  143.5 g 50g 0.35 mol of AgNO 3 1 mole of HCl  36.5 g 50g 1.37 mol of HCl Since the no. of moles of AgNO 3 (0.35 mol ) is less than 1 mol , AgNO 3 is the L.R & HCl is the E.R  

97. For the combustion of sucrose: 1 C 12 H 22 O 11 (342g) + 12 O 2  (32g)  12CO 2  + 11H 2 O there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent? Follow the below steps: Step-1: calculate the number of moles(n=m/M) Step-2: Divide the number of moles by the coefficients Step-3: The lower value is the L.R

98. Example-2 For the combustion of sucrose: 1 C 12 H 22 O 11 (342) + 12 O 2  (32)  12CO 2  + 11H 2 O there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent? Step-1: calculate the number of moles(m/M) Step-2: Divide the number of moles by the coefficients Step-3: The lower value is the L.R Step:1- n(sucrose)= m/M 10/3420.029 mol n(oxygen)=m/M=10/320.312 mol Step-2: Sucrose 0.029/10.029 Oxygen 0.312/12 0.026 Oxygen is the limiting reagent.

99. Method-2 Step-1: calculate the number of moles Step-2: Divide the number of moles by the coefficients Step-3: The lower value is the L.R

100. Example-3 Given data: Question-1: To use all of the Mg, how many moles of O 2 do we need? 3.9 mol of O 2 Question-2: To use all of the O 2 , how many moles of Magnesium do we need? 9.4 mol of Mg The first reactant that runs out is known as the Limiting reagent

101. Conversion factor method Question-1: To use all of the Mg, how many moles of O 2 do we need? 7.8 mol of Mg x 1 mol of Oxygen/ 2 mol of Mg  3.9 mol of O 2 To use all of the Mg, 3.9 mol of O 2 Question-2: To use all of the O 2 , how many moles of Magnesium do we need? 4.7 mol of O 2 x 2 mol of Mg/ 1 mol of O 2  9.4 mol of mg To use all of the O 2 , 9.4 mol of Mg To use all of Mg, we require 3.9 mol of O 2 but the question has 4.7 mol of O 2 making it an E.R To use all of O 2 , we require 9.4 mol of Mg but the question has 7.8 mol of Mg making it a L.R

102. Rule: To find the greatest amount of the product, you need to consider the limiting reagent. With respect to Magnesium, 7.8 mol of Mg x 2 mol of MgO /2 mol of Mg  7.8 mol of MgO To use all of the Mg, 3.9 mol of O 2 Formula: Excess reactant= Total reactant- reactant used Excess reactant= 4.7 mol of O 2 - 3.9 mol 0f O 2 Excess reactant= 0.8 mol of O 2

103. Example-4: Given: Molar masses of Al 27; Cl 2 =71, AlCl 3 =133.3 No of moles of Al= m/M=114/27 4.22 mol No of moles of Cl 2 =m/M= 186/71 2.62 mol In the equation, we have 4.22 mol of Al and 2.62 mol of cl 2

104. Question-1: To use all the Al, how much Cl 2 do we require? 4.22 mol of Al x 3 mol of Cl 2 /2 mol of Al 6.33 mol of Cl 2 To use all the Al, we require 6.33 mol of Cl 2 Question-2: To use all the Cl 2 , how much Al do we require? 2.62 mol of Cl 2 x 2 mol of Al/3 mol of Cl 2 1.75 mol of Al To use all the Cl 2 , 1.75 mol of Al Limiting reagent=Cl 2 To find the greatest amount of aluminium chloride, we need to consider the L.R 2.62 mol of Cl 2 x 2 mol of AlCl 3 /3 mol of Cl 2 1.75 mol of AlCl 3

105. 1.75 mol of AlCl 3 n=m/M n=1.75 M=133.3 Mass of AlCl 3 =n x M= 1.75 x 133.3233.1 gms Excess reactant= Total reactant- reactant used Excess reactant= 4.22 – 1.75 Excess reactant= 2.47 mol n=m/M; n=2.47 mol , M=27 Excess reagent in grams= n x M= 2.47 x 27=66.69 grams of Al is in excess.

106. Example-4 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N 2 (g) + 3H 2 (g) → 2NH 3 (g) (i) Calculate the mass of ammonia produced if 2 × 10 3 (2000) g dinitrogen reacts with 1×10 3 g( 1000g ) of dihydrogen . (ii) Will any of the two reactants remain unreacted? (excess) (iii) If yes, which one and what would be its mass?

107. Answer:

108. Example-5 Methane, CH 4 , burns in oxygen to give carbon dioxide and water according to the following equation: CH 4 + 2 O 2  CO 2 + 2 H 2 O. In one experiment, a mixture of 0.250 mol of methane was burned in 1.25 mol of oxygen in a sealed steel vessel. Find the limiting reactant. Question-1: To use all the methane, how much oxygen is required? Question-2: To use all the oxygen, how much methane is required?

109. Example-6:   Aluminum chloride, AlCl 3 , can be made by the reaction of aluminum with chlorine according to the following equation: 2 Al + 3 Cl 2  2 AlCl 3 . What is the limiting reactant if 20.0 grams of Al and 30.0 grams of Cl2 are used Same two questions and solve it Calculate the number of moles of Al and oxygen using the given data Al=27; Cl 2 -71

110. Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol. Take the reaction: NH 3  + O 2    NO + H 2 O. In an experiment, 3.25 g of NH 3  are allowed to react with 3.50 g of O 2 . aWhich reactant is the limiting reagent? b. How many grams of NO are formed? c. How much of the excess reactant remains after the reaction?

111. 1.10.2: Reactions in solutions A majority of reactions in the laboratories are carried out in solutions. Therefore, it is important to understand as how the amount of substance is expressed when it is present in the form of a solution. The concentration of a solution is the amount of substance present in its given volume It can be expressed in any of the following ways. 1. Mass per cent or weight per cent (w/w %) 2. Mole fraction (Chi X) 3. Molarity (M) 4. Molality( m)

112. Mass Percentage The formula: A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute. Solution: Mass of solute= 2g Mass of solvent-18 g Solution= Solute + Solvent  2+18=20g Mass % of Solute= 2/20 x 100 10 %

113. Mole fraction( Chi X) If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are n A and n B respectively; then the mole fractions of A( x A ) and B( x B ) are given as X A + X B =1

114. Numerical 23 g of ethyl alcohol(C 2 H 5 OH) is dissolved in 54 g of water-(H 2 O). Calculate the mole fraction of ethyl alcohol & water.

115. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.

116. What is the concentration/molarity of sugar (C 12 H 22 O 11 ) in mol L  -1  if its 20 g are dissolved in enough water to make a final volume up to 2 L?

117. Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g mL -1   and the mass percent of nitric acid in it is being 69%.

118. Molarity It is defined as the number of moles of the solute in 1 litre of the solution, It is denoted by M. It can also be written as, M= no of moles x 1000/ V(mL) M= m/ M.M x 1000/ V(mL) Since, no. of moles=m/ M.Mass Units- mol /L Note: Molarity depends upon temperature as volume is involved.

119. Numerical Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g of it in enough water to form 250 mL of the solution. M= m/ M.mass x 1000/ V(mL) M= 4/40 x 1000/250 M= 0.1 x 4= 0.4 mol/L (OR) 0.4 M M stands for Molar

120. Molality: It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m. Molality (m) =No. of moles of solute/Mass of solvent in kg Numerical: The density of 3 M solution of NaCl is 1.25 g mL –1 . Calculate molality of the solution. 3M means 3 moles of NaCl in 1000mL/1 L Molar mass of NaCl = 58.5 g 3 moles of NaCl  1 litre Mass of NaCl in 1 litre= 3 x 58.5= 175.5 g Density= mass/Volume Mass= Density x volume= 1.25 x 1000=1250 g Mass of solvent= 1250-175.5=1074.5 g Molality= 3 x 1000/1074.5= 2.79 m

121. The density of 3 M solution of NaCl is 1.25 g mL –1 . Calculate molality of the solution. Molality= no of moles of solute/ mass of solvent in KG density=1.25( d=m/V); V=1000mL Mass of solution= d x V= 1.25 x 1000= 1250 g Molar mass of NaCl = 35.5+23=58.5 Molarity 3M  3 moles of solute in 1L of solution(1000 mL) n= mass of solute/ M.mass of solute 3= mass of solute/ 58.5 mass of solute= 3 x 58.5=175.5 g Mass of solution= mass of solute + mass of solvent 1250 = 175.5 = mass of solvent Mass of solvent= 1250-175.5=1074.5 g Molality= no of moles of solute/ mass of solvent in KG 3 x 1000/1074.5= 2.79 molal

122. Calculate the mass of sodium acetate (CH 3 COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol -1

123. If the density of methanol is 0.793 kg L  -1 , what is its volume needed for making 2.5 L of its 0.25 M solution?

124. Recap States of matter Definitions of atomic no, & atomic masses Properties of matter Homogeneous & heterogeneous Chemical bonding Scientific notation( uncertainity ) Accuracy & precision( Significant figures) Empirical formula & molecular formula Stoichiometry Laws of chemical combination(5 laws) Dimensional analysis Dalton’s atomic theory

125. Formula mass unit Law of definite proportions Average mass & molar mass % composition Avogadro number – 6.022 x 10 23 Mole concept Yield of chemical reactions Basic SI units Concentration Molarity, molality, mole fraction & mass % Limiting reagent & excess reagent

126. Exam point of view Scientific notation m x 10 n Significant figures Empirical formula & molecular formula E.F- It is the simplest ratio of the atoms present in a molecule Molecular formula = Empirical formula x n Limiting & excess reagent L.R- The substance that is in lesser quantity E.R-The substance that is not completely used up Reactions in solutions Molarity, molality, mole fraction & mass %

127. Take the reaction: 4NH 3  + 5O 2    4NO + 6H 2 O. In an experiment, 3.25 g of NH 3  are allowed to react with 3.50 g of O 2 . Which reactant is the limiting reagent? How many grams of NO are formed? How much of the excess reactant remains after the reaction? Step-1: Calculate the no of moles No of moles of ammonia  3.25/170.191 No of moles of oxygen3.5/320.109 Step-2: Divide it by the coefficients Ammonia0.191/40.047(ER) Oxygen0.109/50.021(LR)

128. If 4.95 g of ethylene (C 2 H 4 ) are combusted with 3.25 g of oxygen. What is the limiting reagent? A reaction container holds 5.77 g of P 4  and 5.77 g of O 2 .The following reaction occurs: P 4  + O 2  P 4 O 6 Find the limiting reagent?

129. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g/ mol Mass=? Molar mass= 82.0245 Volume= 500 mL Molarity 0.375 M 0.37s moles 1000mL  500mL Number of moles of sodium acetate0.1875 moles Answer: 15.38g

130. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL –1 and the mass per cent of nitric acid in it being 69%. Molarity=? Mass % 69gms in 100g Molar mass 63 g No of moles? 1.09 mol Volume? Density mass/volume Volume mass/density100/1.4170.92mL Molarity formula

131. What is the concentration of sugar (C12H22O11) in mol /L if its 20 g are dissolved in enough water to make a final volume up to 2 L? Volume 2 L Mass of sucrose20 g Molar mass 342 g Molarity= n/V(L) Molarity= m/M.M X V(L) Molarity= 20/342 x 2 1000/342 0.029 mol /L

132. If the density of methanol is 0.793 kg/L , what is its volume needed for making 2.5 L of its 0.25 M solution?

133. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in percent by mass. (ii)Determine the molality of chloroform in the water sample. 1 ppm 1 part of 10 6 parts Mass percent for 15 ppm 15/10 6 x 1001.5 x 10 -3 % Molality= mass of solute/ molar mass of solvent in Kg 1 Kg= 1000g

134. 100 g of the sample contains 1.5 × 10 –3 g of CHCl3. ⇒ 1000 g of the sample contains 1.5 × 10 –2 g of CHCl3 Mass of solvent 119.5 g Mass of solute 1.5 x 10 -2 Molality mass of solute/ mass of solvent in Kg Molality 1.5 x 10 -2 / 119.5 g

135. If the speed of light is 3.0 ×10 8 m s-1 , calculate the distance covered by light in 2.00 ns. 1 nano sec  10 -9 sec

136. Questions Approximate atomic weight of an element is 26.89. If its equivalent weight is 8.9, the exact atomic weight of element would be (a) 26.89 (b) 8.9 (c) 17.8 (d) 26.7 Atomic weight = (Equivalent weight × Valency ) =(8.9 × 3) = 26.7 ( Valency /factor n = (26.89)/(8.9) ≈ 3).

137. The number of moles present in 6 gms of carbon is: (a) 2 (b) 0.5 (c) 5 (d) 1 n= mass/molar mass 6/12=0.5

138. What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO 3  and 0.1 M NaCl are mixed together (a) 0.1 N (b) 0.2 M (c) 0.05 M (d) 0.25 M When equal volumes of reactants mix, the volume gets doubled. 0.1 AgNO3+ 0.1 NaCl  NaNo3 + AgCl 0.1/2=0.05 M

139. The total number of ions present in 111 g of CaCl 2  is (a) One Mole (b) Two Mole (c) Three Mole (d) Four Mole An organic compound contains carbon , hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be (a) CHO (b) CH 4 O (c) CH 3 O (d) CH 2 O