* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Some basic concepts of chemistry
Resonance (chemistry) wikipedia , lookup
Freshwater environmental quality parameters wikipedia , lookup
Electron configuration wikipedia , lookup
Water splitting wikipedia , lookup
Nuclear chemistry wikipedia , lookup
Drug discovery wikipedia , lookup
Artificial photosynthesis wikipedia , lookup
Inorganic chemistry wikipedia , lookup
Analytical chemistry wikipedia , lookup
Hypervalent molecule wikipedia , lookup
Abundance of the chemical elements wikipedia , lookup
Chemical element wikipedia , lookup
Electrolysis of water wikipedia , lookup
Isotopic labeling wikipedia , lookup
Computational chemistry wikipedia , lookup
Hydrogen atom wikipedia , lookup
Metalloprotein wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Biochemistry wikipedia , lookup
Chemistry: A Volatile History wikipedia , lookup
Molecular dynamics wikipedia , lookup
History of chemistry wikipedia , lookup
Size-exclusion chromatography wikipedia , lookup
IUPAC nomenclature of inorganic chemistry 2005 wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Stoichiometry wikipedia , lookup
C-1 Vidyasagar Classes SOME BASIC CONCEPTS OF CHEMISTRY INTRODUCTION Matter can be classified as mixtures and pure substances Mixtures contain two or more substances in any ratio. These substances are called components of the mixture and can be separated by physical or mechanical means without loosing their identity. A mixture may be heterogeneous or homogeneous. Heterogeneous mixtures do not have uniform composition through out for example charcoal in water, sugar in salt etc. Homogeneous mixtures have uniform composition for example air, salt in water, alloys etc. Pure Substances have same composition and properties. Their components cannot be separated by simple physical methods for e.g. iron, gold, water, glucose etc. These are subdivided into elements and compounds. Elements are pure substances that cannot be decomposed into simpler substances by chemical changes. It consists of only one type of atoms. e.g. sodium (Na), silver (Ag), oxygen (O2 ), Chlorine (Cl 2 ) etc. Presently 118 different elements are known, out of which 92 are naturally occurring. Compounds are formed when two or more atoms of different elements combine together in a fixed ratio e.g. carbon dioxide (CO 2 ) water (H 2 O) ammonia (NH 3 ) etc. The properties of the constituent atoms are completely different from that of the formed compound. For example water is formed by the reaction of hydrogen and oxygen at very high temperature. Hydrogen is combustible whereas oxygen helps in combustion but the product water is neither combustible nor helps in combustion. It is used for extinguishing fire. MATTER (has mass, occupies space) MIXTURES • Variable composition • Components retain their characteristic properties. • May be separated into pure substances by physical methods • Mixtures of different compositions may have widely different properties. HOMOGENEOUS • Have same composition throughout i.e. one phase • Components indistinguishable are HETEROGENEOUS • Do not have same composition throughout i.e. two or more phase • Components are distinguishable Properties of Matter: Properties are the characteristic quantities of different kinds of matters by which they can be easily recognized. These properties can be classified as physical properties and chemical properties. Physical properties of a substance can be measured or observed without changing their identity or composition. e.g. colour, odour, boiling point, melting points, density etc. Chemical properties of substances are those in which they undergo change in composition to form new substances. e.g. reaction of different substances. 2H 2 + O 2 → 2H 2 O Chemistry -1(Some Basic Concepts of Chemistry) PURE SUBSTANCES • Fixed composition • Cannot be separated into simpler substances by physical methods • Can only be changed in identity and properties by chemical methods. • Properties do not vary. COMPOUNDS • Can be decomposed into simpler substances by chemical changes, always in a definite ratio ELEMENTS • Cannot be decomposed into simpler substances by chemical changes FUNDAMENTAL AND DERIVED UNITS The quantitative scientific observation generally requires the measurement of one or more physical quantities such as mass, length, density, volume, pressure, temperature, etc. A physical quantity is expressed in terms of a number and a unit. Without mentioning the unit, the number has no meaning. Characteristic of a good unit 1. It should be invariable with place and time. 2. It should be easily reproducible. 3. It should be universally acceptable. We teach success XI + XII + Entrance - 2015 C-2 Vidyasagar Classes Common system of units based upon three basic units of length, mass and time are tabulated below. Physical quantity Length Mass Time CGS centimeter gram second Units FPS foot pound second MKS Meter kilogram second The three basic units, i.e., units of mass, length and time are independent units and cannot be derived from any other units, hence they are called fundamental units. The three fundamental units cannot describe all the physical quantities such as temperature, intensity of luminosity, electric current and the amount of the substance. Thus, seven units of measurement, namely mass, length, time, temperature, electric current, luminous intensity and amount of substance are taken as basic units. All other units can be derived from them and are, therefore, called derived units. The units of area, volume, force, work, density, velocity, energy, etc., are all derived units SI Units of Measurement: The SI system has seven basic units. The various fundamental quantities that are expressed by these units along with their symbols are tabulated below. Basic physical quantity Length Mass Time Temperature Electric current Luminous intensity Amount of substance Unit Metre Kilogram Second Kelvin Ampere Candela Mole Symbol m kg s K A cd mol Sometimes sub multiples and multiples are used to reduce or enlarge the size of the different units. Prefix deci centi milli micro nano pico femto atto Symbol d c m µ n p f a Value 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18 Multiples Prefix Symbol deca da hecto h kilo k mega M giga G tera T peta P exa E Chemistry -1(Some Basic Concepts of Chemistry) (b) Volume = length × breadth × height = m × m × m = m3 [cubic metre] (c) Density = (d) Speed = (e) Acceleration = (f) Force = mass × acceleration = kg × ms−2 = kg ms−2 [newton (N)] Pressure = force per unit area kg ms −2 = = kg m−1s−2 or Nm−2 (pascal−Pa) 2 m Energy = force × distance travelled = kg ms−2 × m = kg m2 s−2 (joule − J) (g) (h) mass kg = 3 = kg m−3 volume m distance covered metre = = ms−1 time time ms −1 change in velocity = = ms−2 time taken s (i) Electric charge = Current × Time = A × s = Coulomb (C) (j) Electric potential = (k) Concentration = (l) Electrochemical equivalent (Z) = Energy 2 –2 –1 = kg × m × s A Charge = JA–1s–2 = volt (V) or JC–1 Mole –3 = mol m 3 m E = kg ⋅ C–1 F LAWS OF CHEMICAL COMBINATION The use of SI system is slowly growing; however, older systems are still in use. Sub multiples SI Units for Some Common Derived Quantities 2 (a) Area = length × breadth = m × m = m [square metre] Value 10 102 3 10 106 109 1012 1015 1018 Experimental results point out that chemical combination is governed by following laws. LAW OF CONSERVATION OF MASS (LAVOISIER) (LAW OF INDESTRUCTIBILITY OF MATTER) Statement: This law states that during a chemical change no gain or loss of mass can be measured. OR Mass can neither be created nor be destroyed in a chemical reaction i.e. total mass of the system remains constant. Thus when reactants are completely converted into products, then Total mass of reactants = Total mass of products In case reactants are not consumed completely the above relationship will be Total mass of reactants = Total mass of products + Masses of unreacted reactants We teach success XI + XII + Entrance - 2015 C-3 Vidyasagar Classes SOLVED EXAMPLE 1. LAW OF MULTIPLE PROPORTIONS (JOHN DALTON) What weight of sodium chloride would be decomposed by 4.9 g of sulphuric acid, if 6 g of sodium bisulphite and 1.825 g hydrogen chloride were produced in the reaction? Solution: NaCl + H 2 SO 4 → NaHSO 4 + HCl According to law of conservation of mass W (NaCl) + W ( H 2SO 4 ) = W ( NaHSO4 ) + W (HCl) W (NaCl) + 4.9 = 6 + 1.825 LAW OF RECIPROCAL PROPORTIONS OR EQUIVALENT PROPORTIONS (RICHTER) W (NaCl) = 6 + 1.825 – 4.9 = 2.925 ≈ 2.9 g LAW OF CONSTANT PROPORTIONS OR DEFINITE COMPOSITION (JOSEPH PROUST) Statement: A given chemical compound always contains the same elements combined together in definite proportion by mass, i.e., it has a fixed composition and does not depend on the method of its preparation or the source from which it has been obtained. This law is also known as ‘Law of definite proportions’. E.g. pure water obtained from different sources such as river, well, spring, sea or in a chemical reaction etc. always contains hydrogen and oxygen combined together in the ratio 1 : 8 by weight. SOLVED EXAMPLE 2. In an experiment, 2.4g of iron oxide on reduction with hydrogen yield 1.68g of iron. In another experiment 2.9g of iron oxide give 2.03g of iron on reduction with hydrogen. Show that the above data illustrate the law of constant proportions. Solution : In the first experiment W (FeO) = 2.4 g, W (Fe) = 1.68 g W ( O 2 ) = W (FeO) − W (Fe) = (2.4 − 1.68) = 0.72 g Ratio of oxygen and iron = 0.72 : 1.68 = 1 : 2.33 In the second experiment W (FeO) = 2.9 g, W (Fe) after reduction = 2.03 g W ( O 2 ) = (2.9 − 2.03) = 0.87 g Ratio of oxygen and iron = 0.87 : 2.03 = 1 : 2.33 Thus, the data illustrate the law of constant proportions, as in both the experiments the ratio of oxygen and iron is the same. Chemistry -1(Some Basic Concepts of Chemistry) Statement: When two elements combine to form two or more than two compounds, the weight of one of the elements which combines with a fixed weight of the other, bears a simple whole number ratio. Explanation : Hydrogen and oxygen combine to form two compounds water (H 2 O) and hydrogen peroxide (H 2 O 2 ). H2O H2O2 2 : 16 2 : 32 Ratio of the masses of O which combine with fixed mass of H is 16 : 32 or 1:2 Statement: The weight ratio of two elements A and B which combine with the fixed weight of C separately is either the same or some simple whole number multiple of the weight ratio in which A and B combines together. Explanation: Consider three 2 H2 elements sulphur, oxygen and hydrogen. Both sulphur and H2S H2O oxygen separately combine to form H 2 S and H 2 O respectively. 32 16 They also combine with each O S 32 SO2 32 other to form SO 2 as shown below In H 2 S, 2 parts by weight of hydrogen combine with 32 parts by weight of sulphur. In H 2 O, 2 parts by weight of hydrogen combine with 16 parts by weight of oxygen. ∴ The ratio of the weights of sulphur & oxygen which, combine separately with the fixed weight i.e. 2 parts of hydrogen is, S :O 32 : 16 2 : 1 ……… (i) Now the ratio of sulphur and oxygen in SO 2 S:O 32 : 32 1 : 1 ………(ii) The two ratios (i) and (ii) are related to each other as 2 1 : or 2 : 1 (Simple multiple of each other) 1 1 Sulphur and oxygen also react to form SO 3 which is also in accordance with the law. Now the ratio of sulphur and oxygen in SO 3 S:O 32 : 48 2 : 3 ………(iii) The two ratios (i) and (iii) are also related to each other as 2 2 : or 3 : 1 (Simple multiple of each other) 1 3 We teach success XI + XII + Entrance - 2015 C-4 Vidyasagar Classes SOLVED EXAMPLE 3. Copper sulphide contains 66.5% Cu, copper oxide contais 79.9% Cu and sulphur trioxide contains 40% S. show that the date illustrates the law of reciprocal proportions. Solution : In CuS, % of Cu = 66.5 % of S = 100 – 66.5 = 33.5 66.5 g of Cu combine with 33.5 g of sulphur In CuO, % of Cu = 79.9 % of O = 100 – 79.9 = 20.1 79.9 g of Cu combine with 20.1 g of oxygen Let us fix the mass of copper as 1 g In CuS, 66.5 g of Cu combine with S = 33.5 g 33.5 = 0.503 g 1 g Cu will combine with S = 66.5 In CuO, 79.9 g of Cu combine with O = 20.1 g 20.1 = 0.251 g 1 g of Cu will combine with O = 79.9 The ratio of weights of S and O which combine with the fixed mas sof copper (1 g) S : O 0.503 : 0.251 2 : 1 … (i) In SO 3 , % of S = 40, % of O = 60 Thus, S : O 40 : 60 2 : 3 … (ii) The ratio (i) and (ii) are 2 2 : or 3 : 1 1 3 This is a simple ratio. Hence law of reciprocal proportions is illustrated. Formation of Ammonia: N 2(g) + 3H 2(g) → 2NH 3(g) 1 volume 3 volumes The ratio of volumes is N 2 : H 2 : NH 3 :: 1 : 3 : 2. When the volumes of gases are measured under similar conditions of temperature and pressure it is found in the above reaction that 1 volume of nitrogen gas combines with three volumes of H 2 gas to form two volumes of ammonia gas. SOLVED EXAMPLE 4. 1 volume 1 volume 2 volumes 2C 2 H 2(g) + 5O 2(g) → 4CO 2(g) 2 vol 5 vol 4 vol 40 mL 5 ×40 mL 2 4 ×40 mL 2 40 mL 100 mL 80 mL + 2H 2 O () So, for complete combustion of 40 mL of acetylene, 100 mL of oxygen are required and 80 mL of carbon dioxide is formed. PROBLEMS 1. 0.44 g of a hydrocarbon on complete combustion with oxygen gave 1.8 g water and 0.88 g carbon dioxide. Show that these results are in accordance with the law of conservation of mass. 2. 10 mL of hydrogen combine with 5 mL of oxygen to yield water. When 200 mL of hydrogen at NTP are passed over heated CuO, the CuO loses 0.144 g of its mass. Do these results correspond to the law of constant proportions? 3 Carbon and oxygen are known to form two compounds. The carbon content in one of these compounds is 42.9% while in the other, it is 27.3%. Show that this data is in the agreement with the law of multiple proportions. 4. Water contains 88.90% oxygen and 11.10% of hydrogen, ammonia contains 82.35% of nitrogen and 17.65% of hydrogen and dinitrogen trioxide contains 63.15% oxygen and 36.85% of nitrogen. Show that these data illustrate law of reciprocal proportions. For the above reaction when volumes are measured under similar conditions of temperature and pressure, one volume of hydrogen combines with one volume of chlorine to form two volumes of hydrogen chloride gas. Thus the proportion of H 2(g) and Cl 2(g) and HCl (g) by volume is 1: 1 : 2 Chemistry -1(Some Basic Concepts of Chemistry) How much volume of oxygen will be required for complete combustion of 40 mL of acetylene (C 2 H 2 ) and how much volume of carbon dioxide will be formed? All volumes are measured at NTP. Solution: GAY LUSSAC’S LAW OF COMBINING VOLUMES OF GASES Statement: Under same conditions of temperature and pressure, the volumes of the gaseous reactants and products are in simple ratio of small whole numbers, when the gases participate in chemical reactions. Explanation: Formation of HCl gas: H 2(g) + Cl 2(g) → 2 HCl (g) 2 volumes We teach success XI + XII + Entrance - 2015 C-5 Vidyasagar Classes Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-6 Vidyasagar Classes DALTON’S ATOMIC THEORY AVOGADRO’S HYPOTHESIS Dalton’s atomic theory convincingly explained the laws of chemical combination. Postulates of this theory are i. Matter is made up of minute, indivisible particles called atoms and has independent existence. ii. Atoms of same element have identical properties such as same mass, size, shape and chemical properties. iii. Atoms of different elements differ in their properties and have different masses and size. iv. Compounds are formed when atoms of different elements combine with each other in simple and fixed numerical ratio such as 1 : 1, 1 : 2, 2 : 3 etc. v. Atom cannot be created, destroyed or transformed into atom of other element in the chemical process. A chemical reaction simply change the way in which atoms are grouped together. Statement: Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of gaseous molecules, regardless their chemical nature and physical properties. For example, if we enclose equal volumes of three gases hydrogen (H 2 ), oxygen (O 2 ) and chlorine (Cl 2 ) in different flasks of the same capacity under similar conditions of temperature and pressure, we find that all the flasks have the same number of molecules. However, these molecules may differ in size and mass. Gay-Lussac and Avogadro’s laws can be illustrated as follows: Gay-Lussac’s law Avogadro’s law + 2H 2(g) 2H 2 O (g) 2 vol 2 mol → O 2(g) 1 vol 1 mol 2 vol 2 mol Limitations and Drawbacks: Dalton’s atom still has its significance as the unit participating in chemical reaction, but later researchers proved that Dalton’s theory was not wholly correct. His theory has following drawbacks. Avogadro’s law : It may be stated as, “At constant pressure and temperature, volume of a gas is directly proportional to the number of molecules”. i. V∝N ii. iii. iv. With the discovery of sub-atomic particles i.e. electrons, protons, neutrons the atom can no longer be indivisible. Discovery of isotopes indicated that all atoms of the same elements are not perfectly identical. They may differ in their masses. Atoms of different elements may posses the same mass (isobar) but they always have different atomic numbers and differ in chemical properties. It could not explain why certain substances all containing atoms of the same element are differ in their properties. For example charcoal, graphite and diamond all are made up of only C-atom, but still their properties are quite different. Mathematically, we can say or where N is the number of molecules. V =k N V = volume, N is proportional to number of moles n ∴ V =k n n= Mass of gas Molar mass of gas At STP we can calculate the volume per mol i.e. V n According to general gas equation PV = nRT Vm = V RT = n P v. Atoms of one element can be transmuted into atoms of other element (nuclear reactions). P = 101.325 kPa, vi. In certain organic compounds like proteins, starch cellulose etc. the ratio in which the atoms combine cannot be regarded as simple. Vm = Chemistry -1(Some Basic Concepts of Chemistry) k = proportionality constant We teach success –1 –1 R = 8.314 J mol K , T = 273 K 8.314 × 273 = 22.41 dm3 mol–1 = 22.41 L 101.325 XI + XII + Entrance - 2015 C-7 Vidyasagar Classes 3 Thus, the volume 22.4 Litres or 22.4 dm at NTP corresponds to the molecular weight of the gas, and is referred as gram molecular volume. 6. ATOMICITY: “The number of atoms present in one molecule of a substance is called atomicity”. Relationship between molecular weight and vapour density : Example: The atomicity of oxygen can be calculated by considering the reaction between hydrogen and oxygen to form water vapour. It has been experimentally found that 2 volumes of hydrogen react with 1 volume of oxygen to form 2 volumes of water vapour. Vapour density is defined as the ratio of the weight of a certain volume of a gas to the weight of same volume of hydrogen gas, under similar conditions of temperature and pressure, is called vapour density (V.D). Hydrogen 2 volumes + Oxygen → Water vapour 1 volume 2 volumes n molecules 2n molecules 2 molecules 1 molecule 2 molecules 1 molecule 1/2 molecule Weight of certain volume of gas Weight of same volume of hydrogen If the given gas contains ‘x’ molecules in a certain volume, then according to Avogadro’s Law: V.D = Weight of x molecules of the gas Weight of x molecules of hydrogen gas V.D = Weight of 1 molecule of the gas Weight of 1 molecule of hydrogen gas 1 molecule Thus, one molecule of water contains one molecule of hydrogen (2 atoms) and ½ molecule of oxygen. The molecular mass of water has been found to be 18 amu. Since one molecule of hydrogen contains two atoms of hydrogen, therefore, weight of oxygen in one molecule of water is 18 – 2 = 16 amu. This corresponds to the weight of one atom of oxygen. Thus, one atom is present in half molecule of oxygen. Therefore, 1/2 molecule of O 2 = 1 atom or 1 molecule of O 2 = 2 atoms Thus, atomicity of oxygen is 2. Importance of Avogadro’s Law In this law, the word ‘molecule’ was used for the first time. 2. The law suggested the matter consists of two types of particles namely atoms and molecules. Avogadro’s law made a clear demarcation between atoms and molecules Hydrogen is diatomic, V.D = Weight of 1 molecule of the gas Weight of 2 atoms of Hydrogen ∴ V.D. = 1 Weight of 1 molecule of the gas × 2 Weight of 1 atom of Hydrogen ∴ V.D. = Molecular wt. of gas 2 1. 3. Thus, V.D = Applying Avogadro’s hypothesis 2n molecules It helped in the determination of actual number of molecules (Avogadro’s number) in gram molecular volume of a gas. Weight of 1 molecule of the gas = molecular wt. Weight of 1 atom of Hydrogen ∴ Molecular weight of a gas = 2 × Vapour density. This formula can be used for the determination of molecular weight of volatile substances. CONCEPT OF ATOMS & MOLECULES 4. It explained Gay−Lussac’s law of combining volumes which led to the following conclusions: a. Molecular weight of a gas = 2 × Vapour density. b. Gram molecular volume of all gases is equal to −3 3 22.4 × 10 m or 22.4 dm3 at S.T.P or N.T.P. Hydrogen atom is the smallest with mass 1.667 × 10 c. Atomicity of elements, determination of molecular weights, atomic weights, equivalent weight, molecular formula etc can be obtained. 5. It helped in modification of Dalton’s atomic theory. Molecule: It is the smallest particle of a substance (element or compound), which has free or independent existence and possesses all characteristic properties of the substance. A molecule of an element is composed of like atoms while Chemistry -1(Some Basic Concepts of Chemistry) Atom: The smallest particle of an element that can take part in chemical change and may or may not exist freely, as such. Every atom of an element has a definite mass of the order of 10–26 kg and radius of the order of 10–15 m. –26 We teach success kg. XI + XII + Entrance - 2015 C-8 Vidyasagar Classes a molecule of a compound contains fixed number of atoms of two or more different elements. A molecule may be broken down into its constituent atoms but the atom is indivisible during a chemical change. ATOMIC AND MOLECULAR MASSES Atomic mass (mass of one mole of atom): The present system of atomic masses is based on carbon-12 as the standard. Thus, atomic mass of an element is defined as, “the number which indicates how many times the mass of one atom of the element is heavier than that of 1/12th part of the mass of one atom of carbon-12”. Atomic mass of element Mass of one atom of the element = th (1 / 12) part of the mass of one atom of C - 12 Mass of one atom of the element × 12 = Mass of one atom of carbon - 12 x y ×a+ × b +……… 100 100 xa + yb + ....... = 100 x, y = percentage abundance of the isotopes. SOLVED EXAMPLES Average isotopic mass = 5. Relative abundances and masses of three isotopes of carbon are listed below. Isotope Relative abundance (%) Atomic mass (amu) 98.892 1.108 2 × 10–10 12 13.00335 14.00317 12 C C 14 C 13 Calculate average atomic mass of carbon. Solution: Atomic masses of some elements on the basis of C-12 Avg. at. mass of C H – 1.008 u Mg – 24.305 u Na – 22.989 u C × % 13 C + 14 C × % 14 C 100 98 . 892 12 1 . 108 13.00335 + 2 × 10 −10 × 14.00317 × + × = 100 O – 16.00 u Cu – 63.546 u Zn – 65.38 u Cl – 35.453 u Fe – 55.847 u Ag – 107.868 u In this system 12C is assigned a mass of exactly 12 atomic mass unit (amu). One atomic mass unit is defined as “a mass exactly equal to one-twelfth the mass of one C-12 atom”. 12 = 12 C+ 13 = 12.001 u 6. Boron has two isotopes boron-10 and boron-11, whose percentage abundance are 19.6 and 80.4 respectively. What is the average atomic mass of boron? Actual mass of one atom of C = 1.9924 × 10–23g 1.9924 ×10 −23 –24 = 1.66056 × 10 g 12 Today amu has been replaced by ‘u’ (Unified Mass) C×% 1 amu = Solution: Thus, Actual mass of an atom of an element = At. Mass in amu × 1.66 × 10–24 g Contribution of boron-10 = 10.0 × 0.196 = 1.96 amu Contribution of boron-11 = 11.0 × 0.804 = 8.844 amu Adding both = 1.96 + 8.844 = 10.804 amu Thus, the average atomic mass of boron is 10.804 amu Hence, actual mass of H-atom = 1.008 ×1.66 × 10–24 = 1.6736× 10–24 g 7. Boron occurs in nature in the form of two isotopes having atomic mass 10 and 11. What are the percentage abundances of two isotopes in a sample of boron having average atomic mass 10.8? Using Dulong and Petit’s law approximate atomic weight can be calculated. Atomic weight × specific heat ≈ 6.4 Solution: Average Atomic Mass: In calculations, we actually use average atomic masses of elements instead of atomic mass. Many naturally occurring elements exist as a constant mixture of isotopes (atoms of the same element having same atomic number but different mass number). Let the % abundance of B10 isotope = x The average relative atomic mass depends upon the isotopic composition of that particular element. Thus, Chemistry -1(Some Basic Concepts of Chemistry) 11 Then % abundance of B isotope = 100 – x x × 10 + (100 − x ) × 11 100 But, average atomic mass = 10.8 The average atomic mass = ∴ x × 10 + (100 − x ) × 11 = 10.8 100 We teach success XI + XII + Entrance - 2015 C-9 Vidyasagar Classes Equivalent weight : The number of parts by weight of a substance which combines with or displaces one part by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine. Or 10x + 1100 – 11x = 10.8 × 100 – x = – 1100 + 1080 Or x = 20 10 11 Thus, percentage abundance; B = 20, B = 80 Molecular Mass: It may be defined as,” the mass of a molecule of a substance relative to the mass of one atom of carbon taken as 12 or it is a number which indicates how many times one molecule of a substance is heavier in comparison to 1/12th of the mass of one atom of carbon 12”. Molecular mass (M) = Mass of one molecule of the substance th 1 mass of one atom of carbon - 12 12 Gram equivalent weight: The equivalent weight of a substance expressed in grams is called gram equivalent weight of that substance. METHODS OF DETERMINATION OF EQUIVALENT WEIGHT a. Eq.wt. of metal = b. The mass of a molecule is equal to the sum of the masses of the atoms present in a molecule. One molecule of water consists of 2 atoms of hydrogen and one atom of oxygen. Thus, molecular mass of water = (2 × 1.008) + 16.00 = 18.016 amu. Wt. of metal × 1.008 Wt. of H 2 displaced Equivalent weight of metal by chloride formation method: Eq.wt. of metal = Wt. of metal × 35.5 Wt. of chlorine combined c. Equivalent weight of metal by oxide formation method: Weight of metal ×8 Eq. wt. of metal = Weight of oxygen combined d. Equivalent weight = Gram-molecular mass or Gram molecule: Molecular mass of a substance expressed in grams is called gram-molecular mass or gram molecule or 1 mole. For example, the molecular mass of chlorine is 71 and, therefore, its gram-molecular mass or gram molecule is 71 g. Hydrogen displacement method: Mol.wt. / At. wt. / Formula wt. Valence Approximate At. wt. Eq. wt. Similarly, molecular mass of oxygen (O 2 ) is 32, i.e., 2 × 16 = 32 amu. Valency = Gram molecular mass of oxygen = 32 g Correct At. wt. = Eq. weight × corrected valency. Gram molecular mass should not be confused with the mass of one molecule of the substance in grams. The mass of one molecule of a substance is known as its actual mass. For example, the actual mass of one molecule of oxygen is equal to 32 × 1.66 × 10–24 g = 5.32 × 10–23 g. SOLVED EXAMPLE 8. SOLVED EXAMPLES 9. Solution: Calculate the mass of 1.5 gram molecule of sulphuric acid. Mass of the chlorine in the metal chloride = 49.5% Mass of metal = 100 – 49.5 = 50.5 Solution: Molecular mass of H 2 SO 4 = 2 × 1 + 32 + 4 × 16 = 98.0 amu Equivalent mass of the metal = Gram molecular mass of H 2 SO 4 = 98 g = Mass of 1.5 gram molecule of H 2 SO 4 = 98.0 × 1.5 = 147.0 g EQUIVALENT WEIGHT Chemistry -1(Some Basic Concepts of Chemistry) A chloride of an element contains 49.5% chlorine. The specific heat of the element is 0.056. Calculate the equivalent mass, valency and atomic mass of the element. Mass of metal × 35.5 Mass of chlorine 50.5 × 35.5 = 36.21 49.5 According to Dulong and Petit’s law Approximate atomic mass of the metal = We teach success 6.4 Specific heat XI + XII + Entrance - 2015 C-10 Vidyasagar Classes = 6.4 0.056 = 114.285 ≈ 114.3 Valency = Approximate atomic mass 114.3 = = 3.1 = 3 Equivalent mass 36.21 Hence, exact atomic mass = 36.21 × 3 = 108.63 Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-11 Vidyasagar Classes 10. One gram of the chloride was found to contain 0.835 g of chlorine. Its vapour density is 85. Calculate its molecular formula. Solution: Mass of metal chloride = 1 g The mole is the amount of a substance which contains Avogadro’s number of particles namely 6.023 × 1023. Mass of chloride = 0.835 g −3 Avogadro’s number: The number present in 12 × 10 kg 23 (12g) of 12 6 C isotope which is found to be 6.023 × 10 particles is called Avogadro’s number. These units can be atoms, molecules, ions, electrons etc. Mass of metal = 1 – 0.835 = 0.165 g Equivalent mass of metal = Valency of the metal = = Mole: The quantity of matter which contains as many −3 particles as there are in exactly 12g (12× 10 kg) of 12 6 C isotope. Or 0.165 × 35.5 = 7.01 0.835 2 × V.D. E + 35.5 SOLVED EXAMPLE 11. A piece of copper weighs 0.635 g. How many atoms of copper does it contain? Solution: Gram atomic mass of copper = 63.5 g 2 × 85 170 = = 3.99 ≈ 4 7.01 + 35.5 42.51 Formula of the chloride = MCl 4 PROBLEMS 5. 6. 7. Calculate the molecular mass of glucose (C 6 H 12 O 6 ) molecule. Given At. masses of H = 1.008 amu, C = 12.011 amu, O = 16.0 amu. (180.162 amu) Thallium has two isotopes 203Tl and 205Tl. Knowing that the atomic weight of thallium is 204.4, which isotope is more abundant of the two? 205 ( Tl = 70%) The oxide of an element contains 32.33 percent of the element and the vapour density of its chloride is 79. Calculate the atomic mass of the element. (15.28) No. of moles in 0.635 g of copper = No. of copper atoms in one mole = 6.023 × 1023 No. of copper atoms in 0.01 moles = 0.01 × 6.023 × 1023 = 6.023 × 1021 8. 23 No. of moles = iii. No. of moles = (0.1) What is the mass of 3.01 × 1022 molecule of (0.85 g) ammonia? (At. wt. N = 14, H = 1) 10. Calculate the mass of (2.32 × 10–23 g) i. 1 atom of C14 ii. 1 molecule of N 2 iii. 1 molecule of water iv. 100 molecules of sucrose (C 12 H 22 O 11 ). (5.68 × 10–20 g) (4.65 × 10–23 g) 1 mole = 6.023 × 10 particles = Molecular mass ii. PROBLEMS Calculate the gram atoms in 2.3 g of sodium. [Atomic weight of sodium = 23 g] 9. MOLE CONCEPT i. 0.635 = 0.01 63.5 Mass in grams Mass of one mole in grams (2.99 × 10–23 g) No. of particles 6.023 × 10 23 PERCENTAGE COMPOSITION OF COMPOUNDS Percentage composition of the compound is the relative mass of the each of the constituent element in 100 parts of it. Gram atom of an element iv. Mass of one atom = v. Mass of one molecule = vi. No. of molecules = 6.023 × 10 23 Gram molecular mass 6.023 × 10 23 % of element = Volume of gas (in L) × 6.023 × 1023 22.4 L Chemistry -1(Some Basic Concepts of Chemistry) X × 100 M Where, X = Mass of an element in 1 mole of the compound We teach success XI + XII + Entrance - 2015 C-12 Vidyasagar Classes M = Molecular mass of the compound Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-13 Vidyasagar Classes SOLVED EXAMPLE 12. It is found that 16.5 g of metal combines with oxygen to form 35.60 g of metal oxide. Calculate the percentage of metal and oxygen in the compound. Solution: Mass of oxygen in oxide = (35.60 – 16.50) = 19.10 g 16.50 × 100 = 46.3 35.60 19.10 % of oxygen = × 100 = 53.7 35.60 % of metal = 11. PROBLEMS Calculate the percentage composition of carbon in the following alkanes. (At. no. of C = 12, H = 1) 12. (i) C 2 H 6 (ii) CH 4 (iii) C 3 H 8 (iv) C 4 H 10 (i. 80%, ii. 75%, iii. 85.7%, iv. 83.7%) Determine the percentage of water of crystallisation, iron, sulphur, oxygen in pure ferrous sulphate (FeSO 4 ⋅ 7H 2 O). (At. no. of Fe = 56, S = 32, O = 16, H = 1) (45.32%, 20.14%, 11.51%, 23.02%) EMPIRICAL AND MOLECULAR FORMULAE Calculation of empirical formula :Empirical formula is calculated from the percentage composition. The steps involved in the calculation are : Step 1 : The percentage of each element is divided by its atomic mass. This gives the relative number of different atoms present in the molecule. Step 2 : The relative numbers of different atoms obtained in Step 1 are divided by the lowest one amongst them as to get simple ratio of atoms present in the molecule. Step 3 : The values obtained in Step 2 may or may not be whole numbers. In case one or more values are fractional, these are multiplied by a suitable integer to get simplest ratio in whole numbers. Minor fractions are neglected. Step 4 : The symbols of each element present are written side by side in a line with the number of atoms as determined in Step 2 or Step 3 as subscripts to the lower corner of each. This gives the empirical or simplest formula. Calculation of molecular formula: Knowing the empirical formula, the molecular formula can be ascertained if the molecular mass of the substance is known. It may be the same as the empirical formula of the substance or an exact multiple of it. Molecular formula = n × (Empirical formula) The value of ‘n’ can be determined if molecular mass of the substance is known. Empirical Formula (E.F) : The empirical formula of a compound is the simplest formula which expresses the simple whole number ratio of the atoms of constituent elements present in the molecule. The simplest formula of substance capable of expressing its percentage composition can be called its empirical formula. For example, CH 2 O is the empirical formula of acetic acid. It expresses that the simplest whole number ratio between carbon, hydrogen and oxygen atoms present in one molecular of acetic acid is 1 : 2 : 1. n= SOLVED EXAMPLES 13. Element Water Carbon Methane Hydrogen Peroxide Acetylene Glucose Diborane Tetraphosphorus decoxide M.F H2O C CH 4 H2O2 C2H2 C 6 H 12 O 6 B2 H6 P 4 O 10 Chemistry -1(Some Basic Concepts of Chemistry) % At. mass Relative no. of atoms Simplest ratio 26.6 0.68 K 26.6 39.1 39.1 0.68 = 0.68 =1 35.4 0.68 Cr 35.4 52.0 52 0.68 = 0.68 =1 38.1 2.38 O 38.1 16.0 0.68 16 = 2.38 = 3.5 Therefore, empirical formula is K 2 Cr 2 O 7 . Empirical and Molecular formula of some compounds E.F. H2O C CH 4 HO CH CH 2 O BH 3 P2O5 Calculate the empirical formula for a compound that contains 26.6% potassium, 35.4% chromium and 38.1% oxygen. [Given At. no. of K = 39.1; Cr = 52; O = 16] Solution : Molecular Formula (M.F) : The formula which gives the actual number of atoms of various elements present in the molecule of the substance is termed as molecular formula. Compound Molecular mass Empirical formula mass We teach success Simplest whole number ratio 1×2=2 1×2=2 3.5 × 2 = 7 XI + XII + Entrance - 2015 C-14 Vidyasagar Classes 14. An organic compound contains 40% C, 6.66% H and rest oxygen. Its molar mass is 60. Calculate its empirical and molecular formulas. Solution : Element % C 40.0 At. mass 12 H 6.66 1 O 53.34 (by 16 difference) Relative no. Simplest of atoms ratio 3.33 40 = 3.33 =1 12 3.33 6.66 6.66 = 6.66 =2 1 3.33 53.34 3.33 = 3.33 =1 16 3.33 Empirical formula = CH 2 O Empirical formula mass = (12 + 2 + 16) = 30 Mol. mass 60 n= = =2 30 Emp. mass 14. 1 vol. N 2 + 3 vol. H 2 → 2 vol. NH 3 Thus, calculations based on chemical equations are of three types: i. Calculations based on mass−mass relationship. ii. Calculations based on mass−volume relationship. iii. Calculations based on volume−volume relationship. Calculations using Chemical Equations: A balanced chemical equation gives quantitative information in terms of molecules, masses, moles and volumes between the different reactants and products involved in the reaction. This is called stoichiometry. The quantitative information conveyed by a chemical equation helps in a number of calculations. Complete the following blanks for the reaction indicated Molecular formula = 2 × (Empirical formula) = 2 × (CH 2 O) = C 2 H 4 O 2 13. Volume interpretation: CaH 2(s) + 2H 2 O (g) → Ca(OH) 2(s) + 2H 2(g) PROBLEM A compound of carbon, hydrogen and nitrogen contains the elements in the ratio of 18 : 2 : 7. Calculate its empirical formula. If the molecular mass is 108, what is its molecular formula? (C 6 H 8 N 2 ) Calculate the empirical formula of a mineral which has the following percentage composition CuO = 44.82%, SiO 2 = 34.83% and water = 20.35 % (at. wt. of Cu = 63.5, Si = 28). (CuO. SiO 2 . 2 H 2 O) STOICHIOMETRY QUANTITATIVE RELATIONS IN CHEMICAL REACTIONS: Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the chemical equation and on the relationship between masses and moles. A chemical equation can be interpreted as follows : N 2(g) + 3H 2(g) → 2NH 3(g) i. Moles: 2 Moles + ……. → ……. + ……. ii. Grams: 42 g + ……. → ……. + ……. iii. Hydrogen atoms: 6.02 × 1023 + … → … + … i. SOLVED EXAMPLE 15. Step I: Write the formula of reactants with plus sign on the left hand side. Draw arrow from left to right and write the formula of products right side of the arrow. MgO + HCl → MgCl 2 + H 2 O Step II: Balance the masses of each atom in the chemical formula of reactants and products by proper whole number coefficients to get the balance chemical equation. MgO + 2HCl → MgCl 2 + H 2 O 1 mol 1 molecule N 2 + 3 molecules H 2 → 2 molecules NH 3 Molar interpretation: Mass interpretation: 28 g N 2 + 6 g H 2 → 34 g NH 3 Chemistry -1(Some Basic Concepts of Chemistry) Calculate the number of grams of magnesium chloride that could be obtained from 17.0 g of HCl when HCl is reacted with excess of magnesium oxide. Solution: Molecular interpretation: 1 mol N 2 + 3 mol H 2 → 2 mol NH 3 Mass-mass relationship: Here the weight of the reactants and products are considered. 2 mol 2 × 36.5 g 73 g 1 mol (24 + 71) g 95 g 1 mole 73 g of HCl produce MgCl 2 = 95 g 1 g of HCl produce MgCl 2 = We teach success 95 g 73 XI + XII + Entrance - 2015 C-15 Vidyasagar Classes 17 g of HCl will produce MgCl 2 = 95 × 17 = 22.12 g 73 Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-16 Vidyasagar Classes ii. Mass-Volume relationship: Here such problems are considered which involve both weights and volumes. Weight of the solid substances may be compared with the volume of gases. Gram-molecular volume law, according to which a gram molecule (mol. wt in gram) of every gas at NTP occupies 22.4 litres is used in finding solutions to such problems. SOLVED EXAMPLE 16. Calculate the volume of chlorine that can be obtained at S.T.P. by reaction of 1.58 g of KMnO 4 and excess of hydrochloric acid. Solution: LIMITING REAGENT: When a reaction is carried out in such a way, that the reactants are simultaneously consumed completely, we can say that the reactants are in stoichiometric proportions. The reactant or reagent that is completely consumed determines the quantity of products formed and is called the limiting reactant. The reactant which is taken in excess than the limiting reactant is the excess reactant. SOLVED EXAMPLE 18. Balanced equation 2KMnO 4 + 16HCl 2KCl + 2MnCl 2 + 8H 2 O + 5Cl 2 2 mol (316 g) 5 mol (5 × 22.4 L at S.T.P.) Thus, volume of Cl 2 produced at S.T.P. 5 × 22.4 × 1.58 = = 0.560 L or 560 mL. 316 Carbon monoxide reacts with oxygen to form carbon dioxide according to the equation, 2CO + O 2 → 2CO 2 . In an experiment 400 ml of carbon monoxide and 180 ml of oxygen were allowed to react to produce carbon dioxide. All the volumes were measured under the same conditions of temperature and pressure. Find out the composition of the final mixture. Identify the limiting reagent. Solution: iii. Volume-Volume relationship : Gay-Lussac law of gases supplemented by Avogadro hypothesis is made use of in this type of calculations in which, reactions among gases are studied. Molecules of reacting gases & the products formed thereof are compared to their volumes i.e. the ratio amongst the molecules is taken as the ratio amongst their volumes, according to the reverse of Avogadro's law. SOLVED EXAMPLE 17. Calculate the volume of O 2 at STP required to completely burn 200 ml of acetylene. Solution: The stoichiometric reaction is represented as + Moles of CO = O 2(g) 2CO 2(g) Moles of O 2 400 = 0.179 22400 = 180 = 8.04 × 10–3 22400 According to above equation 2 moles of CO require 1 mole of O 2 for the reaction. Hence for 0.179 mol of CO, the moles of O 2 required will be 0.179 × 1 = 0.0895 moles 2 But we have only 8.04 × 10–3 mole of O 2 . Hence, O 2 is the limiting reagent in this case. CO 2 would be formed only from that amount of available O 2 i.e. 8.04 × 10–3. Since 1 mole of O 2 gives 2 mole of CO 2 . 2CH ≡ CH + 5O 2 → 4CO 2 + 2H 2 O 2 mol 5 mole 2 × 22.4 litre 5 × 22.4 L 200 ml x ml ∴x= 2CO (g) 8.04 × 10 mol O 2 × 2 = 0.0161 mole CO 2 –3 1 mol CO 2 = 22.4 L 0.0161 mole CO 2 = 22.4 × 0.0161 5 × 22.4lit × 200 ml = 500 ml 2 × 22.4lit = 0.361 L = 361 ml Volume of O 2 = 500 ml So the composition of the final mixture is (Note: Experimental conditions for the volume of C 2 H 5 are not mentioned. So, it is assumed to be at STP) CO = 400 – 361 = 39 ml Chemistry -1(Some Basic Concepts of Chemistry) CO 2 = 361 ml We teach success (Excess reagent) XI + XII + Entrance - 2015 C-17 Vidyasagar Classes 19. In a reaction A + B 2 → AB 2 , identify the limiting reagent, if any, in the following reaction mixtures: (i) 300 atoms of A + 200 molecules of B (ii) 2 mol A + 3 mol B (iii) 100 atoms of A + 100 molecules of B (iv) 5 mol A + 2.5 mol B (v) 2.5 mol A + 5 mol B 19. (Mol. wt. of PCl 3 = 137.5, Cl 2 = 71, P 4 O 7 = 236, POCl 3 = 153.5) (PCl 3 ) 1 litre of oxygen and 3 litres of carbon monoxide measured at STP are reached to get carbon dioxide. Calculate the volume of the gases after the reaction and also the weight of carbon dioxide formed in the reaction. (0 litre, 1 litre, 2 litre, 44 g, 3.928 g) Solution: (i) The given reaction is A + B 2 → AB 2 Here 300 atoms of a requires 300 molecules of B, since there are only 200 molecules of B provided ∴ B is the limiting reagent. (ii) 3 mol B requires 3 mol A. Since only 2 mol of A are provided, therefore A is the limiting reagent. (iii) 100 atom of A + 100 molecules of B constitute a Stoichiometric mixture. Neither A nor B is the limiting reagent. (iv) B is the limiting reagent as 5 mol A requires 5 mol B but only 2.5 mol B are given. (v) A is the limiting reagent as 5 mol B requires 5 mol A, but only 2.5 mol A are provided. PROBLEMS 15. Calculate the volume of carbon dioxide at NTP evolved by strong heating of 20g calcium carbonate. (4.48 litres) 16. The hydrated salt Na 2 SO 4 ⋅ nH 2 O, undergoes 55.9 % loss in weight on heating and becomes anhydrous. (10) Calculate the value of n. 17. The reaction, 2C + O 2 → 2CO is carried out by taking 24 g of carbon and 96 g of oxygen. Find out i. ii. iii. 18. Which reactant is left in excess and how much? How many moles of CO are formed? How many gram of other reactant should be taken so that nothing is left at the end of the reaction? (i. 64 g O 2 , ii. 56 g, iii. 72 g) Phosophoryl chloride, POCl 3 , is used in the manufacture of fire retardants, gasoline additives and hydraulic fluids. One reaction by which it can be prepared is 6 PCl 3() + 6 Cl 2(g) + P 4 O 10(s) → 10 POCl 3() If 1.00 kg each of PCl 3 , Cl 2 and P 4 O 10 are allowed to react, which is the limiting reactant? Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-18 Vidyasagar Classes Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-19 Vidyasagar Classes KEY IDEAS 1. 2. 3. 4. 5. 6. Matter is anything that occupies space has mass. Law of conservation of mass (Lavoisier in 1774). During any physical or chemical change, the total mass of the products formed is equal to the total mass of reactants consumed. Law of constant composition (Proust in 1798). A chemical compound always contains same elements combined together in same proportion by mass. The smallest particle of an element that takes part in chemical reactions is atom. 8. The smallest particle of a substance that has independent existence is molecule. 9. One twelfth (1/12) of the mass of an atom of C-12 is atomic mass unit. It is equal to 1.66 × 10–27 kg. 10. The average relative mass of an atom of element as compared with mass of a carbon atom (C-12) taken as 12 a.m.u. is atomic mass. Law of multiple proportions (John Dalton in 1803). When two elements combine together to form two or more than two compounds then the masses of one of elements that combine with fixed mass of the other bear a simple whole number ratio to one another. 11. The average relative mass of a molecule of the substance as compared with mass of a carbon atom (C-12) taken as 12 a.m.u. is molecular mass. 12. One mole is 6.022 × 1023 specified particles. Gay Lussac’s law (1808). When gases react with each other they do so in volumes which bear a simple whole number ratio to one another and to the volume of the products, if they are also gases, provided all volumes are measured under similar condition of temperature and pressure. Avogadro’s law: Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules. 6.022 × 1023 particles 7. 13. Mass percentage composition of a compound gives the mass of each element expressed as percentage of the total mass. 14. The formula which gives the simplest whole number ratio of atoms of different elements present in the molecule of a compound is empirical formula. 15. The reagent that is completely consumed in the reaction is called limiting reagent. in terms of particles 1 mole of O-atoms = 6.022 × 1023 atoms 1 mole of O2 molecules = 6.022 × 1023 molecules in terms of volume 1 MOLE in terms of mass 1 gram atom of element 1 gram molecule of substance 12 g C 18 g H2O Mass of substance 23 g Na 44 g CO2 Multiplied by Multiplied by MOLE Divided by 22.4 L of a gas at N.T.P. Divided by 1 gram formula weight of substance 58.5 g NaCl 100 g CaCO3 Number of particles AVOGADRO NUMBER MOLAR MASS Multiplied by 22.4 L Divided by 22.4 L Vol. at N.T.P. in litres Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-20 Vidyasagar Classes SOLVED EXAMPLES ON MULTIPLE CHOICE QUESTIONS Example 1: The number of gram molecules of oxygen in 6.022 × 1024 molecules of CO is (A) (C) 10 g-moles 5 g-moles (B) (D) 1 g-moles 0.5 g-moles Example 5: In a compound C, H & N are present in the ratio 9 : 1 : 3.5 (by weight). Molecular mass of the compound is 108. Formula of the compound is (A) (C) C2H6N2 C3H4N Elament 6.022 × 10 = 10 6.022 × 10 23 24 Mole of O in CO = ∴ C 10 = 5 g-mole 2 H N The correct alternative is (C). Example 2: Tap water is (A) (C) (B) (D) a mixture none of these N= NaOH & KOH H2O & D2O 1 = (C 3 H 4 N) 2 (B) (D) KCl & KI SO 2 & SO 3 ∴ The correct alternative is (B). Example 6: The reaction of calcium with water is represented by the equation: Ca + 2H 2 O → Ca(OH) 2 + H 2 The volume of hydrogen measured at S.T.P. that would be liberated when 8 g calcium completely reacts with water is (A) (C) 4480 cm3 0.2 cm3 Solution: 4.6 × 1022 atoms of certain element weights 13.8 g, the atomic mass of the element is Ca + 2H 2 O → 290.5 34.4 (B) (D) 40 g 180.6 10.4 (B) (D) 22.42 = 22400 cm3 (1 mL = 1 cm3) 40 g Ca produces = 22400 cm3 of H 2 6.02 × 1023 = 1 mole 4.6 × 1022 atom = 13.8 g 8 g Ca will produce = 13.8 × 6.02 × 10 23 = 180.6 4.6 × 10 22 2240 cm3 0.4 cm3 Ca(OH) 2 + H 2 Solution: ∴ 4 = C6H8N2 The correct alternative is (D). 6.02 × 1073 atom = 3 Mol.wt. 108 = =2 E.F.W. 54 Example 4: (A) (C) Simple ratio M..F. = (E.F.) n The correct alternative is (B). Solution: In SO 2 and SO 3 two same elements (S and O) combine to form two different compounds (SO 2 and SO 3 ). ∴ At. ratio 66.6 = 5.5 12 7.4 = 7.4 1 26.0 = 1.86 14 E.F.W. = 36 + 4 + 14 = 54 a compound an element Example 3: Which of the following pairs of compounds illustrates the law of multiple properties? (A) (C) % Elament 9 = 66.6 13 1 = 7.4 13 3.5 = 26.0 13 Empirical formula = C 3 H 4 N Solution: Tap water contains soluble impurities hence it is a homogeneous mixture. ∴ C6H8N2 C 9 H 12 N 3 Solution: Solution: Mole of CO = (B) (D) ∴ 22400 × 8 = 4480 cm3 of H 2 40 The correct alternative is (A). The correct alternative is (B). Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-21 Vidyasagar Classes THEORY QUESTIONS 1. Briefly discuss the importance of studying chemistry. OR 17. What is gram molar volume? 18. Define the terms element, symbol, compound and formula. 19. Distinguish the following What is the need of studying chemistry? (i) Atom from molecule (ii) Atom from gram–atom (iii) Mole from molecule. 2. What are fundamental units? List their SI units of measurement. 3. What are derived units? 4. What is meant by SI units? Name the seven basis SI units. (i) Why can atomic masses be referred to as relative numbers? 5. List the SI units for the following derived quantities. (ii) Why the atomic masses of most of the elements are not whole numbers? 20. (i) Force (ii) Pressure (iii) Energy Give reasons 21. Explain the need of term average atomic mass. 22. What do you understand by the empirical and molecular formulae of a substance? How two are related to each other? 23. What is atomic mass unit? State and explain Gay Lussac’s Law of gaseous volumes and mention its important application. 24. Write the postulates of Dalton’s theory. 25. Give the limitations of Dalton’s theory. 7. State and explain Avogadro’s hypothesis with a suitable example. What is the importance of this law? 26. Write a note on chemical stoichiometry. 8. Show that the volume of 1 mole of any gas at STP is 22.4 L. 27. What are limiting reagents? 9. Briefly explain the mole concept. 6. State and explain the Gay Lussac’s law of combining volumes of gases with a suitable example. OR OR Define a mole what is its importance in chemistry? 10. Define : (i) (ii) (iii) (iv) 11. Mole Empirical formula Molecular formula Avogadro’s number Describe the following laws (i) Conservation of mass (ii) Conservation of energy (iii) Conservation of mass and energy 12. State the law of constant proportions. Illustrate with an example. 13. State the law of multiple proportions. Illustrate with an example. 14. Define atomic mass and molecular mass. 15. Define atom and molecule. 16. What is Avogadro’s number? Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-22 Vidyasagar Classes 8. EXERCISE 1. The % composition of four hydrocarbons is as follows: (i) (ii) (iii) (iv) %C 75 80 85.7 91.3 %H 25 20 14.3 8.7 2. Boron has two isotopes B and B whose relative abundances are 20% and 80% respectively. Atomic weight of boron is 22.4 lit 16.8 lit (B) (D) 11.2 lit 5.6 lit 13. M 3M (B) (D) 2M 4M 2 8 (B) (D) 15. 16g of O 3 48g of SO 3 3 gram molecules of CO 2 18 × 1023 CO 2 molecules 6 × 1023 oxygen atoms 6 × 1023 carbon atoms Chemistry -1(Some Basic Concepts of Chemistry) 2.88 × 10–3 4.54 × 10–3 (B) (D) 1.66 × 10–3 1.66 × 10–2 C3H3 C2H4 (B) (D) C2H2 C6H6 70 100 (B) (D) 140 65 4 2 (B) (D) 6 3 Four oxides of nitrogen are given below. (ii) (iv) NO 2 N2O5 Oxides having 30.5% nitrogen are (A) (C) One mole of CO 2 contains (A) (B) (C) (D) One mole of oxygen One molecule of sulphur trioxide 100 amu of uranium 44 g of carbon dioxide (i) NO (iii) N 2 O 4 (B) 16g of SO 3 (D) 1g of hydrogen 1.084 × 1020 4.84 × 1017 Caffeine has a molecular weight of 194. If it contains 28.9 % by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is (A) (C) 4 16 (B) (D) An organic compound made of C, H and N contains 20 % nitrogen. What will be its molecular mass if it contains only one nitrogen atom in it? (A) (C) 14. 6.023 × 1019 6.023 × 1023 A compound contains 92.3% of carbon, and the rest hydrogen. The molecule of the compound is 39 times heavier than hydrogen molecule. The molecular formula of the compound is (A) (C) The gas having same number of molecules as 16g of oxygen is (at. wt S = 32) (A) (C) 7. 12. 11 10.8 If molecular mass and atomic mass of sulphur are 256 and 32 respectively then its atomicity is (A) (C) 6. (B) (D) M gram of a substance when vaporised occupy a volume of 5.6 L at NTP. The molecular masses of the substance will be (A) (C) 5. 10.2 10.5 3.125 × 10–2 2.5 × 10–2 If 1021 molecules are removed from 200 mg of CO 2 then the number of moles of CO 2 left will be (A) (C) The total volume of mixture of 2 g of helium and 7 g of nitrogen under S.T.P. conditions is (A) (C) 4. 11. 11 (B) (D) Which among the following is the heaviest? (A) (B) (C) (D) constant proportion conservation of mass multiple Proportions reciprocal Proportions. 0.02 1.25× 10–2 How many water molecules are there in one drop of water (volume 0.0018 ml) at room temperature? (A) (C) 10. 10 (A) (C) 3. (A) (C) 9. The data illustrates the law of (A) (B) (C) (D) How many moles of magnesium phosphate, Mg 3 (PO 4 ) 2 , will contain 0.25 mole of oxygen atoms? 16. (B) (D) (ii) and (iii) (i) and (iv) 20 g of sulphur on burning in air produced 11.2 L of SO 2 at STP. The percentage of unreacted sulphur is (A) (C) We teach success (i) and (ii) (iii) and (iv) 80 60 (B) (D) 20 40 XI + XII + Entrance - 2015 C-23 Vidyasagar Classes 17. 30 g of Mg (At. wt. = 24) combines with 30 g of oxygen in a sealed tube. The composition of residual mixture will be (A) (B) (C) (D) 18. 19. CH CH 4 The limiting reagent for this reaction is NaOH 0.08 moles of Fe(OH) 3 is formed 0.12 mole of FeCl 3 is left unreacted all of the above X2Y3 XY 2 (B) (D) (B) (D) 155.8 55.8 In the reaction 4A + 2B + 3C → A 4 B 2 C 3 , what will be the number of moles of product formed, starting from one mole of A, 0.6 mole of B and 0.72 mole of C? (A) (C) 25. 15.58 5.58 0.25 0.24 (B) (D) 0.3 2.32 Sulphur trioxide is prepared by the following two reactions: S 8(s) + 8O 2(g) → 8SO 2(g) 2SO 2(g) + O 2(g) → 2SO 3(g) How many grams of SO 3 are produced from 1 mole S 8 ? (At. wt. S = 32, O = 16) (A) (C) 1280 960 (B) (D) 640 320 XY X3Y4 What is/are true about calcium carbonate? (At. Wt. Ca = 40) (a) (b) (c) (d) (A) (C) 22. (B) (D) 24. Two elements X (atomic mass = 50) and Y (atomic mass = 16) combine to give a compound having 32 % Y. The formula of the compound is (A) (C) 21. CH 3 CH 2 A metal oxide has the formula X 2 O 3 . It can be reduced by hydrogen to give free metal and water. 0.1596 g of metal oxide requires 6 mg of hydrogen for complete reduction. The atomic mass of metal in amu is (A) (C) A solution containing 0.2 mole of ferric chloride is allowed to react with 0.24 mole of sodium hydroxide. The correct statement for this reaction is that (A) (B) (C) (D) 20. 40g MgO + 20g O 2 45g MgO + 15g O 2 50g MgO + 10g O 2 60g MgO only 3.0 g of a hydrocarbon on combustion gives 8.8 g of CO 2 and 5.4 g of water. The empirical formula of the compound is (A) (C) 23. The percentage of carbon atom is 20 % The percentage of oxygen is four times the percentage of carbon. 10 g of calcium carbonate contains 4 g of calcium. 10 g of calcium carbonate contains 6.02 × 1022 CO 32− ions. All are true a, c and d (B) (D) b, c and d c and d 254 g of iodine and 142 g of chlorine are made to react completely to give a mixture of ICl and ICl 3 . The moles of each one formed is (At. wt. I = 127, Cl = 35.5) (A) (B) (C) (D) 0.1 ICl and 0.1 mole ICl 3 1.0 mol ICl and 1.0 mol ICl 3 0.5 ICl and 0.1 mol ICl 3 0.5 mol ICl and 1.0 mol ICl 3 Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015 C-24 Vidyasagar Classes 8. EXERCISE : HOMEWORK 1. A pure substance can only be (A) x+y (B) A symbol not only represents the name of the element but also its x y y (C) xy (D) x At room temperature and pressure, two flasks of equal volumes are filled with H2 and SO2 separately. Particles which are equal in number, in the two flasks are (A) (C) (A) (C) atoms molecules (B) (D) (A) (B) (C) (D) 2. 3. (C) (D) atomic number atomic volume All elements are homogeneous Compounds always contain two or more different elements A mixture is not always heterogeneous Air is a heterogeneous mixture 10. 11. kg m2 s−2 kg m2 s−1 (B) (D) 12. reciprocal proportions multiple proportions constant proportions conservation of mass elements consist of impurities these are mixtures of allotropes these are mixtures of isobars these are mixtures of isotopes If the weight of 5.6 litres of a gas at S.T.P is 11 g, the gas is (Given at.wt. of P = 31, C = 12, N = 14, O = 16, Cl = 35.5) (A) (C) phosphine nitric oxide (B) (D) phosgene nitrous oxide 290 34.4 (B) (D) 180 10.4 XY XY 2 (B) (D) X2Y X2Y3 5 mL, 10 mL 5 mL, 15 mL (B) (D) 10 mL, 5 mL 10 mL, 10 mL If 6 litre of H 2 and 5.6 litre of Cl 2 are mixed and exploded in an eudiometer, the volume of HCl formed is (A) (C) 15. 315 g 342 g If 5 mL of methane is completely burnt, the volume of oxygen required and the volume of CO 2 formed under the same conditions are (A) (C) 14. (B) (D) The simplest formula of a compound containing 50% of element X (At. mass = 10) and 50% of the element Y (At. mass = 20) is (A) (C) 13. 150 g 327 g 4.6 × 1022 atoms of an element weigh 13.8 g. The atomic mass of the element is (A) (C) kg m−1 s2 kg m2 s2 electrons neutrons What is the mass of one mole of aluminium sulphate Al 2 (SO 4 ) 3 ? (A) (C) The atomic masses of the elements are usually fractional because (A) (B) (C) (D) 7. (B) (D) Different proportions of oxygen in the various oxides of nitrogen, prove the law of (A) (B) (C) (D) 6. atomic mass atomicity 9. SI unit of energy is (A) (C) 5. a compound an element an element or a compound a heterogeneous mixture Which one of the following statements is incorrect? (A) (B) 4. The atomicity of a species is x and its atomic weight is y. The molecular weight of the species is 6.0 litre 11.2 litre (B) (D) 5.6 litre 11.6 litre 23 g if sodium will react with ethyl alcohol to give (A) (B) (C) (D) one mole of hydrogen one mole of oxygen one mole of NaOH 1/2 mole of hydrogen ANSWERKEY 1. (C) 2. (A) 3. (D) 4. (A) 5. (B) 6. (D) 7. (D) 8. (C) 9. (C) 10. (D) 11. (B) 12. (B) 13. (B) 14. (C) 15. (D) Chemistry -1(Some Basic Concepts of Chemistry) We teach success XI + XII + Entrance - 2015