# Fundamental Concepts of Chemistry

August 4, 2017 | Author: Fatehullah Khan | Category: Stoichiometry, Mole (Unit), Significant Figures, Molecules, Logarithm

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Chapter 1

Introduction to Fundamental Concepts of Chemistry SIGNIFICANT FIGURES •

―Significant figures are the reliable digits in a number or measurement which are known with certainty.‖ • Significant figures show the accuracy in measurements. • We can understand the precision of a measurement if we know exactly the significant figures in the measurement. • A measurement that contains more number of significant figures is more accurate than a measurement that contains less number of Significant figures. • For example: Radius of a bob is 3.3679 cm and that of the other is 3.36 cm. • In this situation the first measurement is the most accurate as it has more number of significant figures. USE OF SIGNIFICANT FIGURES IN ADDITION AND SUBTRACTION • We consider the significant figures on the right side of decimal point. • This means that only as many digits are to be retained to the right side of decimal point as the number with fewest digits to the right of the decimal point. • 4.345 + 23.5 =27.845 Answer after rounding off: 27.8 USE OF SIGNIFICANT FIGURES IN MULTIPLICATION AND DIVISION • The number obtained after calculation of two or more numbers must have NO more significant figure than that number used in multiplication or division. • For example: 4.3458 x 2.7 =11.73366 • Answer after rounding off: 12 • Because 2.7 has only two significant figures) RULES OF SIGNIFICANT FIGURES • (1) All the non-zero digits are significant figures. For Example: • 3.456 has four significant figures. • 12.3456 has six significant figures.

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0.34 has two significant figures. (2) Zeros between non-zero digits are significant. For Example: 2306 has four significant figures. 200894 has six significant figures (3) Zeros locating the position of decimal in numbers of magnitude less than one are not significant. For Example: 0.0004 has only one significant figures. 0.0000034 has two significant figures. (4) Final zeros to the right of the decimal point are significant. 3.0000 has five significant figures. 1002.00 has six significant figures. (5) Zeros that locate decimal point in numbers greater than one are not significant. For Example: 30000 has only one significant figure. 120000 has two significant figures

ROUNDING OFF THE DATA • • • • • • • • •

―The procedure of dropping digits from a number or measurement so as to acquire greatest or desired significant value is known as rounding off the data.‖ Certain rules are followed to round off the given data. Rule # 1: If the digit to be dropped is greater than 5, then add "1" to the last digit to be retained and drop all digits farther to the right. For example: 3.677 is rounded off to 3.68 if we need three significant figures in measurement. 3.677 is rounded off to 3.7 if we need two significant figures in measurement. Rule # 2: If the digit to be dropped is less than 5, then simply drop it without adding any number to the last digit. For example: 6.632 is rounded off to 6.63 if we need three significant figures in measurement. 6.632 is rounded off to 6.6 if we need two significant figures in measurement. Rule # 3: If the digit to be dropped is exactly 5 then:

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(A) If the digit to be retained is even, then just drop the "5". For example: 6.65 is rounded off to 6.6 if we need two significant figures in measurement. 3.4665 is rounded off to 6.466 if we need four significant figures in measurement. (B) If the digit to be retained is odd, then add "1" to it. For example: 6.35 is rounded off to 6.4 if we need two significant figures in measurement. 3.4675 is rounded off to 6.468 if we need four significant figures in measurement. Zero is an even number 3.05 is rounded off to 3.0 if we need two significant figures in measurement.

• EXPONENTIAL NOTATION • • • • • • • • • • • •

The method of writing numbers as multiples or powers of 10 are known as exponential or scientific notation. In scientific work sometimes very large or very small numbers are confronted. Such numbers are conveniently expressed as powers of 10. The number of particles in one mole of any substance is 6,02,000,000,000,000,000,000,000. This number is expressed as 6.02 x 1023. 0.000000000000000000000000000000911kg which is the mass of electron can be expressed as 9.11 x 10-31 kg 0.0023=2.3 x 10-3 (negative power=number < 1) 2530 = 2.53x103 (positive power=number > 1) Decimal is given after first non-zero digit. The number of digits between the two decimal points is shown as the exponent. In scientific notations, numbers are more conveniently added, subtracted, multiplied & divided. This method is used for all kinds of numbers, e.g. There are two numbers= 32000 x 0.0023 Converting into exponential form= 3.2 x 104 x 2.3 x 10—3

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Multiplying coefficients & powers are added = (3.2x2.3) x 104-3 Hence= 7.36 x 101 or 73.6

LOGARITHMS • In an expression ax=y, x is the log of y to the base a. • 32=9, • 2 is the log of 9 to the base 3. • 24=16 • 4 is the log 16 to the base 2. • The are two types of log • 1) Natural log, base is 10, invented by Alkhawarzmi • 2) Scientific log, base is e, invented by John Nepeir. Natural Log • 100=1 (log of 1 is ―0‖) 10-1=0.1 (log of 0.1 is ―1‖) • 101=10 (log of 10 is ―1‖) 10-2=0.01 (log of 0.01 is ―2‖) • 102=100 (log of 100 is ―2‖) 10-3=0.001 (log of 0.001 is ―3‖) • 103=1000 (log of 1000 is ―3‖) Dealing with logs • For example we have to find the log of 5823 • First we shall convert into exponential form • 5.823 x 103 • The decimal fraction (5.823) is called Mantissa, which is always positive & seen from log tables. • The exponent (3) is called characteristics & it can be positive or negative, found by seeing the number. Rules of log • log (a x b) = log a + log b log (a/b) = log a – log b n • log a = n log a • Examples: log (100 x 0.00001)= ?? • = log 100 + log 0.0001= 2 + (-4) = 2 – 4 = - 2 • Example: - log (100 x 1000 x 0.001) = ? • = log 100 + log 1000 + log 0.001

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= 2 + 3 + (-3) = 2+3-3 =2 Example: - log (0.00001/100) = ? = log 0.00001 – log 100 =–5–2 =–7 Example: - log (0.001)4 = 4 x log 0.001 = 4 x -3 = -12 Example : - log 10 x 100 x (1000)3 100 x 0.001 x (0.01)2 = (1+2+9) – (2-3-4) = 12 – (– 5) = 12 + 5 = 17

ATOMIC MASS • • • • •

ACCURACY AND PRECISION • • • • • • • • • •

How close is the actual value to the expected value is called accuracy. How close are the several replicate measurement of same quantity is known as precision. One measurement can, at the same time, be accurate but not precise. & otherwise. One measurement can, at the same time, be accurate as well as precise. Sometimes there is a difference between the accepted value and the experimental value. This difference is known as error. Error = accepted value – experimental value Error can be positive or negative depending on whether the experimental value is greater than or less than the accepted value. Often it is useful to calculate relative error, or percent error. Percent error = = (error/expected value) x 100%

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"The mass of one atom of the element compared with the mass of one atom of C12" Atomic mass is a ratio therefore it has no unit. Generally atoms mass is expressed in (a.m.u). One atomic mass unit is equal to 1/12 of the mass of a C12 atom, e.g. Atomic Number - Symbol - Name - Atomic Weight 1 H - Hydrogen - 1 2 He - Helium 4 3 Li - Lithium 7 4 Be - Beryllium - 9 5 B - Boron 11 6 C - Carbon - 12 7 N - Nitrogen - 14 8 O - Oxygen - 16 9 F - Fluorine - 19 10 Ne - Neon 20 11 Na - Sodium - 23 12 Mg - Magnesium 24 13 Al - Aluminium 27 14 Si - Silicon 28 15 P - Phosphorus - 31 16 S - Sulfur 32 17 Cl - Chlorine - 35.5 18 Ar - Argon 40 19 K - Potassium 39 20 Ca - Calcium - 40. 21 Sc - Scandium - 45 22 Ti - Titanium - 48 23 V - Vanadium 51 24 Cr - Chromium 52 25 Mn - Manganese - 55 26 Fe - Iron 56 27 Co - Cobalt 59 28 Ni - Nickel 59 29 Cu - Copper - 64 30 Zn - Zinc 65

MOLECULAR MASS • • • • •

―The sum of atomic masses of all the atoms present in a molecule of any compound is called its molecular mass.‖ A vast number of compounds are composed of molecules (covalent compounds) & they are assigned with molecular masses, e.g. Molecular mass of H2O= 1x2+16x1=18 a.m.u Molecular mass of CO2= 12+16x2= 44 a.m.u Glucose C6H12O6 = 12x6+1x12+16x6=72+12+96=180 a.m.u

FORMULA MASS

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―The sum of atomic masses of atoms present in the simple empirical formula of a compound is called its formula mass.‖ There are many compounds that are not composed of molecules, i.e. ionic compounds. Such compounds are composed of charged particle called ions. Ions do not exist individually that’s why ionic compounds are network solids & they are assigned formula masses, e.g. Formula mass of NaCl= 23x1+35.5x1= 58.5 a.m.u Formula mass of MgO= 24+16=40 a.m.u

EMPIRICAL FORMULA • • • • • • • •

"The formula of a compound which expresses the ratio in which atoms of different elements are combined in a compound" Empirical formula only indicates atomic ratios but it does not indicate actual number of atoms of different kinds present in the molecule of a compound. Two or more compound may have same empirical formula. Empirical formula is ONLY determined by experiment. The empirical formula of benzene is CH which is indicating that carbon & hydrogen atoms are combined in a ratio of 1:1. Generally, ionic compounds are represented by their empirical formula. KCl, NaCl, MgO, KMnO4, K2Cr2O7 etc are all empirical or simple formula. A compound contains 20.00% C, 6.71% H, 46.65% N, and 26.64% O by mass. Determine its empirical formula Element= %age ÷ atm mass = mol ratio ÷ HCF=Ratio C = 20% ÷ 12 = 1.667 ÷ 1.667 = 1 H = 6.71% ÷1 = 6.71 ÷ 1.667 = 4 N = 46.65 ÷ 14 = 3.332 ÷ 1.667 = 2 Hence C1H4N2 is the empirical formula

MOLECULAR FORMULA

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"The formula of a compound which not only expresses the relative number of atoms of each kind but also expresses the actual number of atoms of each element present in one molecule". • Molecular formula and empirical formula of a compound are related as: • MOLECULAR FORMULA = (EMPIRICAL FORMULA)n • Where "n" is an integer and is given by: • n = molecular mass of compound • empirical formula mass of compound • Molecular formula of propane = C3H8. • Molecular formula of sugar = C12H22O11. Problem solving • A compound contains 20.00% C, 6.71% H, 46.65% N, and 26.64% O by mass. Determine its empirical formula. • a. CH4N2O b. C2H7N3O2 c. C3HN7O4 d. C2H8N4O2 e. C4HN2O4 • In case if molecular formula is required, n is calculated. MOLE • • • • • • •

The gram atomic mass or gram molecular mass or gram formula mass of a substance that contains 6.02 x 1023 particles is called one mole. . Carbon = 12 a.m.u. So 12 gram of carbon = 1 mole of carbon. Nitrogen molecule = 28 a.m.u. 28 gram of N2 = 1 mole of N2 Formula mass of NaCl = 58.5 a.m.u. Therefore 58.5 gram of NaCl = 1 mole of NaCl. Mole is denoted by "n". Number of moles of substance = Mass of substance (in grams) Molar mass

AVOGADRO'S NUMBER • • • • •

One mole of any substance contains equal number of particles (atoms or molecules or ions). Value of this number is 6.02 x 1023. This constant value or number is referred to as "AVOGADRO'S NUMBER" One mole of hydrogen = 6.02 x 1023 molecule of hydrogen. One mole of sodium = 6.02 x 1023 sodium of hydrogen. One mole of Ca+2 = 6.02 x 1023 ions of Ca+2.

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It is denoted by "NA". It is very useful in determining the number of particles (atoms, ions or molecules) in a given mass of any substance,

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• Mole ratio • It is the ratio of theoretical moles of required substance to the theoretical mole of given substance.

STOICHIOMETRY •

The alphabetical representation of a chemical reaction is known as stoichiometry. • Example • NaCl + AgNO3  NaNO3 + AgCl • Species on the left side are reactants, species which react at the beginning of reaction. • Species on the right side are products, species which produce at the end of reaction. • A chemical equation shows the kind of species that react or are produced but does not determine the exact quantity of the substances involved. • Stoichiometry is the technique based upon the balanced chemical equation by which the quantity of products from the given quantity of reactants or as the vice versa. • The traditional method for determining the quantities is time taking, we can deduce certain formula to perform stoichiometry specially made for doing mcq tests. Stoichiometric relationships • There are three basic relations or stoichiometric calculations. • Mass - Mass relationship • Mass - Volume relationship • Volume - Volume relationships • But there can be more… • Like mole – mole, mole – mass, etc • All follow basically same pattern. • Practice to perfect. Formula for stoichiometry • Required mass = mol ratio x given moles x molar mass • Required vol = mol ratio x given moles x molar volume • Required volume = mol ratio x given volume • Required moles = mol ratio x given moles

Mass to mass relation • Required mass = mol ratio x given moles x molar mass • Required vol = mol ratio x given moles x molar volume • Required volume = mol ratio x given volume • Required moles = mol ratio x given moles

LIMITING REACTANT • • • • • • •

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"The substance which produces least amount of products when it is completely consumed in a chemical reaction". Stoichiometric measurement become more complicated when more than one reactant are used, it is because any one of the reactant is completely converted into products, i.e. limiting reactant. However the other reactant is always leftover as the reaction is completed, i.e. excess reactant. It shall be inappropriate to calculate the quantity of products from that of excess reactant as its given quantity is never completely used up. Determination of limiting reactant, thus, is an important step while executing stoichiometric calculations. Limiting reactant is found by stoichiometric formulas. Nitric oxide is made from the oxidation of ammonia. How many moles of nitric oxide can be made from the reaction of 3.80 mol NH3 with 5.15 mol O2? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Before calculating moles of NO, we shall attempt to find which of the reactant NH3 or O2 is limiting. Lets find that how many moles of oxygen are required to consume 3.8 moles of NH3. Formula… Required moles = mole ratio x given moles 5/4 * 3.80 = 4.75 moles

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All 3.80 moles of NH3 will complete consume when 4.75 moles of O2 are consumed & the reaction will stop proceeding any further. What about remaining O2??? 5.15 – 4.75 = 0.4 moles of O2 are left over. Hence NH3 is the limiting reactant.

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