Chapter 1

INTRODUCTION TO

FUNDAMENTAL CONCEPTS OF CHEMISTRY

CHEMISTRY• It is the science that deal with structure, composition & properties of matter.

• It also deals with the changes that matters undergo & energy associated with such changes.

• At intermediate level, three main branches of chemistry are studied.

• Physical Chemistry, Inorganic Chemistry & Organic Chemistry.• Physical Chemistry: The branch of science which deals with the forces & principal involved in the combination of atoms & molecules.• Inorganic Chemistry: The branch of chemistry that deals with the study of all elements & their compounds except the covalent compounds involving carbon & hydrogen.

• Organic Chemistry is the study of compounds of carbon.

• The exception are CO2, CO, CO3-

2, HCO3-1, CN-1 etc

• The main purpose of chemistry are

• To feed the race,

• To cure the sick &

• To provide facilities to mankind.

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.

• 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:

• (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 is 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

• Multiplying coefficients & powers are added

• = (3.2x2.3) x 104-3

• Hence= 7.36 x 101 or 73.6

DIVISION• 48000/0.00012

• Converting into scientific notation.

• 4.8 x 104/1.2 x 10-4

• Decimals divide where as powers subtract.

• 4.8/1.2 x 104+4

• 4 x 108

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”)

• 101=10 (log of 10 is “1”)

• 102=100 (log of 100 is “2”)

• 103=1000 (log of 1000 is “3”)

• 10-1=0.1 (log of 0.1 is “-1”)

• 10-2=0.01 (log of 0.01 is “-2”)

• 10-3=0.001 (log of 0.001 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

• log an = 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

• = 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

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.

• In this picture, all of the darts land on the bulls-eye which illustrates good precision and accuracy.

• In this picture, all of the darts land near each other, but away from the bulls-eye which illustrates good precision, but poor accuracy.

• In this picture, all of the darts land on the bulls-eye which illustrates good precision and accuracy.

• In this picture, all of the darts land near each other, but away from the bulls-eye which illustrates good precision, but poor accuracy.

• 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%

• The percent error will always be a positive value.

CLASSIFICATION OF MATTERMatter

Pure substances Mixtures

elementscompound

s

Form by bonds between elements

Ionic compounds

Covalent compounds

Composed of atoms

Identified by symbols & atomic

masses

Composed of ionsIdentified by

empirical formula & formula mass

Composed of molecules

Identified by molecular formula &and molecular

mass

ATOMIC MASS• "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.

• Hydrogen= 1

a.m.uCarbon= 12

a.m.u

Helium= 4 a.m.u

Nitrogen= 14 a.m.u

Oxygen= 16 a.m.u

Sodium= 23 a.m.u

Sulphur= 32 a.m.u

Magnesium= 24 a.m.u

Calcium= 40 a.m.u

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• “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

PROBLEM SOLVING • Calculate the formula mass of Fe(NO3)3·9H2O.

• a. 515.695 u

• b. 403.999 u

• c. 241.862 u

• d. 394.928 u

• e. 375.986 u

• H2O= 18 a.m.u…So… 18x9= 162

• NO3=14+3x16=14+48=62…So.. For (NO3)3= 62 x 3 = 186

• Fe = 56

• So 56+171+186= 404 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.

MOLECULAR FORMULA• "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.

• Molecular formulae are generally assigned to compounds that are composed of molecules, i.e. covalent compounds, e.g.

• CH4, H2O, NH3 etc are molecular formula of few covalent compounds.

• Ionic compounds DONOT & CANNOT have molecular formula

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.

element

÷ at. mass

= mole ratio ÷ HCF = ratio

C

H

N

÷ 12

= 6.71÷ 1

= 1.667

÷ 14=

3.332

÷ 1.667

÷ 1.667

= 1

÷ 1.667

= 4

= 2

%age

20%

6.71%46.65

%

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)

Molecular mass or atomic mass or formula 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.

• 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,

PROBLEM SOLVING • How many moles of CO2 are present in 2200

mg?

• A) 5 moles

• B) 0.005 mole

• C) 5000 moles

• D) 10 moles

• E) 50 moles

• How many moles are present in 64 g of methane, CH4?

• A) 3 moles

• B) 4 moles

• C) 5 moles

• D) 6 moles

• E) 2.5 moles

• How many grams are needed for 2 moles of sulfuric acid, H2SO4?

• A) 145 grams

• B) 165 grams

• C) 196 grams

• D) 164 grams

• E) 250 grams

• How many molecules are present in 88 g of carbon dioxide, CO2?

• A) 3.01 x 1023 molecules

• B) 6.02 x 1023 molecules

• C) 9.03 x 1023 molecules

• D) 1.204 x 1024 molecules

• E) 1.204 x 1023 molecules

STOICHIOMETRY• The alphabetical representation of a

chemical reaction is known as chemical equation.

• 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.

Reactants

Products

• 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 is measured 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.

MOLE RATIO• It is the ratio of theoretical moles of

required substance to the theoretical mole of given substance.

• Suppose

• 3NH4ClO4 (s) + 3Al (s) Al2O3 (s) + AlCl3 (s) + 3NO (g) + 6H2O (g)

theoretical 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

• If 5.0 moles of Al react with excess O2, how many moles of Al2O3 can be formed? 4 Al(s) + 3 O2(g) → 2 Al2O3(s)

• a. 1.0 mol

• b. 2.0 mol

• c. 2.5 mol

• d. 5.0 mol

• e. 10.0 mol

• What mass of H2 can be formed from the reaction of 51.3 g of carbon with excess water?

• C(s) + H2O(g) → H2(g) + CO(g)

• a. 4.31 g

• b. 8.61 g

• c. 17.2 g

• d. 102.6 g

• e. 306 g

• What mass of oxygen is consumed by the complete combustion of 23.0 grams of ethylene?

• C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g)

• a. 8.75 g

• b. 26.2 g

• c. 60.5 g

• d. 69.0 g

• e. 78.7 g

• Sodium carbonate reacts with hydrochloric acid as shown below in an unbalanced chemical equation. What mass of CO2 is produced from the reaction of 53 g Na2CO3 with excess HCl?

• Na2CO3(s) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)

• a. 22 g

• b. 44 g

• c. 94 g

• d. 88 g

• e. 80 g