UNIT 1 SOME BASIC CONCEPTS OF CHEMISTRY
SYLLABUS
1.1 Importance of chemistry.
1.2 Nature of matter
1.3 Properties of matter and their Measurement
1.4 Uncertainty in measurement
1.5 Laws of Chemical Combinations
1.6 Dalton’s Atomic Theory
1.7 Atomic and Molecular masses
1.9 Percentage composition
1.10 Stoichiometry and Stoichiometric Calculations
INTRODUCTION
Chemistry is defined as the branch of science which deals with the study of composition, structure and properties of matter and changes which the matter undergoes under different conditions and the laws which govern these changes.
Importance of chemistry
Science can be viewed as a continuing human effort to systematize knowledge for describing and understanding nature. For the sake of convenience science is sub-divided into various disciplines : chemistry, physics, biology , geology etc. Chemistry is the branch of science that studies the composition, properties and interaction of matter. Chemists are interested in knowing how chemical transformations occur. Chemistry plays a central role in science and is often intertwined with other branches of science like physics , biology, geology etc. Chemistry also plays an important role in daily life.
NATURE OF MATTER
Anything which has mass and occupies space is called matter. Everything around us , book, water, air, all living being etc. are composed of matter.
Classification of matter
There are two ways of classifying matter :
(i) Physical classification.
(ii) Chemical classification.
Physical classification
Based on physical state under ordinary conditions of temperature and pressure, matter is classified into the following three types : (i) solids (ii) liquids (iii) gases
A substance is said to be solid if possesses a definite volume and definite shape e.g. sugar, iron, wood etc.
A substance is said to be liquid, if it possesses a definite volume but no definite shape.
A substance is said to be gaseous, if it neither possesses a definite volume nor a definite shape. This is because they fill up the whole vessel in which they are put, e.g. hydrogen, oxygen, carbon dioxide, air etc.
Chemical classification
All kinds of matter can be classified into the following two types :
(i) Homogeneous
(ii) Heterogeneous
The word ‘material’ is commonly used for all kinds of matter whether homogeneous or heterogeneous.
A material is said to be homogeneous, if it has uniform composition and identical properties throughout.
Since any distinct portion of matter that is uniform throughout in composition and properties is called a ‘’phase’’ , hence,
A material is said to be homogeneous, if it consists of one phase. On the other hand, a material is said to be heterogeneous, if it consists of a number of phases.
However, a more scientific classification is called ‘chemical classification of matter’ is briefly described below :
All types of materials are believed to be made up of ‘’substances’’ . A material containing only one substance is called a ‘’pure substance’’. On the other hand , materials containing more than one substance are not pure and are called ‘’mixtures’’. Pure substances are further classified into ‘elements’ and ‘compounds’. Mixtures are also of two types, namely ‘Homogeneous mixtures’ and ‘Heterogeneous mixtures’. Homogeneous mixtures are called ‘’solutions’’. The single phase in which a solution occurs may be gaseous, liquid or solid.
Elements
An element is the simplest form of matter which cannot be further sub-divided into simpler form by chemical or physical methods. Examples : Copper, sulphur, oxygen etc. Elements can be further divided into following types :
i. Metals
These are generally solids and have characteristics such as hardness, ductility, malleability, lustre, high tensile strength and ability to conduct heat and electricity. Examples : copper silver, aluminium etc.
ii. Non-metals
These are generally non-lustrous , brittle, poor conductors of heat and electricity. e.g., sulphur, nitrogen, hydrogen, oxygen etc.
iii. Metalloids
These elements have characteristics common to metals as well as non-metals. e.g., arsenic, bismuth, antimony etc.
Compound
A compound consists of two or more elements, joined together in a fixed ratio by chemical bonds.
For example,
Water is formed from hydrogen and oxygen.
Sodium chloride formed from sodium and chlorine
Sugar formed from carbon, hydrogen and oxygen.
Atom
An atom is the smallest particle of an element which can take part in a chemical reaction. It may or may not exist free in nature, e.g., O represents one atom of oxygen ; Cl represents one atom of chlorine.
Molecule
A molecule is the smallest particle of an element or compound that can exist by itself and also retains the properties of the substance. Example, O2 represents one molecule of oxygen ; MgSO4 represents one molecule of magnesium sulphate.
Mixture
A mixture is a system in which two or more elements and / or compounds are simply mixed together in any proportions. For example, air is a mixture gases like nitrogen, oxygen, carbon dioxide, water vapour, argon, dust particles etc.,. Similarly gun powder containing charcoal, sulphur and potassium nitrate ; brass is a mixture containing copper and zinc.
Mixtures are of two types
i) Homogeneous mixture
ii) Heterogeneous mixture.
Homogeneous Mixture : It is one which has uniform composition throughout. For example, salt solution in water, sulphur solution in carbon disulphide etc.
Heterogeneous mixture
It is one which has non-uniform composition. For example, a suspension of sand in water.
PROPERTIES OF MATTER AND THEIR MEASUREMENT
Every substance has unique or characteristic properties. These properties can be classified in two categories
i) physical properties and
ii) chemical properties
Physical properties
Physical properties are those properties which can be measured or observed without changing the identity or the composition of the substance. Example : colour, odour, melting point, boiling point, density etc.
Chemical properties
The observation or measurement of chemical properties require a chemical change to occur. The examples of chemical properties are characteristic reactions of different substances , like acidity, basicity, combustibility etc.
Physical quantities and their measurement in chemistry
In everyday life , we come across a number measurements, e.g. we buy vegetables in kilograms, milk in litres, cloth in metres. However, during scientific studies, in addition to the measurements of mass, volumes and lengths, we come across a number of other quantities such as temperature, pressure, concentration, force , work, density etc. All such quantities , which we come across during our scientific studies are called physical quantities.
Evidently, the measurement of any physical quantity consists of two parts :
(i) the number , and
(ii) the unit
For example , if an object weighs 4.5 kg, it involves two parts :
(i) 4.5 , i.e., the number and
(ii) kg, i.e. the unit.
The main aim in this section are :
(i) To see how accurately the number has been expressed , i.e. the concept of significant figures.
(ii) To study the units of the measurement, i.e. the S.I units.
(iii) To derive the units of any physical quantity and check the accuracy of any scientific equation by seeing the dimensions of both sides of the equation are same, i.e. the concept of dimensional analysis.
Measurement and S .I. units
A unit is defined as the standard of reference chosen to measure any physical quantity.
Since early times , different types of units of measurements have been very popular in different parts of the world e.g. Sers, pounds etc. for mass ; miles, furlongs, yards etc for distances. However, these units are quite cubersome because no uniformity in the conversion factors involved. e.g. mile = 1760 yards, 1 yard = 3 feet, 1 foot = 12 inches.
In view of the difficulties mentioned above, French Academy of Science, in 1791, introduced a new system of measurements called ‘metric system’ in which the different units of a physical quantity are related to each other as multiples of powers of 10, e.g. 1 km = 103 m , 1 cm = 102 m etc. This system was found to be so convenient that scientists all over the world adopted the system for reporting scientific data . It remained in use till 1960.
In 1960 , the General Conference of Weights and Measures (Conference Gererale des Poids et Measures , CGPM) , adopted the International system of units ( or SI, after the French Le System Internationale Units). This system has seven SI base units ( Table 1). In SI , large and small quantities are are expressed by using appropriate prefix ( Table 2) , with base units.
Table 1 SI Base Units
Physical quantity Name of the unit Symbol
Mass kilogram kg
Length metre m
Time Second s
Temperature Kelvin K
Electric current Ampere A
Luminous intensity Candela cd
Amount of substance Mole mol
Metre
The metre is the length of the path traveled by light in vaccum during a time interval of 1/299 792 458 of a second.
Kilogram
The kilogram is the unit of mass ; it is equal to the mass of the international prototype of the kilogram.
Second
The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of caesium-133 atom.
Ampere
The ampere is that constant current which if maintained in straight parallel conductors of infinite length, of negligible circular cross-section , and placed 1 metre apart in vaccum, would produce between these conductors a force equal to 2 x 107 newton per metre of length.
Kelvin
The Kelvin, unit of thermodynamic temperature, is 1/273.16 of the thermodynamic temperature of the triple point of water.
Mole
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.12 kilogram of carbon-12 ; its symbol is ‘’mol.’’ When the mole is used , the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.
Candela
The candela is the luminous intensity , in a given direction, of a source that emits monochromatic radiation of frequency 540 x 1012 hertz and that has a radiant intensity in that direction 1/1683 watt per steradian.
The SI system allows the use of prefixes to indicate the multiples or submultiples of a unit. These prefixes are listed in Table 2.
Table 2 SI Prefixes
Multiple Prefix Symbol Multiple Prefix Symbol
1024 yotta Y 101 deci d
1021 zetta Z 102 centi c
1018 exa E 103 milli m
1015 peta P 106 micro
1012 tera T 109 nano n
109 giga G 1012 pico p
106 mega M 1015 femto f
103 kilo k 1018 atto a
102 hecto h 1021 zepto z
10 deca da 1024 yocto y
DERIVED UNITS
At first glance, the SI system seems to be a very limited system of units. There are many quantities , such as area and volume, whose units do not appear in Table 1.1. In the SI , units for such quantities are obtained by appropriate combinations (multiplication or division) of the base units and are called derived units.
Table 3 Some common SI Derived Units
Quantity Definition SI unit
Area Length squared m2
Volume Length cubed m3
Density Mass per unit volume kg m3
Velocity Distance traveled per unit time ms 1
Acceleration velocity change per unit time. m s 2
Force Mass times acceleration of object. Kg m s 2
= N(newton)
Pressure Force per unit area kg m 1s 2
= (Pascal, Pa )
= N m 2
Energy
(work, heat) Force times distance traveled kg m2 s 2
= Joule = J
Electric charge Ampere times second A s ( Coulomb , C)
Electrical potential Energy per unit charge J / ( A s)
potential difference ( volt, V)
The way the SI base units are combined in such cases depend on the dimensions of the measured quantities. For instance, if you wished to calculate the area of a rectangular carpet, you would multiply its length by its width. The unit for area is likewise obtained as the product of the unit for length and the unit for width. Since length and width are both distances , for which the SI unit is meter (m) , the SI unit for area is m2(meter squared or square meter).
Length x width = area
m x m = m2
Similarly , speed is the ratio of distance travelled divided by elapsed time. The SI unit for speed is therefore meter / second or or m/s or m s1. Some commonly used SI derived units are given in Table 3
Many quantities like force , pressure , energy etc. are commonly defined in terms of their own units which are expressible in terms of the base units.
Force
The SI unit of force is newton (1 N = kg m s2). To appreciate its size of the unit , imagine holding a mass of 1 kg. The force required to hold is about 10 N, since if 1 kg is released, it will fall downwards with acceleration of 9.8 m s2 or a force of 9.8 kg m s2 ( 9.8 N). In other words, 1 N is the force needed to hold 100 g of mass.
Pressure
The SI unit of pressure is pascal and is defined as a force of 1 N applied to an area of 1 m2. A popular unit is atmosphere (atm) ; 1 atm is equal to 101.325 kPa. For rough estimate, we can take 1 atm 105 N m2. To appreciate the size of the unit, recall that a 760 mm high mercury column exerts a pressure of 1 atm.
Energy
The joule ( 1 J = 1 N m) is the SI unit of energy. It is defined as the energy needed to push against a force of 1 N through 1 m. If we raise 1 kg of mass by 1 m , we are roughly spending 10 J of energy. Energy is also expressed in in kcal mol1 in older literature of chemistry. The data is easily converted into kJ mol1 by multiplying by 4.184.
Note
1. The unit named after a scientist is started with a small letter and not with a capital letter e.g. unit of force is written as newton and not as Newton. Likewise unit of heat or work is written as joule and not as Joule.
2. Symbols of the units do not have a plural ending like ‘s’ . For example we have 10 cm and not 10 cms.
3. Words and symbols should not be mixed e.g. we should write either joules per mole or J mol1. not joules mol1.
4. Prefixes are used with basic units e.g. kilometer means 1000 m ( because meter is the basic unit). Exception : Though kilogram is the basic unit of mass, yet prefixes are used with gram because in kilogram, kilo is already a prefix.
5. A unit written with a prefix and a power for complete unit e.g. cm3 means (centimetre)3 and not centi(metre)3.
Mass and 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. The mass of a substance can be determined very accurately in the laboratory by using an analytical balance.
The SI unit of mass is kilogram. However, its fraction gram (1 kg = 1000 g) , is used in laboratories due to smaller amounts of chemicals used in chemical reactions.
Volume
Volume has units of (length)3. So in SI system, volume has units of m3. But in laboratories , smaller volumes are used. Hence volume is often denoted in cm3 or dm3 units.
A common unit , litre (L) which is not an SI unit, is used for measurement of volume of liquids.
1 L = 1000 mL
1000 cm3 = 1 dm3
In laboratory, volume of liquids or solutions can be measured by graduated cylinder, burette, pipette etc. A volumetric flask is used to prepare a known volume of a solution. These measuring devices are shown in Fig.
Some volume measuring devices
Density
Density of a substance is its amount of mass per unit volume. So SI unit of density can be obtained as follows :
This unit is very large and a chemist often expresses density in g cm3, where mass is expressed in gram and volume is expressed in cm3.
Temperature
There are three common scales to measure temperature ; C (degree celsius) , F ( degeee farenheit and K (Kelvin) . Here , K is the SI unit. The thermometers based on these scales are shown in Fig. Generally , the thermometer with Celsius scale are calibrated from 0 to 100 where these two temperatures are the freezing point and boiling point of water respectively. The farenheit scale is represented between 32 and 212.
The temperatures on two scales are related to each other by the following relationship.
The Kelvin scale is related to Celsius scale as follows :
K = C + 273.15
Thermometers using different temperature scales
The below 0C (i.e., negative values) are possible in Celsius scale but in Kelvin scale, negative temperature is not possible.
Intensive and Extensive properties
A property which independent of the amount of substance taken is called intensive property. Example, density, boiling point, refractive index etc.
A property depends on the amount of substance taken is called extensive properties. Example, mass , volume etc.
Problems
01. Fill in the blanks in the following conversions :
(i) 1 km = ……. mm = ………. pm
(ii) 1 mg = ……. kg = ……… ng
(iii) 1 mL = ……. L = ……… dm3
02. If the speed of light is 3.0 x 108 m s1, calculate the distance covered by light in 2.0 ns.
03. What physical quantities are represented by the following units and what are their most common names ?
(i) kg m s2 (ii) kg m2 s2
04. Deduce the SI unit of :
(i) area (v) accleration
(ii) volume (vi) force
(iii) density (vii) work
(iv) speed (viii) energy
05. Express each of the following in SI units.
(i) 93 million miles ii) 5 feet 2 inches.
(iii) 100 miles per hour iv) 14 pounds per square inch.
(v) 0.74 Å (vi) 46C
(vii) 150 pounds.
06. Vanadium metal is added to steel to impart strength. The density of vanadium is 5.96 g cm3. Express this in SI unit.
07. Match the following prefixes with their multiples.
Prefixes Multiples
(i) micro 106
(ii) deca 109
(iii) mega 106
(iv) giga 1015
(iv) femto 10
MEASUREMENT AND SIGNIFICANT FIGURES
Precision and Accuracy
Every measurement is limited by reliability of the measuring instrument and skill of the person making the measurement. In each mesurement uncertainty should be reported correctly.
If we repeat a particular measurement, we usually do not obtain precisely the same result as each measurement is subjected to experimental error. Different measured values vary slightly from one another. The term precision refers for the closeness of the set of values obtained from identical measurements of a quantity. Accuracy , a related term, refers to the closeness of a single measurement to its true value.
Illustration
Three students were asked to determine the mass of a piece of metal where mass is known to be 0.520 g. The data obtained by each student are recorded in the Table 4.
Table 4 Data to illustrate Precision and Accuracy
Measurements in gram
Student 1 2 3 Average
A 0.521 0.515 0.509 0.515
B 0.516 0.515 0.514 0.515
B 0.521 0.520 0.520 0.520
The data for student A are neither very precise nor accurate ; the individual values differ widely from one another and average value is not accurate. Student B was able to determine the mass of the metal more precisely. The values deviate but little from one another, however the average mass is still not accurate. In contrast , the data for the student C is both precise and accurate.
SCIENTIFIC ( EXPONENTIAL) NOTATION
Very large and very small numbers are common in chemistry. Repeatedly writing down such numbers in the ordinary way ( as in the case of Avogadro constant 602,213,700 000 000 000 000 000 ) would be tedious and inevitably result in errors. A better way to represent such numbers is essential for a convenient and accurate manipulation. Scientific notation offers such a way.
In scientific notation, all numbers , however, large or small , are expressed as a number between 1.000…. and 9.999…. multiplied or divided by 10, an appropriate number of times.
For example,
138.42 = 1.3842 x 10 x 10 = 1.3842 x 102
Here 2 is the power or exponent , to which 10 is raised.
In scientific notation, a number is generally expressed in the form
N x 10n
Where N is a number ( called digit term) between 1.000…. and 9.999….. and n is a number ( not necessarily a single digit) called an exponent. To express a number smaller than 1.000…. in scientific notation, the appropriate number between 1.000…. and 9.999….. is divided by an appropriate number of times.
For example,
To transform a number larger than 9.999…. to scientific notation, the decimal point is moved to the left until there is only one non-zero digit before the decimal point. If the decimal point is moved x places to the left , the exponent n = x . Thus in transforming 138.42 to scientific notation the decimal point is moved to the left two places. So exponent n = 2 and we can write :
To transform a number smaller than 1.000….. to scientific notation, the decimal point is moved to the right until there is one non-zero digit before the decimal point. If the decimal point is moved by y places to the right, the exponent is y. For example, when 0.0001382 is to be transformed to scientific notation, decimal point is moved four places to the right.
So the exponent is n = 4 and we can write
To add and subtract numbers in scientific notation , the power of 10 , that is , the exponent, n must be the same in both the numbers. For example, we have to add 6.234 x 104 and 1.203 x 103. We transform 1.203 x 103 to 0.123 x 104 and then add
6.234 x 104 + 0.1203 x 104
= ( 6.234 + 0.1203 ) 104 = 6.3543 x 104
To multiply two numbers in scientific notation, we make use of the relation,
(10)x (10)y = 10(x+y)
For example,
( 3.025 x 103 ) ( 6.217 x 106 = 18.81 x 10 3 6
= 18.81 x 103 = 1.881 x 102
To divide two numbers in scientific notation, we make use of the relation,
For example,
SIGNIFICANT FIGURES
When we make measurement, we obtain numbers by reading them from a scale of some sort on a measuring device. Because of this, there is always some limitation on the numbers of meaningful digits that can be obtained . For example, in a centimeter scale, the measurement can be done to the nearest centimeter and not further. Similarly, a millimeter scale will make the measurement possible to the nearest millimeter. Thus, we conclude that the measurement of various variables such as weight, volume, density , pressure, temperature etc. can only precise and not exact. Since every measuring scale has its own precision, the uncertainty in measurement can be expressed in terms of certain figures which are often termed as significant figures. The significant figures in a number are all certain digits plus one doubtful digit. In order to understand the implication of the term ‘significant figure’, let us assume the height of a man measured in centimeter scale has reported in three ways i.e., 160 cm, 160.0 cm and 160.00 cm. Apparently, there seems to be no difference in the values but each one has a certain specific information to convey.
(i) The number 160 cm indicates that the measuring scale has the precision in centimeters only and the height of the man is somewhere between 159 and 161 cm. The digits 1 and 6 are certain digits, whereas the digit zero has some doubt or uncertainty. Thus there are three significant figures in the above number.
(ii) The next number 160.0 cm has four digits. The result is more precise and clearly indicates the scale of measurement has been calibrated beyond centimeters . The number lies between 159.9 cm and 160.1 cm. There are four significant figures out of which three are definite while the last one is a doubtful digit.
(iii) Similarly, the third number , i.e., 160.00 cm indicates that the scale has still better precision beyond the first point of decimal. The value lies somewhere between 159.99 cm and 160.01 cm. Thus the number has five significant figures out of which the first four are definite , while the last is doubtful.
Thus we can conclude that the number of significant figures in any number depends upon the precision of the scale employed for the purpose. The more the significant figures there are in a measured quantity, the greater is the precision of measurement.
Rules for Reporting the number of Significant Figures
The following rules should be observed in counting the number of significant figures of a given measured quantity :
(i) All the digits are significant except the zeros at the beginning of a number. In other words leading zeros are never significant.
For example:
0.132 has three significant figures.
0.013 has two significant figures.
0.001 has one significant figure.
(ii) A zero becomes significant in case it comes in between two non-zero numbers.
For example
6.023 has four significant figures.
6.0023 has five significant figures.
(iii) All the zeros placed at the right of a number are significant. In fact they represent the accuracy or precision of the measuring scale.
Example
343.0 has 4 significant figures
343.00 has five significant figures.
343.000 has 6 significant figures.
(iv) Exact numbers have infinite number of significant figures. For example in 2 balls or 20 eggs, there are infinite number of significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e., 2 = 2.000000 or 20 = 20.000000
(v) In exponential notations , the numerical portion represents the number of significant figures.
Example,
0.000054 is expressed as 5.4 x 105 in terms of scientific notations. The number of significant figures in this number is 2. Similarly, the number of significant figures in the Avogadro’s number , 6.023 x 1023 is four.
The different ways of expressing a length of 1200 cm in different number of significant figures is given below:
1.200 x 103 m : four significant figures.
1.20 x 103 m : three significant figures
1.2 x 103 m : two significant figures.
Rounding off
If we wish to round off a number at certain point :
• We simply drop the digits that follow, if the first term is less than 5. Thus 6.2317 rounds to 6.23 if we wish only two decimal places.
• If the digit after the point of round off is larger than five or it is 5 followed by other non-zero digits, then we add 1 to the preceding digit. Thus, 6.236 and 6.2351 both round to 6.24.
• When the digit after the point of round off is 5 and no other digits follow the five, then we drop 5 if the preceding digit is even and we add 1 if it is odd. Thus 8.165 rounds to 8.16, but 8.175 rounds to 8.18.
Note : While solving problems all significant digits used in calculation should be taken up and only the final result should be round off to the desirable number of significant figures.
Calculations involving significant figures
To get the final result of any experiment, usually calculations are required which involve addition, subtraction, multiplication and division of different numbers. These numbers may have different accuracies , i.e., may contain different number of significant figures or decimal places. The final result involving these numbers , therefore, cannot be more accurate or precise than the last precise number involved in a particular calculation. The following rules are applied in determining the number of significant figures in the answer of any particular calculation.
Rule 1 : The result of an addition or subtraction should be reported to the same number of decimal places as that of the term with least number of decimal places. The number of significant figures of different numbers have no role to play.
Example 1
First number has three decimal places, second has one and the third has two. Hence the answer should be reported only up to one decimal place. Note that the significant figures in the three numbers are 4, 2 and 3 respectively.
Example 2
Each number has three decimal places. So the answer is also reported up to three decimal places. Further note that the significant figures in each of the three numbers is 4, but the result has five significant figures.
Example 3
The last number is an exact number ( involving no decimal place ) . Hence answer is reported as an exact number.
Example 4
As the second number has two decimal places only, while the first has four, the answer is reported up to the two decimal places only.
Example 5
As the first number has one decimal places only, while the second has four, the answer is reported up to one decimal place.
Rule 2 : The result of a multiplication or division should be reported to the same number of significant figures as is possessed by the least precise term used in the calculation.
Example 1
The first number has four significant figures, while the second has two. The actual product has been rounded off to give a reported product of 12 i.e., containing two significant figures only. This is because the least precise term in the calculation ( viz. 2.8) has only two significant figures.
Example 2
0.46 15.734 gives
Actual quotient = 0.029236
Reported quotient should be 0.029 containing only two significant figures because the least precise term in the calculations (viz., 0.46) has only two significant figures only.
It may be noted that both the above rules, in fact , may be interpreted as follows :
‘’ The reported answer should not be more precise than the least precise term used in the calculation.’’
This generalization helps to check the reported answer in case where doubt arises. For example, in example 1, the precision of the least precise term is 0.1 part in 2.8 i.e., 1 part in 28 or nearly 35 parts per thousand ( i.e. 35 p.p.t). The precision of the reported answer is 1 part in 12 or nearly 83 p.p.t. Thus the reported result is not more precise than the least precise term. If we take the reported answer as 12.1 , the precision will be 0.1 in 12.1 or 1 part in 121 i.e., nearly 8 p.p.t. which is more precise than the least precise term. Hence the reported result should be 12 and not 12.1.
Rule 3
If a calculation involves a number of steps, the result should contain the same number of significant figures as that of the least precise number involved, other than the exact numbers.
Example
Leaving the exact number 4, the least precise term has two significant figures. Hence after rounding off , the reported result will be 0.77 i.e., containing two significant figures. Alternatively, the above rule is applied as follows :
First the number of significant figures that the answer should contain is decided ( i.e., it should be equal to that of the least precise term, other than the exact number). Before carrying out the mathematical operations, every number is round off to contain one significant figure more than the answer would have. The answer obtained is then round off to contain the required number of significant figures.
Thus in the above example, the answer should have two significant figures. Hence every number is first round off to contain three significant figures.
After rounding off to two significant figures, reported answer should be 0.77.
Note
The rules that have been stated above apply only to non-integral measured quantities because only in these cases the uncertainty in measurement has significance. These do not apply to exact numbers where uncertainty has no significance.
Problems
08. What is the difference between 5.0 g and 5.00 g ?
09. Express the following in the scientific notation :
(i) 0.0048 (ii) 234,000
(iii) 8008 (iv) 500.00
(v) 6.0012
10. How many significant figures are present in the following ?
(i) 0.0025 (ii) 208
(iii) 5005 (iv) 126,000
(v) 500.0 (vi) 2808
11. Convert the following into basic units :
(i) 26.7 pm (ii) 15.15 s (iii) 25365 mg
12. How many significant figures should be present in the answer of following calculations ? (N/24/1.31)
13. (a) Express the following numbers to four significant figures.
(i) 5.607892 (ii) 32.392800 (iii) 1.78986 x 103
(iv) 0.007837
(b) Express the result of the following calculations to appropriate number of significant figures.
(i) 3.24 x 0.08666
5.006
(ii) 0.582 + 324.65
(iii) 943 x 0.00345 + 101
14. How many significant figures are there in each of the following numbers ?
(i) 6.005 (iv) 0.0025
(ii) 6.022 x 1023 (v) the sum of 18.5 + 0.4325
(iii) 8000 (vi) the product of 14 x 6.345
15. Express the following to four significant figures :
(i) 6.45372 (ii) 48.38250 (iii) 70000
(iv) 2.65986 x 103 (v) 0.004687
16. A sample of a metal weighs 6.5425 g and has a density of 8.8 g cm3. What is the volume ? Report the answer to correct decimal place.
17. How many significant figures are there in each of the following numbers ?
(i) 6.200 (ii) 0.052 (iii) 7.5 x 104
(iv) 0.00050 (v) 67.32 6.3
18. What is the number of significant figures in :
(i) Avogadro number (6.0 x 1023 ) and
(ii) Planck’s constant ( 6.62 x 1034)
19. Express the number 45000 in exponential notation to show :
(i) two significant figures.
(ii) four significant figures.
20. Express the following in SI base units using power of 10 notation.
(a) 1.35 mm (b) 1 day (c) 6.45 mL (d) 48 g (e) 0.0426 in
DIMENSIONAL ANALYSIS
It is frequently necessary to convert one set of units to another in calculations. The systematic procedure used for this is called factor label method or unit factor method or dimensional analysis.
Table Conversion factors
Quantity S I Units Conversion factor
Length
Metre 1 m = 100 cm = 1.0936 yards
1cm = 0.3937 inch
1 in = 2.54 cm
= 0.0254 m
1 Angstrom (A) = 1010 m
1 mile = 1.6094 km
Mass
kilogram(kg) 1 kg = 1,000 g
= 2.205 lb
1 lb = 453.6 g
1 atomic mass unit (amu)
= 1.66054 x 1024 g
Temperature
Kelvin(K) 0 K = *2730 C
= *459.670 F
0F = (9/5) 0C + 32
0C = (5/9) 0F * 32
K = 0C + 273
Volume
Cubic metre 1 litre (L) = 103 m3.
1 in3 = 16.4 cm3.
1 cm3 = 1 m L
Force
Newton
(N = kg ms2 )
1 dyne = 105 N
Pressure
Pascal
(Pa = N m2) 1 atmosphere
= 101,325 Pa
= 760 mm Hg
= 14.7 lb in2.
= 1.013 x 106 dyn cm2.
Energy
Joule
(J = N m) 1 calorie = 4.184 J
1 electron volt = 96,485 kJ mol1
1 litre-atmosphere = 101.325 J
1 J = 107 ergs
This is based on the relationship between different units that express the same quantity. We can construct conversion factors from relationships between units. A conversion factor is a fraction that we use to convert a given quantity to a desired quantity by multiplication.
(given quantity) x (conversion factor) = (desired quantity)
A conversion factor has the property of changing units of the given quantity to appropriate units of desired quantity.
Suppose we wished to convert a measured length of 4.0 yd ( 4.0 yards) into the equivalent length measured in feet. To do this, we must use the known relationship between yards and feet,
1 yd = 3 ft
This can be used to construct two conversion factors, which are
Each of these has a numerical equivalence of 1 . If we multiply anything by 1, we don’t change its size. Therefore, if we multiply any quantity by a conversion factor, its magnitude doesn’t change. All we do is change the units.
To convert 4.0 yd to ft, we must multiply by one of the conversion factors that we have constructed. To make the choice , we simply examine the units. We know that the unit ‘’yd’’ must not be the answer , so we choose the factor that allows us to cancel yd. This is the second one.
The units can often help you to detect mistakes , because if you do the wrong arithmetic, you get the wrong units.
For example, suppose we have used the first conversion factor by mistake.
The units are yd2/ft (square yards per foot). These are not the units we want, so we know that we made a mistake. In solving chemistry problems, it is very helpful to know when you have made a mistake.
We can use the concept of an equivalence between quantities, instead of an equality. For example if you have a job that pays 5 rupees per hour, there is an equivalence between the time you work and the rupees you earn. Each hour that you work is equivalent to 5 rupees pay. We express this in a mathematical way by using the symbol to mean ‘’is equivalent to’’. Thus ,
1 hr 5 rupees
Equivalencies such as this are used to make conversion factors. In this case , we can make these two factors,
If you worked for 12 hours, you could use the second factor to calculate your earnings.
Problems
21. A jug contains 2 L of milk. Calculate the volume of milk in m3.
22. How many seconds are there in 2 days ?
23. Using the unit conversion factors, express the given measurements in the designated units.
(a) 25 L to m3
(b) 25 g L1 to mg d L1
(c) 1.54 mm s1 to pms1
(d) 2.66 g cm3 to g m3 (e) 4.2 L h2 to mLs2
24. For a precious stone , ‘carat’ is used for specifying its mass. If 1 carat = 3.168 grains ( a unit of mass) and 1 gram = 15.4 grains, find the total mass in kilogram of the ring that contains a 0.500 carat diamond and 7.00 gram gold.
25. ‘The star of India’ sapphire weighs 563 carats. If one carat is equal to 200 mg, what is the weight of the gemstone in grams.
26. What is the name given to the unit that equals:
i) 109 g ii) 103m iii) 106 L
27. Calculate the volume of 40 kg of carbon tetrachloride whose density is 1.60 g cm3.
28. What is the density of a steel ball which has a diameter of 7.5 mm and a mass of 1.755 g ?
29. (a) Convert the following in kilogram.
(i) 0.91 x 1027 g (mass of electron)
(ii) 1 f g ( mass of human DNA molecule)
(iii) 500 M g (mass of jumbo jet loaded)
(iv) 3.34 x 1024 g (mass of hydrogen molecule)
(b) Convert into metre
(i) 7 nm (diameter of small virus)
(ii) 40 Em (thickness of Milky Way Galaxy)
(iii) 1.4 Gm (diameter of Sun)
(iv) 41 Pm (distance of the nearest star)
30. What is the mass of 1 L of mercury in grams and in kilograms if the density of liquid mercury is 13.6 g cm3 ?
31. Convert litre-atmosphere to Joule (SI system of energy).
LAWS OF CHEMICAL COMBINATIONS
On the basis of various quantitative experiments on the combination of elements to form compounds, Lavoiser and other scientists made certain generalisations known as laws of chemical combinations. The fundamental laws are :
1. Law of conservation of mass
2. Law of constant composition or definite proportions.
3. Law of multiple proportions.
4. Law of reciprocal proportions.
5. Gay-Lussac’s law of gaseous volumes.
The first four laws deals with the mass relationships and fifth deals with the volumes of the reacting gases.
1. LAW OF CONSERVATION OF MASS
This law deals with the masses of the reactants and the products of a chemical reaction ( or a physical change) and was studied by Antonie Lavoisier (1789). This law may be stated as follows :
In all physical and chemical changes, the total mass of the reactants is equal to that of the products.
Thus according to this law , there is no increase or decrease in the total mass of matter during a chemical or physical change. In other words,
Matter can neither be created nor destroyed.
Hence , this law is also called the Law of indestructibility of matter. The following experiments illustrate the truth of the law.
(a) When matter undergoes a physical change.
A piece of ice (solid water) is taken in a small conical flask. It is well corked and weighed. The flask is now heated gently to melt the ice (solid) into water (liquid).
The flask is again weighed. It is found that there is no change in the weight though a physical change has taken place.
(b) When matter undergoes a chemical change
The following chemical changes illustrate the law :
(i) Precipitation reaction : Landolt took solutions of sodium chloride and silver nitrate separately in two limbs of the Landolt’s tube (Fig)
Landolt’s experiment
The tube was sealed and then weighed. After weighing , the two solutions are mixed thoroughly by shaking the tube. The reaction occurred with the formation of a white precipitate of silver chloride as :
AgNO3 + NaCl AgCl + NaNO3
White ppt.
No heat was evolved or absorbed during the reaction. After the reaction, the tube was again weighed and the weight was found to be the same. This confirms the truth of the law.
(ii) Decomposition of mercuric oxide
100g of mercuric oxide when heated in a closed tube, decomposed to produce 92.6 g of mercury and 7.4 g of oxygen gas.
Thus during the above decomposition reaction, matter is neither gained or lost.
Present position of the law
In reactions involving mass energy conversions, the law of conservation of mass has to be modified. Matter can be converted into equivalent amount of energy , i.e., in a transformation , the sum of matter and energy remains unaltered. Matter and energy are related by the expression :
Energy = mass (velocity of light)2
E = m c2
Where E is energy in joules, m is the mass of matter in kilograms and c is the velocity of light ( = 3 x 108 m s1)
It is not generally possible to bring about complete conversion of matter into energy. However, in atomic pile or atomic bomb , matter is converted into energy. In ordinary reactions, mass energy conversion is too small to detect by ordinary chemical analysis. So for practical purposes, the law of conservation of mass holds good for all chemical reactions.
Problems
32. 10 g of calcium carbonate on heating gave 4.4 g of CO2 and 5.6 g of CaO. Show that these observations are in agreement with the law of conservation of mass.
33. (a) When 4.2 g of NaHCO3 is added to a solution of acetic acid weighing 10.0 g , it is observed that 2.2 g of CO2 is released into atmosphere. The residue is found to weigh 12.0 g . Show that these observations are in agreement with the law of conservation of mass.
(b) If 6.3 g of NaHCO3 are added to 15.0 g CH3COOH solution, the residue is found to weigh 18.0 g. What is the mass of CO2 released in the reaction ?
2. THE LAW OF DEFINITE PROPORTIONS OR
CONSTANT COMPOSITION.
This law was given by Joseph Proust and deals with the composition of elements present in a given compound.
The law states that : The same compound always contain the same elements combined together in the same fixed proportions by weight.
This means that the composition of a compound is always the same irrespective of the method by which it is produced.
Illustration
Sodium chloride may be obtained from sea water or from salt mines. It can also be prepared by the reaction between sodium hydroxide and hydrochloric acid. These samples, on analysis are found to contain sodium and chlorine in the proportion 23 : 35.46 by mass.
Note
The law of constant composition is true only if the same compound is obtained from same isotope of an element, because masses of isotopes of an element are different. For example, two samples of sodium chloride (NaCl) obtained from sodium(23Na) and two isotopes of chlorine (viz.., 35Cl and 37Cl ) are 23Na35Cl and 23Na37Cl and the ratio of masses of two elements Na : Cl in these will be 23 : 35 and 23 : 37 respectively.
Moreover , this law does not hold good for non-stiochiometric compounds, which possess variable composition.
Problem
34. 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098 g. In another experiment, 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476 g. Show that these results illustrate the law of constant composition.
3. THE LAW OF MULTIPLE PROPORTIONS (DALTON)
Statement
When two elements combine to form more than one compound, the different masses of one of the elements which combines with a fixed mass of the other element bear a simple ratio.
Illustration
Carbon combines with oxygen to form two different oxides namely, carbon monoxide and carbon dioxide. In carbon monoxide, 12 g of carbon combine with 16 g of oxygen. In carbon dioxide 12 g of carbon combine with 32 g of oxygen. Therefore the weights of oxygen combining with 12 g of carbon in these oxides are in the ratio 16 : 32 i.e., 1 : 2 which is a simple ratio.
Problems
35. Hydrogen and oxygen are known to form two compounds. The hydrogen content in one of these is 5.93 % while in the other it is 11.2 %. Show that this data illustrates the law of multiple proportions
36. Carbon and oxygen are known to form two compounds. The carbon content in one of these is 42.9% , while in the other it is 27.3%. Show that this data is in agreement with the law of multiple proportions.
37. The following data are obtained when dinitrogen and dioxygen react together to form different compounds.
Mass of dinitrogen Mass of dioxygen
(i) 14 g 16 g
(ii) 14 g 32 g
(iii) 28 g 32 g
(iv) 28 g 80 g
Which law of chemical combination is obeyed by the above experimentl data ? Give its statement.
(iv) The Law Of Reciprocal Proportions (Richter)
The law was proposed by Richter (1792).
Statement
If an element A combines with an element B and also with an element C, then if B and C are to combine together the ratio (by mass) in which they combine should be simply related to the different masses of B and C combining with a fixed mass of A.
or
If an element A combines with an element B in the ratio x : y and with another element C in the ratio x : z then if B and C are to combine they do so in the ratio y : z or a simple multiple of it.
Illustration
Hydrogen combine with carbon to form methane in the ratio 1 : 3 . Hydrogen also combine with oxygen to form water in the ratio 1 : 8. If carbon and oxygen are to combine, they do so in the ratio 3 : 8. By experiment it is found that the formation of carbon dioxide is in the ratio 3 : 8.
Problem
38. Phosohorus(atomic mass 31) combines with hydrogen (atomic mass 1) in the mass ratio of 3.1 : 0.3. In water , hydrogen and oxygen (atomic mass 16) combine in the ratio 0.2 : 1.6.
(i) From the above data , calculate the mass ratio of phosphorus and oxygen in the oxide of phosphorus.
(ii) What is the name of the law that is applied in calculating this ratio ?
(iii) What is the valency of phosphorus in an oxide where mass ratio of phosphorus to oxygen is 3.1 : 4 ?
5. LAW OF COMBINING VOLUMES
The French chemist, Gay-Lussac experimented with the reaction of gases and found that the volumes of reactants and products in a large number of chemical reactions are related to each other by small integers, provided the volumes are measured at the same temperature and pressure.
Example :
In the reaction of hydrogen with oxygen to produce water vapour, it was found that two volumes of hydrogen and one volume of oxygen give two volumes of water vapour. This means that 100 mL of H2 gas will combine exactly with 50 L of O2 gas to give exactly 100 mL of water vapour if all the gases are measured at the same temperature and pressure ( say 300 K and 1 atm pressure).
Gay-Lussac’s discovery of integer ratio in volume relationship is actually the law of definite proportion by volume. The Gay-Lussac’s law , which was based on experimental observation , was explained properly by the work of Amedeo Avogadro in 1811.
Avogadro’s law
Avogadro’s law states that : The volume of a gas (at fixed pressure and temperature) is proportional to the number of moles (or molecules of gas present).
Mathematically, we can write :
V n
(where n is the number of moles of gas) . Avogadro’s law can be stated in another simple way : Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.
Both the Gay-Lussac and Avogadro laws can be illustrated as follows :
Problem
39. If 10 volumes of dihydrogen gas reacts with 5 volumes of dioxygen gas , how many volumes of water vapour would be produced ?
Applications of Avogadro’s hypothesis
1. Determining Atomicity of Elementary Gases
Atomicity of an element is the number of atoms present in a molecule of it. Avogadro’s hypothesis helps in determining the atomicity of elementary gases such as hydrogen, oxygen, chlorine etc., Let us determine the atomicity of chlorine molecule. The atomicity of chlorine can be determined by considering the reaction between hydrogen and chlorine to form hydrogen chloride. It is known that at the same temperature and pressure:
Hydrogen + Chlorine * Hydrogen chloride
1 volume 1 volume 2 volumes
Suppose that 1 volume of hydrogen contain ‘n’ molecules. Therefore according to Avogadro’s hypothesis 1 volume of chlorine also contain ‘n’ molecules and 2 volumes of hydrogen chloride would have ‘2n’ molecules. Hence we can say that :
This means that 1 molecule of hydrogen chloride molecule is made up of half a molecule of hydrogen and half a molecule of chlorine. This can be possible only if half a molecule of hydrogen contains at least one atom because a fraction of atom is not possible. Now the molecule of hydrogen chloride contains one atom of hydrogen. This shows that a molecule of hydrogen is actually equal to two atoms. In a similar manner, it can be shown that a molecule of chlorine also contains two atoms of it.
2. Derivation of the relation between vapour density and molecular mass.
Vapour density of a gas is the ratio of the mass of certain volume of gas to the mass of the same volume of hydrogen at the same temperature and pressure.
According to Avogadro’s law equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules. Let ‘n’ is the number of molecules in the volume considered. Then ,
( Molecular mass of a substance is defined as the ratio of the mass of 1 molecule of the substance to the mass of 1 atom of hydrogen).
3. To show that Molar volume ( Gram molecular volume ) of all gases is the same at STP.
The Molar volume ( gram molecular volume ) of a gas is the volume occupied by one mole of gas at standard temperature and pressure and is 22.4 litres or 22,400 cm3.
3. Determination of molecular formula of gaseous compounds
The following example illustrates the deduction of molecular formula of gaseous compounds.
From experiments , it is found that 1 volume of nitrogen combines with 3 volumes of hydrogen to form 2 volumes of ammonia. If 1 volume of a gas contains n molecules, then on the basis of Avogadro’s hypothesis, we have,
Since both nitrogen and hydrogen are diatomic, ½ molecule of nitrogen will contain one atom of it while 3/2 molecules of hydrogen will contain (3/2) x 2 = 3 atoms of it.
This implies that one molecule of ammonia is composed of one atom of nitrogen and three atoms of hydrogen. Hence the molecular formula of ammonia should be NH3.
DALTON’S ATOMIC THEORY
John Dalton , in 1803 published ‘ A New System of Chemical Philosophy’ in which he proposed the following :
1. Matter consists of indivisible atoms.
2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. 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.
MODERN ATOMIC THEORY
Dalton’s atomic theory played a very significant role in the growth and development of chemistry. But later discoveries and knowledge made it necessary to modify his theory. The major modifications are mentioned below:
1. Atom is divisible and consists of sub-atomic particles like protons, electrons and neutrons.
2. Atoms of the same element need not be alike. Isotopes of the same element differ e.g., 126C , 136C and 146C.
3. Atoms of different elements may possess the same mass e.g., 4019K and 4020Ca (such atoms with same mass number but different atomic numbers are called isobars)
4. The ratio in which atoms combine and need not necessarily be simple e.g., molecules of cane sugar(C12H22O11) , stearic acid(C17H35COOH) and many others with complex ratios and formulae.
5. The postulate that atoms can neither be created nor destroyed is not strictly true. New atoms can be created through artificial transmutation. Partial destruction of atoms and matter is the basis for the production of atomic energy, according to Einstein’s equation, E = mc2, E is the energy, m, the mass converted into energy and c , the velocity of light. Indestructibility of atoms is true about chemical reactions.
Problems
40. 4 g of a gas occupies 1400 cm3 at STP. Calculate its molecular mass.
41. What will be the volume occupied at STP by the following ?
i) 7 g carbon monoxide
ii) 32 g of oxygen
iii) 19 g of carbon disulphide.
42. 100 cm3 of a gas at STP weighs 0.152 g. Calculate the molecular mass of the compound.
43. 0.39 g of a gas occupies 112 cm3 at STP. Calculate :
i) Density
ii) Vapour density
iii) Molecular mass
ATOMIC MASS
Since atom is a very small particle, its actual mass cannot be determined. However, it is possible to determine the relative masses of atoms. For this purpose, the mass of some atom is taken as standard and the masses of atoms of all other elements are compared to this standard. The standard for comparison of atomic masses is taken to be the most common isotope of carbon, which has been assigned an atomic mass of 12 amu. Thus a new scale called Atomic Mass Unit ( amu ) has been accepted for expressing the relative atomic masses of elements.
Atomic mass unit (amu) (U)
Atomic mass unit is exactly (1/12)th the mass of a C-12 atom.
Note
The new symbol used is ‘u’ (unified mass) in place of amu.
Average Atomic Mass of an Element
The relative atomic mass of an element is defined as the ratio of the average mass of an atom of the element to (1/12) th mass of a C-12 atom.
The atomic mass of an element tells us to how many times an atom of the element is heavier than (1/12)th mass of an atom of C-12.
For example, atomic mass of magnesium is 24 amu. It means that an atom of magnesium is 24 times heavier than (1/12)th mass of a C-12 atom.
Concept of Average Atomic mass
It has been found that majority of elements occur in nature as a mixture of several isotopes. Isotopes are atoms of the same element with different atomic masses. The best way to define atomic mass of these elements is to determine the atomic masses of each isotope separately and combine them in the ratio of their proportions of occurrence. This is called average atomic mass. For example, chlorine contains two types of atoms having relative atomic masses 35 and 37 amu. The relative abundance of these isotopes in nature is in the ratio 3 : 1. Thus atomic mass of chlorine is the average of these different relative atomic masses as follows.
The naturally occurring carbon is a mixture of three isotopes known as carbon-12, carbon-13 and carbon-14. The average of carbon atom in such a sample is 12.011 amu. The isotopic mass of various isotopes of an element are determined by very accurate measurements and these are used to calculate the atomic mass of the element.
In general,
Atomic mass = fractional abundance x isotopic mass
Fractional abundance of an isotope is the fraction of the total number of atoms that is composed of that particular isotope.
The isotopic masses are very close to the whole numbers , while atomic masses being , the average values, are generally fractional.
The average atomic mass for a naturally occurring element , expressed in atomic mass units is called Atomic mass.
The atomic masses of elements are given in the following Table
Atomic masses of Elements Referred to12C = 12.0000
Element Symbol Atomic number Atomic mass
Actinium Ac 89 (227)
Aluminium Al 13 27
Amwercium Am 95 243
Antimony Sb 51 121.8
Argon Ar 18 39.9
Arsenic As 33 74.9
Astatine At 85 210
Barium Ba 56 137.3
Berkyllium Bk 97 (245)
Beryllium Be 4 9.01
Bismuth Bi 83 209
Boron B 5 10.8
Bromine Br 35 79.9
Cadmium Cd 48 122.4
Caesium Cs 55 132.9
Calcium Ca 20 40.1
Californium Cf 98 (251)
Carbon C 6 12
Cerium Ce 58 140.1
Chlorine Cl 17 35.5
Chromium Cr 24 52
Cobalt Co 27 58.9
Copper Cu 29 63.5
Curium Cm 96 (245)
Dyspronsium Dy 66 162.5
Einstenium Es 99 (254)
Erbium Eb 68 167.3
Europium Eu 63 152
Fermium Fm 100 (254)
Fluorine F 9 19
Francium Fr 87 (223)
Gadolinium Gd 64 157.3
Gallium Ga 31 69.7
Germanium Ge 32 72.6
Gold Au 79 197
Hafnium Hf 72 178.5
Helium He 2 4
Holmium Ho 67 164.9
Hydrogen H 1 1.008
Indium In 49 114.8
Iodine I 53 126.9
Iridium Ir 77 192.2
Iron Fe 26 55.8
Krypton Kr 38 83.8
Lanthanum La 57 138.9
Lawrencium lr 103 (257)
Lead Pb 82 207.2
Lithium Li 3 6.94
Lutetium Lu 71 175.0
Magnesium Mg 12 24.3
Manganese Mn 25 54.9
Mendelevium Md 101 (256)
Mercury Hg 80 200.6
Element Symbol Atomic number Atomic mass
Molybdenum Mo 42 95.9
Neodymium Nd 60 144.2
Neon Ne 10 20.2
Neptunium Np 93 237.0
Nickel Ni 28 58.7
Niobium Nb 41 92.9
Nitrogen N 7 14.0
Nobelium No 102 (254)
Osmium Os 76 190.2
Oxygen O 8 16.0
Palladium Pd 46 106.4
Phosphorus P 15 31.0
Platinum Pt 78 195.1
Plutonium Pu 94 (242)
Polonium Po 84 (210)
Potassium K 19 39.1
Praseodymium Pr 59 140.9
Promethium Pm 61 (145)
Protactinium Pa 91 231.0
Radium Ra 88 226.0
Radon Rn 86 (222)
Rhenium Re 75 186.2
Rhodium Rh 45 102.9
Rubedium Rb 37 85.5
Ruthenium Ru 44 101.1
Samarium Sm 62 150.4
Scandium Sc 21 45.0
Selenium Se 34 79.0
Silicon Si 14 28.1
Silver Ag 47 107.9
Sodium Na 11 23.0
Strontium Sr 38 87.6
Sulphur s 16 32.1
Tantalum Ta 73 180.9
technetium Te 52 127.6
Terbium Tb 65 158.9
Thallium Tl 81 204.4
Thorium Th 90 232.0
Thulium Tm 69 168.9
Tin Sn 50 118.7
Titanium Ti 22 47.9
Tungsten W 74 183.8
Uranium U 92 238.0
Vanadium V 23 50.9
Xenon Xe 54 131.3
Ytterbium Yb 70 173.0
Yttrium Y 39 88.9
Zinc Zn 30 65.4
Zirconium Zr 40 91.2
(Numbers in parentheses give the mass number of most stable isotope.)
Approximate atomic masses of common elements
Element Symbol Atomic number Atomic mass
Aluminium Al 13 27
Argon Ar 18 40
Barium Ba 56 137
Bromine Br 35 80
Calcium Ca 20 40
Carbon C 6 12
Chlorine Cl 17 35.5
Chromium Cr 24 52
Iron Fe 26 56
Fluorine F 9 19
Gold Au 79 197
Helium He 2 4
Lead Pb 82 207
Lithium Li 3 7
Hydrogen H 1 1
Manganese Mn 25 55
Magnesium Mg 12 24
Mercury Hg 80 200
Neon Ne 10 20
Nickel Ni 28 58
Nitrogen N 7 14
Oxygen O 8 16
Phosphorus P 15 31
Potassium K 19 39
Radium Ra 88 226
Silver Ag 47 108
Sodium Na 11 23
Thorium Th 90 232
Tin Sn 50 118
Uranium U 92 238
Zinc Zn 30 65
Problems
44. The element chlorine contains two isotopes3517Cl and , which has a mass of 34.98 amu and 3717Cl , which has a mass of 36.98 amu. The atomic mass of chlorine is 35.46 amu. Calculate the percentage of each of the isotopes in chlorine.
45. Use the data given in the following table to calculate the molar mass of naturally occurring argon.
Isotope Isotopic molar mass Abundance
36Ar 35.96755 g mol1 0.337%
38Ar 37.96272 g mol1 0.063%
40Ar 39.9624 g mol1 99.600%
Determination of Isotopic masses
The isotopic masses are most accurately determined by using an instrument called mass spectrograph (Aston’s mass spectrograph and Dempster’s mass spectrograph)
GRAM ATOMIC MASS
It may be defined as that much quantity of the element whose mass in grams is numerically equal to its atomic mass. Gram atomic mass is also called one gram -atom of the element. For example, atomic mass of Mg is 24 amu.
1 gram-atom of Mg = gram atomic mass of
magnesium = 24 g
2 gram-atom of Mg = 2 x gram atomic mass of Mg
= ( 2 x 24 ) = 48 g
From the above discussion, we get the relationship between mass of the element and its gram atoms as :
MOLECULAR MASS
The actual masses of molecules are very small and cannot be measured directly. Therefore for the purpose of comparison, we depend upon the relative masses of molecules. The standard mass taken for reference is the mass of carbon atom ( C-12 ) taken as 12 amu.
Molecular mass of a substance tells us the number of times a molecule of the substance is heavier than (1/12)th mass of carbon (C-12 atom). For example, molecular mass of water (H2O) is 18 amu. It means that a molecule of water is 18 times heavier than (1/12)th mass of a C-12 atom
Molecular mass of a substance can be obtained by adding the atomic masses of all the atoms present in a molecule of the substance. For example,
Molecular mass of H2O = (2 x At. Mass of H) + ( At. Mass of O )
= 2 x 1 + 16 = 18 amu
Formula mass
In the case of ionic compounds, formula of the compound does not represent its molecules, but only represents the ratio of different ions in the compound. This is called a formula unit of the compound. In such compound, therefore we do not use the term molecular mass. Instead ,we use the term formula mass. Formula mass of an ionic compound is obtained by adding atomic masses of all the atoms in a formula unit of the compound. For example,
Formula mass of sodium chloride (NaCl) =
= Atomic mass of Na + atomic mass of Cl
= 23 + 35.5 = 58.5
Gram Molecular Mass
It may be defined as that much quantity of the substance ( element or compound) whose mass in grams is numerically equal to its molecular mass. Gram molecular mass is also called Gram Molecule of the substance. For example, molecular mass of ammonia is 17 amu, therefore:
Percentage Composition and Molecular Formula
Percentage composition of a compound is the amount of various constituent elements present in 100 grams of the compound. Percentage of an element in a compound can be calculated by dividing the total mass of the element by molecular mass and multiplying the quotient by 100.
The mass percentage of a particular constituent species in ores and minerals can be calculated as :
Problems
46. Calculate the molecular mass of the following :
(i) H2O (ii) CO2 (iii) CH4
47. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).
48. Fe2(SO4)3 is the empirical formula of a crystalline compound of iron. Calculate the mass percentage of iron, sulphur and oxygen in this compound.
49. Calculate the percentage composition of CaCO3.
50. Find the percentage composition of potassium nitrate, whose formula is KNO3.
51. What is the percentage of water of crystallization in Al2(SO4)3 .18 H2O ?
52. Calculate the mass of 2.5 gram atoms of calcium.
53. Calculate the mass of 1.5 gram molecules of water.
54. Calculate the number of gram atoms and gram molecules in 48 g of oxygen.
55. Calculate the mass of 2 gram molecules of sulphuric acid.
56. What will be the mass of one 12C atom in g ?
57. Which one of the following have largest number of atoms ?
(i) 1 g Au(s) (ii) 1 g Na(s) (iii) 1 g Li(s) (iv) 1 g Cl2(g)