3. LANGUAGE OF CHEMISTRY
THE “CRIME CHEMISTRY”
Before starting with any complex calculations or reactions, one needs to know the Language of Chemistryranging from basics of units to balancing of equations.
1. Some Basics to be Revised
1.1 Physical Quantity
The quantities that are measurable are known as physical
quantities.
Eg. Length, mass, time etc.
A physical quantity is represented by number followed by a unit.
The number and its unit collectively represent the magnitude of
the physical quantity.
Figure 3.1: Measurement of physical
quantity
www.plancess.com
Language of Chemistry
3.2
(a) Fundamental physical quantities:
The quantities that are independent of each other are called as physical quantities. Length, mass,
time, electric current, temperature, luminous intensity and amount of substance are taken as
fundamental or basic quantities.
(b) Derived physical quantities:
The quantities, which are obtained from fundamental quantities are known as derived quantities.
The quantities like area, volume, speed, acceleration, force etc. are derived physical quantities.
1.2 Unit
A unit is the standard of comparison for measurements. Units are essential for presenting a measured
quantity correctly.
Modern System of Units
Table 3.1: Modern system of units
MKS or Metric system:
C.G.S. System:
F.P.S. or British system:
Length
meter (m)
centimeter (cm)
foot (ft)
Mass
kilogram (kg)
gram (gm)
pound (lb)
Time
second (s)
second (s)
second (s)
According to SI system there are seven fundamental units and two supplementary units.
1.3 Basic or Fundamental SI units
Supplementary SI Units
All other quantities (for example speed, volume, pressure etc.) can be derived from these fundamental
quantities.
The metric system is a decimal system. Different units for physical quantities are expressed in powers of ten.
Table 3.2: Supplementary SI unit
Prefix
Abbreviation
Multiple
Example
Yotta
Y
1024
1 yattameter(Ym)= 1 × 1024 m
Zeta
Z
1021
1 zetameter(Zm)= 1 × 1021 m
Exa
E
1018
1 exameter(Em)= 1 × 1018 m
www.plancess.com
Foundation for Chemistry
Prefix
Abbreviation
Multiple
Peta
P
1015
1 petameter(Pm)= 1 × 1015 m
Tera
T
1012
1 terameter(Tm)= 1 × 1012 m
Giga
G
109
1 gigameter(Gm)= 1 × 109 m
Mega
M
106
1 megameter(Mm)= 1 × 106 m
Kilo
k
103
1 kilometer(km)= 1 × 103 m
Hecto
h
102
1 hectometer(hm)= 1 × 102 m
Deka
dam
101
1 dekameter(dam)= 1 × 101 m
Deci
da
10 −1
1 decimeter(dm)= 1 × 10−1 m
Centi
c
10 −2
1 cetimeter(cm)= 1 × 10−2 m
Milli
m
10 −3
1 millimeter(mm)= 1 × 10−3 m
Micro
µ
10−6
1 micrometer( µ m)= 1 × 10−6 m
Nano
n
10−9
1 nanometer(nm)= 1 × 10−9 m
Pico
p
10−12
1 picometer(pm)= 1 × 10−12 m
Femto
f
10−15
1 femtometer(fm)= 1 × 10−15 m
Atto
a
10−18
1 attometer(am)= 1 × 10−18 m
Zepto
z
10−21
1 zeptometer(1zm)= 1 × 10−21 m
Yocto
y
10−24
1 petameter(1ym)= 1 × 10−24 m
1.
3.3
Example
Mass and weight: Mass of a system represents the amount of matter present in a system, while
weight represents the force that gravity exert on that system (or object). These terms are often
used interchangeably, although strictly speaking, they are different quantities.
www.plancess.com
Language of Chemistry
3.4
The SI base unit of mass is kilogram (kg) but in chemistry unit gram (g) is more convenient and
frequently used. 1 kg = 1000 g = 1 × 103 g
2.
Volume: Volume of a body = (length)3
Because in SI system, unit of length is meter, so SI unit of volume is m3.
1 cm3 = (1 × 10–2 m)3 = 1 × 10–6 m3
1 dm3 = (1 × 10–1 m)3 = 1 × 10–3 m3
3.
Density: Density (d) i.e., density of a substance is the mass of its unit volume. Density is an
intensive property. Intensive properties are those properties which does not change when we
divide the system,for example temperature after dividing the system the temperature of the two
new formed system will remain the same but mass is an extensive properties ,if we divide the
system then mass of the two systems changes.
The SI unit of density is kg/m3 but g/cm3 or g/mL are more commonly used, for expressing the
densities of solids and liquids.
1 g/cm3 = 1 g/mL = 1000 kg/m3
1 g/L = 0.001 g/mL
4.
Temperature: Temperature of a body represents the intensity of heat energy associated with it.
Three scales are most common for measuring the temperature.
(a) Degree Farenheit (°F) : On this scale, freezing and boiling points of water are 32°F and
212°F respectively. These two limits are divided into 180 divisions.
(b) Degree Celsius (°C) : On this scale, freezing and boiling points of water are 0°C and 100°C
respectively and this range between freezing and boiling points of water is divided into 100
parts.
(c) Kelvin (K and not °K) : It is absolute temperature scale and is SI unit of temperature. 0 K is
the lowest temperature that can be attained theoretically.
Following relations can be used directly to interconvert temperature in different units.
Here C and F are temperatures in °C and °F respectively.
Also,
K = C + 273.15
1.4 Conventions for Writing Units
Accuracy and Precision:
Accuracy represents closeness to the true value whereas Precision is agreement between a groups of
measurement.
In simple words, Accuracy refers to how close a measurement is to the true value of the quantity that was
measured while precision tells us how closely individual measurements agree with each other.
www.plancess.com
Foundation for Chemistry
3.5
Suppose for example our true value is 20 units. In two sets of experiments let us suppose our values
are(18,18.5,19,19.3,19.7) and (12,12.5,12.7,13,11.9) then first set is accurate and second set is more precise.
1.5 Derived Units
The unit which are derived from the fundamental units or basic units are called derived units. Some
example of derived units are listed below:
Non-metric Units:
(a) Non-metric Units of Length:
Units of length for large distances:
Light year (ly):
It is equal to the distance travelled by light in vacuum in one year. i.e. 1 ly = 9.46 ×1015m.
Astronomical Unit (A.U.):
One astronomical unit is equal to the mean distance between the earth and the sun.
1 A.U. = 1.496 × 1011 metres
Parsec: 1 Parsec = 3.26 ly = 3.08 ×1016 m
Units of length for small distances.
Angstrom (Å) :
1Å = 10–10 m
Fermi (fm) :1fm = 10–15 m
(b) Non-metric Unit of Mass:
Atomic mass unit (amu):
It is defined as of mass of one carbon-12 atom, thus 1 amu = 1.67 ×10–27 kg
www.plancess.com
3.6
Language of Chemistry
(c) Non-metric Units of Time:
Examples of these are day, month, lunar month, year, decade, century and millennium etc.
1 decade = 10 years
1 century = 100 years
1 millenium = 1000 years.
1.6 Significant Figures
Significant figures are meaningful digits which are known with certainty. The uncertainty is indicated by
writing the certain digits and the last uncertain digit. Thus, if we write a result as 11.2 mL, we say the 11
is certain and two is uncertain and the uncertainty would be +1 in the last digit.
Rules for determination of significant figures:
1.
All non-zero digits are significant.
For example:
3.14 has three significant figures; two certain +1 uncertain.
5.153 has four significant figures; three certain +1 uncertain.
0.5153 has four significant figures; three certain +1 uncertain.
2.
The zeros to the right of the decimal point are significant.
For example:
3.0 has two significant figures; one certain +1 uncertain digit.
3.10 has three significant figures; two certain +1 uncertain digit.
3.140 has four significant figures; three certain +1 uncertain digit.
3.
The zero to the left of the first non-zero digit in a number are not significant. The merely
represent the position of a decimal point.
For example:
0.02 g has one significant figure.
0.002 g has one significant figure.
4.
The zeros between two non-zero digits are also significant.
For example:
6.01 has three significant figures; two certain and last digit uncertain.
6.001 has four significant figures; three certain and last digit uncertain.
www.plancess.com
Foundation for Chemistry
5.
3.7
Leading zero before the decimal point are never significant.
For example:
0.618 has three significant figures; two certain and last digit uncertain.
6.
When a number ends in zeros that are not to the right of decimal point, the zeros are not
necessarily significant.
For example:
180 cm has two or three significant figures, 18600 g has three or four or five significant figures.
This ambiguity has been removed by using exponential notation.
7.
The use of exponential notation avoids the potential ambiguity of whether the zeros at the end
of a number are significant (rule six) or not. For example, a mass of 19400 g can be written in
exponential notations showing three, four or five significant figures as:
1.94 × 104
(three significant figures)
1.940 × 104
(four significant figures)
1.9400 × 104
(five significant figures)
Thus, all significant figure came before the exponent and the exponential term does not add
to number of significant figures. Similarly significant numbers in numerical value of Avogadro
number (6.023 × 1023) are four and in Planck’s constant (6.626 × 10-34Js) are four.
8.
Exact numbers can be treated as if they have an infinite numbers of significant figures.
This rule applies to many definitions between units. Thus when we say one foot has 12 inches,
the number 12 is exact we need not worry about the number of significant figures in it.
9.
Rounding off the uncertain digit:
The rounding off of uncertain digit of significant figure is made as follows:
(a) If the digit to be rounded off is more that five, the preceding number is increased by one.
For example:
2.16 is rounded off to 2.2
3.58 is rounded off to 3.6
(b) If the digit to be rounded off is less than five, the preceding number is retained.
For example:
2.14 is rounded off to 2.1
4.13 is rounded off to 4.1
www.plancess.com
3.8
Language of Chemistry
(c) If the digit of be rounded off is equal to five, the preceding number is not changed if it is even
and increased by one, if it is odd.
For example:
3.25 is rounded off to 3.2
(Preceding digit 2 is even)
2.35 is rounded off to 2.4
(Preceding digit 3 is odd)
1.6.1 Calculations Involving Significant Figures
(a) Rule 1: In addition and subtraction, the result should be reported to the same number of decimal
places as that of the term with the least number of decimal places.
(b) Rule 2: In multiplication and division, the results should be reported to the same number of
significant figures as the least precise term used in calculation.
One should not calculate to a degree of accuracy greater than required.
2. Chemical Equation
An atom of an element is represented by symbol, the molecule of a compound by a formula, and a
chemical change is represented by an equation. Thus, chemical equation is the symbolic representation
of an actual chemical change taking place where reactant species are written on the left hand side and the
products formed are written on the right hand side.
Rules for writing an equation –
1.
Substances which take part in chemical reaction (i.e., starting material, reactants) are usually
written on left hand side and substances formed as a result of chemical reaction (i.e., products)
are written on right hand side. The reactants and products are separated by an arrow head mark
(→) which starts from reactant and points towards the products. Sign of equal to (=) may also be
used in place of arrow head mark.
2.
Each reactant or the product is separated from one another by using plus sign (+) in between
them.
3.
Now the number of atoms of the substances on both sides are equalized. This is known a balanced
equation.
Example: (i) Magnesium bums with oxygen and form magnesium oxide.
Note: Magnesium burns in air with a brilliant white light, and for this reason. It is often used in
flares and fireworks.
www.plancess.com
Foundation for Chemistry
3.9
2Mg(s) + O2(g) → 2MgO(s)
Magnesium also bums in an environment of carbon dioxide such as in a beaker full of dry ice:
2Mg(s) + CO2(g) → 2MgO(s) + C(s)
(ii) Nitrogen and hydrogen combines together to give ammonia. (Haber synthesis)
N2 + H2 → NH3 (Skeleton equation)
N2 + 3H2 → 2NH3 (Balanced equation)
PLANCESS CONCEPTS
A Breathalyzer test for alcohol makes use of the oxidation of
alcohol by dichromate ion. The dichromate ion is orange, but its
reduction product is green Cr3+ ion. The breath of a person who
is intoxicated contains alcohol vapor, which passes through an
acidic solution containing when the person blows through the
Breathalyzer device.
Figure 3.2: Breathalyzer
Any alcohol in the exhaled air is oxidized, which is signaled by
3+
the appearance of green Cr . The more alcohol the person has consumed, the greater the
intensity of the green color.
Vipul Singh
AIR 1, NSO
PLANCESS CONCEPTS
The explosive decomposition of sodium azide releases nitrogen gas, which
rapidly inflates an air bag during a crash.
Figure 3.3: Inflation of air bag
Vipul Singh
AIR 1, NCO
(iii) When potassium chlorate is heated it is decomposed to give potassium chloride and oxygen.
KClO3 → KCl + O2 (Skeleton)
www.plancess.com
3.10
Language of Chemistry
2KClO3 → 2KCl + 3O2 (Balanced)
(iv) When hydrochloric acid reacts with calcium carbonate, calcium chloride and water are formed
and carbon dioxide gas is given out.
CaCO3 + HCl → CaCl2 + CO2 + H2O(Skeleton)
CaCO3 + 2HCl → CaCl2 + H2O + CO2 (Balanced)
A chemical equation should be balanced so as to satisfy the requirements of the law of
conservation of mass, as no matter is destroyed or created during a chemical reaction.
Features of the equation:
(1) The reactants appear on the left, and the products appear on the right. The arrow joining them
indicates the direction of reaction.
(2) An integer precedes the formula of each substance. These numbers are known as stoichiometric
coefficients. The quantitative relation among reactants and products is known as stoichiometry.
The stoichiometric coefficient for H2 in this equation is three and that of NH3 is two. When no
number appears, we understand that the stoichiometric coefficient is one. The stoichiometric
coefficient for N2 in this equation is one.
(3) The stoichiometric coefficients in a chemical equation are the set of smallest integers that gives a
balanced equation.
(4) Charge is conserved. In this equation, all participants are neutral species, so charge is conserved
regardless of the stoichiometric coefficients. For reactions that include ions, however, charge
conservation must be explicitly addressed.
(5) Stoichiometric coefficients indicates relative numbers of molecules involved in the reaction.
Note: When the Haber synthesis is carried out industrially, immense numbers of molecules are
involved, but for every one molecule of N2 that reacts, three of H2 react and two of NH3 are
produced.
Theoretical yield is always more than experimental yield.
www.plancess.com
Foundation for Chemistry
3.11
The amount of product actually obtained in a reaction is called actual yield.
Balancing a chemical equation refers to establishing the mathematical relationship between
the quantity of reactants and products. The quantities, are expressed as grams or moles.
2.1 Balancing of a Chemical Equation
The number of atoms on the both sides of arrow head marks should be equal in a chemical equation.
Such equation is called balanced equation. The equation can be balanced by two methods.
(1) Hit and trial method
1.
(2) Partial equation method
Hit and Trial Method:
There is no definite rule for balancing an equation by this method and, therefore, it requires a
good deal of skill and practice. The equation is balanced by hit and trial method in the following
steps:
(i) First of all write the skeleton equation for the reaction and count the number of atoms of each
substance on either side of the equation.
(ii) Balance the atoms of the elements which occurs at minimum number of places on both the sides
of equation.
(iii) When elementary gases like oxygen, nitrogen, hydrogen etc. appear as a reactant or a product,
these gases are first kept in atomic state. Later on the whole equation is multiplied by two and
the atomic equation is made molecular.
Example:
Acetylene gas burns in air producing carbon dioxide and water.
The skeleton equation of the above reaction:
C2H2 + O → CO2 + H2O
Here first of all balance carbon, then hydrogen and in the end balance oxygen atom.
(a) To balance carbon atoms, multiply the molecule CO2 by two
C2H2 + O → 2CO2 + H2O
(b) Number of hydrogen atoms are same on both sides, but there are five oxygen atom on right hand
side and there is only one oxygen atom on left hand side. Hence multiply the atom O by five on
left hand side.
C2H2 + 5O → 2CO2 + H2O
www.plancess.com
3.12
Language of Chemistry
Now it is balanced atomic equation.
(c) To convert balanced atomic equation into balanced molecular equation, multiply the whole
equation by 2.
2C2H2 + 5O2 → 4CO2 + 2H2O
ANSWER TO THE “CRIME CHEMISTRY”
As you know, when someone picks up an object with their
bare hands, they leave fingerprints on it. Crime investigators
call them latent fingerprints, which means they cannot be seen
unless developed in some way. (latent means “hidden”.)
There’s a number of ways latent fingerprints can be made
visible, including dusting with a fine powder, as seen on TV
crime shows. Another method makes use of an ionic reaction
of the kind discussed in this section.
When you touch something, among the substances present in the fingerprint you leave
behind is NaCl. This comes from the small amount of sweat present on your fingers. As the
sodium chloride in the fingerprint dissolves in the solution, the silver nitrate reacts with it to
give a precipitate of silver chloride, AgCl.
Silver chloride is white, but when exposed to sunlight it decomposes to give a black deposit
of metallic silver, which reveals the lines and swirls of the fingerprint.
2. Partial Equation Method
The balancing of equations by this method involve the following steps:
(i) It is assumed that the reaction has taken place in two or more stages. A separate balanced equation
is written for each stage.
(ii) If necessary, these partial equations are multiplied by a suitable integer. This will cancel out the
products which do not appear in the final equation.
Example
When chlorine reacts with hot solution of sodium hydroxide sodium chloride, sodium chlorate
and water are formed.
(i) The skeleton equation of the above reaction is
Cl2 + NaOH → NaCl + NaClO3 + H2O
www.plancess.com
Foundation for Chemistry
3.13
(ii) The reaction may be assumed to take place in the following stages
………….(1)
Cl2 + H2O → HCl + HClO
Sodium hydroxide reacts with the acids formed in stage (1)
NaOH + HCl → NaCl + H2O
………… (2)
NaOH + HClO → NaClO + H2O
………… (3)
Sodium hypochlorite is converted into sodium chlorate at higher temperature
3NaClO + HClO → 2NaCl + NaClO3 …………. (4)
(iii) In get order to cancel the intermediate products which do not appear in final equation multiply
equation (1), (2) and (3) by three.
3Cl2 + 3H2O → 3HCl + 3HClO
………….(5)
3NaOH + 3HCl → 3NaCl + 3H2O
………… (6)
3NaOH + 3HClO → 3NaClO + 3H2O
………… (3)
(iv) Add equation (4), (5), (6) and (7)
3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2 O
This is the desired balanced chemical equation.
Illustration 1: Write balanced chemical equations from the following word equations. Include the
physical state of each element or compound.
(a) Sodium metal plus water yields hydrogen gas and an aqueous sodium hydroxide solution.
Sol.
2Na(s) + 2H2O(l)
H2(g)
+
2NaOH(aq)
(Sodium) (Water)
(Hydrogen Gas) (Sodium Hydroxide)
(b) Potassium chlorate when heated yields potassium chloride plus oxygen gas. (Ionic compounds
are solids.)
Sol.
2KClO3(s)
2KCl(s)
+
3O2(g)
(Potassium Chlorate) (Potassium Chloride)
www.plancess.com
Language of Chemistry
3.14
(c) An aqueous sodium chloride solution plus an aqueous silver nitrate solution yields a silver
chloride precipitate (solid) and a sodium nitrate solution.
Sol.
Cu(s) + 2H2SO 4 → CuSO 4 (s) + SO2 (g) ↑ + 2H2O(l)
copper
(Conc)
Copper sulphate
Sulphur dioxide
water
(d) An aqueous phosphoric acid solution and an aqueous calcium hydroxide solution yields water
and solid calcium phosphate.
Sol.
NaCl(aq)+ AgNO3 (aq) → AgCl(s)+ NaNO3 (aq)
Sodium
Chloride
Silver
Nitrate
Silver
Chloride
Sodium
Nitrate
Illustration 2: Either the following chemical equations presents correct balanced chemical equation.
(1) H3PO4 + Ca(OH)2
CaCO3(PO4)2 + 6H2O
(2) Cu(NO3 )2 → 2CuO + 4NO2 + O2
Copper nitrate
Sol.
(1) No. First reaction is not correct. Here it is not specified that the acid used is dilute or
concentrated and moreover physical state of reactant is also not mentioned.
Thus, correct balanced equation will be :
3H3PO 4 (aq)+ 3Ca(OH)2 (aq) → CaCO3 (PO 4 )2 (s)+ 6H2 O(l)
Phosphoric
Acid
Calcium
Hydroxide
Calcium
Phosphate
Water
(2) No. Second equation is also not correctly balanced as atoms of different elements are not same
in reactants and products. Moreover Physical state of reactants and products are not specified.
Thus, correct balanced equation will be
Cu(NO3 )2 →
2CuO + 4NO2 (g) ↑ + O2 (g) ↑
Copper (II) oxide Nitrogen dioxide
oxygen
Copper nitrate
www.plancess.com
Foundation for Chemistry
3.15
SUMMARY
Terminologies
Basic concepts
Symbol
Symbol is the short that stands for the atom of a specific element.
Valency
Valency is the combining capacity of an atom or a radical. It is equal
to the number of electron(s) lost/gained or shared combining with
another atom or radical. Some elements, like iron, mercury, lead,
show variable valencies.
Radical
Radical is an atom or a group of atoms of the same or different
elements that behave as single unit and has positive or negative charge.
A radical with positive charge is a cation, e.g. NH+4 (ammonium ion),
Na+ (sodium ion) and a radical with negative charge is an anion, e.g.
Cl− (chloride), CO32 (carbonate).
Molecular formula
Molecular formula is a shorthand notation for the molecule of
a substance in terms of symbols and numbers of atoms of each
element present in it.
Atomic mass unit (amu)
Atomic mass unit (amu) is equal to one twelfth the mass of an
atom of an atom of carbon-12 (atomic mass of carbon taken as 12).
Molecular mass or relative
molecular mass
Molecular mass or relative molecular mass of a substance is the
relative mass of its molecules as compared with the mass of a carbon
– 12 atom, taken as 12 units.
www.plancess.com
3.16
Language of Chemistry
Chemical equation
Chemical equation is the symbolic representation of a chemical
reaction using symbols and formulae of the substance involved in
the reaction. Since matter is neither created nor destroyed in the
course of a chemical reaction so every equation needs to be balanced.
Balanced chemical equation
Balanced chemical equation tell us (i) which substance take part
is a chemical reaction (reactants) and which substances are formed
(products) and (ii) the number of molecules of each substance
involved.
www.plancess.com
Foundation for Chemistry
3.17
SOLVED EXAMPLES
Example.1. What is a symbol? What information does it convey?
Sol.
Symbol: A symbol is an abbreviation or short representation of an element.
It gives following information about the element:
(i) It represents specific element.
(ii) It represents one atom of an element.
(iii) A symbol represents how many atoms are present in its one gram (gm) atom.
(iv) It represents no. of times an atom is heavier than one atomic mass unit (a.m.u.) taken as a
standard.
Example.2. Why is the symbol S for sulphur, but Na for sodium and Si for silicon?
Sol.
(i) Generally the first letter of the name of element is taken as the symbol for that element and
written in capitals.
(ii) As in case of sulphur we use symbol S. But in cases where the first letter has already adopted
there we use symbol derived from Latin names as in case of sodium we use symbol Na
(Natrium).
(iii) In some cases we take initial letter in capital together with small letter from its name as in
case of silicon we use symbol Si.
Example.3. (a) Explain the term ‘valency’ and ‘variable valency’.
(b) How are the elements with variable valency named?
Sol. (a) Valency:
(i) The valency of an element is a measure of its combining power with other atoms when it
forms chemical compounds or molecules.
(ii) It is the no. of electrons that are shared, lost or gained during the course of a chemical
reaction.
(iii) The minimum no. of electrons it should lose or gain to attain stability or cobine with other
elements .
e.g. (a) Valency of K is +1 because it can lose one electron.
(b) Valency of Cl is -1 because it can accept one electron.
www.plancess.com
Language of Chemistry
3.18
Variable valency:
Certain metals exhibit more than one valency this is called as variable valency.
It is the combining capacity of an element in which the metal loses more electrons from a shell
next to valence shell in addition to electrons of valence shell.
E.g. (i) Iron when added to hydrochloric acid form iron (II) chloride Fe (II) Cl2. (FeCl2 )
(ii) Iron when heated with chloride forms iron (III) chloride (FeCl3)
If an element exhibits two different positive valencies then
(i) For the lower valency, use the suffix – OUS at the end of the name of the metal.
(ii) For the higher valency use a suffix – IC at the end of the name of the metal.
e.g.
Element
Lower valency
Higher valency
Ferrum (Iron)
Ferrous (Fe2+)
Ferric (Fe3+)
Example.4. Give the formula and valency of:
(a) Aluminate
(b) Chromate
(c) Aluminium
(d) Cupric
Sol.
Name
Formula
Valency
Aluminate
AlO2
-2
Chromate
CrO4
-2
Aluminium
Al
+3
Cupric
Cu
+2
Example.5. What is a chemical formula? What is the rule for writing a formula correctly?
Sol. Chemical formula;
1. It is the symbolic representation of actual number of atoms present in one molecule of that
substance.
2. It also indicates the fixed proportion by weight in which atoms combine.
Rules:
(i) The positive and negative radicals are written side by side (+ve first) with their charge as super
script on the right side.
www.plancess.com
Foundation for Chemistry
3.19
(ii) The charges are then interchange and written as subscript.
(iii) The final formula is written without the sign of charge e.g. Hg2O
(a) Hg+1O2-
(b) Hg2O
Example.6. What do you understand from the following?
(a) Acid radical
(b) Basic radical
Sol.
(a) Acid radical: Negatively charged radical is called acid radical. e.g. Cl–, O2–
(b) Basic Radical: positively charged radical is called basic radical. e.g. K+, Na+
Example.7. Write the chemical names of the following compounds:
(a) Ca3 (PO 4 )2
(b) K 2 CO3
(c) K 2 MnO 4
(d) Mn3 (BO3 )2
(e) Mg(HCO3 )2
(f ) Na4 Fe(CN)6
(g) Ba(ClO3 )2
(h) Ag2S O3
(i) (CH3COO)2 Pb
(j) Na2SiO3
Sol.
(a)
Ca3 (PO 4 )2
Calcium phosphate
(b)
K 2 CO3
Potassium carbonate
(c)
K 2 MnO 4
Potassium manganate
(d)
Mn3 (BO3 )2
Maganese borate
(e)
Mg(HCO3 )2
Magnesium bicarbonate
(f )
Na4 Fe(CN)6
Sodium ferrocyanide
www.plancess.com
Language of Chemistry
3.20
(g)
Ba(ClO3 )2
Barium chlorate
(h)
Ag2S O3
Silver sulphite
(i)
(CH3COO)2 Pb
Lead acetate
(j)
Na2SiO3
Sodium silicatate
Example.8. (a) What is polyatomic ions? Give two examples.
(b) Name the fundamental law which is involved in every equation.
Sol.
(a) A charged ion composed of two or more atoms covalently bounded that is known as Poly
atomic ion. That can be carbonate ( CO32− ) and sulphate ( SO24− )
(b) The fundamental laws which are involved in every equation are:
(i) A chemical equation consists of formulae of reactants connected by plus sign (+) and
arrow ( → ) followed by the formulae of products connected by plus sign (+).
(ii) The sign of an arrow ( → ) is to read ‘to form’. It also shows the direction in which the
reaction is predominant.
Example.9. Why an equation should be balanced? Explain your answer by giving a simple equation?
Sol.
This can be explained by considering law of mass conservation, “matter can neither be created
nor can it be destroyed.”
This law will be applicable only if total no. of atoms on the reactants side is equal to total no. of
atoms on products side or the mass on the right side is equal to the mass on left side.
Thus, a chemical reaction should be always balanced.
e.g. KNO3 → KNO2 + O2
The balanced form of this equation will be
2KNO3 → 2KNO2 + O2
www.plancess.com
Foundation for Chemistry
3.21
Example.10. (a) Define the atomic mass unit.
(b) Calculate the molecular mass of the following:
(i) CuSO4. 5H2O
(ii) (NH4)2CO3
(iii) (NH2)2CO
(iv) Mg3N2
(Given atomic mass of Cu = 63.5,
H = 1, O = 16, C = 12, N = 14, Mg = 24, S = 32.)
Sol.
(a) The atomic mass unit (amu) is defined as 1 th of the mass of an atom of carbon.
12
1 a.m.u. = 1.67 × 10−24 gm
= 1.67 × 10−27 kg
1 gm mass = 6.02 × 1023 amu. And mass 1 kg mass = 6.02 × 1026 amu.
(b) (i) The molecular mass of CuSO4. 5H2O
= 63.5+32+ (16 × 4) + 5(2+16)
= 159.5+90
=249.5 gm
(ii) The molecular mass of (NH4)2CO3
=N2H8CO3
= 14 × 2 + 1 × 8 + 12 + 3 × 16 = 28+8+12+48
=96 gm
(iii) The molecular mass of (NH2)2CO
= N2H4CO
= 2 × 14 + 1 × 4 + 12 + 16 = 28+4+12+16
=60 gm
(iv) The molecular mass of Mg3N2
= 3 × 24 + 2 × 14 = 28+72=100 gm
www.plancess.com
3.22
Language of Chemistry
Example.11. What are the limitations of a chemical equation?
Sol.
Limitations of chemical equation
(i) It does not tell about physical state of reactants and products.
(ii) It does not tell about rate of reaction.
(iii) It does not tell about the progress of the reaction i.e. it does not tell whether reaction will
complete or not.
(iv) It does not tell about conditions necessary for reaction.
(v) It does not tell whether energy is evolved or absorbed.
(vi) It does not tell about the changes in colour, precipitation etc.
www.plancess.com
Foundation for Chemistry
3.23
EXERCISE 1 – For School Examinations
Q.1.
Write the basic and acid radicals of these compounds and then write the chemical formula.
(a) Barium sulphate
(b) Bismuth nitrate
(c) Calcium bromide
(d) Ferrous sulphide
(e) Chromium sulphate
(f ) Calcium silicate
(g) Potassium ferrocyanide
(h) stannic oxide
(i) Calcium silicate
(j) Sodium zincate
(k) Magnesium phosphate
(l) Stannic phosphate
(m) Sodium thiosulphate
(n) Potassium manganate
(o) Nickel bisulphate
Q.2.
Complete the following statements by selecting the correct option:
(a) The formula of a compound represents
(i) an atom,
(ii) a particle
(iii) a molecule
(iv) a combination
(b) The correct formula of aluminium oxide is
(i) AlO3
(ii) AlO2
(iii) Al2O3
(c) The valency of nitrogen in nitrogen dioxide NO2 is
Q.3.
(i) one
(ii) two
(iii) three
(iv) four
Fill in the blanks
(a) Dalton used symbol
(b) Symbol represents
for oxygen
for hydrogen.
atoms of an element.
(c) Symbolic expression for molecule is called
(d) Sodium chloride has two radicals. Sodium is
radical.
(e) Valency of carbon in CH4 is
and in C2H2 is
(f ) Valency of iron in FeCl2 is
radical while chloride is
in C2H6
in C2H4
and FeCl3 it is
(g) Formula for iron (III) carbonate is
www.plancess.com
3.24
Q.4.
Language of Chemistry
Complete the following table.
Acid Radicals
→
Chloride
Nitrate
Sulphate
Carbonate
Hydroxide
Phosphate
MgCl2
Mg(NO3)2
MgSO4
MgCO3
Mg(OH)2
Mg3(PO4)2
Basic Radicals ↓
Magnesium
Sodium
Zinc
Silver
Ammonium
Calcium
Iron II
Potassium
Q.5.
Sodium chloride reacts with silver nitrate to produce silver chloride and sodium nitrate.
(a) Write the equation.
(b) Check whether it is balanced, if not balance it.
(c) Find the weight of reactants and products.
(d) State the law which this equation satisfies.
Q.6.
What information does the following chemical equation convey?
(a) Zn + H2SO 4 → ZnSO 4 + H2
Q.7.
(b) Mg + 2HCl → MgCl2 + H2
Balance the following equations.
(a) Fe + H2O → Fe3O 4 + H2
(b) Ca + N2 → Ca2N2
(c) Zn + KOH → K 2 ZnO2 + H2
(d) Fe2O3 + CO → Fe + CO2
(e) PbO + NH3 → Pb + H2O + N2
(f ) Pb3O 4 → PbO + O2
(g) PbS + O2 → PbO + SO2
(h) S + H2SO 4 → SO2 + H2O
(i) S + HNO3 → H2SO 4 + NO2 + H2O
(j) MnO2 + HCl → MnCl2 + H2O + Cl2
(k) C + H2SO 4 → CO2 + H2O + SO2
(l) KOH + Cl2 → KCl + KClO + H2O
www.plancess.com
Foundation for Chemistry
Q.8.
3.25
(m) NO2 + H2O → HNO2 + HNO3
(n) Pb3O 4 + HCl → PbCl2 + H2O + Cl2
(o) H2O + Cl2 → HCl + O2
(p) NaHCO3 → Na2CO3 + H2O + CO2
(q) HNO3 + H2S → NO2 + H2O + S
(r) P + HNO3 → NO2 + H2O + H3PO 4
Write the balanced chemical equations of the following reactions.
(a) Sodium hydroxide + sulphuric acid → sodium sulphate + water
(b) Potassium bicarbonate + sulphuric acid → potassium sulphate + carbon dioxide + water
(c) Iron + sulphuric acid → ferrous sulphate + hydrogen
(d) Chlorine + sulphur dioxide + water → sulphuric acid + hydrogen chloride
(e) Silver nitrate → silver + nitrogen dioxide + oxygen
(f ) Copper + nitric acid → copper nitrate + nitric oxide + water
(g) Ammonia + oxygen → nitric oxide + water
(h) Barium chloride + sulphuric acid → barium sulphate + hydrochloric acid
(i) zinc sulphide + oxygen → zinc oxide + sulphur dioxide
(j) Aluminium carbide + water → aluminium hydroxide + methane
(k) Iron pyrites + oxygen → ferric oxide + sulphur dioxide
(l) potassium permanganate + hydrochloric acid → potassium chloride + manganese chloride
+ chlorine + water
(m) Aluminium sulphate + sodium hydroxide → sodium sulphate + sodium metaaluminate
+ water
(n) Aluminium + sodium hydroxide + water → sodium meta aluminate + hydrogen
(o) Potassium dichromate + sulphuric acid → Potassium sulphate + chromium sulphate +
water + oxygen
(p) Potassium dichromate + hydrochloric acid → potassium chloride + chromium chloride
+ water + chlorine
(q) Sulphur + nitric acid → sulphuric acid + nitrogen dioxide + water
(r) sodium chloride + manganese dioxide + sulphuric acid → Sodium hydrogen sulphate +
manganese sulphate + water + chlorine
www.plancess.com
3.26
Q.9.
Language of Chemistry
Choose the correct answer from the options given below.
(a) Modern atomic symbols are based on the method proposed by
(i) Bohr
(ii) Dalton
(iii) Berzelius
(iv) Alchemist
(b) The number of carbon atoms in hydrogen carbonate radical is
(i) One
(ii) Two
(iii) Three
(iv) Four
(c) The formula of iron (III) sulphate is
(i) Fe3SO4
(ii) Fe(SO4)3
(iii) Fe2(SO4)3
(iv) FeSO4
(d) In water, the hydrogen – to – oxygen mass ratio is
(i) 1:8
(ii) 1:16
(iii) 1:32
(iv) 1:64
(e) Sodium carbonate is Na2CO3, that of calcium hydrogencarbonate will be
(i) CaHCO3
(ii) Ca(HCO3)2
(iii) Ca2HCO3
(iv) Ca(HCO3)3
Q.10. Correct the following statements
(a) A molecular formula represents an element.
(b) The molecular formula of water (H2O) represents 9 parts by mass of water.
(c) A balanced equation obeys the law of conservation of mass and so does an unbalanced
equation.
(d) A molecule of an element is always monoatomic.
(e) CO and Co both represents cobalt.
Q.11. Balance the following equations:
(i) ZnCO3 (s) + HCl(aq) → ZnCl2 (aq) + H2O(l) + CO2 (g)
(ii) NH4 Cl(s) + Ca(OH)2 (s) → CaCl2 (s) + NH3 (g) + H2 O(l)
(iii) Cu + H2 SO 4 (s) → CuSO 4 (aq) + SO2 (aq) + H2 O(l)
www.plancess.com
Foundation for Chemistry
3.27
Q.12. Balance the following equations.
(i) Na(s) + H2O(l) → NaOH(aq) + H2 (g)
(ii) Na2O2 (s) + H2O(l) → NaOH(aq) + O2 (g)
∆
(iii) NaNO3 (s) 
→ NaNO2 (aq) + O2 (g)
∆
(iv) Cu(NO3 )2 (s) 
→ CuO(s) + NO2 (g) + O2 (g)
∆
(v) Hg(NO3 )2 (s) 
→ Hg(l) + NO2 (g) + O2 (g)
Q.13. Write the formulae and balance following equations:
(i) Zinc + dil. Sulphuric acid → Zinc sulphate + Hydrogen
(ii) Ammonium sulphate + Calcium hydroxide → Calcium sulphate + Ammonia + water
(iii) Lead dioxide + Hydrochloric acid → Lead chloride + Water + Chlorine
(iv) Aluminium oxide + Sulphuric acid → Aluminium sulphate + Water
(v) Iron + conc. Sulphuric acid → Iron (II) sulphate + Sulphur dioxide gas + Water.
Q.14. Balance the following equations.
(a) CaCO3 (s) + HCl(aq) → CaCl2 (aq) + H2O(l) + CO2 (g)
(b) Zn(s) + HCl(aq) → ZnCl2 (aq) + H2 (g)
(c) MnO2 (s) + HCl(aq) → MnCl2 (aq) + H2O(l) + Cl2 (g)
(d) Na2CO3 (s) + H2SO 4 (aq) → Na2SO 4 (aq) + H2O(l) + CO2 (g)
(e) Mg3N2 (s) + H2O(l) → Mg(OH)2 (s) + NH3 (g)
(f ) NH3 (g) + O2 (g) → N2 (g) + H2O(l)
(g) MgCl2 (aq) + Na2CO3 (aq) → MgCO3 (s) + NaCl(aq)
www.plancess.com
3.28
Language of Chemistry
SOLUTIONS
Exercise 1 – For School Examinations
1.
Compounds
Chemical formula
Basic Radicals
Acidic Radicals
(a) Barium sulphate
Ba(SO 4 )
Ba2+
SO24−
(b) Bismuth nitrate
Bi(NO3 )3
Bi3+
NO3−
(c) Calcium bromide
CaBr2
Ca2+
Br −
(d) Ferrous sulphide
FeS
Fe2+
S2 −
(e) Chromium sulphate
Cr2 (SO 4 )3
Cr3+
SO24−
(f ) Calcium silicate
CaSiO3
Ca2+
SiO32−
(g) Potassium ferrocyanide
K 4 Fe(CN)6
K+
Fe(CN)64 −
(h) Stannic oxide
SnO2
Sn4 +
O2 −
(i) Calcium silicate
CaSiO3
Ca2+
SiO32−
(j) Sodium zincate
Na2 ZnO2
Na+
ZnO22−
(k) Magnesium phosphate
Mg3 (PO 4 )2
Mg2+
PO34−
(l) Stannic phosphate
Sn3 (PO 4 )4
Sn4 +
PO34−
(m) Sodium thiosulphate
Na2S2O3
Na+
S2O32−
(n) Potassium manganite
K 2MnO 4
K+
MnO24−
(o) Nickel bisulphate
Ni(HS O 4 )2
Ni2+
HSO −4
www.plancess.com
3.29
Foundation for Chemistry
2.
(a) a molecule
(b) Al2O3
(c) Four
3.
(a) [O],[H]
(b) Gram
(c) Molecular formula
(e) 4, 3, 2, 1
(f ) 2,3
(g) Fe2[CO3]3
(d) Basic, Acid
4.
Acid
Radicals →
Chloride
Nitrate
Sulphate
Carbonate
Hydroxide
Phosphate
Magnesium
MgCl2
Mg(NO3)2
MgSO4
MgCO3
Mg(OH)2
Mg3(PO4)2
Sodium
NaCl
NaNO3
Na2SO4
Na2CO3
NaOH
Na3PO4
Zinc
ZnCl2
Zn(NO3)2
ZnSO4
ZnCO3
Zn[OH]2
Zn3(PO4)2
Silver
AgCl
AgNO3
Ag2SO4
Ag2CO3
AgOH
Ag3PO4
Ammonium
NH4Cl
NH4NO3
(NH4)2SO4
(NH4)2CO3
NH4OH
(NH4)3PO4
Calcium
CaCl2
Ca(NO3)2
CaSO4
CaCO3
Ca(OH)2
Ca3(PO4)2
Iron II
FeCl2
Fe(NO3)2
FeSO4
FeCO3
Fe(OH)2
Fe3(PO4)2
Potassium
KCl
KNO3
K2SO4
K2CO3
KOH
K2PO4
Basic
Radicals ↓
5.
Sodium chloride + Silver nitrate → Silver chloride + Sodium nitrate
(a) Chemical Equation:
NaCl + AgNO3 → AgCl + NaNO3
this is a displacement reaction in which anion of first compound replaces the cation of second
element.
(b) Yes, it is balanced
(c) Molecular mass of NaCl = (23 + 35.5)
= 58.5
Molecular mass of AgNO3 = (108+14+48)
= 170
Molecular Mass of AgCl= (108+35.5)
143.5
Molecular Mass of NaNO3 = (23+14+48) = 85
www.plancess.com
3.30
Language of Chemistry
NaCl +
AgNO3
58.5 +
170
→
AgCl+
NanO3
143.5 +
85
228.5 g
Wt. of reactants=228.5 g
228.5 g
Wt. of products 228.5 g
(d) This equation satisfies the “Law of Conservation of Matter.”
Law of Conservation of Matter: “Matter is neither created nor destroyed in course of chemical
reaction.”
6.
(a) Zn + H2SO 4 → ZnSO 4 + H2
This equation conveys following information
(i) The actual result of chemical change.
(ii) All reactant take part in reaction and product formed as a result of reaction.
(iii) Here one molecule of zinc, one molecule of sulphuric acid react to give one molecule of Zinc
sulphate and one molecule of Hydrogen.
(iv) Composition of respective molecules i.e. one molecule of sulphuric acid contains two atoms
of hydrogen, one atom of sulphur and four atoms of oxygen.
(v) Relative molecular masses of different substances
i.e. molecular mass of
Zn = 65,
H2SO4 = (2+32+64) = 98,
ZnSO4 (65+32+64) =161
and H2 = 2
(vi) 22.4 litres of hydrogen are formed at S.T.P.
(b) Mg + 2HCl → MgCl2 + H2
(i) That magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen
gas.
(ii) 24g of magnesium react with 2(1+35.5) =73g of Hydrochloric acid to produce (24 + 71)
i.e. 95g of magnesium chloride.
(iii) That hydrogen produced at STP is 22.4 litres.
www.plancess.com
Foundation for Chemistry
3.31
7.
(a) 3Fe + 4H2O → Fe3O 4 + 4H2 ↑
(b) 3Ca + N2 → Ca3N2
(c) Zn + 2KOH → K 2 ZnO2 + H2 ↑
(d) Fe2O3 + 3CO → 2Fe + 3CO2
(e) 3PbO + 2NH3 → 3Pb + 3H2O + N2 ↑
(f ) 2Pb3O 4 → 6PbO + O2 ↑
(g) PbS + 3O2 → 2PbO + 2SO2 ↑
(h) S + 2H2SO 4 → 3SO2 + 2H2O
(i) S + 6HNO3 → H2SO 4 + 6NO2 + 2H2O
(j) MnO2 + 4HCl → MnCl2 + 2H2O + Cl2 ↑
(k) C + 2H2SO 4 → CO2 + 2H2O + 2SO2 ↑
(l) 6KOH + 3Cl2 → 5KCl + KClO + 3H2O
(m) 2NO2 + H2O → HNO2 + HNO3
(n) Pb3O 4 + 8HCl → 3PbCl2 + 4H2O + Cl2 ↑
sunlight
(o) H2O + 2Cl2 
→ 4HCl + O2 ↑
(p) 2NaHCO3 → Na2CO3 + H2O + CO2 ↑
(q) 2HNO3 + H2S → 2NO2 + 2H2O + S
(r) P + 5HNO3 → 5NO2 + H2O + H3PO 4
8.
(a) 2NaOH + H2SO 4 → Na2SO 4 + 2H2O
(b) 2KHCO3 + H2SO 4 → K 2SO 4 + 2CO2 + 2H2O
www.plancess.com
3.32
Language of Chemistry
(c) Fe + H2SO 4 → FeSO 4 + H2
(d) Cl2 + SO2 + 2H2O → H2SO 4 + 2HCl
(e) 2AgNO3 → 2Ag + 2NO2 + O2
(f ) 3Cu + 8HNO3 → 3Cu(NO3 )2 + 2NO + 4H2O
pt.800°C
(g) 4NH3 + 5O2 →
6H2O + 4NO ↑ +Heat
(h) BaCl2 + H2SO 4 → BaSO 4 + 2HCl
(i) 2ZnS + 3O2 → 2ZnO + 2SO2
(j) Al4 C3 + 12H2O → 4 Al(OH)3 + 3CH4
(k) 4FeS2 + 11O2 → 4Fe2O3 + 8SO2
(l) 2KMnO 4 + HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O
(m) Al2 (SO 4 )3 + 8NaOH → 3Na2SO 4 + 2NaAlO2 + 4H2O
(n) 2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2
(o) 2K 2 Cr2 O7 + 8H2SO 4 → 2K 2SO 4 + 2Cr2 (SO 4 )3 + 8H2O + 3O2
(p) K 2 Cr2 O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2
(q) S + 6HNO3 → H2SO 4 + 6NO2 + 2H2O
(r) 2NaCl + MnO2 + 3H2SO 4 → 2NaHSO 4 + MnSO 4 + 2H2O + Cl2
9.
a. (iii) Berzelius
b. (i) one
d. (i) 1:8
e. (ii) Ca(HCO3)2
c. (iii) Fe2 (SO4)3
10.
(a) A molecular formula represent the molecule of an element or of a compound.
(b) The molecular formula of water (H2O) represents 18 parts by mass of water.
(c) A balanced equation obeys the law of conservation of mass but unbalanced equation does not
obey this law.
(d) A molecule of a noble element is always monoatomic.
(e) CO represents carbon monoxide and Co represents Cobalt.
www.plancess.com
Foundation for Chemistry
11.
3.33
(i) ZnCO3 + 2HCl → ZnCl2 + H2O+ CO2 ↑
(ii) 2NH4 Cl+ Ca(OH)2 → CaCl2 + 2NH3 + 2H2 O
(iii) Cu + H2 SO 4 → CuSO 4 + H2 O+ SO2 ↑
12.
(i) 2Na + 2H2O → 2NaOH + H2 ↑
(ii) 2Na2O2 + 2H2O → 4NaOH + O2 ↑
∆
(iii) 2NaNO3 
→ 2NaNO2 + O2 ↑
∆
(iv) 2Cu(NO3 )2 
→ 2CuO + 2NO2 + 3O2 ↑
∆
(v) Hg(NO3 )2 
→ Hg + 2NO2 + O2 ↑
13.
(i) Zn + dil.H2SO 4 → ZnSO 4 + H2
(ii) (NH4 )2 SO 4 + Ca(OH)2 → CaSO 4 + 2NH3 + 2H2O
(iii) PbO2 + 4HCl → PbCl2 + 2H2O + Cl2 ↑
(iv) Al2O3 + 3H2SO 4 → Al2 (SO 4 )3 + 3H2O
(v) Fe + 2H2SO 4 (conc.) → FeSO 4 + 2H2O + SO2 ↑
14.
(a) CaCO3 + 2HCl → CaCl2 + H2O+ CO2 ↑
(b) Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2 (g) ↑
(c) MnO2 + 4HCl → MnCl2 + 2H2O + Cl2 ↑
(d) Na2CO3 + H2SO 4 → Na2SO 4 + H2O+ CO2 ↑
(e) Mg3N2 + 6H2O → Mg(OH)2 + 2NH3 ↑
(f ) 4NH3 + 3O2 → 2N2 + 6H2O
(g) MgCl2 + Na2CO3 → MgCO3 + 2NaCl
www.plancess.com