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Showing posts with label 9th Atoms and Molecules. Show all posts
Showing posts with label 9th Atoms and Molecules. Show all posts

Thursday, December 5, 2019

Atoms and molecules class 9 solved question

QUESTIONS FROM Exam PAPERS

Questions: 5 g of calcium combine with 2 g of oxygen to form a compound. Find the molecular formula of the compound. (Atomic mass of Ca = 40 u ; O =16 u)

Solution: Number of moles in 5g of calcium = mass / molar mass = (5 / 40) = 0.125

Number of moles in 2g of oxygen = mass / molar mass = (2 / 16) = 0.125

Now we will calculate the simplest ratio of the element by dividing the number of moles of each element by the smallest value.

Since number of moles of each element is 0.125, therefore calcium and oxygen are present in a ratio of 1 : 1. Thus the empirical formula of the compound is CaO.

For calculating the molecular formula, we need the molecular mass of the compound. However, because a compound with the formula CaO is known,

Therefore the molecular formula of the compound is CaO.

Questions: When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is produced.
What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.00 g of oxygen ?
Which law of chemical combination will govern your answer ? State the law.

Solution: The mass of the carbon dioxide formed will be only 11 grams. The remaining oxygen is not used up.

This indicates the law of definite proportions which says that in compounds, the combining elements are present in definite proportions by mass.

Class 9 Atoms and molecules solved CBSE Test Papers New addition

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Saturday, October 19, 2013

Dalton’s atomic theory,Atomic mass and Relative Atomic mass (9th CBSE Ch.Atoms and Molecules)

In 1803, A British school teacher John Dalton provided the basic theory about the nature of matter which provides explanation for the law of conservation of mass and the law of definite proportions.
According to Dalton’s atomic theory, all matter, whether an element, a compound or a mixture is composed of small particles called atoms

The postulates of this theory may be stated as follows:
 (i) All matter is made of very tiny particles called atoms.
(ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
(iii) Atoms of a given element are identical in mass and chemical properties.
(iv)  Atoms of different elements have different masses and chemical properties.
(v)  Atoms combine in the ratio of small whole numbers to form compounds.
(vi) The relative number and kinds of atoms are constant in a given compound.

Drawbacks of Dalton’s Atomic Theory:
ð  The atom is further subdivided into protons, neutrons and electrons.
ð  The atoms of same elements are not similar in all respect. They may vary in mass and density. These are known as isotopes. For example: chlorine has two isotopes having mass numbers 35 a.m.u and 37 a.m.u.
ð  Atoms of different elements are not different in all respects. Atoms of different elements that have the same atomic mass are called isobar.
ð  According to Dalton atoms of different elements combine in simple whole number ratio to form compounds. This is not seen in complex organic compounds like sugar C12H22O11.
ð  The theory fails to explain the existence of allotropes like Diamond and Graphite which having different properties even these are made up of same kind ot atom namely Carbon.
The introduction of matter wave concept by de Broglie, the principle of uncertainty by Heisenberg etc., paved the way for modern atomic theory [Modification in Dalton’s atomic theory]

Modifications in Dalton’s atomic theory - Modern atomic theory are as follows.

ð  Atom is considered to be a divisible particle.
ð  Atoms of the same element may not be similar in all respects. eg: Isotopes (17Cl35,17Cl37 )
ð   Atoms of different elements may be similar in some respects eg. Isobars (18Ar 40 , 20Ca 40 )
ð   Atom is the smallest particle which takes part in chemical reactions.
ð  The ratio of atoms in a molecule may be fi xed and integral but may not be simple e.g., C12H22O11 is not a simple ratio (Sucrose)
ð  Atoms of one element can be changed into atoms of other element by transmutation.
ð  The mass of an atom can be converted into energy. This is in accordance with Einstein’s equation E = mc2

Atom: It is the smallest particle of an element which may or may not have independent existence. The atoms of certain elements such as hydrogen, oxygen, nitrogen, etc .do not have independent existence whereas atoms of helium, neon, argon, etc. do have independent existence. Thus we can say that all elements are composed of atoms.

Q. How do we know the presence of atoms if they do not exist independently for most of the elements?
Answer: Atom join in different way to form matter (neutral molecules or ion) that we are able to touch, feel and see.

How big are atoms?
Atoms are extremely small. They are so small, that you cannot see them with most microscopes. Now, Scan tunneling Microscope (STM) is the modern instrument that made it possible to take photograph of atom. The size of an isolated atom can 't be measured because we can 't determine the location of the electrons that surround the nucleus. We can estimate the size of an atom, however, by assuming that the radius of an atom is half the distance between adjacent atoms in a solid. This technique is best suited to elements that are metals, which form solids composed of extended planes of atoms of that element. The results of these measurements are therefore often known as metallic radii.

Q. What is the unit of measurement of atomic radius?
Ans: Picometers (pm) or Angstroms (Å)

Q.The size of sodium atom is bigger than that of hydrogen atom. Why?
Answer: Size of atom is the distance between the nucleus and outermost shell ( valence shell ) of an atom. The atomic number of sodium is greater than that of hydrogen. So, it needs more number of shells to fill electrons and hence will have more number of shells than hydrogen. Hence, atomic size of sodium is bigger than that of sodium.

Naming of an element
Dalton was the first scientist to use the symbols for elements in a very specific sense.

Q. Why are Dalton’s symbols not used in chemistry?
Answer: Dalton was the first scientist to use the symbol for the name of the elements a specific sense but it was difficult to memorize and in uses so Dalton's symbol are not used in chemistry
Berzelius suggested that the symbols of elements be made from one or two letters of the name of the element.

IUPAC (International Union of Pure and Applied Chemistry) approves names of elements. Many of the symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (uppercase) and the second letter as a small letter (lowercase)
For example:  (i) hydrogen, H (ii) aluminum, Al and not AL (iii) cobalt, Co and not CO.
Symbols of some elements are formed from the first letter of the name and a letter, appearing later in the name. Examples are: (i) chlorine, Cl, (ii) zinc, Zn etc.
Other symbols have been taken from the names of elements in Latin, German or Greek. For example, the symbol of iron is Fe from its Latin name ferrum, sodium is Na from natrium, and potassium is K from kalium. Therefore, each element has a name and a unique chemical symbol.

Molecule: A molecule is the smallest or the simplest structural unit of an element (or) a compound which contains one (or) more atoms. It retains the characteristics of an element. A molecule can exist freely and it is a combined form of bonded units whereas an atom is a singular smallest form of non bonded unit.
Molecules are of two types, namely homo atomic molecules and hetero atomic molecules.

Homo atomic molecules: These are the molecules which are made up of atoms of the same element. For example hydrogen gas consists of two atoms of hydrogen (H2).Similarly oxygen gas consists of two atoms of oxygen (O2).

HETERO ATOMIC MOLECULES : The hetero atomic molecules are made up of atoms of different elements. They are also classified as diatomic, triatomic, or polyatomic molecules depending upon the number of atoms present. H2O, NH3, CH4, etc., are the examples for hetero atomic molecules.

Atomicity:The number of atoms present in one molecule of an element is called the atomicity of an element. Depending upon the number of atoms in one molecule of an element, molecules are classified into monoatomic, diatomic, triatomic or poly atomic molecules containing one, two, three, or more than three atoms respectively.
Mono atomic molecules: Helium (He) Neon (Ne) Metals ;  Di atomic molecules:   Hydrogen H2 Chlorine Cl2
Tri atomic molecules: Ozone (O3) ; Poly atomic molecules: phosphorous P4 Sulphur S8

Atomicity = Molecular Mass/Atomic mass

MORE TO KNOW:
Isotopes ⇒ These are the atoms of same element with same atomic number (Z) but different mass number (A). Example (17Cl35,17Cl37 )
Isobars These are the Atoms of the different element with same mass number but different atomic number. Example (18Ar40, 20Ca 40 )
 Isotones ⇒ These are the atoms of different elements with same number of neutrons Example ( 6C13, 7N14 )

AVOGADRO’S HYPOTHESIS: Amedeo Avogadro put forward hypothesis and is based on the relation between number of molecules and volume of gases that is “volume of a gas at a given temperature and pressure is proportional to the number of particles”.

Avogadro’s Law: Equal volumes of all gases under the same conditions of temperature and pressure. contain the equal number of molecules.

TEST YOUR UNDERSTANDING SKILL
 (a) Find the atomicity of chlorine if its atomic mass is 35.5 and its molecular mass is 71 
(b) Find the atomicity of ozone if its atomic mass is 16 and its molecular mass is 48

WHAT IS AN ION? 

An ion is a charged particle and can be negatively or positively charged.
A negatively charged ion is called an ‘anion’ and the positively charged ion, a ‘cation’. For example, sodium chloride (NaCl). Its constituent particles are positively charged sodium ions (Na+) and negatively charged chloride ions (Cl–).
Ions may consist of a single charged atom or a group of atoms that have a net charge on them. A group of atoms carrying a charge is known as a polyatomic ion e.g. Calcium oxide (Ca+2 O-2)

Atomic mass and Relative Atomic mass (RAM):

Q. Each element had a characteristic atomic mass even then we are using Relative Atomic mass. Give reason?

Answer: Since determining the mass of an individual atom was a relatively difficult task due to extremely smaller size, relative atomic masses were determined using the laws of chemical combinations and the compounds formed.

Relative Atomic mass (RAM): In 1961 IUPAC selected an isotope of carbon (12C) as a slandered for comparing atomic and molecular mass of element and compound.

Relative atomic mass of an element is the ratio of mass of one atom of element to the 1/12th part of mass of one atom of carbon. Relative atomic mass is a pure ratio and has no unit. If the atomic mass of an element is expressed in grams, it is known as gram atomic mass. e.g. 

Gram atomic mass of hydrogen = 1g where as gram atomic mass of carbon = 12g

Atomic mass is expressed in atomic mass unit (amu). One atomic mass unit is defined as 1/12th part of the mass of one atom of carbon.

Q. The atomic mass of an element is in fraction .What does it mean?
Ans If the atomic mass of an element is in fraction, this mean that it exists in the form of isotopes. The atomic mass is the average atomic mass and is generally fractional.

Chemical Formulae: The chemical formula is a symbolic representation of a compound of its composition.
For writing Chemical Formulae the name or symbol of the metal is written first then non-metals with their valencies. Then we must crossover the valencies of the combining atoms. For example:
 (a) Formula for aluminium oxide: Al3+ O-2 Þ Al2O(b) calcium hydroxide :  Ca+2 OH-1 Þ Ca(OH)2

Valency: The combining power (or capacity) of an element is known as its valency. Valency can be used to find out how the atoms of an element will combine with the atom(s) of another element to form a chemical compound.

RELATIVE MOLECULAR MASS (RMM) : The relative molecular mass of an element or a compound is the ratio of mass of one molecule of the element or a compound to the mass of 1/12 th part of mass of one atom of carbon. Relative Molecular mass is a pure ratio and has no unit. If the molecular mass of a given substance is expressed in gram, it is known as gram molecular mass of that substance.
Molecular mass is the sum of the masses of all the atoms present in one molecule of the compound or an element.

Test your numerical skill:
Problem:    Find the gram molecular mass of water (H2O)
Solution:  Þ 2(H) = 2 x 1 = 2  and   1(O) = 1 x 16 = 16    ;  Gram molecular mass of H2O = 2 + 16 = 18g

Problem:  Find the gram molecular mass of carbon dioxide
Solution:  Þ  (CO2) 1(C) = 1 x 12 = 12 and  2(O) = 2 x 16 = 32;  Gram molecular mass of CO2 = 12 + 32 = 44 g

Calculate the percent by mass (weight) of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl)
Relative molecular mass of NaCl  = 22.99 + 35.45 = 58.44
The total mass of Na present:
1 Na is present in the formula, mass = 22.99
Tthe percent by mass (weight) of Na in NaCl: 
%Na = (22.99 ÷ 58.44) x 100 = 39.34%
The total mass of Cl present:
Cl is present in the formula, mass = 35.45
The percent by mass (weight) of Cl in NaCl:
%Cl = (35.45 ÷ 58.44) x 100 = 60.66%
 Here,   39.34 + 60.66 = 100.
The answers above are probably correct if %Na + %Cl = 100, that is, 

Empirical formula: The empirical formula is the simplest formula for a compound in which atoms of different elements are present in simple ratio. It shows the relative number of atoms of each element. For example CH2O is the empirical formula of Glucose C6H12O6

Molecular formula: It is the formula in which the actual number of atoms of  different element are present. For example, if the empirical formula of benzene is CH where as molecular formula is C6H6, etc.


9th CBSE Science Chap 3 Atoms and Molecules Self ...: 1. Fill in the blanks (a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called —...

Related post to IX(9th)  Atoms and Molecules

9th(CBSE) Ch. Atoms and Molecules - Mole Concept and Related Problems

MOLE CONCEPT [CBSE Class IX Chemistry]
While performing a reaction, to know the number. of atoms (or) molecules involved, the concept of mole was introduced. The quantity of a substance is expressed in terms of mole.
Definition of mole:  Mole is defined as the amount of substance that contains as many specified elementary particles as the number of atoms in 12g of carbon-12 isotope.
One mole is also defined as the amount of substance which contains Avogadro number (6.023 x 1023) of particles.
MORE TO KNOW
Avogadro number: Number of atoms or molecules or ions present in one mole of a substance is called Avogadro number. Its value is 6.023x1023.
Therefore, one mole of any substance = 6.023 x 1023 particles may be atoms, molecules, ions
For example: One mole of oxygen atoms represents 6.023 x 1023 atoms of oxygen and 5 moles of oxygen atoms contain 5 x 6.023x1023 atoms of oxygen.
Questions based on mole concept:
1. When the mass of the substance is given:  Use this formula: Number of moles = given mass/ atomic mass
(a). Calculate the number of moles in  (i) 81g of aluminium ii) 4.6g sodium  (iii) 5.1g of Ammonia  (iv) 90g of water  (v) 2g of NaOH 
Solution: (i) Number of moles of aluminum = given mass of aluminium / atomic mass of aluminium = 81/27 = 3 moles of aluminum [Rest Question do yourself]
(b) Calculate the mass of 0.5 mole of iron
Solution: mass = atomic mass x number of moles = 55.9 x 0.5 = 27.95 g
Do yourself:  Find the mass of 2.5 mole of oxygen atoms [ Mass = molecular mass x number of moles]
2. Calculation of number of particles when the mass of the substance is given:
Number of particles =( Avogadro number x given mass)/gram molecular mass
Problem: Calculate the number. of molecules in 11g of CO2
Solution: gram molecular mass of CO2 = 44g
Number of molecules = (6.023 x 1023 x 11) / 44 = 1.51 x 1023 molecules
Do yourself: Calculate the number of molecules in 360g of glucose
3. Calculation of mass when number of particles of a substance is given:
Mass of a substance = (gram molecular mass x number of particles)/6.023 x 1023
Problem: Calculate the mass of 18.069 x 1023 molecules of SO2
Solution: Gram molecular mass SO2 = 64gm
The mass of 18.069 x 1023 molecules of SO2 = (64x18.069 x 1023)/ (6.023 x 1023) = 192 g
Do yourself: (a) Calculate the mass of glucose in 2 x 1024 molecules (b) Calculate the mass of 12.046 x 1023 molecules in CaO
4. Calculation of number of moles when you are given number of molecules:
Problem:  Calculate the number moles for a substance containing 3.0115 x 1023 molecules in it.
Solution: Number of moles = [Number of molecules/(6.023 x 1023)]  = ( 3.0115 x 1023)/( 3.0115 x 1023) =0.5 moles
Do yourself: (a) Calculate number of moles in 12.046x 1022 atoms of copper (b) Calculate the number of moles in 24.092 x 1022 molecules of water.
Problem: Calculate the number of aluminum ions present in 0.051 g of aluminum oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al=27 u)
Solution: Mass of the 1 mole of Al2 O3  = 2x27 + 3x16 = 102gm
The number of ions present in 102 gm of aluminum oxide  = 6.023 x 1023 ion
The number of ions present in 0.051g of aluminum oxide= (6.023 x 1023 ion x 0.051g)/ 102 gm 
=  6.023 x 1023 ion x0.0005 = 3.0115 x 1020 ions
In Al2 O3, Aluminium and oxygen are in ratio 2:3
So, The number of aluminum ions present(Al3+) in 0.051g of aluminum oxide = 2 x 3.0115 x 1020 ions =6.023 x 1020 ion
MORE TO KNOW
Volume occupied by one mole of any gas at STP is called molar volume. Its value is 22.4 litres 22.4 litres of any gas contains 6.023 x 1023 molecules.
Problem: Calculate the volume occupied at STP by :- (a) 64 gram of oxygen gas (b) 6.02 x1022 molecules of CH4 (c) 5 moles of nitrogen gas
Solution: (a) One mole of a gas occupies 22.4 L volume at STP
Now, number of moles in 64 g oxygen gas = 64/32 = 2
Therefore, volume occupied by 2 moles(64 g) of oxygen gas = 2 x 22.4 L = 44.8 L
(b) 1 mole = 6.02 x 1023 molecules
Therefore, 1 mole (6.02 x 1023 molecules) of CH4 gas occupies 22.4 L volume at STP.
(c) One mole of a gas occupies 22.4 L volume at STP
Therefore, volume occupied by 5 moles of nitrogen gas = 5 x 22.4 L = 112 L
Problem: Calculate the volume at STP occupied by 1021 molecules of Oxygen?
Solution: The molar volume that is the volume occupied by one mole of gas is 22.4 L. We know there are 6.022 X 1023 particles of a substance in one mole of that substance. Thus
volume occupied by 6.022 X 1023 molecules of oxygen = 22.4 L
volume occupied by 1021 molecule of oxygen = [(22.4 X 1021) / (6.022 X 1023)] 
                                                                              = 3.72 X 10-2 L = 37.2 ml
Q. Calculate the mass and volume of oxygen at STP, which will be evolved on electrolysis of 1 mole of(18 g) of water.
Answer: Electrolysis of water will break it down in its component as Hydrogen and Oxygen
Balanced chemical reaction:
2H2O → 2H2 + O2
From above equation, 2 mole of water will evolve 1 mole of oxygen gas upon electrolysis. Therefore 1 mole of water will produce 1/2 mole of Oxygen.
Mass of Oxygen evolved = number of moles of Oxygen evolved × Molecular wt. of Oxygen =1/2 × 32 = 16 g
At STP 1 mole of any gas will occupy 22.4l volume.
Volume of Oxygen evolved = No. of moles × 22.4l = 1/2 × 22.4l = 11.2l.
Q. How many molecules are present in 1 ml of water?
Answer: we know that density of water is 1gm/ml.
Hence, 1 gm water will = 1 ml water.
Now, we have molecular mass of water H2O = 1x2 + 16 = 18 gm
18 gm of water contain 6.022 x 1023 molecules
1 gm of water will contain = (6.022 x 1023)/18 molecules = 0.33 x 1023 molecules
So, the no. of molecules of water in 1ml of water = 3.3 x 1022

Friday, October 18, 2013

Laws of chemical combination and Landolt Experiment ( IX Atoms and Molecules)

How and why elements combine and what happens when they combine. Antoine L. Lavoisier laid the foundation of chemical sciences by establishing two important laws of chemical combination.

(a) The Law of conservation of mass was stated by Antoine L. Lavoisier in 1785 as” Mass can neither be created nor destroyed in a chemical reaction” [The Law of conservation of mass is the 2nd  postulate of Dalton's atomic theory. It states that Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.]
Example:
ð  Water forms by the union of hydrogen and oxygen. If we weigh the reactants (hydrogen and oxygen) before the chemical reaction, we find the weight of the product (water) equal to the combined weight of reactants.
ð  The weight of iron increases on rusting. The increase in weight is equal to the weight of oxygen added to iron.
ð  Carbon combines with Sulphur to form Carbon disulphide. The mass of reactants i.e. carbon and sulphur is same mass of products (carbon disulphide).
 Carbon + Sulphur   -----   Carbon DiSulphide
 C          +   S            -----     C2S  
1g         +  5. 34g            =       6.34 g
LHS                                =     RHS

The verification of the Law of conservation of mass by Landolt Experiment

H. Landolt was German Chemist. He proved the law of conservation of mass by using an H-shaped glass tube. He filled silver nitrate in limb A and hydrochloric acid in limb B. The tube was sealed and weighed before the chemical reaction. The reactants were mixed by inverting and shaking the tube. A white precipitate of silver chloride was formed along with Sodium nitrate . The tube was weighed again. He found that there was no change in weight during the following chemical reaction. 
AgNO3 + NaCl      ® AgCl2 (white precipitate) + NaNO3

In all the chemical reactions, energy is evolved or absorbed which would be at the expense of change in mass. In ordinary chemical reactions, this change in mass is so small that it cannot be registered on the most sensitive balance. This suggests that some matter of the reaction mixture gets converted into energy such as light, heat etc. Thus mass and energy are interconvertible. The mass is converted to energy by Einstein’s relation E = mc2.

(b) The law of constant proportions which is also known as the law of definite proportions was stated by Proust in 1799 as “In a chemical substance the elements are always present in definite proportions by mass”.
[The Law of constant proportions is the 6th postulate of Dalton's atomic theory. The relative number and kinds of atoms are constant in a given compound.]
E.g. In a compound such as water, the ratio of the mass of hydrogen to the mass of oxygen is always 1:8, whatever the source of water. Thus, if 9 g of water is decomposed, 1 g of hydrogen and 8 g of oxygen are always obtained.
if the element ‘A’ and ‘B’ combine chemically to form the compound AB, then in whatever manner AB is formed, it is always composed of same two elements ‘A’ and ‘B’ combined together in the same fixed ratio or proportion by weight.
For example:  Sulphur dioxide can be obtained b following sources:
         (i).     Sulphur is burnt in air, 
                   S               +    O2                    ---------
®   SO2
         (ii).    Copper is heated with conc. sulphuric acid
                   Cu            +   2H2SO4       ---------
®  CuSO4   +   2H2O   +   SO2
         (iii).   Dilute hydrochloric acid is added to sodium bisulphate
                   NaHSO3   +   HCl            ----------
®   NaCl   +   H2O   +   SO2
In each case, The ratio of sulphur and oxygen in the sulphur dioxide obtained is of 32 : 32 or 1 : 1 by mass.
The Law of Multiple Proportions
when two elements A and B combine to form more than one compounds, then the weight of one is constant and the other has a simple ratio.  [The Law of Multiple Proportions is the third postulate of Dalton's atomic theory. It states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers.]
E.g. Two different compounds are formed by the elements carbon and oxygen.
C  (12gm)        +         ½ O2   (16gm)                         ---------®         CO (28gm)
C  (12gm)        +         O2       (32gm)            ---------®         CO2 (44gm)

Here, 12 gm of carbon combine with 16g and 32gm of Oxygen to form Carbon monoxide and Carbon dioxide respectively. The ratio of oxygen combining with 12 gm of Carbon is 16: 32 or, 1:2 which is in a simple ratio.

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CBSE 9th [Atoms and molecules]Empirical formula and molecular formula

Empirical formula :The empirical formula is the simplest formula for a compound in which atoms of different elements are present in simple ratio. It shows the relative number of atoms of each element. For example CH2O is the empirical formula of Glucose C6H12O6


Calculating Empirical Formula from Percentage Composition

Assume 100g of sample

Convert all percentages to a mass in grams, eg, 21% = 21g, 9% = 9g

Find the relative atomic mass (r.a.m) of each element present using the Periodic Table

Calculate the moles of each element present: n = mass ÷ r.a.m

Divide the moles of each element by the smallest of these to get a mole ratio

If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula

If the numbers in the mole ratio are NOT whole numbers, you will need to further manipulate these until the 
mole ratio is a ratio of whole numbers (integers)

A compound is found to contain 47.25% copper and 52.75% chlorine. Find the empirical formula for this 
compound.

element
Cu
Cl
mass in grams
47.25
52.75
r.a.m
63.6
35.5
moles = mass ÷ r.a.m
47.25 ÷ 63.6 = 0.74
52.75 ÷ 35.5 = 1.49
divide throughout by lowest number
0.74 ÷ 0.74 = 1
1.49 ÷ 0.74 = 2.01 = 2


Empirical formula for this compound is CuCl2

Molecular formula  : It is the formula in which the actual number of atoms of  different element are present. For example, if the empirical formula of benzene is CH where as molecular formula  is C6H6 , etc.

An empirical formula is often calculated from elemental composition data. The weight percentage of each of the elements present in the compound is given by this elemental composition. 
Let's determine the empirical formula for a compound with the following elemental composition:40.00% C, 6.66% H, 53.34% O....Read full Post
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