Nitrogen Family

Introduction:
In periodic table elements are arranged in the ascending order of their atomic number.  They are arranged in such a way that elements with same electronic configuration fall under the same group.  As electronic configuration decides chemical property of an element, elements with in a group will have same chemical property.  Each group is called as a family with the name of the first element in that group.  Hence elements in the group under Boron are said to belong Boron family.  Similarly elements under the nitrogen are called as nitrogen family.

Position of Nitrogen Family in Periodic Table
In periodic table nitrogen family can be found in p Block elements.  Elements in which the valence electrons can be found in  p orbitals are called as p block elements.  They have general electronic configuration ns² np^1-6 . These elements occupy right corner of periodic table
  
Nitrogen family comes next to carbon in periodic table and this family consists of Nitrogen, phosphorus, Arsenic, Antimony and Bismuth

General Properties of Nitrogen Family:

This family elements have the general electronic configuration ns²,np³ . As the valence shell contains five electrons these elements have a maximum valency of five.
But as we move down the group of nitrogen family, the electrons in the S orbital will not take part in valency due to inert pair effect.  Thus elements in the bottom of the group will have a maximum valency of three.
As we down the group the atomic size increases and hence the metallic nature increases.  So Bismuth is metal while nitrogen is non metal
As the metallic nature increases the basic nature of the oxides increases.  So the Bismuth oxide is basic while nitrogen oxides are acidic

Inert Nature of Nitrogen:

In nitrogen family all other elements except nitrogen are highly reactive.  But nitrogen is chemically inert.  This is due to strong triple bond formed in molecular nitrogen.  But nitrogen is an important element in proteins and is required by animation for continuous survival.  Hence the nitrogen in atmosphere is fixed by nitrogen fixation to earth

Electro negativity and the periodic table

what is electronegativity ?
Electronegativity is the measure of the capacity of an atom of an element to attract the shared electron pair in a covalent bond towards itself.The electronegatiivty is different from the electron gain enthalpy in the sense that while electron gain enthalpy calculate the measure of attracting the electrons towards it self in the isolated gaseous state, the electronegativity measures the  power to attract the electrons pair in combined state.

The concept of electronegativity was introduced by Pauling in 1932,the trends of electronegativity in the periodic table is quite evident. In a period from left to right , the value of the electronegativity  increases while in a group from top to bottom , the value of electronegativity decreases.

For example :
Period 2nd   Li < Be< B< C< N<O<F 
Period 3rd   Na<Mg<Al<Si<P<S<Cl
Period 4th   K<Ca<Ga<Ge<As<Se<Br
Similarly  in group we find the electronegativity goes on decreasin from top to bottom:

for example :
Group 1A     Li>Na>K>Rb>Cs

Group  2A     Be>Mg>Ca>Sr>Ba  etc.
The electronegativity of an element is not constant and depends on the following factors:
1.State of hybridization : sp> sp2>sp3
2.Oxidations state of an element .Fe 3+> Fe2+.

In the periodic table the elements with low value of electronegativity are metals and with high values are non metals,the fluorine is the most electronegative element of the periodic table and the Cesium with lowest electronegativity.

Problems on Electronegativity Periodic Table

Question for practice :

• Question 1. Arrange the following elements in the decreasing order of electronegativity  :  F,Br,I Cl

Answer : as these are the halogens  or the elements of 7A group of the periodic table and as we know the electronegatiivty decreases down the group so the required order is F>Cl>Br>I

• Question2 .arrange the following in the increasing order of electronegativity : Na,Mg,K, Al,C,O ,F ,Cl

Answer : the required order is  K<Na<Mg<Al<Cl<O<F
NOTE : In periodic table the four most electronegative elements are  F>O >Cl >N.

Law of Conservation of Mass

Introduction

The law of conservation of mass is also termed as principle of mass/matter conservation. The law of conservation of mass states that mass cannot be created/destroyed, but it may be changed from one form to another. This concludes that for any chemical process in a closed system, the mass of the reactants must be equal the mass of the products.

Experiment for Law of the Conservation of Mass:

Discussion:

In this experiment, aqueous solutions of three different compounds will produce two separate and distinct chemical reactions. Balanced chemical equations for the two reactions are:

Na₂CO₃(aq) + CaCl₂(aq) ----2NaCl(aq) + CaCO₃(aq) (Eq.1)

CaCO₃(s) + H₂SO₄(aq) ------CaSO₄(s) + H₂O + CO₂(g) (Eq.2)

The combined masses of the three solutions will be measured before and after each reaction is completed.

PURPOSE

To recognize the law of conservation of mass and to determine the masses of reactants and products.

EQUIPMENT

Laboratory balance corks (to fit test tubes) (2), Erlenmeyer flask: 125-Ml, labels rubber stopper (for flask), safety goggles, and graduated cylinder: 10-mL, lab apron and coat test tubes 13 x 100-mm (2)

MATERIALS

1 M aqueous solutions of: Na₂CO₃, CaCl₂ and H₂SO₄

PROCEDURE

1. Measure exactly 10.0 mL of sodium carbonate (Na₂CO₃) solution in a graduated cylinder. Pour Na2CO3 into a clean, dry 125-mL Erlenmeyer flask. Stopper the flask.

2. Measure exactly 3.0 mL of 1 M calcium chloride (CaCl₂) solution by a pipette and pour into a clean, dry test tube. Put cork and label the tube.

3. Measure exactly 3.0 mL of 1 M sulfuric acid (H₂SO₄) solution by a pipette and pour into a clean, dry test tube. Put cork and label the tube.

4. Put the stoppered flask and the corked test tubes together on the pan of the laboratory balance. Tilt the test tubes to avoid liquids from touching the corks. Measure the total mass of these containers, stoppers, and solutions.

5. Now, remove the flask and the test tube containing the CaCl₂ solution from the balance pan. Pour the CaCl₂ solution into the Na₂CO₃ solution in the flask. Shake the flask to mix the two solutions thoroughly.

6. Replace the stopper and cork in their initial containers. Once again, measure the combined mass of the three containers, stoppers, and contents.

7. Remove the flask and the test tube containing H₂SO₄ from the balance pan. Pour the H₂SO₄ solution into the flask with care. Shake the flask until all bubbling stops. Note the readings.

8. Replace the stopper and cork and the solutions in their initial containers. Again measure the mass of the three containers, stoppers, and contents.

Result:

We will observe that the mass of the all three solutions will be conserved. Hence, this experiment proves the law of conservation of mass.

Molecules calculation

Mass of a molecule is sum of the masses of the constituent atoms.
The mass of an atom is sum of masses of neutrons and protons.
In turn, masses of certain number of molecules too would depend on the masses of the constituent atoms.
Using this logic, Avogadro devised a special term 'mole' which is a certain mass equivalent to the molar mass expressed in grams.

Introduction :

Avogadro was the first person to establish that elements exist as molecules.
Thus a mole can be defined as a certain number of molecules. This number was found out to be 6.022 x 10²³ molecules.
Thus a mole of oxygen, hydrogen, carbon dioxide, carbon monoxide or any substance would contain 6.022 x 10²³ molecules.
It can be further illustrated as under
If two moles of hydrogen combines with one mole of oxygen what is the number of molecules of water formed?
One mole= 6.022 x 10²³ molecules.
Two moles= 12.044x 10²³ molecules
                  2H₂            +                                O₂ ---------->                 2H₂O
12.044x 10²³ molecules + 6.022 x 10²³ molecules.---->          2 moles
So the answer is 12.044x 10²³ molecules of water. It may be mentioned here that it is not 2+1=3, as in mathematics.

Illustrations of Molecules Calculation

Let us consider a few problems.
Question 1;- How many molecules are there in 56 g of carbon monoxide?
Ans: First let us find the number of moles

number of moles = mass in gram/molar mass
=56/28
=2 moles
Now by Avogadro's hypothesis,
One mole of a gas contains 6.022 x 10²³ molecules.
so two moles would contain
2 x 6.022 x 10²³ molecules.
12.044x 10²³ molecules.

Question 2 :-
Find the moles in12.044x 10²³ molecules of hydrogen at STP.
Ans:  One mole of hydrogen = 6.022 x 10²³ molecules.
so 12.044x 10^23 /6.022 x 10²³ 
=2 moles
Question 3
A gas has volume of 44.8 L. What is the number of molecules in it?
22.4 L volume is occupied by one mole i.e. 6.022 x 10²³ molecules.
So, 44.8 L would be occupied by   6.022 x 10²³ x 2
= 12.0444 x 10²³ molecules.

Empirical and molecular formula

Introduction :

Empirical Formula

The empirical formula for a compound is the simplest ratio of the numbers of atoms of each element present in the compound, e.g. hydrogen peroxide H₂O₂, its empirical formula is HO.

Empirical formula mass of a compound: It refers to the sum of the atomic masses of the elements present in the empirical formula. We can calculate the empirical Formula using the composition of elements in a compound.

Find Molecular Formula

A molecular formula represents the number of atoms of each element in a molecule. Therefore, the multiples of empirical formula gives the molecular formula of a, e.g. hydrogen peroxide H₂O₂. The Molecular Mass (molecular weight) of a compound is calculated with the atomic masses of each element and times with the number of atoms present in a compound.

    Molecular formula mass = n x empirical formula mass

The Molecular formula shows each element by its chemical symbol and indicates the number of atoms of each element found in each molecule of that compound. For example, methane, a small molecule consisting of one carbon atom and four hydrogen atoms, has the chemical formula CH₄. The sugar molecule glucose has six carbon atoms, twelve hydrogen atoms and six oxygen atoms, so its chemical formula is C₆H₁₂O₆.

Examples of Empirical and Molecular Formula:

• Example 1: 

Common sugar: glucose

Ratio of carbon, hydrogen and oxygen are in the ratio of 1:2:1. So, the empirical formula is CH₂O.

Empirical formula mass of glucose: 12.0 + (2 x 1.0) + 16.0 = 30g/mol.

Molar mass or molecular weight of glucose is: 180.16g/mol 

Molecular formula mass = n x empirical formula mass

180.16g/mol = n x 30g/mol

n = 6

Therefore, the molecular formula for a compound is 6 x empirical formula, i.e.., 6 x CH₂O = C₆H₁₂O₆   

• Example 2:  Ethane 

Ratio of carbon, hydrogen and oxygen are in the ratio of 1:2. So,  the empirical formula is CH₃ 

Empirical formula mass is: 12.0 + (3 x 1.0) = 15.0 g/mol

Molar mass or molecular weight of  Ethane is 30.0g/mol 

Molecular formula mass = n x empirical formula mass

30.0 = n x 15.0

n = 2

Therefore, the molecular formula for a compound is 2 x empirical formula, i.e.., 2 x (CH₃) which is C₂H₆

Comparing Molecular and Empirical Formulas

Compound                                     Molecular formula                        Empirical formula

Water                                                 H₂O                                               H₂O

Hydrogen Peroxide                              H₂O₂                                              HO

Glucose                                             C₆H₁₂O₆                                        CH₂O

Methane                                            CH₄                                                CH₄

Ethane                                              C₂H₆                                              CH₃