Monday, July 4, 2016

Chapter 2 - Mass of Protons, Neutrons and Electrons

In the previous section, we saw the details about Chemical equations. In this chapter, we will see Details of Atoms.

We have seen elements, compounds, molecules, atoms etc., The objects that we learn about, are becoming smaller and smaller. Compare this with the research of astronomers: They explore more and more by studying higher and higher levels. Like, upper atmosphere, outer space, deep space etc., The objects that they deal with, become larger and larger. Like planets, stars, etc.,

We are going in the opposite direction. The objects that we study, become smaller and smaller. The comparison is shown in the fig.2.1 below:
Fig.2.1
We have seen molecules and atoms. Now we go still deeper. We want to know what is inside the atoms. At the center of the atom, there is the nucleus. Different types of particles are present inside the nucleus. One such particle is the Proton. It is a positively charged particle. Each atom of a particular element will have a particular number of protons. For example,
• an atom of Sodium will have 11 protons in it’s nucleus
• an atom of Nitrogen will have 7 protons in it’s nucleus
The number of protons in the atom of a particular element is a unique value, that will never change.

Even when the nucleus carries positively charged protons, the atom as a whole, is electrically neutral. Why is that so? The answer is that, the positive charge is neutralised by another type of particle called the Electron. This electron is a negatively charged particle. And the number of electrons in an atom will be equal to the number of protons. So the charges get neutralised. The electrons revolve around the nucleus. Just like planets revolve around the sun. The electrons revolve around the nucleus in fixed orbits. These orbits are also called shells. This is shown in the fig.2.2 below:
Fig.2.2
The shells around the nucleus are given fixed numbers. The numbering is done starting from the shell which is closest to the nucleus. So the shell which is closest to the nucleus will be numbered 1. The next shell 2, and so on. Larger the number of a shell, greater is it’s distance from the nucleus. In addition to numbering, the shells are also given names. K, L, M, N etc., So the shell number 1 is called the K-shell, shell number 2 is called the L-shell, shell number 3 is called the M-shell and so on.

The above model was put forward by Danish scientist Neils Bohr, and is called the Bohr model of atom.

Mass of particles
If we take a piece of iron in our hand, we will feel some weight. This is because, the iron piece that we took has ‘mass’. If we take a piece of wood of the same size, we will not feel the same weight. This is because, iron has more mass than wood of the same size. All substances have mass, even if they are very small. Molecules, atoms, protons, electrons.., all have mass.

When scientists studied about the mass of atoms, protons, electrons etc., they noticed an important difference:
Consider the following sum:
• Total mass of all the protons in an atom + Total mass of all the electrons in that atom.
The above sum must be equal to the ‘mass of the atom as a whole’. This is because, the atom consists of protons in the central nucleus, and the electrons revolving around the nucleus.

But it was found that, the above sum was very less than the mass of the atom. This is shown schematically in the fig. 2.3 below:
Fig.2.3
Based on this difference, scientists predicted that, protons and electrons are not the only components of an atom. Some thing else is also present.

In 1932, British scientist James Chadwick discovered the unknown particle. It was named as Neutrons. They are present in the nucleus of the atom. They are neither positively charged nor negatively charged. That is., they are electrically neutral. So we can include neutrons also, in the calculations related to mass.

In day to day life, we require various units to make various measurements. For example, 
• We use the unit ‘meter’ to measure distances. 
• We use the unit kilogram to measure mass (or weight)
In the same way, we now need an ‘appropriate unit’ to measure very small masses. Because, electrons, protons etc., have very small masses. The unit that we use for this purpose is called the Atomic mass unit. It is abbreviated as u, just as meter is abbreviated as m and kilogram is abbreviated as kg.

Now we must discuss an important point: 
■ One meter is the distance travelled by light in 1/299,792,458 of a second. So we have a condition: The distance travelled by light during a certain time. Such a condition puts a ‘fix’ on the meter. The advantage of having a ‘fix’ is that, different people will not take different values for ‘one meter’.
■ One gram also have a fix. It is the mass of 1 cm3 of water. From this, the mass corresponding to 1 kg can be easily calculated because, 1 kg = 1000 g
■ In the same manner, we want a fix on ‘u’ also. It is done as follows:
• One u is 1/12 of the mass of a carbon-12 atom
• One carbon-12 atom has a mass of 1.992646547 × 10 -27 kg
• 1/12 of that is 1.6605 × 10 -27 kg
• So, one u = 1.6605 × 10 -27

What is this ‘carbon-12 atom’ ? Why not just say ‘carbon atom’? 
Carbon-12 is a close relative of carbon. ‘12’ is strictly specified so that, there will not be any ambiguity while choosing from among close relatives who are ‘look-alikes’. We will learn about the relation later in this chapter.
Any way, it is not important for our present discussion. All we need to know at present is that 1 u = 1.6605 × 10 -27

So we have a 10, raised to the power of -27. let us analyse it further:
1 kg mass is known to us
• 0.1 kg = 10 -1 = one tenth of a kg
• 0.01 kg = 10 -2 = one hundredth of a kg. A smaller quantity.
• 0.001 kg = 10 -3 = It is a very small quantity. It comes into use when we measure the mass of precious metals like gold
• 0.0001 kg = 10 -4 . A still smaller quantity.

Notice that when the number of zeros (on the right side of the decimal point) increases, the quantity becomes smaller and smaller. 1.6605 × 10 -27 will have 26 zeros on the right side of the decimal point. It is an extremely small quantity. 

Now we need to express the masses of electrons, protons etc., in terms of u. So that we can just write ‘u’ instead of 1.6605 × 10 -27. How do we do that​?

It is an ordinary case of ‘conversion from one unit to another’. Let us see a common example.
Suppose we have Rs. 2500/- with us. How many Euros is that? We can easily calculate it if we know ‘how much is one Euro in Rs.’ At the present rate, One Euro = 74.44 Rs. So Rs. 2500 = 2500/74.44 = 33.58 Euros. In this way we can convert the 'mass of a proton in kg', into 'mass of a proton in u'. The following comparison will help to understand the process better:

So we find that the procedure is to simply divide. 
• If we have Rs 2500/-, it is equivalent to having 33.58 Euros. 
• If we have 1.6605 × 10 -27 kg, it is equivalent to having 1.00727 u
In this way we can express the mass of neutron and electron also in terms of u. The steps are shown below:

For easy comparison, let us write the results together:
• Mass of Proton = 1.00727 u
• Mass of Neutron = 1.00866 u
• Mass of Electron = 0.005484 u

Consider the mass of proton. There are 2 zeros after the decimal point. That means, the quantity after the decimal point is small. So, for practical purposes, this portion after the decimal point is discarded. That leaves ‘1’ on the left side of the decimal point. That means, for practical purposes, the mass of one proton is taken as 1u

Next consider the mass of neutron. Here also there are 2 zeros after the decimal point. That means, the quantity after the decimal point is small. So, for practical purposes, this portion after the decimal point is discarded. That leaves ‘1’ on the left side of the decimal point. That means, for practical purposes, the mass of one neutron is taken as 1u


Now consider the mass of electron. Here also there are 2 zeros after the decimal point. That means, the quantity after the decimal point is small. So, for practical purposes, this portion after the decimal point is discarded. That leaves ‘0’ on the left side of the decimal point. That means, for practical purposes, the mass of neutron is taken as 0 u. So the neutron is given 'zero mass' in calculations.

The above calculations tallies with the experimental results. Experimentally, it is proved that most of the mass of an atom is concentrated in the nucleus. That is., total mass of the atom is due mainly to the nucleus. The electrons out side the nucleus does not contribute much to the total mass of the atom. This is shown schematically in the fig.2.4 below. 
Fig.2.4
So it is appropriate to consider electrons to have zero mass.

In the next section we will see 'Mass number'.

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Sunday, July 3, 2016

Chapter 1.2 - Chemical Equations of Reactions

In the previous section, we saw the details about Chemical formula. In this section we will see Chemical Equations.
Consider the reaction between zinc and hydrochloric acid. As a result of the reaction, we will get two substances: zinc chloride and hydrogen. Substances taking part in a reaction are called Reactants. Substances formed as a result of the reaction are called products. In the above reaction:
• Reactants are zinc and hydrochloric acid
• Products are zinc chloride and hydrogen

We have seen that the molecules can be represented by chemical formulae. Using them, we can write 'equations to represent a reaction'. For the reaction between zinc and hydrochloric acid, the equation will be:
Zn + HCl → ZnCl2 + H2
Let us analyse the above equation:
• On the left side we have two items. Zn and HCl  
   ♦ Zn represents Zinc, and HCl represents Hydrochloric acid. These are the reactants
• On the right side we have two items. ZnCl2 and H2 
   ♦ ZnCl2 represents zinc chloride, and H2 represents hydrogen. These are the products
• So a 'chemical equation' gives us an idea about 'what happens during a reaction'. But some details are missing:
   ♦ On the left side, there is only one chlorine atom. But on the right side, there are two chlorine atoms
   ♦ On the left side, there is only one hydrogen atom. But on the right side, there are two hydrogen atoms
• So the equation is not balanced. It is our job to balance a given equation

• In the above equation, zinc chloride is present. It is a compound containing Zn (zinc) and Cl (chlorine). Why is zinc chloride represented as ZnCl2 and not ZnCl ? 
• The reason is that, a molecule of zinc chloride will always consist of:
    ♦ One atom of zinc and
    ♦ Two atoms of chlorine
• So why is it that 'one molecule of  zinc chloride will always consist of One atom of zinc and Two atoms of chlorine'?

• Similarly, in the above equation, hydrochloric acid is present. It is a compound containing H2 (hydrogen) and Cl (chlorine). Why is hydrochloric acid represented as HCl ? Why not HCl2 or H2Cl ?  
• The reason is that, a molecule of hydrochloric acid will always consist of:
    ♦ One atom of hydrogen and
    ♦ One atom of chlorine
• So why is it that 'one molecule of  hydrochloric acid will always consist of One atom of hydrogen and One atom of chlorine'?

Such a question can arise in the case of many compounds. We will learn the answers in later chapters. At present, all we need to learn is, how to balance a 'given equation'. Later, we will learn to write equations on our own, and then balance them.

To get 2 hydrogen atoms and 2 chlorine atoms on the right side, can we write H2Cl2 (instead of HCl) on the left side ? No. because hydrochloric acid is HCl.  H2Cl2 represents something else. 
While balancing chemical equations, we must never alter the subscripts. Because altering them will alter the compound. We must alter only the coefficients.

We can follow a definite procedure to balance any chemical equation. We will discuss the procedure with the help of a few examples. First we will take the above equation:
• Write the 'number of atoms' of each element on the reactants side and products side. In the above table, in step 1, we have written the number of atoms according to the 'given equation'. These numbers must be same on both sides. Other wise the equation is not a balanced one. In our case, 
    ♦ Hydrogen has 1 on the reactants side, and 2 on the products side
    ♦ Chlorine has 1 on the reactants side and 2 on the products side
Such differences should be under lined. We do not have to pay attention to the numbers which are not underlined. Among those which are under lined, it is always better to consider elements other than hydrogen and oxygen first. 
• So we consider chlorine. It's number on the left side must become 2. 
• Remember that we can change only the coefficients. So, in the given equation, put 2 in front of HCl on the left side
• Now write the number of each element again in step 2. We find that hydrogen also becomes 2 on the left side
• There is nothing to underline. The equation is balanced

Another example:
For a chemical equation to be balanced, the number of atoms of each element on both sides of the equation must be the same.
• Write the 'number of atoms' of each element on the reactants side and products side. In the above table, in step 1, we have written the number of atoms according to the 'given equation'. These numbers must be same on both sides. Other wise the equation is not a balanced one. In our case, 
    ♦ Oxygen has 2 on the reactants side, and 1 on the products side
This difference is under lined. We do not have to pay attention to the numbers which are not underlined.  
• So we consider oxygen. It's number on the right side must become 2. 
• Remember that we can change only the coefficients. So, in the given equation, put 2 in front of H2O on the right side
• Now write the number of each element again in step 2. We find that oxygen is balanced, but hydrogen is 2 on the left side, and 4 on the right side
• This 2 on the left side must become 4. So put 2 in front of H2 on the left side
• Now write the numbers for each element in step 3. There is nothing to underline. The equation is balanced

Another example:
• Write the 'number of atoms' of each element on the reactants side and products side. In the above table, in step 1, we have written the number of atoms according to the 'given equation'. These numbers must be same on both sides. Other wise the equation is not a balanced one. In our case, 
    ♦ Oxygen has 2 on the reactants side, and 1 on the products side
This difference is under lined. We do not have to pay attention to the numbers which are not underlined.  
• So we consider oxygen. It's number on the right side must become 2. 
• Remember that we can change only the coefficients. So, in the given equation, put 2 in front of MgO on the right side
• Now write the number of each element again in step 2. We find that oxygen is balanced, but magnesium is 1 on the left side, and 2 on the right side
• This 1 on the left side must become 2. So put 2 in front of Mg on the left side
• Now write the numbers for each element in step 3. There is nothing to underline. The equation is balanced

Another example:
• Write the 'number of atoms' of each element on the reactants side and products side. In the above table, in step 1, we have written the number of atoms according to the 'given equation'. These numbers must be same on both sides. Other wise the equation is not a balanced one. In our case, 
    ♦ Carbon has 3 on the reactants side, and 1 on the products side
    ♦ Hydogen has 8 on the reactants side and 2 on the products side
    ♦ Oxygen has 2 on the reactants side, and 3 on the products side
Such differences should be under lined. We do not have to pay attention to the numbers which are not underlined. Among those which are under lined, it is always better to consider elements other than hydrogen and oxygen first. 
• So we consider carbon. It's number on the right side must become 3 
• Remember that we can change only the coefficients. So, in the given equation, put 3 in front of CO2 on the right side
• Now write the number of each element again in line with step 2. We find that
    ♦ Carbon is balanced: 3,3
    ♦ Hydrogen remains in the same unbalanced state: 8,2
    ♦ Oxygen's numbers have changed. But it is still unbalanced: 2,7
• We kept hydrogen and oxygen for the last. So we took carbon first. Now carbon is balanced. Out of hydrogen and oxygen, we must take hydrogen first.
• For hydrogen, the present numbers are: 8,2. The right side also must become 4
• For that, we put 4 in front of H2O on the right side
• Now write the number of each element again in step 3. We find that
    ♦ Carbon is balanced: 3,3
    ♦ Hydrogen is balanced: 8,8
    ♦ Oxygen's numbers have changed. But it is still unbalanced: 2,10
• For oxygrn, the 2 must become 10 on the left side
• For that, we put 5 in front of O2 on the left side
• Now write the number of each element again in line with step 4. We find that
    ♦ Carbon is balanced: 3,3
    ♦ Hydrogen is balanced: 8,8
    ♦ Oxygen is balanced: 10,10
• There is nothing to underline. The equation is balanced.

Some more solved examples can be seen here. In the next section we will see the 'Details of Atoms'.

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Sunday, June 19, 2016

Chapter 1.1 - Chemical Formula of a Compound

In the previous section, we saw the molecules of Elements. In this section we will see the molecules of Compounds. We know that a compound contains more than one element. So there will be more than one type of atoms. For example, in water, there will be atoms of hydrogen and oxygen. 

The different types of atoms in a compound will not exist independently. The atoms will always be in groups. Groups cannot be formed in any manner. There are strict rules for the formation of groups:
• Each group will have atoms of different elements
• The number of atoms of each element will be the same in any given group 
• In the case of compounds also, the name of such groups is Molecule. 
• For example, in water, there will be atoms of two elements: hydrogen and oxygen
• In any given molecule of water, there will be two atoms of hydrogen and one atom of oxygen. This is shown in the fig.1.3 below:
Fig.1.3 Molecule of Water

Representing Atoms and Molecules

The method of representing atoms and molecules can be demonstrated with the help of some examples:
We have seen that the symbol for the element sulphur is S. One atom of sulphur is also represented by the letter S. So if we see the letter S, it indicates one atom of sulphur. But no one atom of sulphur can exist independently. It is always found as groups of 8 sulphur atoms. We have seen that such groups are called molecules. So a group of 8 sulphur atoms is one molecule of sulphur. (sulphur is a polyatomic molecule). One atom of sulphur is represented as S. How will we represent one molecule of sulphur? For that we write the number 8 as a subscript. So one molecule of sulphur is represented as S8.

If there are more than one sulphur molecule, that number is written on the left side. So, if a sample of sulphur contains 3 molecules of sulphur, it is written as 3S8. Such a notation will also help us to calculate the total number of sulphur atoms in that sample. The calculation is as follows:
• S8 indicates that, one Sulphur molecule has 8 sulphur atoms
• 3S8 indicates 3 molecules of sulphur. 
• So the total number of sulphur atoms in 3 molecules = 3 × 8 = 24
• If the sample contains 2 sulphur molecules, then the total number of sulphur atoms will be 2 × 8 = 16 

Another example: The symbol for the element helium is He. When we write He, it represents one atom of helium. Each of the helium atoms are able to exist independently. So a molecule of helium will consist of only one helium atom. (Helium is a monoatomic molecule). So when we write He, it will represent one molecule of helium also. 

If there are more than one helium molecule, that number is written on the left side. if a sample of helium contains 5 helium molecules, it is written as 5He. As we saw above in the case of sulphur, this type of representation will give us the number of atoms also:
• He indicates that, one helium molecule has 1 helium atom.
• 5He indicates 5 molecules of helium 
• So the total number of helium atoms in 5 molecules = 5 × 1 = 5
• If we see 7He, it will indicate  7 molecules of helium, which is equal to 7 atoms of helium

One more example: We have seen that the symbol for the element hydrogen is H. One atom of hydrogen is also represented by the letter H. So if we see the letter H, it indicates one atom of hydrogen. But no one atom of hydrogen can exist independently. It is always found as groups of 2 hydrogen atoms. We have seen that such groups are called molecules. So a group of 2 hydrogen atoms is one molecule of hydrogen. (hydrogen is a diatomic molecule). One atom of hydrogen is represented as H. How will we represent one molecule of hydrogen? For that we write the number 2 as a subscript. So one molecule of hydrogen is represented as H2.

If there are more than one hydrogen molecule, that number is written on the left side. So, if a sample of hydrogen contains 3 molecules of hydrogen, it is written as 3H2. This will also give us the total number of hydrogen atoms in that sample. It can be calculated as follows:
• H2 indicates that, one hydrogen molecule has 2 hydrogen atoms
• 3H2 indicates 3 molecules of hydrogen 
• So the total number of hydrogen atoms in 3 molecules = 3 × 2 = 6
• If the sample contains 5 hydrogen molecules, then the total number of hydrogen atoms will be 5 × 2 = 10


We will now see some solved examples:
Solved example 1.1
Write the number of atoms and molecules in each of the following:
(i) 3N2 (ii) O3 (iii) 6Cl2
Solution: (i) 3N2
• The subscript 2 indicates that there are two atoms in one molecule
• 3 on the left side indicates that there are a total of 3 molecules
• So the total number of atoms = 3 × 2 = 6
(ii) O3
• The subscript 3 indicates that there are 3 atoms in one molecule
• There is no number on the left side. It indicates 1 molecule
• So the total number of atoms = 1 × 3 = 3
(iii) 6Cl2
• The subscript 2 indicates that there are two atoms in one molecule
• 6 on the left side indicates that there are a total of 6 molecules
• So the total number of atoms = 6 × 2 = 12

So we have seen how the molecules of elements are represented. Now we will see the representation of the molecules of compounds. Consider the example of carbon dioxide. It is a compound, formed when carbon burns in oxygen. It can also form during the decomposition of calcium carbonate. But the ‘method of formation’ of carbon dioxide is not important for our present discussion. We want to know the elements present inside the carbon dioxide, and also quantity of each of those elements.

In one molecule of carbon dioxide:
• Number of carbon atoms = 1
• Number of oxygen atoms = 2

The above numbers will be true for every molecule of carbon dioxide. That is., what ever be the method by which we obtain carbon dioxide, or from which ever source, we obtain the carbon dioxide, every molecule of it will contain one atom of carbon and two atoms of oxygen. We can write in in the form of ratio:
• In every carbon dioxide molecule, the ratio of carbon to oxygen is 1:2. 
If this ratio is not maintained, we will not get carbon dioxide. For example, if 2 atoms of carbon combine with 3 atoms of oxygen, the ratio of carbon to oxygen is 2:3. It will not be carbon dioxide.

As this ratio is always maintained, we can represent carbondioxide as CO2. It indicates one atom of carbon and 2 atoms of oxygen. This type of representation is called the Chemical formulae of the compound. 

Another example: Any one water molecule will have two atoms of hydrogen and one atom of oxygen. So the chemical formula of water is H2O

• A chemical formula can have natural numbers (1, 2, 3 etc.,) in two forms: Regular form and subscript form
• Subscripts are found on the right of each element. It indicates the number of atoms of that element in the molecule. If there is only one atom of an element, it’s subscript ‘1’ is not written
• Numbers in the Regular form is always seen on the extreme left side of the chemical formula. it indicates the total number of molecules. If the total number of molecules is 1, it is not written. Such numbers in the regular form will never be seen in side a chemical formula. it will be only seen on the extreme left side

With this knowledge, we can analyse any given chemical formula, and write the number of atoms of each element present in it. The following solved examples will demonstrate the procedure
Solved example 1.2:
Determine the number of atoms of each element from the following chemical formulae:
(i) H2SO4 (ii) 5H2O (iii) ZnCl2 (iv) 7NH3
Solution: (i) H2SO4
■ There is 'no number' on the extreme left. This indicates '1'. So the given chemical formula indicates one molecule. 
• The first element is hydrogen (indicated by H). 
   ♦ It's subscript is 2. So there are 2 atoms of hydrogen
   ♦ Total number of hydrogen atoms = Total number of molecules × number of atoms per molecule = 1 × 2 = 2
• The second element is sulphur (indicated by S)
   ♦ There is 'no subscript'. This indicates '1' atom
   ♦ Total number of sulphur atoms = Total number of molecules × number of atoms per molecule = 1 × 1 = 1
• The final element is oxygen (indicated by O). 
   ♦ It's subscript is 4. So there are 4 atoms of oxygen
   ♦ Total number of oxygen atoms = Total number of molecules × number of atoms per molecule = 1 × 4 = 4
(ii) 5H2O
■ There is '5' on the extreme left. So the given chemical formula indicates 5 molecules. 
• The first element is hydrogen (indicated by H). 
   ♦ It's subscript is 2. So there are 2 atoms of hydrogen
   ♦ Total number of hydrogen atoms = Total number of molecules × number of atoms per molecule = 5 × 2 = 10
• The second element is oxygen (indicated by O)
   ♦ There is 'no subscript'. This indicates '1' atom
   ♦ Total number of Oxygen atoms = Total number of molecules × number of atoms per molecule = 5 × 1 = 5
(iii) ZnCl2
■ There is 'no number' on the extreme left. This indicates '1'. So the given chemical formula indicates one molecule.
• The first element is zinc (indicated by Zn)
   ♦ There is 'no subscript'. This indicates '1' atom
   ♦ Total number of Zinc atoms = Total number of molecules × number of atoms per molecule = 1 × 1 = 1
• The second element is chlorine (indicated by Cl). 
   ♦ It's subscript is 2. So there are 2 atoms of chlorine
   ♦ Total number of chlorine atoms = Total number of molecules × number of atoms per molecule = 1 × 2 = 2
(iv) 7NH3
■ There is '7' on the extreme left. So the given chemical formula indicates 7 molecules
• The first element is nitrogen (indicated by N)
   ♦ There is 'no subscript'. This indicates '1' atom
   ♦ Total number of nitrogen atoms = Total number of molecules × number of atoms per molecule = 7 × 1 = 7
• The second element is hydrogen (indicated by H). 
   ♦ It's subscript is 3. So there are 3 atoms of hydrogen

   ♦ Total number of hydrogen atoms = Total number of molecules × number of atoms per molecule = 7 × 3 = 21

In the next section we will see Chemical Equations.

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