Chapter 10.3 - Gram Molecular Mass

In the previous section, we saw that 18 grams of water will give water molecules. In this section we will see two more examples like that:

■ Consider a molecule of carbon dioxide (CO₂). How many grams of carbon dioxide should we take so that, there will be N_A molecules of CO₂?
Solution:

1. If there is to be N_A molecules of CO₂, there must be:
• N_A carbon atoms and
• (N_A × 2 = 2N_A) oxygen atoms
2. What is the mass of N_A carbon atoms?
• Mass of N_A atoms of any element is it's GAM
• So mass of N_A carbon atoms is GAM of carbon which is equal to 12 grams
3. What is the mass of N_A oxygen atoms?
• Mass of N_A atoms of any element is it's GAM
• So mass of N_A oxygen atoms is GAM of oxygen which is equal to 16 grams
• So mass of 2N_A oxygen atoms = 2 × 16 = 32 grams 
4. It seems that 12 grams of carbon and 32 grams of oxygen will give us N_A carbon dioxide molecules. Let us check: 
5. Let us consider (12+32) = 44 grams of water. Let this '44 grams' consist of 12 grams of carbon  and 32 grams of oxygen  
6. We can group it in the following way:
• The N_A carbon atoms (obtained from 12 grams of carbon) should be split into N_A groups, so that, each group has 1 carbon atom
• The 2N_A oxygen atoms (obtained from 32 grams of oxygen) should also be split into N_A groups, so that, each group has 2 oxygen atoms
7. Thus the 12 grams of carbon is now split into N_A groups. Let this be Set 1
• The 32 grams of oxygen is also now split into N_A groups. Let this be Set 2
The two sets are shown in fig.10.4 below:

Fig.10.4

8. Both the sets have the same number (N_A) of groups
• So each group from Set 1 will get a partner from Set 2
9. When a group from Set 1 (consisting of one carbon atom) enter into partnership with a group from Set 2 (consisting of two oxygen atoms), we get a molecule (CO₂) of carbon dioxide
10. Since there are N_A groups in each set, we will get N_A carbon dioxide molecules
11. Thus we can conclude: 44 grams of carbon dioxide will give N_A carbon dioxide molecules

One more example: Consider a molecule of glucose (C₆H₁₂O₆). How many grams of glucose should we take so that, there will be N_A molecules of C₆H₁₂O₆?
Solution:
1. If there is to be N_A molecules of C₆H₁₂O₆, there must be:
• (N_A × 6 = 6N_A) carbon atoms
• (N_A × 12 = 12N_A) hydrogen atoms
• (N_A × 6 = 6N_A) oxygen atoms
2. What is the mass of N_A carbon atoms?
• Mass of N_A atoms of any element is it's GAM
• So mass of N_A carbon atoms is GAM of carbon which is equal to 12 grams
• So mass of 6N_A carbon atoms = 6 × 12 = 72 grams 
3. What is the mass of N_A hydrogen atoms?
• Mass of N_A atoms of any element is it's GAM
• So mass of N_A hydrogen atoms is GAM of oxygen which is equal to 1 gram
• So mass of 12N_A oxygen atoms = 12 × 1 = 12 grams 
4. What is the mass of N_A oxygen atoms?
• Mass of N_A atoms of any element is it's GAM
• So mass of N_A oxygen atoms is GAM of oxygen which is equal to 16 grams
• So mass of 6N_A oxygen atoms = 6 × 16 = 96 grams 
5. It seems that 72 grams of carbon, 12 grams of hydrogen and 96 grams of oxygen will give us N_A carbon glucose molecules. Let us check: 
6. Let us consider (72+12+96) = 180 grams of glucose. Let this '180 grams' consist of 72 grams of carbon, 12 grams of hydrogen  and 96 grams of oxygen  
7. We can group it in the following way:
• The 6N_A carbon atoms (obtained from 72 grams of carbon) should be split into N_A groups, so that, each group has 6 carbon atoms
• The 12N_A hydrogen atoms (obtained from 12 grams of hydrogen) should also be split into N_A groups, so that, each group has 12 hydrogen atoms
• The 6N_A oxygen atoms (obtained from 96 grams of oxygen) should also be split into N_A groups, so that, each group has 6 oxygen atoms
8. Thus the 72 grams of carbon is now split into N_A groups. Let this be Set 1
• The 12 grams of hydrogen is also now split into N_A groups. Let this be Set 2
• The 96 grams of oxygen is also now split into N_A groups. Let this be Set 3
The three sets are shown in fig.10.5 below:

Fig.10.5

9. All the three sets have the same number (N_A) of groups
• So each group from Set 1 will get a partner from Set 2 and set 3
10. When a group from Set 1 (consisting of six carbon atoms) enter into partnership with a group from Set 2 (consisting of twelve hydrogen atoms) and a group from Set 3 (consisting of six oxygen atoms), we get a molecule (C₆H₁₂O₆) of glucose
11. Since there are N_A groups in each set, we will get N_A glucose molecules
12. Thus we can conclude: 180 grams of glucose will give N_A  glucose molecules

Let us write a summary of the above discussion on molecules:
■ Molecules of elements:
• To get N_A hydrogen molecules, we must take 2 grams of hydrogen atoms 
• To get N_A oxygen molecules, we must take 32 grams of oxygen atoms 
• To get N_A ozone molecules, we must take 48 grams of oxygen atoms
• To get N_A sulphur molecules, we must take 256 grams of sulphur atoms 
• To get N_A helium molecules, we must take 4 grams of helium atoms
■ Molecules of compounds
• To get N_A water molecules, we must take 18 grams of water
• To get N_A carbon dioxide molecules, we must take 44 grams of carbon dioxide
• To get N_A glucose molecules, we must take 180 grams of glucose

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Is there an easier method to get the above masses? Let us try:

■ Consider a molecule of hydrogen. 
1. The molecule is H₂. It has two hydrogen atoms. 
2. So total mass of a molecule of hydrogen will be twice the mass of a hydrogen atom
3. We know that, the mass of a hydrogen atom is 1 u
4. So mass of a hydrogen molecule = (2 × 1 u) = 2 u
5. Above we found that, to get N_A hydrogen molecules, we must take 2 grams of hydrogen atoms   
• In (4) we have 2 u. 
• In (5) we have 2 grams
• The numeric parts are the same 

■ Consider a molecule of oxygen. 
1. The molecule is O₂. It has two oxygen atoms. 
2. So total mass of a molecule of oxygen will be twice the mass of a oxygen atom
3. We know that, the mass of a oxygen atom is 16 u
4. So mass of a oxygen molecule = (2 × 16 u) = 32 u
5. Above we found that, to get N_A oxygen molecules, we must take 32 grams of oxygen atoms   
• In (4) we have 32 u. 
• In (5) we have 32 grams
• The numeric parts are the same

■ Consider a molecule of ozone. 
1. The molecule is O₃. It has three oxygen atoms. 
2. So total mass of a molecule of ozone will be thrice the mass of a oxygen atom
3. We know that, the mass of a oxygen atom is 16 u
4. So mass of a ozone molecule = (3 × 16 u) = 48 u
5. Above we found that, to get N_A ozone molecules, we must take 48 grams of oxygen atoms   
• In (4) we have 48 u. 
• In (5) we have 48 grams
• The numeric parts are the same

■ Consider a molecule of sulphur. 
1. The molecule is S₈. It has eight sulphur atoms. 
2. So total mass of a molecule of sulphur will be eight times the mass of a sulphur atom
3. We know that, the mass of a sulphur atom is 32 u
4. So mass of a sulphur molecule = (8 × 32 u) = 256 u
5. Above we found that, to get N_A sulphur molecules, we must take 256 grams of sulphur atoms   
• In (4) we have 256 u. 
• In (5) we have 256 grams
• The numeric parts are the same

■ Consider a molecule of helium. 
1. The molecule is H. It has only one helium atom. 
2. So total mass of a molecule of helium will be same as the mass of a helium atom
3. We know that, the mass of a helium atom is 4 u
4. So mass of a helium molecule = (1 × 4 u) = 4 u
5. Above we found that, to get N_A helium molecules, we must take 4 grams of helium atoms   
• In (4) we have 4 u. 
• In (5) we have 4 grams
• The numeric parts are the same

■ Consider a molecule of water. 
1. The molecule is H₂O. It has two hydrogen atoms and one oxygen atom. 
2. So total mass of a molecule of water will be the sum of:
• twice the mass of a hydrogen atom
• mass of  an oxygen atom
3. We know that, the mass of a hydrogen atom is 1 u and that of oxygen atom is 16 u
4. So mass of a water molecule = (2 × 1 u) + (1 × 16 u)   = 2 u + 16 u = 18 u
5. Above we found that, to get N_A water molecules, we must take 18 grams of water 
• In (4) we have 18 u. 
• In (5) we have 18 grams
• The numeric parts are the same

■ Consider a molecule of carbon dioxide. 
1. The molecule is CO₂. It has one carbon atom and two oxygen atoms. 
2. So total mass of a molecule of carbon dioxide will be the sum of:
• mass of a carbon atom
• twice the mass of an oxygen atom
3. We know that, the mass of a carbon atom is 12 u and that of oxygen atom is 16 u
4. So mass of a carbon dioxide molecule = (1 × 12 u) + (2 × 16 u)   = 12 u + 32 u = 44 u
5. Above we found that, to get N_A carbon dioxide molecules, we must take 44 grams of carbon dioxide
• In (4) we have 44 u. 
• In (5) we have 44 grams
• The numeric parts are the same

■ Consider a molecule of glucose. 
1. The molecule is C₆H₁₂O₆. It has six carbon atoms, twelve hydrogen atoms and six oxygen atoms. 
2. So total mass of a molecule of glucose will be the sum of:
• six times the mass of a carbon atom
• twelve times the mass of a hydrogen atom
• six times the mass of an oxygen atom
3. We know that, the mass of a carbon atom is 12 u, mass of a hydrogen atom is 1 u and that of an oxygen atom is 16 u
4. So mass of a glucose molecule = (6 × 12 u) + (12 × 1 u) + (6 × 16 u) = 72 u + 12 u + 96 u = 180 u
5. Above we found that, to get N_A glucose molecules, we must take 180 grams of glucose
• In (4) we have 180 u. 
• In (5) we have 180 grams
• The numeric parts are the same

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So we can write:

• Each molecule (it may be 'a molecule of an element' or 'a molecule of a compound') will have a unique value for the 'mass in grams' 

• If we take that much mass of that element/compound, there will be N_A number of molecules

• This unique mass is called 'Gram Molecular Mass' (GMM) of that molecule
Some examples:
• GMM of element oxygen is 16 grams
    ♦ In 16 grams of oxygen, there will be N_A molecules of oxygen
• GMM of element helium is 4 grams
    ♦ In 4 grams of helium, there will be N_A molecules of helium
• GMM of compound glucose is 180 grams
    ♦ In 180 grams of glucose, there will be N_A molecules of glucose

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The GMM of any molecule can be written using the following steps

Step 1: Calculate the total mass of one molecule. It's unit will be in 'u'
Step 2: Write the numeric part
Step 3: Write 'grams' on the right side
This will give the GMM of that molecule

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Some examples:

1. Find the GMM of ammonia (NH₃)
Solution:
Step 1: Total mass of one molecule:
Mass of one nitrogen atom = 14 u
Mass of one hydrogen atom = 1 u
So total mass = (1 × 14 u) + (3 × 1 u) = 14 u + 3 u = 17 u
Step 2: Write the numeric part: 17
Step 3: Write 'grams' on the right side. We get: 17 grams
So GMM of ammonia is 17 grams

1. Find the GMM of HCl
Solution:
Step 1: Total mass of one molecule:
Mass of one hydrogen atom = 1 u
Mass of one chlorine atom = 35.5 u
So total mass = (1 × 1 u) + (1 × 35.5 u) = 1 u + 35.5 u = 36.5 u
Step 2: Write the numeric part: 36.5
Step 3: Write 'grams' on the right side. We get: 36.5 grams
So GMM of ammonia is 36.5 grams

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Once we understand the basics, there will not be any need to write the steps. Just write the total mass of one molecule, and write 'grams' instead of 'u'

We will now see some solved examples

Solved example 10.1
How many water molecules are present in 90 grams of water?
Solution:
1. GMM of water (H₂O) = (2 × 1) + (1 × 16) = 2 + 16 = 18 grams
2. No. of GMMs in 90 grams of water = ⁹⁰⁄₁₈  = 5
3. 1 GMM will contain N_A molecules. 
4. So no. of molecules in 5 GMMs = 5N_A = 5 × 6.022×10²³

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Let us do the above problem by another method, which uses some very basic information:

1. We have:
No. of molecules ×  Mass (in grams) of one molecule = Given total Mass (in grams)
⇒ No. of molecules = ^{Given total Mass (in grams)}⁄_{Mass (in grams) of one molecule}.  
2. Given total mass = 90 grams
3. Mass (in u) of one molecule of water = (2 × 1) + (1 × 16) = 2 + 16 = 18 u
4. But 1 u = 1.6605×10^-24 grams  (Details here)
So 18 u = 18 × 1.6605×10^-24 grams
5. Substituting this value in (1) we get:

■ We get the same result 5 × 6.022×10²³ that we got earlier.  
In these types of problems, it is worthy to always remember that, the reciprocal of 'u in grams' gives the Avogadro number N_A. That is:      

We can check it using a calculator

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Solved example 10.2

Samples of some compounds are given:
(i) 85 grams of ammonia (NH₃) 
(ii) 90 grams of glucose (C₆H₁₂O₆)
(iii) 88 grams of carbon dioxide (CO₂)
(iv) 50 grams of hydrogen (H₂)  
• Find the number of molecules in each sample. (Hint: GAMs are: H = 1g, C = 12 g, N = 14 g, O = 16 g)
Solution (i):
1. GMM of ammonia (NH₃) = (1 × 14) + (1 × 3) = 14 + 3 = 17 grams
Note: For this step we require the atomic masses (in u) of nitrogen and hydrogen. But they are given to us in the form of GAM. We know that, GAM is numerically same as u. 
2. No. of GMMs in 85 grams of water = ⁸⁵⁄₁₇  = 5
3. 1 GMM will contain N_A molecules. 
4. So no. of molecules in 5 GMMs = 5N_A = 5 × 6.022×10²³

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Let us do the problem by another method, which uses some very basic information:

1. We have:
No. of molecules × Mass (in grams) of one molecule = Given total Mass (in grams)
⇒ No. of molecules = ^{Given total Mass (in grams)}⁄_{Mass (in grams) of one molecule}.  
2. Given total mass = 85 grams
3. Mass (in u) of one molecule of water = (1 × 14) + (1 × 3) = 14 + 3 = 17 u
4. But 1 u = 1.6605×10^-24 grams  (Details here)
So 18 u = 18 × 1.6605×10^-24 grams
5. Substituting this value in (1) we get:

■ We get the same result 5 × 6.022×10²³ that we got earlier.

Solution (ii):
1. GMM of glucose (C₆H₁₂O₆) = (6 × 12) + (1 × 12) + (6 × 16) = 72 + 12 + 96 = 180 grams
2. No. of GMMs in 90 grams of water = ⁹⁰⁄₁₈₀  = 0.5
3. 1 GMM will contain N_A molecules. 
4. So no. of molecules in 0.5 GMMs = 0.5N_A = 0.5 × 6.022×10²³

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Let us do the above problem by the other method:

1. We have:
No. of molecules × Mass (in grams) of one molecule = Given total Mass (in grams)
⇒ No. of molecules = ^{Given total Mass (in grams)}⁄_{Mass (in grams) of one molecule}.  
2. Given total mass = 90 grams
3. Mass (in u) of one molecule of glucose = (6 × 12) + (1 × 12) + (6 × 16) = 72 + 12 + 96 = 180 u
4. But 1 u = 1.6605×10^-24 grams  (Details here)
So 180 u = 180 × 1.6605×10^-24 grams
5. Substituting this value in (1) we get:

■ We get the same result 0.5 × 6.022×10²³ that we got earlier.  

Solution (iii):
1. GMM of carbon dioxide (CO₂) = (1 × 12) + (16 × 2) = 12 + 32 = 44 grams
2. No. of GMMs in 88 grams of water = ⁸⁸⁄₄₄  = 2
3. 1 GMM will contain N_A molecules. 
4. So no. of molecules in 2 GMMs = 2N_A = 2 × 6.022×10²³

---

Let us do the above problem by the other method:

1. We have:
No. of molecules × Mass (in grams) of one molecule = Given total Mass (in grams)
⇒ No. of molecules = ^{Given total Mass (in grams)}⁄_{Mass (in grams) of one molecule}.  
2. Given total mass = 88 grams
3. Mass (in u) of one molecule of carbon dioxide = (1 × 12) + (16 × 2) = 12 + 32 = 44 u
4. But 1 u = 1.6605×10^-24 grams  (Details here)
So 44 u = 44 × 1.6605×10^-24 grams
5. Substituting this value in (1) we get:

■ We get the same result 2 × 6.022×10²³ that we got earlier.

Solution (iv):
1. GMM of hydrogen (H₂) = (2 × 1) = 2 grams
2. No. of GMMs in 50 grams of water = ⁵⁰⁄₂  = 25
3. 1 GMM will contain N_A molecules. 
4. So no. of molecules in 25 GMMs = 25N_A = 25 × 6.022×10²³

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Let us do the above problem by the other method:

1. We have:
No. of molecules × Mass (in grams) of one molecule = Given total Mass (in grams)
⇒ No. of molecules = ^{Given total Mass (in grams)}⁄_{Mass (in grams) of one molecule}.  
2. Given total mass = 50 grams
3. Mass (in u) of one molecule of glucose = (2 × 1) = 2 u
4. But 1 u = 1.6605×10^-24 grams  (Details here)
So 2 u = 2 × 1.6605×10^-24 grams
5. Substituting this value in (1) we get:

■ We get the same result 2 × 6.022×10²³ that we got earlier.

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Based on the above discussion, we can now take up 'mole'. We will see it in the next section. 

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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.

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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.

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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. 

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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.

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

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