In the previous section, we saw that, in any chemical reaction, we must take a specified number of molecules of each reactants. Other wise the reaction will not be completed.
So we have to answer a question:
■ Molecules are very minute particles. How can we count them?
• To answer this question, we will consider the example of coins.
1. Suppose in a bank, it is required to make up an amount of Rs. 100000/- (Rs. One lakh). The amount must consist of only one rupee coins.
2. It will be very difficult and time consuming to count one lakh coins. So an easier method is adopted:
3. Let the mass of one 1 Re coin be 5 grams. Then:
• Mass of two 1 Re coins = Number of coins × mass of one coin = 2 × 5 = 10 grams
• Mass of five 1 Re coins = Number of coins × mass of one coin = 5 × 5 = 25 grams
4. Conversely,
• If a 'group of 1 Re coins' has a total mass of 10 grams, then:
♦ the number of coins in that group = Total mass of the group⁄Mass of one coin
= 10⁄5 = 2 coins
• If a 'group of 1 Re coins' has a total mass of 280 grams, then:
♦ the number of coins in that group = Total mass of the group⁄Mass of one coin
= 280⁄5 = 56 coins
5. This gives us an easy method to make up 1 lakh rupees:
Step 1: No. of 1 Re coins in 1 lakh = 100000⁄1 = 100000
Step 2: Mass of this much 1 Re coins = Number of coins × mass of one coin
= 100000 × 5 grams = 500000 grams = 500 kilograms
Step 3: Make a pile of 1 Re coins in such a way that total mass of that pile is 500 kilograms
That pile will be worth 1 lakh
Or make 10 piles each with a total mass of 50 kilograms. The ten piles taken together will be worth 1 lakh
Another example:
Mass of a Rs. 2 coin is 8 grams. Make up an amount of Rs. Two lakhs and Seventy five thousand (Rs. 2,75,000/-)
Step 1: Number of Rs. 2 coins in 2,75,000 = 275000⁄2 = 1,37,500
Step 2: Mass of 137500 coins = Number of coins × mass of one coin
= 137500 × 8 grams = 1100000 grams = 1100 kilograms
Step 3: Make a pile of Rs. 2 coins in such a way that total mass of that pile is 1100 kilograms
That pile will be worth Rs. Two lakhs and Seventy five thousand
Or make 10 piles each with a total mass of 110 kilograms. The ten piles taken together will be worth Two lakhs and Seventy five
■ So we see that, large numbers can be easily counted by making use of mass.
[It may be noted that, this method will work only if:
• All the 1 Re coins are of the same mass
• All the Rs.2 coins are of the same mass]
■ Now the next question arises:
How can we find the mass of very minute particles like atoms and molecules?
[To get a rough idea about 'how minute the atoms and molecules are', let us see an example:
• One drop (0.05 ml) of water contain 1.67×1027.
♦ That is., 1670000000000000000000000000 molecules.
• If this huge number of molecules are to be accommodated in a single drop, they will have to be very very small.
• Now consider atoms. Each water molecule will have three atoms. Two of hydrogen and one of oxygen.
• So the total number of atoms is three times 1670000000000000000000000000.
• This much atoms are accommodated in a single drop. So the atoms are still smaller than molecules]
Coming back to our main discussion, we want the mass of atom and molecules. For that, we have to revisit the topic where we derived mass number A. Details here.
■ Let us write the important points that we learned there:
• 1 u = 1.6605×10-27 kg = 1.6605×10-24 grams
• Mass of a proton is 1 u
• Mass of a neutron is 1 u
• Mass of an electron is 0 u
• Mass of an atom
= Sum of the masses of all protons and neutrons in the nucleus of that atom
= [(No. of protons × 1 u) + (No. of neutrons × 1 u)]
= [(No. of protons) + (No. of neutrons)]
So the mass of an atom (when expressed in u) = sum of the number of protons and neutrons in the nucleus
• Consider the words: 'number of protons and neutrons'
• That means we have to count the numbers. So they are counting numbers. Counting numbers are whole numbers. So the mass of any atom will be a whole number. But this is not always the case. Let us see an example:
1. Consider the element neon (20Ne). It has 10 protons and 10 neutrons. So it's mass will be 20 u.
• But there is an isotope of neon (21Ne). It has 10 protons and 11 neutrons. So it's mass will be 21 u
• There is yet another isotope of neon (22Ne). It has 10 protons and 12 neutrons. So it's mass will be 22 u
2. Thus we have three masses for neon: 20 u, 21 u and 22 u
Which one will we take?
The answer is that, we take the average. The average 'based on abundance'
If we take a sample of neon from nature,
• 90.48% of that sample will be 20Ne
• 0.27% of that sample will be 21Ne
• 9.25% of that sample will be 22Ne
So the average is calculated as:
(20 × 0.9048) + (21 × 0.0027) + (22 × 0.0925) = 20.18 u
• 20.18 is not a whole number.
• In fact, the atomic mass of most elements are calculated in this way and so it is difficult to obtain whole numbers
For example,
average atomic mass of nitrogen is 14.0067
average atomic mass of phosphorus is 15.9994
• The following table shows the results:
3. However, for practical purposes and calculations, most of these values are rounded off to whole numbers. The following table shows the modified values:
• Thus we can easily obtain the mass of one atom of any element
Step 2: 'M grams' have a numeric part 'M'
Step 3: The numeric part 'M' should be same as the atomic mass of the element
Then 'M grams' is the GAM of that element
Solution:
Step 1: Consider a certain mass (in grams) of nitrogen. Let it be 'M grams'
Step 2: 'M grams' have a numeric part 'M'
Step 3: The numeric part 'M' should be same as the atomic mass of nitrogen, which is 14
Then '14 grams' is the GAM of nitrogen
2. Find the GAM of oxygen
Solution:
Step 1: Consider a certain mass (in grams) of oxygen. Let it be 'M grams'
Step 2: 'M grams' have a numeric part 'M'
Step 3: The numeric part 'M' should be same as the atomic mass of oxygen, which is 16
Then '16 grams' is the GAM of oxygen
3. Find the GAM of helium
Solution:
Step 1: Consider a certain mass (in grams) of helium. Let it be 'M grams'
Step 2: 'M grams' have a numeric part 'M'
Step 3: The numeric part 'M' should be same as the atomic mass of helium, which is 4
Then '4 grams' is the GAM of helium
■ Once we understand the basics, there will be no need to write the steps. Just write the atomic mass (with out 'u') and write 'grams' on the right side.
• 14 grams of nitrogen contains 6.022×1023 atoms of nitrogen
♦ That is., 14 grams of nitrogen contains NA atoms of nitrogen
2. GAM of hydrogen = 1 grams
• 1 gram of hydrogen contains 6.022×1023 atoms of hydrogen
♦ That is., 1 gram of hydrogen contains NA atoms of hydrogen
3. GAM of chlorine = 35.5 grams
• 35.5 grams of chlorine contains 6.022×1023 atoms of chlorine
♦ That is., 35.5 grams of chlorine contains NA atoms of chlorine
This GAM gives us an easy method to find the 'total number of atoms in a sample'. We will see it in the next section
So we have to answer a question:
■ Molecules are very minute particles. How can we count them?
• To answer this question, we will consider the example of coins.
1. Suppose in a bank, it is required to make up an amount of Rs. 100000/- (Rs. One lakh). The amount must consist of only one rupee coins.
2. It will be very difficult and time consuming to count one lakh coins. So an easier method is adopted:
3. Let the mass of one 1 Re coin be 5 grams. Then:
• Mass of two 1 Re coins = Number of coins × mass of one coin = 2 × 5 = 10 grams
• Mass of five 1 Re coins = Number of coins × mass of one coin = 5 × 5 = 25 grams
4. Conversely,
• If a 'group of 1 Re coins' has a total mass of 10 grams, then:
♦ the number of coins in that group = Total mass of the group⁄Mass of one coin
= 10⁄5 = 2 coins
• If a 'group of 1 Re coins' has a total mass of 280 grams, then:
♦ the number of coins in that group = Total mass of the group⁄Mass of one coin
= 280⁄5 = 56 coins
5. This gives us an easy method to make up 1 lakh rupees:
Step 1: No. of 1 Re coins in 1 lakh = 100000⁄1 = 100000
Step 2: Mass of this much 1 Re coins = Number of coins × mass of one coin
= 100000 × 5 grams = 500000 grams = 500 kilograms
Step 3: Make a pile of 1 Re coins in such a way that total mass of that pile is 500 kilograms
That pile will be worth 1 lakh
Or make 10 piles each with a total mass of 50 kilograms. The ten piles taken together will be worth 1 lakh
Another example:
Mass of a Rs. 2 coin is 8 grams. Make up an amount of Rs. Two lakhs and Seventy five thousand (Rs. 2,75,000/-)
Step 1: Number of Rs. 2 coins in 2,75,000 = 275000⁄2 = 1,37,500
Step 2: Mass of 137500 coins = Number of coins × mass of one coin
= 137500 × 8 grams = 1100000 grams = 1100 kilograms
Step 3: Make a pile of Rs. 2 coins in such a way that total mass of that pile is 1100 kilograms
That pile will be worth Rs. Two lakhs and Seventy five thousand
Or make 10 piles each with a total mass of 110 kilograms. The ten piles taken together will be worth Two lakhs and Seventy five
[It may be noted that, this method will work only if:
• All the 1 Re coins are of the same mass
• All the Rs.2 coins are of the same mass]
■ Now the next question arises:
How can we find the mass of very minute particles like atoms and molecules?
[To get a rough idea about 'how minute the atoms and molecules are', let us see an example:
• One drop (0.05 ml) of water contain 1.67×1027.
♦ That is., 1670000000000000000000000000 molecules.
• If this huge number of molecules are to be accommodated in a single drop, they will have to be very very small.
• Now consider atoms. Each water molecule will have three atoms. Two of hydrogen and one of oxygen.
• So the total number of atoms is three times 1670000000000000000000000000.
• This much atoms are accommodated in a single drop. So the atoms are still smaller than molecules]
■ Let us write the important points that we learned there:
• 1 u = 1.6605×10-27 kg = 1.6605×10-24 grams
• Mass of a proton is 1 u
• Mass of a neutron is 1 u
• Mass of an electron is 0 u
• Mass of an atom
= Sum of the masses of all protons and neutrons in the nucleus of that atom
= [(No. of protons × 1 u) + (No. of neutrons × 1 u)]
= [(No. of protons) + (No. of neutrons)]
So the mass of an atom (when expressed in u) = sum of the number of protons and neutrons in the nucleus
• Consider the words: 'number of protons and neutrons'
• That means we have to count the numbers. So they are counting numbers. Counting numbers are whole numbers. So the mass of any atom will be a whole number. But this is not always the case. Let us see an example:
1. Consider the element neon (20Ne). It has 10 protons and 10 neutrons. So it's mass will be 20 u.
• But there is an isotope of neon (21Ne). It has 10 protons and 11 neutrons. So it's mass will be 21 u
• There is yet another isotope of neon (22Ne). It has 10 protons and 12 neutrons. So it's mass will be 22 u
2. Thus we have three masses for neon: 20 u, 21 u and 22 u
Which one will we take?
The answer is that, we take the average. The average 'based on abundance'
If we take a sample of neon from nature,
• 90.48% of that sample will be 20Ne
• 0.27% of that sample will be 21Ne
• 9.25% of that sample will be 22Ne
So the average is calculated as:
(20 × 0.9048) + (21 × 0.0027) + (22 × 0.0925) = 20.18 u
• In fact, the atomic mass of most elements are calculated in this way and so it is difficult to obtain whole numbers
For example,
average atomic mass of nitrogen is 14.0067
average atomic mass of phosphorus is 15.9994
• The following table shows the results:
| Element | Average atomic mass (u) |
|---|---|
| Hydrogen | 1.0079 |
| Helium | 4.0026 |
| Carbon | 12.0111 |
| Nitrogen | 14.0067 |
| Oxygen | 15.9994 |
| Phosphorus | 30.9738 |
| Chlorine | 35.453 |
| Element | Average atomic mass (u) |
Atomic mass considered for practical purposes (u) |
|---|---|---|
| Hydrogen | 1.0079 | 1 |
| Helium | 4.0026 | 4 |
| Carbon | 12.0111 | 12 |
| Nitrogen | 14.0067 | 14 |
| Oxygen | 15.9994 | 16 |
| Phosphorus | 30.9738 | 31 |
| Chlorine | 35.453 | 35.5 |
We have already learned the above points in a previous chapter. We can continue our present discussion on 'how to count the number of minute particles' based on the above points.
Gram Atomic Mass and Avogadro Number
■ Consider the element carbon
1. It has an atomic mass 12 u. In '12 u', the unit of mass is 'u'
2.Let us convert this unit to 'grams'
3. We know that 1 u = 1.6605×10-24 grams
So mass of one carbon atom in grams = 12×1.6605×10-24
4. Let us take a sample of some carbon. Let that sample have a total mass of 12 grams.
5. How many carbon atoms will be present in that carbon sample?
Obviously,
• Note that, the '12' in both numerator and denominator cancels each other
■ Consider the element helium
1. It has an atomic mass 4 u. In '4 u', the unit of mass is 'u'
2.Let us convert this unit to 'grams'
3. We know that 1 u = 1.6605×10-24 grams
So mass of one helium atom in grams = 4×1.6605×10-24
4. Let us take a sample of some helium. Let that sample have a total mass of 4 grams.
5. How many helium atoms will be present in that helium sample?
Obviously,
• Note that, the '4' in both numerator and denominator cancels each other
■ Consider the element phosphorus
1. It has an atomic mass 31 u. In '31 u', the unit of mass is 'u'
2.Let us convert this unit to 'grams'
3. We know that 1 u = 1.6605×10-24 grams
So mass of one phosphorus atom in grams = 31×1.6605×10-24
4. Let us take a sample of some phosphorus. Let that sample have a total mass of 31 grams.
5. How many phosphorus atoms will be present in that phosphorus sample?
Obviously,
• Note that, the '31' in both numerator and denominator cancels each other
We saw three examples. In all of them, we got the same number of atoms. But this 'same number of atoms' was due to the 'particular mass' of each element that we took. Let us write more details about this 'particular mass':
1. Take a sample of any element
2. Let the total mass of that sample be 'M grams'
3. In 'M grams', the numeric part is 'M' and units part is 'grams'
4. This numeric part 'M' should be same as the 'atomic mass' of that element
For example,
'M' should be 16 for oxygen
'M' should be 35.5 for chlorine
'M' should be 4 for helium
5. Then the sample will contain 6.022×1023 number of atoms
This is true for all elements
■ This number '6.022×1023' is called the Avogadro number. It's symbol is NA
• Each element will have a unique value for the 'mass in grams'
• If we take that much mass of that element, there will be NA number of atoms
• This unique mass is called 'Gram Atomic Mass' (GAM) of that element
The GAM of any element can be written using the following steps
Step 1: Consider a certain mass (in grams) of the element. Let it be 'M grams'Step 2: 'M grams' have a numeric part 'M'
Step 3: The numeric part 'M' should be same as the atomic mass of the element
Then 'M grams' is the GAM of that element
Some examples:
1. Find the GAM of nitrogenSolution:
Step 1: Consider a certain mass (in grams) of nitrogen. Let it be 'M grams'
Step 2: 'M grams' have a numeric part 'M'
Step 3: The numeric part 'M' should be same as the atomic mass of nitrogen, which is 14
Then '14 grams' is the GAM of nitrogen
2. Find the GAM of oxygen
Solution:
Step 1: Consider a certain mass (in grams) of oxygen. Let it be 'M grams'
Step 2: 'M grams' have a numeric part 'M'
Step 3: The numeric part 'M' should be same as the atomic mass of oxygen, which is 16
Then '16 grams' is the GAM of oxygen
3. Find the GAM of helium
Solution:
Step 1: Consider a certain mass (in grams) of helium. Let it be 'M grams'
Step 2: 'M grams' have a numeric part 'M'
Step 3: The numeric part 'M' should be same as the atomic mass of helium, which is 4
Then '4 grams' is the GAM of helium
■ Once we understand the basics, there will be no need to write the steps. Just write the atomic mass (with out 'u') and write 'grams' on the right side.
• So now we know how to write the GAM of any given element.
• Also we know the speciality of GAM. The speciality is 'same number of atoms'.
Let us see some examples:
1. GAM of nitrogen = 14 grams• 14 grams of nitrogen contains 6.022×1023 atoms of nitrogen
♦ That is., 14 grams of nitrogen contains NA atoms of nitrogen
2. GAM of hydrogen = 1 grams
• 1 gram of hydrogen contains 6.022×1023 atoms of hydrogen
♦ That is., 1 gram of hydrogen contains NA atoms of hydrogen
3. GAM of chlorine = 35.5 grams
• 35.5 grams of chlorine contains 6.022×1023 atoms of chlorine
♦ That is., 35.5 grams of chlorine contains NA atoms of chlorine
This GAM gives us an easy method to find the 'total number of atoms in a sample'. We will see it in the next section
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