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**Atom and Atomic Weight**

- The atom is the
**smallest**particle that can not be divided into its constituents.

- The smallest particle of an element that takes part in a chemical reaction is called an
**atom**. - An atom is so minute that it cannot be detected even with the most powerful microscope, let alone placed on a balance pan and weighed. So, there is no question of determining the absolute weight of an atom.
- So, chemists decided to determine the relative masses of atoms (i.e., how many times one atom of an element is heavier than another).
**Hydrogen****atom**was first selected as standard.- The atomic weight of an element is:
- When we state that the atomic weight of chlorine is 35.5, we mean that an atom of chlorine is 35.5 times heavier than an atom of hydrogen. It was later felt that the standard for reference for atomic weight may be oxygen, the advantage being that the atomic weights of most other elements became close to whole numbers.
- The atomic weight of an element is:
- The modern reference standard for atomic weights is the
**carbon isotope**of mass number 12. - The atomic weight of an element is:
- On this basis, atomic weight of oxygen 16 was changed to 15.9994.
- Nowadays atomic weight is called
**relative atomic mass**and denoted by**amu**(atomic mass unit). The standard for atomic mass is C^{12}. - Atomic weight is not a weight but a number. Atomic weight is not absolute but relative to the weight of the standard
**reference****element**(C^{12}). **Gram****atomic****weight**is atomic weight expressed in grams, but it has a special significance with reference to a mole.- One
**atomic mass unit**is defined as a mass exactly equal to one-twelfth of the mass of one carbon-12 atom and 1 amu = 1.66056 × 10^{–24}g

__Relationship between gram and amu:__

⇒ 1 amu = 1/12 wt of one C-12 atom

⇒ For C, 1 mole C = 12 gm = 6.023 × 10^{23} atoms or wt. of 6.023 × 10^{23} atoms = 12 gm

⇒ wt. of 1 atom of C = 12/N_{A} gm (NA → Avogadro's number = 6.23 × 10^{23})

⇒ 1 amu = 1/12 wt of one C-12 atom

⇒ 1 amu = 1/12 × 12/N_{A }gm

⇒ 1 amu = 1/N_{A} g

So, **1 amu = 1.66 × 10 ^{-24} g**

Today, ‘amu’ has been replaced by ‘u’ which is known as

When we use atomic masses of elements in calculations, we actually use average atomic masses of elements which are explained below.

Mass of an atom of hydrogen = 1.6736 × 10

- Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (percent occurrence), the average atomic mass of that element can be computed as:

(exact weight of isotope #1) * (% abundance of isotope #1) + (exact weight of isotope #2) * (abundance of isotope #2) = Average atomic weight of the element**Example:****Carbon has the following three isotopes with relative abundances and masses as shown against each of them.**

From the above data, the average atomic mass of carbon will come out to be:(0.98892) (12 u) + (0.01108) (13.00335 u) + (2 ×10^{–12}) (14.00317 u) =**12.011 u** **Elements**are found in different isotopic forms (atoms of same elements having different atomic mass), so the atomic mass of any element is the average of all the isotopic mass within a given sample.

**Solved Examples**

**Example 1.** **Use the date given in the following table to calculate the molar mass of naturally occurring argon.**

**Solution.** Molar mass of Ar = 35.96755 × 0.071 + 37.96272 × 0.163 + 39.96924 × 0.766

= 39.352 g mol^{−1}**Example 2. Carbon occurs in nature as a mixture of carbon 12 and carbon 13. The average atomic mass of carbon is 12.011. What is the percentage abundance of carbon 12 in nature?****Solution. **Average atomic mass is:

1201.1 = 12x + 1300 – 13x

x = 1300 – 1201.1 = 98.9%**Example 3. Nitrogen is made up of two isotopes, N-14 and N-15. Given nitrogen's atomic weight of 14.007, what is the percent abundance of each isotope?****Solution.** (14.003074) (x) + (15.000108) (1 - x) = 14.007

Notice that the abundance of N-14 is assigned 'x' and the N-15 is 'one minus x.' This is the "trick" refered to above. The two abundances always add up to one (or, if you prefer, 100%)__For example, I might have this:__

(14) (x) + (15) (1 - x) = 14.007

14x + 15 - 15x = 14.007

x = 15 - 14.007 = 0.993 & 1 - x = 0.007**Example 4. Copper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu). Given copper's atomic weight of 63.546, what is the percent abundance of each isotope?****Solution.** (62.9296) (x) + (64.9278) (1 - x) = 63.546

Once again, notice that 'x' and 'one minus x' add up to one.__Now solve for x:__

x = 0.6915 (the decimal abundance for Cu-63)

**Dulong and Petit Law **

- They measured the specific heat of a number of metals and found that the product of the specific heat and the atomic weight is a constant, having an approximate value of 6.4.
**Specific heat (cal/g-deg) × atomic weight (g/g–atom) = 6.4 (cal/deg.g.atom)** - This correlation has been used to ‘correct’ the atomic weights of some elements in the periodic table. Dulong and Petit’s law is applicable only to
**metals.**

- Avogadro (1811) suggested that the fundamental chemical unit is not an atom but a molecule, which may be a cluster of atoms held together in some manner causing them to exist as a unit. The term
**molecule**means the smallest particle of an element or a compound that can exist free and retain all its properties. - The
**molecular mass**of a substance is an additive property and can be calculated by adding the atomic masses of all the atoms of different elements present in one molecule.

mic masse Relative Molar Mass or relative Molar Mass is defined as the smallest mass unit of a compound with one twelfth of the mass of one carbon – 12 atom.

**Molecular Mass of Sugar (Sucrose)**

Molecular mass C_{12}H_{22}O_{11} = 12(mass of C) + 22(mass of H) + 11(mass of O)

Molecular mass C_{12}H_{22}O_{11} = 12(12) + 22(1) + 11(16)

Molecular mass C_{12}H_{22}O_{11} = 342g/mol__One can use the following methods to find the Molecular weight:__**(a) Vapour Density Method**

Molecular mass = 2 × V.D.**(b) Diffusion Method:** According to Graham’s law of diffusion, rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

**(c) Victor Meyer Method:** This method applies to volatile organic liquids. Suppose vapour of an organic liquid having mass W g occupies a volume of V mL at STP. Then its molecular mass is:

- Some substances such as sodium chloride do not contain
**discrete molecules**as their constituent units. In such compounds, positive (sodium) and negative (chloride) entities are arranged in a three-dimensional structure. It may be noted that in sodium chloride, one Na^{+}is surrounded by six Cl^{–}and vice-versa. - The formula such as NaCl is used to calculate the formula mass instead of molecular mass as in the solid-state sodium chloride does not exist as a single entity.

Thus, formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine

= 23.0 u + 35.5 u

= 58.5 u

Try yourself:Given that the abundances of isotopes ^{54}Fe, ^{56}Fe and ^{57}Fe are 5%, 90% and 5% respectively, the atomic mass of Fe is:

View Solution

Try yourself:The percentage of naturally isotopes ^{35}CI and ^{37}CI that accounts for atomic mass of chlorine as 35.45 is:

View Solution

Try yourself:12 g of an alkaline earth metal gave 14.8 g of its nitride Atomic weight of that metal is:

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Try yourself:The specific heat of a metal is 0.11 and its equivalent weight is 18.61 its exact atomic weight is:

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Try yourself:The sulphate of metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO_{4.} 7 H_{2}O. The atomic of M is:

View Solution

**Mole Concept**

- A mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.
**One mole**is defined as the amount of substance containing the same number of**discrete entities**(atoms, molecules, ions, etc.) as the number of atoms in a sample of pure 12 C weighing exactly 12 g.C-12 Isotope- In order to determine this number precisely, the mass of a carbon-12 atom was determined by a
**mass spectrometer**and found to be equal to 1.992648 × 10^{–23}g. Knowing that one mole of carbon weighs 12 g, the number of atoms in it: - The number of entities composing a mole has been experimentally determined to be 6.0221367 × 10
^{23}, a fundamental constant named**Avogadro’s number**(NA) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. - This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being
**6.022×10**^{23}/mol. - It may be emphasized that the mole of a substance always contains the same number of entities, no matter what the substance may be.

1 mole = collection of 6.02 × 10^{23}species

6.02 × 10^{23}= N_{A}= Avogadro's No. - 1 mole of atoms is also termed as
**1 gm-atom**, 1 mole of ions is termed as**1 gm-ion**and 1 mole of a molecule termed as**1 gm-molecule**.

**(a)** If no. of some species is given, then no. of moles is:**(b)** If the weight of a given species is given, then no of moles is:

or **(c)** If the volume of a gas is given along with its temperature (T) and pressure (P)

use:

where R = 0.0821 liter-atm/mol-K (when P is in atmosphere and V is in litre.)

Note:1 mole of any gas at STP (0°C & 1 bar) occupies

22.7 litres.

1 mole of any gas at STP (0°C & 1 atm) occupies22.4 litres.

__A mole of any substance is related to:__

- Number of particles
- Mass of a substance
- Volume of the gaseous substance

- A mole is a collection of 6.023 × 10
^{23}particles, ions, atoms etc. - Avogadro Number (NA): The number of carbon atoms present in one gram-atom (1-mole atom) of C-12 isotope is called
**Avogadro’s****number**. - One gram-atom (12 grams) of C–12 contains
**6.02 × 10**^{23}atoms. Thus the numerical value of Avogadro’s number (NA) is 6.02 × 10^{23}per mol. - It should be noted that 1 a.m.u.= 1/12
^{th}of mass of a Carbon-12 atom.**(i)**6.023 × 10^{23}atoms of Na constitute one-mole atom of Na.**(ii)**6.023 × 10^{23}molecules of oxygen constitute 1 mole of oxygen molecules.**(iii)**6.023 × 10^{23}electrons constitute one mole of electrons.**(iv)**No. of moles = Number of particles/6.023 × 10^{23 }

- One mole of every substance weighs equal to the gram atomic weight of the substance or to the gram molecular weight of the substance.
**Example:****(i)**1 mole of sodium weighs 23 g of Na.**(ii)**1 mole of CaCO_{3}weighs 100 g.

Number of moles = Mass of substance in grams/Mass of weight in grams

- One mole of every gas occupies 22.4 lit. of volume at STP i.e. 1 mole of O
_{2}occupies 22400 ml of volume at STP. - 1 mole of He occupies 22400 ml of volume at STP.
- Number of moles = 'V' of gas in litres at STP/22.4Mole Concept

- For 'n' mole of a compound (C
_{3}H_{7}O_{2}) - Moles of C = 3n
- Moles of H = 7n
- Moles of O = 2n

__Density is of two types:__

**(a) **Absolute density**(b)** Relative density

- It is defined only for gas. It is a density of gas with respect to
**H**_{2}gas at the same temp & pressure. - The density of Cl
_{2}gas with respect to O_{2}gas is:

**Solved Examples**

**Example 1. Find the weight of water present in 1.61 g of Na _{2}SO_{4}. **

Moles of water = 10 × moles of Na

wt. of water = 0.5 × 18 = 0.9 gm

= 234.18 / N

Number of moles = Number of particles/6.023 × 10

**Example 4. What will be the number of moles in 13.5 g of SO _{2}Cl_{2}?**

**Example 5. What will be the number of moles of oxygen in one litre of air containing 21% oxygen by volume under STP conditions?****Solution.** 100 ml of air at STP contains 21 ml of O_{2}

1000 ml of air at STP contains = 21/100 × 1000 = 210 ml O_{2}

∴ No. of moles = Volume in ml under STP conditions/22400 litre

= 210/22400 = 9.38 × 10^{-3 }mole.**Example 6. Calculate the mass of (i) an atom of silver (ii) a molecule of carbon dioxide.****Solution.** **(i)** 1 mole of Ag atoms = 108 g

(∴ Atomic mass of silver = 108u)

= 6.022 × 10^{23} atoms

6.022 × 10^{23} atoms of silver have mass = 108g

∴ Mass of one atom of silver

**(ii)** 1 mole of CO_{2} = 44 g

(Molecular mass of CO_{2} = 1 x 12 + 2 x 16 = 44u)

= 6.022 × 10^{23} molecules

Thus, 6.022 × 10^{23} molecules of CO_{2} has mass = 44 g

∴ 1 molecule of CO_{2} has mass

**Example 7. Calculate the number of molecules present ****(i) In 34.20 grams of cane sugar (C _{12}H_{22}O_{11}).**

[Molecular mass of cane sugar C

= 12 x 12 + 22 × 1 + 11 × 16 = 342 amu

= 6.022 × 10

Now 342 g of cane sugar contain 6.022 × 10

∴ 34.2 g of cane sugar will contain

**(ii)** 1 mole of water = 18 g = 6.022 × 10^{23} molecules.

Mass of 1 litre of water = Volume × density = 1000 × 1 = 1000 g

Now 18 g of water contains = 6.022 x 10^{23} molecules.

∴ 1000 g of water will contain

= 3.346 x 10^{25} molecules**(iii)** 1 mole of H_{2}O = 18 g = 6.022 × 10^{23} molecules.

Mass of 1 drop of water = 0.05 g

Now 18 g of H_{2}O contain = 6.022 × 10^{23} molecules.

∴ 0.05 g of H_{2}O will contain =**Example 8. Calculate the number of moles in each of the following:(i) 392 grams of sulphuric acid(ii) 44.8 litres of carbon dioxide at STP(iii) 6.022 × 10**

∵ Molecular mass of H_{2}SO_{4} = 2 × 1 + 32 + 4 × 16 = 98u)

Thus 98 g of H_{2}SO_{4} = 1 mole of H_{2}SO_{4}

∴ 392 g of H_{2}SO_{4}

**(ii) **1 mole of CO_{2} = 22.4 litres at STP

i.e. 22.4 litres of CO_{2} at STP = 1 mole

∴ 44.8 litres of CO_{2} at STP

= 1/22.4 × 44.8

= 2 moles CO_{2}

**(iii)** 1 mole of O_{2} molecules = 6.022 × 10^{23} molecules

6.022 × 10^{23} molecules = 1 mole of oxygen molecules

**(iv)** 1 mole of Al = 27 g of Al

(∴ Atomic mass of aluminium = 27 g)

i.e. 27 g of aluminium = 1 mole of Al

**(v)** 1 metric ton of Fe = 10^{3} kg = 10^{6} g

1 mole of Fe = 56 g of Fe

∴ 10^{6} g of Fe = 10^{6}/56 molecules

= 1.786 × 10^{4} moles

**(vi)** 7.9 mg of Ca = 7.9 × 10^{−3} g of Ca

**(vii)** 65.5 µg of C = 65.5 10^{-6} g of C

**Try Yourself!**

**Q.1. Find the weight of 6023 molecules of CO _{2} in grams. **

**Ans. **4.4 × 10^{–19} gm

**Q.2. Calculate the total number of electrons present in 1.6 g of methane. (Molecular wt. of methane = 16 a.m.u.)**

**Ans. **6.023 × 10^{23} electrons

**Q.3. 0.24 gm of a volatile substance upon vaporization gives 45 ml vapours at STP. What will be the vapour density of the substance?**

**Ans. **59.73**Q.4. What is the mass of carbon present in 0.5 mole of K _{4}[Fe(CN)_{6}]? **

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