__Formulas in brief__* –*

**Mass = Molar mass x Number of moles**

**Number of moles = Given number of particles / Avogadro number****Number of particles = (Given mass / molar mass) x Avogadro number**(from 1 and 2)- If one carbon atom has a mass of 12 atomic mass units and one magnesium atom has a mass of 24 atomic mass units, then as a magnesium atom is twice as heavy as a carbon atom. It follows that this ratio will be maintained for any number of atoms.
- Equal masses of carbon and magnesium contain different numbers of atoms.

6 g of carbon contains 6 / 12 moles of carbon = 0.5 moles

6 g of magnesium contains 6 / 24 moles of magnesium =0.25 moles

- 1 mole = 6.022 x 10
^{23}atoms or molecules or formula units of that substance.

1 mole of water (H_{2}O)contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms. It is a simple matter of multiplying the moles of the compound by the atoms or ions that make it up.

**Read more –**

What is Matter? What is its Characteristics?

States of Matter

Law of Conservation of Mass

Problems based on Law of Conservation of Mass

Law of Definite Proportions / Constant Composition

Problems based on law of definite proportions or constant composition

Mole Concept – Importance and formulas

Quiz – Mole Concept and Stoichiometry – 1

** Problem based on formula no. – 1
Mass = Molar mass x Number of moles
**Calculation of mass from mole of any

*fundamental unit like atom, molecule*and vice – versa.

**Note that if it is not mention atom or molecule before mole, it always means one mole of that substance in its natural form. **

** **

**Example 1- **How many grams are there in 5.5 mol of sulphur?

**Solution – **For converting mass into mole and vice versa, we always need the molar mass.

Molar mass of sulphur = 32.0 g mol^{–1}

Number of moles of sulphur = 5.50 mol** **

Therefore, mass of sulphur (in grams) = molar mass x number of moles

= 5.50 mol × 32.0 *g* mol^{–1}= **176.0 g sulphur**

**Example 2- **Calculate number of moles present in 64 g of oxygen.

**Solution – **Molar mass of oxygen = 32 g mol^{–1}

Oxygen in natural form will be molecular oxygen, O2

Therefore, number of moles of oxygen = 64 g / 32 g mol^{–1} = **2.0 moles**

** **

**Example 3- **Calculate the number of moles present in 108 g of aluminium.

**Solution – ** Atomic mass of Al = 27gm

So, 27g of aluminium = 1 mole of aluminium

Or, by using formula , Number of moles = given mass / atomic mass

Or, Number of moles = 1/27 x 108 = ** **4 moles of aluminium** **

Thus, 108g of aluminium = **4 moles of aluminium **

**Example 4 – **Calculate the mass of 0.5 mole of iron.

**Solution – **Atomic mass of iron = 55.9 g

Mass of the 1 mole of iron = 55.9 g

Or, by using formula, mass = atomic mass x number of moles

Mass of the 0.5 mole of iron = 55.9 x 0.5 g = **27.95 g**

**Now, Check your understanding –**

**1. Atomic number (Z) of Iron, Fe is 26, Atomic mass 56.**

Using the above information answer the following-

(a) 1 mole of iron weighs __________ grams.

(b) 7.4 moles of iron weighs ________________ grams.

2. 1 litre of an aqueous solution of NaCl contains 117g of the dissolved salt. Calculate the number of moles of the solute in the solution. (Na = 23u , Cl = 35.5)

3. Two bottles A and B contain 1000 ml. water solutions of 60g NaOH and 117g NaCl respectively. Which bottle has more concentrated solution?

** Problem based on formula no. – 2
Number of moles = Given number of particles / Avogadro number
**Calculation of mole from number of particles and vice – versa.

**Example 5 – **Calculate the number of moles of 24.088 x 10^{23} numbers of sodium atom.

**Solution – **Given number of particles = 24.088 x 10^{23}

Avogadro number = 6.022 x 10^{23}

Number of moles = Given number of particles / Avogadro number**
= **24.088 x 10

^{23}/ 6.022 x 10

^{23}

** = 4 moles**

Example 6 – Calculate the number of particles in 0.5 moles of N atom.

**Solution – ** Number of moles = 0.5 moles

Avogadro number = 6.022 x 10^{23}

Number of particles = Number of moles x Avogadro number**
= **0.5 x 6.022 x 10

^{23}

** = 3.011 x 10**^{23}

**Now, Check your understanding –**

**Atomic number (Z) of Iron, Fe is 26, Atomic mass 56.**

Using the above information answer the following-

(a) 7.4 moles of iron contains ________________ atoms.

(b) 1.2 x 10^{24 }atoms of iron would be _____________ moles of iron.

*Problem based on formula no. – 3*

**Number of particles = (Given mass / molar mass) x Avogadro number
**Calculation of number of particles from mass and vice – versa.

**Example 7 – **Calculate the number of molecules in 22 g of *CO*_{2}

**Solution – ** Gram molecular mass or molar mass of CO_{2} = 44 g

Number of particles = (Given mass / molar mass) x Avogadro number = ( 22 g / 44 g ) x 6.022 x 10^{23} = **3.011 x 10**^{23}** ****molecules**

**Example 8 – **Calculate the mass of 18.066 x 10^{23 }molecules of SO_{2}

**Solution – ** Gram molecular mass or molar mass of SO_{2} = 64 g

Mass = (Number of particles x molar mass) / Avogadro number

= (18.066 x 10^{23} x 64) / 6.022 x 10^{23}

** = 192 g**

**Example 9 –** Calculate the mass of glucose in 2 x 10^{24} molecules.

**Solution – ** Gram molecular mass or molar mass of glucose = 180 g

** **Mass = (Number of particles x molar mass) / Avogadro number

= (2 x 10^{24} x 180) / 6.022 x 10^{23}

** = 597.8 g
**

**Example 10 –**How many atoms are there in 24 g of carbon?

**Solution – ** 24 g of carbon = 24 / 12 moles = 2 moles

1 mole of atoms = 6.022 x 10^{23}

Therefore 2 moles of carbon contains 2 x 6.022 x 10^{23} atoms = **1.204 x 10**^{24}** atoms. **

** ****Now, Check your understanding –**

**Atomic number (Z) of Iron, Fe is 26, Atomic mass 56.**

Using the above information answer the following-

(a) 1 gram of iron contains ________________ atoms.

(b) 1 atom of iron weighs _________________ grams.

**Thus, One mole of oxygen stands for –
**1. 6.022 x 10

^{23}molecules of oxygen

2. 2 x 6.022 x 10

^{23}atoms of oxygen

3. 1 gram molecule of oxygen

4. 2 gram atoms of oxygen

5. 32 grams oxygen

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adminThanks. Please try – Quiz based on Mole Concept and Stoichiometry – 1

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