Let us consider if we provide a name for a quantity of things taken as a whole is common in everyday life. Some examples are dozen, ream and a gross. Each word represents a specific quantity of items and not dependent on the commodity. For example, a ream of paper always measures 500 identical sheet. Similarly a dozen of oranges and bananas always represents 12 items.
In chemistry we have a unit that describes a quantity of particles. It is called the mole. A mole is 6.023 $\times$ 10 The particles can be atoms, molecules, ions, electrons and so forth. Because particles are so small in chemistry the mole is a very convenient unit.^{23} particles. |

**The chemist counting unit is the mole.**What is unusual about the mole in chemistry is its magnitude. A mole is 6.02 $\times$ 10

^{23}objects. The extremely large size of the mole unit is necessitated by the extremely small size of atoms and molecules.

To the chemist, one mole always means 6.02 $\times$ 10

^{23}objects, just as one dozen means 12 objects. Two moles of objects is two times 6.02 $\times$ 10

^{23}objects, and five moles of objects is five times 6.02 $\times$ 10

^{23}.

The number 6.02 $\times$ 10

^{23}is often referred to as

**Avogadro's number**in honor of the Italian scientist whose hypothesis led to its determination.

**Definition of Mole:**

The mole is defined such that

**"a sample of a natural element with a mass equal to the element's atomic mass expressed in grams contains 1 mole of atoms."**

The mole chemistry definition in SI unit is the amount of a substance that contains as many entities as there are in exactly 0.012kg of carbon-12. For our purpose it is most convenient to define the mole as the number equal to the number of carbon atoms in exactly 12 grams of pure 12C.

**Mole Ratio Definition**:

The ratio between the amounts in moles of reactants and the amounts in moles of products are called the mole ratios.

Mole is the formula weight of a substance expressed in grams. The mole equation is given below. **Mole = $\frac{grams}{molecular\ mass}$**

**Mole formula = 6.02 $\times$ 10**

^{23}formula unitsIn mole formula chemistry the formula for mole is given by

**One mole of a compound = Relative formula mass in grams**

The mole is a unit the measures amount of substance in such a way that equal amounts of elements consist of equal number of atoms. The ratio in which individual atoms combine to form molecules are exactly the same as the ratios in which moles of these atoms combine.

The formula of a compound implies a mole ratio formula. It can be used to determine the number of moles, the masses and the percent composition of each element in a compound.

The formula of a compound implies a mole ratio formula. It can be used to determine the number of moles, the masses and the percent composition of each element in a compound.

Mole calculations starts by determining the amount in moles of the solution for which we know the molar concentration and have measured the volume used.

The conversion from "moles" of one substance to "moles" of another substance in a reaction will be

The stoichiometry mole to mole problem is given below.

The conversion from "moles" of one substance to "moles" of another substance in a reaction will be

**governed by the values of the coefficients.**Mole to mole calculations (conversion) are one step calculations that are always based on balanced chemical equations.The stoichiometry mole to mole problem is given below.

### Solved Example

**Question:**How many moles of CO

_{2}are produced in the combustion of 0.750 mole of hexane C

_{6}H

_{14}?

The equation is 2C

_{6}H

_{14}(l) + 19O

_{2}(g) $\rightarrow$ 12CO

_{2}(g) + 14H

_{2}O(l)

**Solution:**

First make certain equation is balanced. It is in this case. Assembling the facts.

Coefficient of CO

_{2}= 12

Coefficient of C

_{6}H

_{14}= 3

Moles of C

_{6}H

_{14}= 0.750 mole

Moles CO

_{2 }= $\frac{12\ moles\ CO_{2}}{2\ moles\ C_{6}H_{14}} \times [0.750\ mole C_{6}H_{14}]$ = 4.50 moles CO

_{2}

Combustion of 0.750 mole of C

_{6}H

_{14}produces 4.50 moles of CO

_{2}.

The example problems shown below explains the conversion of moles, masses and volumes.

### Solved Example

**Question:**Consider the reaction

4NH

_{3}(g) + 5O

_{2}(g) $\rightarrow$ 4NO(g) + 6H

_{2}O(g)

How many moles of water are formed in this reaction?

**Solution:**

Coefficient of H

_{2}O = 6

Coefficient of 1.35 moles of O

_{2}= 5

Moles of H

_{2}O = $\frac{6\ moles H_{2}O}{5\ moles\ O_{2}} \times [1.35\ moles\ O_{2}]$

= 1.62 moles of water.

The molar mass of an element is the mass in grams of 1 mole of that element. One molar mass of hydrogen is 1.008g of hydrogen. One molar mass of nitrogen is 14.01g of nitrogen. The term molar mass also applies to compounds.

One mole of an element is an amount of that element equal to its molar mass (an amount equal to its atomic mass in grams).

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One mole of an ideal gas occupies the same volume under the same conditions of temperature and pressure. This assertion leads to the idea of the molar volume, the volume per mole of any gas under stated conditions.One mole of an element is an amount of that element equal to its molar mass (an amount equal to its atomic mass in grams).

The numerical value of molar volume depends on the temperature and pressure of the gas. Doubling the number of molecules present doubles the volume of the system.

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Below are the examples for chemistry mole problems given which shows how to solve mole problems, mole conversion problems and mole to mole problems.

Mole to mole problems are nothing but the quantity of starting substance in moles and the quantity of desired substance is requested in moles.

Mole to mole problems are nothing but the quantity of starting substance in moles and the quantity of desired substance is requested in moles.

### Solved Examples

**Question 1:**How many moles of carbon dioxide will be produced by the complete reaction of 2.0 mol of glucose (C

_{6}H

_{12}O

_{6}) according to the following equation?

C

_{6}H

_{12}O

_{6}+ 6O

_{2}$\rightarrow$ 6CO

_{2}+ 6H

_{2}O

**Solution:**

**C**

_{6}H_{12}O_{6}+ 6O_{2}$\rightarrow$ 6CO_{2}+ 6H_{2}OThe balanced equation states that 6 mol of CO

_{2}will be produced from 1 mol C

_{6}H

_{12}O

_{6}.

moles of C

_{6}H

_{12}O

_{6}= moles of CO

_{2 }

The mole ratio from the balanced equation is $\frac{6mol\ CO_{2}}{1mol\ C_{6}H_{12}O_{6}}$

(2.0 mol of C

_{6}H

_{12}O

_{6})[$\frac{6mol\ CO_{2}}{1mol\ C_{6}H_{12}O_{6}}$]

**= 12 mol of CO**

_{2 }**Question 2:**Find the mass of 1 mole of potassium alum KAl(SO

_{4})2.12H

_{2}O?

**Solution:**

1 K = 39

1 Al = 27

2 (SO

_{4}) = 192 = 2 (32 + 16 $\times$ 4)

12 H

_{2}O = 216 = 12 (2 + 16)

Therefore 1 mole of potassium alum is = 474g