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Mole Unit

Since it is not possible to calculate the weight of particles individually, a collection of such particles called mole is taken for all practical purposes. Avogadro discovered that under standard conditions of temperature and pressure, (1atm and 273K) a sample of gas occupies a volume of 22.4 L.

Avogadro Number

It was discovered that the number of atoms present in 12g of carbon is 6.023 x 1023 atoms. This is referred to as Avogadro number.

A mole of a gas is the amount of a substance containing 6.023 x 1023 particles.

Avogadro number is the number of atoms present in C12 isotope = 12g of C.

One mole of any gas at STP will have a volume of 22.4 L.

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Standard Temperature and Pressure

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Volumes of gases change with changes of temperature and pressure. Therefore standard temperature and pressure (abbreviated STP) have been chosen of comparing volumes of gases. Standard temperature is 0oC and standard pressure is 1 atm or 760mm. Volumes of gases are corrected to STP prior to most stoichiometric calculations by performing a combined Boyle's law-Charles law calculation.

The number of moles of gas is required to perform stoichiometric calculation using the mole concept. It has been found experimentally that one mole of any ideal gas at STP occupies 22.4 liters. Thus 1.00 mole H2 (2.02 grams) at STP occupies 22.4 liters; 0.500 mole H2 (1.01 gram) at STP occupies 11.2 liters.

Example Problem

Example 1: A sample of hydrogen gas has a volume of 1.10 liters at -40oC and 0.520atm. What will be its volume at STP.

Convert the required temperature to K.

K = C + 273 = -40 + 273 = 273K (T1)

K = C + 273 = 0 + 273 = 273K (T2)

P1 = 0.520atm, V1 = 1.10 liters

P2 = 1.00atm, V2 = ?

Using the combined gas equation

P1V1 / T1 = P2V2 / T2

Then V2 (Volume of H2 at STP) = (1.10 liters) {(0.520atm / 1.00atm) (273K / 273K)}

Ans: V2 = 0.670 liters

Example 2: Calculate the number of moles of hydrogen collected in the above example problem.

The volume of H2 at STP is 0.670 liters. Determine the number of moles of H2 by dividing the volume by the molar volume of an ideal gas.

Number of moles of H2 = {0.670 liters / (22.4 liter / mole)}

Ans: Number of moles of H2 = 0.0299 moles of H2

Avogadro Law

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The relationship between the mass of gas present and its volume is described by Avogadro law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Thus if the temperature, pressure and volume of two gases are the same then the two gases contain the same number of molecules regardless of their identity.

The actual temperature and pressure at which we compare two or more gases do not matter. It is convenient however to select one temperature and one pressure as standard and chemists have chosen 1atm as the standard pressure and 0oC as the standard temperature. These conditions are called Standard temperature and pressure (STP).

Avogadro law: Equal volumes of gases measured at the same temperature and pressure contain equal number of molecules.
Avogadro law allows us to write a gas that is valid not only for any pressure,volume and temperature but also for any quantity of gas. This law called the ideal gas law expressed as

PV = nRT

P = Pressure of the gas in atmosphere (atm)
V = Volume of gas in liters (L)
n = Amount of gas in moles (mol)
T = Temperature of gas in kelvin (K)
R = Ideal gas constant

Mole Unit Problems

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Problem 1: In the equation 2C2H6 + 7O2 4CO2 + 6H2O, how many molecules of C2H6 are needed to produce 27.0 grams of H2O?Solution:

The solution map for this problem is given below.
The problem is solved by factor label method (FLV)

Problem 2: In the equation 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l) how many liters of NH3(g) at STP are needed to react with 200 liters of O2(g) at STP?Solution:

To solve this problem we need to convert between volume and numbers of moles. Our solution map is

The solution is given below

Volume problems are particularly easy to solve because the factor 22.4 always cancels in the calculations.

Problem 3: How many grams of O2 are required to burn completely 114g of octane.Solution:

The combustion reaction is written as follows

Molecular weight of octane = 114g
Number of moles of octane = 114 / 114 = 1 mole.
According to the equation one mole of octane reacts with 25 / 2 moles of O2.
Weight of O2 = Number of moles x Molecular weight
= (25 / 2) x 32
= 400g

Gram molar volume

The volume occupied by one mole of a gas at STP is known as its gram molar volume.
1 mole of gas = 6.023 x 1023 molecules = 22.4 L in STP
Moles of a gas = Volume of the gas at STP in L / 22.4 L

Problem 4: Calculate the volume of O2 at STP produced by heating 12,25g of KClO3Solution:

Molecular weight of KClO3 = 122.5g

2KClO3 2KCl + 3O2

Number of moles of KClO3 taken = 12.25 / 122.5 = 1/10 = 0.1mole
According to the equation
2 moles of KClO3 = 3 moles of O2 = 3 x 22.4 L of O2 at STP
0.1 mole of KClO3 = {(3 x 0.1) / 2} moles of O2
= {(3 x 0.1) x 22.4 / 2} L of O2 at STP
= 3.36 L of O2 at STP

Problem 5: Calculate the amount of water formed when a mixture of 1g H2 and 2g O2 is sparked.Solution:

1 gm H2 = 1 / 2 mole H2; 2gm O2 = 2 / 32 mole O2 or 1 / 16 mole O2

Now the equation for the formation of water is

2H2 + O2 2H2O
(2mole) (1mole) (2mole)

From the above equation 1 mole H2 requires 1 / 2 mole oxygen therefore 1 / 2 mole H2 requires 1 / 4 mole O2 for complete conversion to water. Since number of moles of O2 available is much less therefore all of the oxygen will be converted to water but not all of the hydrogen.

Therefore from the equation 1 / 16 mole O2 will produce 2 / 16 mole H2O, that is

(2 x 18) / 16
(or) 2.25 gm H2O

Problem 6: 56ml of NH3 at STP is passed over heated cupric oxide to produce nitrogen. How many grams of magnesium nitride can be obtained from the liberated N2.Solution:

The reaction are

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