Chemistry

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Empirical Formula and Molecular Formula

Elemental analysis of a compound gives the mass percentage of atoms present in the compound. Using the mass percentage, we can determine the empirical formula of the compound. Molecular formula of the compound can be arrived at from the empirical formula using the molar mass of the compound.

Empirical formula of a compound is the formula written with the simplest ratio of the number of different atoms present in one molecule of the compound as subscript to the atomic symbol. Molecular formula of a compound is the formula written with the actual number of different atoms present in one molecule as a subscript to the atomic symbol.

Let us understand the empirical formula by considering acetic acid as an example.

The molecular formula of acetic acid is C₂H₄O₂

The ratio of C : H : O is 1 : 2 : 1 and hence the empirical formula is CH₂O.

Determination of Empirical Formula from Elemental Analysis Data:

Step 1:

Since the composition is expressed in percentage, we can consider the total mass of the compound as 100 g and the percentage values of individual elements as mass in grams.

Step 2:

Divide the mass of each element by its atomic mass. This gives the relative number of moles of various elements in the compound.

Step 3:

Divide the value of relative number of moles obtained in the step 2 by the smallest number of them to get the simplest ratio.

Step 4:

(only if necessary) in case the simplest ratios obtained in the step 3 are not whole numbers then they may be converted into whole number by multiplying by a suitable smallest number.

Example:

1. An acid found in Tamarind on analysis shows the following percentage composition: 32 % Carbon; 4 % Hydrogen; 64 % Oxygen. Find the empirical formula of the compound.

The empirical formula is C₂H₃O₃

2. An organic compound present in vinegar has 40 % carbon, 6.6 % hydrogen and 53.4 % oxygen. Find the empirical formula of the compound.

The empirical formula is CH₂O

Molecular formula of a compound is a whole number multiple of the empirical formula. The whole number can be calculated from the molar mass of the compound using the following expression

Calculation of Molecular Formula from Empirical Formula:

Let us understand the calculations of molecular formula from the following example.

Two organic compounds, one present in vinegar (molar mass: 60 g mol^-1), another one present in sour milk (molar mass: 90 g mol^-1) have the following mass percentage composition. C-40%, H-6.6%; O-53.4%. Find their molecular formula.

Since both compounds have same mass percentage composition, their empirical formula are the same as worked out in the example problem no 2. Empirical formula is CH₂O. Calculated empirical formula mass (CH₂O) = 12 + (2×1) + 16 = 30 g mol^-1.

Formula for the compound present in vinegar

∴ Molecular formula = (CH₂O)₂
= C₂H₄O₂ (acitic acid)

Caluculation of molecular formula for the compound present in sour milk.

Molecular formula = (CH₂O)₃
= C₃H₆O₃ (lactic acid)

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Gram Equivalent Mass | Formula, Definition, Diagrams

Similar to mole concept gram equivalent concept is also widely used in chemistry especially in analytical chemistry. In the previous section, we have understood that mole concept is based on molecular mass. Similarly gram equivalent concept is based on equivalent mass.

Definition:

Gram equivalent mass is defined as the mass of an element (compound or ion) that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.

Consider the following reaction:

Zn + H₂SO₄ → ZnSO₄ + H₂

In this reaction 1 mole of zinc (i.e. 65.38 g) displaces one mole of hydrogen molecule (2.016 g).

Mass of zinc required to displace 1.008 g hydrogen is

= \(\frac{65.38}{2.016}\) × 1.008
= \(\frac{65.38}{2}\)

The equivalent mass of zinc = 32.69

The gram equivalent mass of zinc = 32.69 g eq^-1

Equivalent mass has no unit but gram equivalent mass has the unit g eq^-1

It is not always possible to apply the above mentioned definition which is based on three references namely hydrogen, oxygen and chlorine, because we can not conceive of reactions involving only with those three references. Therefore, a more useful expression used to calculate gram equivalent mass is given below.

On the basis of the above expression, let us classify chemical entities and find out the formula for calculating equivalent mass in the table below.

Equivalent Mass of Acids, Bases, Salts, Oxidising Agents and Reducing Agents

Mole concept requires a balanced chemical reaction to find out the amount of reactants involved in the chemical reaction while gram equivalent concept does not require the same. We prefer to use mole concept for non-redox reactions and gram equivalent concept for redox reactions.

For example,

If we know the equivalent mass of KMnO₄ and anhydrous ferrous sulphate, without writing balanced chemical reaction we can straightaway say that 31.6 g of KMnO₄ reacts with 152 g of FeSO₄ using gram equivalent concept.

The same can also be explained on the basis of mole concept. The balanced chemical equation for the above mentioned reaction is

10FeSO₄ + 2KMnO₄ + 8H₂SO₄
↓
K₂SO₄ + 2 MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O

i.e. 2 moles (2 × 158 = 316 g) of potassium permanganate reacts with 10 moles (10 × 152 = 1520 g) of anhydrous ferrous sulphate.

∴ 31.6 g KMnO₄ reacts with

\(\frac{1520}{316}\) × 31.6 = 152 g of FeSO₄

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Avogadro’s number | Definition & Units

The total number of entities present in one mole of any substance is equal to 6.022 × 10²³. This number is called Avogadro number which is named after the Italian physicist Amedeo Avogadro who proposed that equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules. Avogadro number does not have any unit.

In a chemical reaction, atoms or molecules react in a specific ratio. Let us consider the following examples

Reaction 1:

C + O₂ → CO₂

Reaction 2:

CH₄ + 2O₂ → CO₂ + 2 H₂O

In the first reaction, one carbon atom reacts with one oxygen molecule to give one carbon dioxide molecule. In the second reaction, one molecule of methane burns with two molecules of oxygen to give one molecule of carbon dioxide and two molecules of water.

It is clear that the ratio of reactants is based on the number of molecules. Even though the ratio is based on the number of molecules it is practically difficult to count the number of molecules.

Because of this reason it is beneficial to use ‘mole’ concept rather than the actual number of molecules to quantify the reactants and the products. We can explain the first reaction as one mole of carbon reacts with one mole of oxygen to give one mole of carbon dioxide and the second reaction as one mole of methane burns with two moles of oxygen to give one mole of carbon dioxide and two moles of water. When only atoms are involved, scientists also use the term one gram atom instead of one mole.

Molar Mass:

Molar mass is defined as the mass of one mole of a substance. The molar mass of a compound is equal to the sum of the relative atomic masses of its constituents expressed in g mol^-1.

Examples:

• Relative atomic mass of one hydrogen atom = 1.008 u
• Molar mass of hydrogen atom = 1.008 g mol^-1
• Relative molecular mass of glucose = 180 u
• Molar mass of glucose = 180 g mol^-1

Molar Volume:

The volume occupied by one mole of any substance in the gaseous state at a given temperature and pressure is called molar volume.

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Mole Concept – What is a Mole? Related Formulae and Examples

Often we use special names to express the quantity of individual items for our convenience. For example, a dozen roses means 12 roses and one quire paper means 24 single sheets. We can extend this analogy to understand the concept of mole that is used for quantifying atoms and molecules in chemistry. Mole is the SI unit to represent a specific amount of a substance.

To understand the mole concept, let us calculate the total number of atoms present in 12 g of carbon – 12 isotope or molecules in 158.03 g of potassium permanganate, 294.18 g of potassium dichromate and 180 g of glucose.

Table 1.2 Calculation of Number of Entities in One Mole of Substance.

From the calculations we come to know that 12 g of carbon-12 contains 6.022 × 10²³ carbon atoms and same numbers of molecules are present in 158.03 g of potassium permanganate and 294.18 g of potassium dichromate. Similar to the way we use the term ‘dozen’ to represent 12 entities, we can use the term ‘mole’ to represent 6.022 × 10²³ (atoms or molecules or ions)

One mole is the amount of substance of a system, which contains as many elementary particles as there are atoms in 12 g of carbon-12 isotope. The elementary particles can be molecules, atoms, ions, electrons or any other specified particles.

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Molar Mass – Definition, Formula

Similar to relative atomic mass, relative molecular mass is defined as the ratio of the mass of a molecule to the unified atomic mass unit. The relative molecular mass of any compound can be calculated by adding the relative atomic masses of its constituent atoms.

For example,

(i) Relative molecular mass of hydrogen molecule (H₂)
= 2 × (relative atomic mass of hydrogen atom)
= 2 × 1.008 u
= 2.016 u.

(ii) Relative molecular mass of glucose (C₆H₁₂O₆)
= (6 × 12) + ( 12 × 1.008) + (6 × 16)
= 72 + 12.096 + 96
= 180.096 u

Relative Atomic Masses of Some Elements