We know that atoms are very small units that combine to form molecules. Because of their small size, atoms are not countable units. Therefore we must have some measurement criteria to measure the mass of an atom or molecule. We have to determine the number of atoms in a molecule to calculate the molecular mass. The first method was purposed by *John Dalton* and *Thomson* to determine the atomic weights of atoms. We cannot calculate the atomic weight directly by taking that atom, but we have to take the weight of an atom relative to some atom. For example; hydrogen is the lightest element with atomic weight 1.00. Similarly we can consider 1/16th of mass of 1 atom of O as a unit of atomic mass.

Today we are using the carbon element as standard element for the measurement of atomic mass whose atomic mass is considered as 12 amu. Hence, 1 atomic mass unit is equal to 1/12th of the mass of one atom of C-12. The unit of atomic mass is equal to the mass of one H-atom. The relative weights of H, C and Nare considered as 1.01, 12.01 and 14.01. We can check the atomic mass of other elements with respect to these standard elements.

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The atom is very small in size so the mass of an atom is very small. Now the atomic mass unit is represented as ‘u’.

**
Use below widget to calculate the molar mass**

Molarity =$\frac{`(Number\ of\ mol)}{(volume\ of\ solution)`}$

A molecular shows all the physical and chemical properties of that matter. Molecules can be obtained by same or different combination of elements. Molecule formula is related to empirical formula.**Molecule formula = n x empirical formula**

= 33.6g N_{2} x {(1 mole N_{2}) (28.02g N_{2}) x (2 mol NH_{3}) (1 mol N_{2})}

= 2.40 mol of NH_{3}

Studying chemistry is much easier once we learn to visualize what reaction equations are indicating symbolically. Some terminology required for this module is shown in the following diagram.

Stoichiometric coefficients are necessary to balance chemical equations. Balancing insures that equal numbers of atoms of each element are present in both the products and reactants. Otherwise the reaction would violate the law of conservation of mass indicating that new atoms have been made in the reaction rather than redistributing the atoms into new chemical compounds.

**Example**

1. Molar mass of NaCl

[22.989 + 35.453] $\times$ $\frac{1g}{mole}$ = $\frac{ 58.443g}{mole}$

2. Molar mass of Sugar C_{12}H_{22}O_{11}

([12 × 12.0107] + [22 ×1.008] + [11 ×15.9994]) $\times$ $\frac{1g}{mole}$ = $\times$ $\frac{342.29g}{mole}$

**Average Molar Mass of Mixtures**

The average molar mass of any mixture $\bar{M}$ may be calculated from the mole fractions x_{i} of the components and their molar masses M_{i}:

$\sum_{i}xiMi$
### Solved Examples

**Question 1: **How many kg of potassium are contained in 10 kg of KCl?

** Solution: **

**Question 2: **Find the mass% of Nitrogen in ammonia.

** Solution: **

**Question 3: **How many grams of KCl will contain 10 g of potassium?

** Solution: **

Capacity of mass equivalent to $\frac{1}{12}$ of the mass of an atom of carbon-12 is called atomic mass.

1 u = $\frac{mass\ of\ C^{12}_{6}\ atom}{12}$

Absolute mass of a is 1.9945 x 10^{-23} g

1 u = $\frac{1.9945 x 10^-23 g}{12}$

1 u = $\frac{ 1.66 x 10^-24 g}{12}$

1 u =1.66 x 10^{-27} kg.

Molarity =$\frac{`(Number\ of\ mol)}{(volume\ of\ solution)`}$

Representation of molarity is M. Molar can also be computed from the strength.

The following formula represents how to get molar from strength.

molarity =$\frac{(potential\ (gram\ or\ litre))}{(mol\ mass\ of\ the\ solute)}$

A molecular shows all the physical and chemical properties of that matter. Molecules can be obtained by same or different combination of elements. Molecule formula is related to empirical formula.

Where,

n = $\frac{(molecular\ mass)}{(empirical\ formula\ mass)}$

**Example for molecules**

**C _{2}H_{6}**

The molecule benzene consists of two carbon atoms and six hydrogen atoms. Here the molecule benzene is the combination of carbon and hydrogen elements.

**Formula for molecular mass**

The molecular mass of a substance can be computed by adding the atomic masses of all atoms present in one molecule of the substance. The average mass value of a molecule of that substance is represented in atomic mass unit.

**Example**

Find the molecular mass of H_{2}O

**Solution**

Molecular mass of H_{2}O = (2*1.0u) + (1*16.0u)

=18u

**Relative atomic mass**

Relative atomic mass of any species is always compared to the mass of a carbon-12 atom and is always compared with its relative value of 12.

→ Read MoreChemistry involves the making and breaking of bonds. So, calculating the mass per mole of a complex molecule is essentially no different from finding the mass per mole of a mono atomic element. For example, one molecule of glucose (C_{6}H_{12}O_{6}) is assembled from 6 carbon atoms. 12 hydrogen atoms and 6 oxygen atoms.

To calculate the number of grams per mole of a complex molecule simply follow the steps given below.

To calculate the number of grams per mole of a complex molecule simply follow the steps given below.

- Multiply the number of atoms per mole of the first element by its atomic mass.
- Multiply the number of atoms per mole of the second element by its atomic mass.
- Multiply the number of atoms per mole of the third element by its atomic mass. Keep going until we have covered all the elements in the molecule.
- Finally add the masses together.

**Example Problem**

How many moles of NH_{3} are produced from 33.6g of N_{2}.

**Solution**

Before we can convert moles of N_{2} to moles of NH_{3} we must first convert the mass of N_{2} to moles of N_{2}. This means a two-step conversion.

- In the first step the mass of N
_{2}is converted to moles using the molar mass of N_{2}as the conversion factor. - In the second step moles of N
_{2}are converted to moles of NH_{3}using mole ratio.

**The complete unit map for this problem is illustrated below**

= 33.6g N

= 2.40 mol of NH

The significance of chemical formula will be obvious.

The true formula of a substance expresses the quantitative relations in which the body stands to one, two, or more others.

The formula of sugar expresses the sum of all its elements which combines with an equivalent oxide of lead and it also represents the sum of the weights of carbonic acid and alcohol into which it is resolved by fermentation.

Chemical formula is a symbolic representation of what happens when chemicals come in contact with one another. A balanced chemical equation shows how many of what molecules react, how many of what molecules result, and sometimes the state of the substances.

The symbols that chemists use to depict chemical reactions are designed to specify what the atoms are doing when a reaction occurs. For example, which atoms or ions are breaking chemical bonds and then forming new bonds? The true formula of a substance expresses the quantitative relations in which the body stands to one, two, or more others.

The formula of sugar expresses the sum of all its elements which combines with an equivalent oxide of lead and it also represents the sum of the weights of carbonic acid and alcohol into which it is resolved by fermentation.

Chemical formula is a symbolic representation of what happens when chemicals come in contact with one another. A balanced chemical equation shows how many of what molecules react, how many of what molecules result, and sometimes the state of the substances.

Studying chemistry is much easier once we learn to visualize what reaction equations are indicating symbolically. Some terminology required for this module is shown in the following diagram.

Stoichiometric coefficients are necessary to balance chemical equations. Balancing insures that equal numbers of atoms of each element are present in both the products and reactants. Otherwise the reaction would violate the law of conservation of mass indicating that new atoms have been made in the reaction rather than redistributing the atoms into new chemical compounds.

Molecules are the smallest particles of matter which can have independent existence. Like atoms, molecules are so small that their absolute mass can’t be determined by actual weighing. Therefore, like atomic masses, molecular masses are expressed relatively.

Atomic weight

The atomic mass of an element is the average relative mass of an atom of the element as compared with the mass of 12C atom taken as 12 units. The atomic mass of an element is the number of times one atom of the element is heavier than 1/12 of a 12C atom. Therefore, on Carbon scale,

Atomic mass = $\frac{Mass\ of\ one\ single\ atom\ of\ the\ element}{Mass\ of\ 1/12th\ atom\ of\ carbon}$

**Relative Molecular Mass**

The relative molecular mass of an element or compound is the mass of one molecule of the element or compound compared with the mass of an atom of _{6}**C**^{12} which is arbitrarily assigned as 12.000.

Molecular weights are mere numbers and have no units like atomic weights as these are relative weights. The molecular weight is not the actual weight of the molecule, it is a number which indicates how many times the molecule of a particular substance is heavier than one-twelfth of an atom of** **_{6}C^{12} isotope.

**Example**

The molecular weight of oxygen is 32. This means that one molecule of oxygen is 32 times heavier than one-twelfth of an atom of_{6}**C**^{12} isotope.

**Gram Molecular Weight:** The molecular weight expressed in grams is called gram molecular weight.

Example – Molecular weight of H_{2}O is 18, **therefore the gram molecular weight of H**_{2}O is 18 g.

Atomic weight

Atomic mass = $\frac{Mass\ of\ one\ single\ atom\ of\ the\ element}{Mass\ of\ 1/12th\ atom\ of\ carbon}$

Molecular weights are mere numbers and have no units like atomic weights as these are relative weights. The molecular weight is not the actual weight of the molecule, it is a number which indicates how many times the molecule of a particular substance is heavier than one-twelfth of an atom of

The molecular weight of oxygen is 32. This means that one molecule of oxygen is 32 times heavier than one-twelfth of an atom of

Example – Molecular weight of H

- Calculation of Molecular weight by the sum of the standard atomic weights of the atoms which form the compound.
- Multiply the molecular weight by molar mass constant $\frac{1g}{mole}$

[22.989 + 35.453] $\times$ $\frac{1g}{mole}$ = $\frac{ 58.443g}{mole}$

2. Molar mass of Sugar C

([12 × 12.0107] + [22 ×1.008] + [11 ×15.9994]) $\times$ $\frac{1g}{mole}$ = $\times$ $\frac{342.29g}{mole}$

The average molar mass of any mixture $\bar{M}$ may be calculated from the mole fractions x

$\sum_{i}xiMi$

A chemical formula or molecular formula **is one of the methods of providing information about the atoms that constitutes a particular chemical compound. **

**Following information is given by chemical formula:**

**Example**

**Determination of Molecular weights **

The methods of determining molecular weights are given below

**1. Molar Volume Method**

By this method we find the weight of known volume of the gas at STP. Molar volume of any gas means that volume of one mole of any gas is 22.4L at STP.

**Example:** 10L of a gas at 14 degree C and 729 mm pressure weight 17.925g. Calculate the molecular weight of the gas.

First we calculate the volume of the gas at STP

Volume of the gas at STP = $\frac{(10\times 273\times 729)}{287\times 760}$

= 9.125 L

Weight of 9.125 L = 17.925g

Weight of 22.4 L = $\frac{17.925\times 22.4}{9.125g}$

= 44 g

**The molecular weight of the gas = 44.**

**2. Vapor Density Method **

VD= $\frac{Weight\ of\ certain\ volume\ of\ a\ gas\ or\ vapor\ under\ certain\ temperature\ and\ pressure}{Weight\ of\ same\ volume\ of\ a\ hydrogen\ under\ same\ temperature\ and\ pressure}$

** Molecular weight = 2 x VD **

**Example: **1L of hydrogen at STP weighs 0.09g. If 2 L of a gas at STP weighs 2.880g. Calculate the molecular weight of the gas.

1L of hydrogen at STP weighs = 0.09 g

2L of hydrogen at STP weighs = 0.09x2= 0.18g

VD = $\frac{2.88g}{0.18g}$ = 16

** Molecular Weight = 2x16 = 32 **

**3. **By** the ideal gas equation **

Ideal gas equation is ** PV=nRT**

Where n is the number of moles = $\frac{w}{M}$ , w is the weight of the gas in gram and M is the molecular weight.

**Example:** What is the molecular weight of a gas, if 0.0866 g sample in 60 mL bulb has pressure of 400 mm at 20 degree C?

T = 273+20=293k

P = 400/760 atm = 0.526 atm

V= 60 / 1000 L = 0.06 L

R= 0.082 L atm Mole^{-1} K^{-1}

M= $\frac{w\times R\times T}{P\times V}$ = $\frac{0.0866\times 0.082\times 293}{0.526\times 0.006}$

M = 65.92

- The chemical formula gives out information for all constituent elements by its chemical symbol and specifies the number of atoms of each element found in each of these discrete molecules of the compound.
- If a molecule is having more than one atom of a particular element, this number is indicated by using a subscript after the chemical symbol of the atom.

- In a molecule of
**carbon dioxide**, one atom of**carbon**is combined with two atoms of**oxygen**. The formula is CO_{2}. - Sucrose has twelve carbon atoms, twenty-two hydrogen atoms and eleven oxygen atoms, and so the chemical formula stands as C
_{12}H_{22}O_{11}.

The methods of determining molecular weights are given below

By this method we find the weight of known volume of the gas at STP. Molar volume of any gas means that volume of one mole of any gas is 22.4L at STP.

First we calculate the volume of the gas at STP

Volume of the gas at STP = $\frac{(10\times 273\times 729)}{287\times 760}$

= 9.125 L

Weight of 9.125 L = 17.925g

Weight of 22.4 L = $\frac{17.925\times 22.4}{9.125g}$

= 44 g

VD= $\frac{Weight\ of\ certain\ volume\ of\ a\ gas\ or\ vapor\ under\ certain\ temperature\ and\ pressure}{Weight\ of\ same\ volume\ of\ a\ hydrogen\ under\ same\ temperature\ and\ pressure}$

1L of hydrogen at STP weighs = 0.09 g

2L of hydrogen at STP weighs = 0.09x2= 0.18g

VD = $\frac{2.88g}{0.18g}$ = 16

Where n is the number of moles = $\frac{w}{M}$ , w is the weight of the gas in gram and M is the molecular weight.

T = 273+20=293k

P = 400/760 atm = 0.526 atm

V= 60 / 1000 L = 0.06 L

R= 0.082 L atm Mole

M= $\frac{w\times R\times T}{P\times V}$ = $\frac{0.0866\times 0.082\times 293}{0.526\times 0.006}$

M = 65.92

Since the formula of a compound expresses the ratio of the numbers of its constituent atoms, a formula also conveys information about the relative masses of the elements it contains. In order to make this connection work, we have to know the relative masses of the different elements.

**Example:** Masses of carbon, hydrogen and oxygen in one mole of ethanol (C_{2}H_{5}OH).

Carbon: (2 mol)(12.0 g mol^{–1}) = 24 g of C

Hydrogen: (6 mol)(1.01 g mol^{–1}) = 6 g of H

Oxygen: (1 mol)(16.0 g mol^{–1}) = 16 g of O

**Mass fraction**: Mass fraction of an element in a compound is the ratio of the mass of that element to the mass of the entire formula unit or molecular weight of the compound. Mass fractions are always between 0 and 1, but most of the time is expressed as percent.

**Mass% = Mass fraction x 100 %****Example **

Mass fraction and mass percentage of oxygen in ethanol (C_{2}H_{5}OH)

The molar mass of ethanol is (24 + 6 + 16) g mol^{–1} = 46 g mol^{–1}.

Of this, 16 g is due to oxygen, so its mass fraction in the compound is

$\frac{16g}{46g}$ = 0.35

Mass% = 0.35 x100 % = 35%.

Carbon: (2 mol)(12.0 g mol

Hydrogen: (6 mol)(1.01 g mol

Oxygen: (1 mol)(16.0 g mol

Mass fraction and mass percentage of oxygen in ethanol (C

The molar mass of ethanol is (24 + 6 + 16) g mol

Of this, 16 g is due to oxygen, so its mass fraction in the compound is

$\frac{16g}{46g}$ = 0.35

Mass% = 0.35 x100 % = 35%.

The mass fraction of K in KCl is $\frac{39.1}{74.6}$ =.524;

10 Kg of KCl contains(39.1/74.6) × 10 Kg of K

= 5.24 Kg of K

Molar mass of Ammonia (NH3) = $\frac{(14+1\times 3)g}{mole}$

= $\frac{17g}{mole}$

Mass ratio of Nitrogen = $\frac{(14g/mole)}{17g/mole}$ = 0.823

Mass % of Nitrogen = 0.823 x 100 % = 82.3% .

The mass ratio of KCl/K is 74.6 ÷ 39.1 = 1.908

10 g of potassium will be present in 1.908 x 10 grams of KCl

=**19.08 grams. KCl**

10 g of potassium will be present in 1.908 x 10 grams of KCl

=

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