**Important formula**

Density of a substance tells us about how closely its particles are packed. If density is more, it means particles are more closely packed.

Density = Mass / Volume

SI unit of density = SI unit of mass/ SI unit of volume

= kg / m³ or kg m^{-3}

There are three common scales to measure temperature — °C (degree celsius), °F (degree fahrenheit) and K (kelvin).

Freezing point of water 0°C

Boiling point of water 100 °C

The temperatures on two scales are related to each other by the following relationship: °F = (9/5) °C + 32

The kelvin scale is related to celsius scale as follows: K = °C + 273.15

It is interesting to note that temperature below 0 °C (i.e., negative values) are possible in Celsius scale but in Kelvin scale, negative

temperature is not possible.

One atomic mass unit is defined as a mass exactly equal to one-twelfth of the mass of one carbon - 12 atom.

1 amu = 1.66056 × 10^{-24 }g

Mass of an atom of hydrogen = 1.6736×10^{-24} g

Thus, in terms of amu, the mass of hydrogen atom = 1.0080 amu

**Problem**

Calculate the molecular mass of glucose C_{6}H_{12}O_{6 }molecule.

**solution**

Molecular mass of glucose (C_{6}H_{12}O_{6}) = 6(12.011 u)+12(1.008 u)+6(16.00 u)

= (72.066 u) + (12.096 u) + (96.00 u)

= 180.162 u

One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.

This number of entities in 1 mol is so important that it is given a separate name and symbol. It is known as ‘**Avogadro constant**’, or Avogadro number denoted by N_{A }= 6.022×10^{23}

1 mol of water molecules = 6.022 × 10^{23} water molecules

1 mol of sodium chloride = 6.022 × 10^{23} formula units of sodium chloride

The mass of one mole of a substance in grams is called its molar mass.

The molar mass in grams is numerically equal to atomic/molecular/formula mass in u.

Molar mass of water = 18.02 g mol^{-1}

Molar mass of sodium chloride = 58.5 g mol^{-1}

Example of water (H_{2}O).

**Problem **

what is the percentage of hydrogen and oxygen in water.

**Solution**

Mass % of an element = (mass of that element in the compound X 100) / (molar mass of the compound)

Molar mass of water = 18.02 g

Mass % of hydrogen = (2 X 1.008) X 100 / (18.02) =11.18

Mass % of Oxygen = (16.00) X 100 / (18.02) = 88.79

An **empirical formula** represents the simplest whole number ratio of various atoms present in a compound, whereas, the **molecular formula** shows the exact number of different types of atoms present in a molecule of a compound.

**Problem**

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

**Solution**

** Step 1**. Conversion of mass per cent to grams

Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present.

** Step 2**. Convert into number moles of each element Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound

Moles of hydrogen = 4.07 g / 1.008g = 4.04

Moles of carbon = 24.27g / 12.01g= 2.021

Moles of chlorine = 71.65g / 35.453g =2.021

** Step 3**. Divide each of the mole values obtained above by the smallest number amongst them

Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl . In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

** Step 4**. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements

CH_{2}Cl is, thus, the empirical formula of the above compound.

** Step 5**. Writing molecular formula

- Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula. For CH
_{2}Cl, empirical formula mass is 12.01 + (2 × 1.008) + 35.453 = 49.48 g - Divide Molar mass by empirical formula mass = 98.96g / 49.48g = 2 = (n)
- Multiply empirical formula by n obtained above to get the molecular formula

Empirical formula = CH_{2}Cl, n = 2. Hence molecular formula is C_{2}H_{4}Cl_{2}.

**Reactions in Solutions**

The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways.

- Mass per cent or weight per cent (w/w %)
- Mole fraction
- Molarity
- Molality
**Mass per cent**

It is obtained by using the following relation:

Mass per cent = (Mass of the solute / Mass of the Solution) X 100

**Mole Fraction**

It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB , respectively, then the mole fractions of A and B are given as:

Mole fraction of A = No of moles of A / No of moles of solution = nA / (nA + nB)

Mole fraction of B = No of moles of B / No of moles of solution = nB / (nA + nB)

**Molarity**

It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution. Thus,

Molarity (M) = no of moles of solute / volume of solution in liters

Suppose, we have 1 M solution of a substance, say NaOH, and we want to prepare a 0.2 M solution from it.1 M NaOH means 1 mol of NaOH present in 1 litre of the solution. For 0.2 M solution, we require 0.2 moles of NaOH dissolved in 1 litre solution. Hence, for making 0.2M solution from 1M solution, we have to take that volume of 1M NaOH solution, which contains 0.2 mol of NaOH and dilute the solution with water to 1 litre. Now, how much volume of concentrated (1M) NaOH solution be taken, which contains 0.2 moles of NaOH can be calculated as follows: If 1 mol is present in 1L or 1000 mL solution

then, 0.2 mol is present in

(1000 mL / 1 mol) X 0.2 mol solution

= 200 mL solution

Thus, 200 mL of 1M NaOH are taken and enough water is added to dilute it to make it 1 litre.

Note that the number of moles of solute (NaOH) was 0.2 in 200 mL and it has remained the same, i.e., 0.2 even after dilution ( in 1000 mL) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to NaOH. But keep in mind the concentration.

**Molality**

It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.

Thus, Molality (m) = No of moles of solute / mass of solvent in Kg

**Problem**

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

**Solution**

Step 1. Conversion of mass per cent to grams

Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present.

Step 2. Convert into number moles of each element

Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound

Moles of hydrogen = 4.07 g/1.008g = 4.04

Moles of carbon = 24.27 g/12.01g = 2.021

Moles of chlorine = 71.65g/35.453g= 2.021

Step 3. Divide each of the mole values obtained above by the smallest number amongst them

Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl . In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements

CH_{2}Cl is, thus, the empirical formula of the above compound.

Step 5. Writing molecular formula

(a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula.

For CH_{2}Cl, empirical formula mass is

12.01 + (2 × 1.008) + 35.453 = 49.48 g

(b) Divide Molar mass by empirical formula mass

Molar Mass / Empirical Formula Mass = 98.96 g / 48.49 g = 2 = (n)

(c) Multiply empirical formula by n obtained above to get the molecular formula

Empirical formula = CH_{2}Cl, *n *= 2. Hence molecular formula is C_{2}H_{4}Cl_{2}.

Combustion of methane. A balanced equation for this reaction is as given below:

CH_{4} (g) + 2O_{2} (g) → CO_{2} (g) + 2 H_{2}O (g)

Here, methane and dioxygen are called reactants and carbon dioxide and water are called products.

Mass/Volume = Density

**Balancing a chemical equation**

According to the law of conservation of mass, a balanced chemical equation has the *same number of atoms of each element on both sides *of the equation.

Many chemical equations can be balanced by *trial and error*.

Let us take the reactions of a few metals and non-metals with oxygen to give oxides

* *

4 Fe(s) + 3O_{2}(g) → 2Fe_{2}O_{3}(s) (a) balanced equation

2 Mg(s) + O_{2} (g) → 2MgO(s) (b) balanced equation

P_{4}(s) + O_{2} (g) → P_{4}O_{10} (s) (c) unbalanced equation

Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on each side of the equations.

However equation (c) is not balanced. In this equation, phosphorus atoms are balanced but not the oxygen atoms.

To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation.

P_{4}(s) + 5O_{2} (g) → P_{4}O_{4}(s) balanced equation

Now, let us take combustion of propane, C_{3}H_{8}. This equation can be balanced in steps.

**Step 1**

Write down the correct formulas of reactants and products.

Here, propane and oxygen are reactants, and carbon dioxide and water are products.

C_{3}H_{8}(g) + O_{2} (g) → CO_{2} (g) + H_{2}O(l) unbalanced equation

**Step 2**

*Balance the number of C atoms: *Since 3 carbon atoms are in the reactant, therefore, three CO_{2} molecules are required on the right side.

C_{3}H_{8} (g) + O_{2} (g) → 3CO_{2} (g) + H_{2}O (l)

** **

**Step 3**

*Balance the number of H atoms*: on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side.

C_{3}H_{8} (g) + O_{2} (g) → 3CO_{2} (g)+4H_{2}O (l)

**Step 4**

*Balance the number of O atoms*: There are 10 oxygen atoms on the right side (3 × 2 = 6 in CO_{2} and 4 × 1= 4 in water).

Therefore, five O_{2} molecules are needed to supply the required 10 oxygen atoms.

C_{3}H_{8} (g) +5O_{2} (g) → 3CO_{2} (g) + 4H_{2}O (l)

**Step 5**

*Verify that the number of atoms of each element is balanced in the final equation*.

The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side.

All equations that have correct formulas for all reactants and products can be balanced.

Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation.

**Problem**

Calculate the amount of water (g) produced by the combustion of 16 g of methane.

**Solution**

The balanced equation for the combustion of methane is :

CH_{4} (g) +2O_{2} (g) → CO_{2} (g) + 2H_{2}O (g)

(i) 16 g of CH_{4} corresponds to one mole.

(ii) From the above equation, 1 mol of CH_{4} (g) gives 2 mol of H_{2}O (g).

2 mol of water (H_{2}O) = 2 × (2+16) = 36g

1 mol H_{2}O = 18 g H_{2}O

Hence, 2 mol H_{2}O = 2 × 18 g H_{2}O = 36 g H_{2}O

**Problem**

How many moles of methane are required to produce 22g CO2 (g) after combustion?

**Solution**

According to the chemical equation CH_{4} (g) +2O_{2} (g) → CO_{2} (g) + 2H_{2}O (g)

44g CO_{2} (g) is obtained from 16 g CH_{4} (g).

[1 mol CO_{2} (g) is obtained from 1 mol of CH_{4}(g)]

Number of moles of CO_{2} (g) = [22 g CO_{2} (g)] X [1 mol CO_{2} (g) / 44 g CO_{2} (g)] = 0.5 mol CO_{2} (g)

Hence, 0.5 mol CO_{2} (g) would be obtained from 0.5 mol CH_{4} (g) or 0.5 mol of CH_{4} (g) would be required to produce 22 g CO_{2} (g).

**Problem**

50.0 kg of N_{2} (g) and 10.0 kg of H_{2} (g) are mixed to produce NH_{3} (g). Calculate the amount of NH_{3} (g) formed. Identify the limiting reagent in the production of NH_{3} in this situation.

**Solution**

A balanced equation for the above reaction is written as follows :

N_{2} (g) +3H_{2} (g) ⇔ 2NH_{3} (g)

Calculation of moles :

Number of moles of N_{2} = [50.0 kg N_{2}] X [1000 g N_{2 }/ 1 kg N_{2}] X [1 mol N_{2 }/ 28.0 g N_{2}]

=17.86 X 10^{2} mol

Number of moles of H_{2} = [10.0 kg H_{2}] X [1000 g H_{2 }/ 1 kg H_{2}] X [1 mol H_{2 }/ 2.016 g H_{2}]

=4.96 X 10^{3 }mol

According to the above equation, 1 mol N_{2 }(g) requires 3 mol H_{2} (g), for the reaction. Hence, for 17.86×102 mol of N_{2}, the moles of H_{2} (g) required would be

= [17.86 X 10^{2} mol] X [3 mol H_{2} (g)/ 1 mol N_{2 }(g)]

= 5.36 X 10^{3 }mol H_{2}

But we have only 4.96×10^{3 }mol H_{2}. Hence, dihydrogen is the limiting reagent in this case. So, NH_{3} (g) would be formed only from that amount of available dihydrogen i.e., 4.96 × 10^{3} mol

Since 3 mol H_{2}(g) gives 2 mol NH_{3}(g)

[4.96 X 10^{3 }mol H_{2}(g) ] X [2 mol NH_{3} (g)/ 3 mol H_{2 }(g)] = 3.30 X 10^{3 }mol NH_{3 }(g) is obtained

If they are to be converted to grams, it is done as follows :

1 mol NH_{3} (g) = 17.0 g NH_{3} (g)

[3.30 X 10^{3 }mol NH_{3 }(g)] X [17.0 g NH_{3} (g) / 1 mol NH_{3} (g) ]

=3.30 X 10^{3 }X 17g NH_{3 }= 56.1 X 10^{3 } NH_{3}

= 56.1 kg NH_{3}

Mass per cent = [mass of solute / mass of solution] X 100

**Problem**

A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.

**Solution**

Mass per cent of A = [mass of A / mass of the solution] X 100

we know mass of the solution = 2g of A + 18 g of water = 20 g

Mass per cent of A = [2g / 20 g] X 100 = 10%

Mole fraction of A = No of moles of A / No of moles of solution = n_{A} / (n_{A} + n_{B})

Mole fraction of B = No of moles of B / No of moles of solution = n_{B} / (n_{A} + n_{B})

Molarity (M) = no of moles of solute / volume of solution in liters

**Problem**

Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.

**Solution**

since Molarity (M) = no of moles of solute / volume of solution in liters

[Mass of NaOH/ Molar mass of NaOH] / 0.250 L

=[4 g/40 g] /0.250 L = 0.1 mol / 0.250 L

= 0.4 mol^{-1 }= 0.4 M

Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent.

**Problem**

The density of 3 M solution of NaCl is 1.25 g ml^{ -1} Calculate the molality of the solution.

**Solution**

M = 0.3 mol^{-1}

Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g

Mass of 1L solution = 1000 × 1.25 = 1250 g

(since density = 1.25 g mL^{-1})

Mass of water in solution = 1250 –75.5 = 1074.5 g

Molality (m) = No of moles of solute / mass of solvent in Kg

3 mol /1074.5 g = 2.79 m

Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that the molality of a solution does not change with temperature since mass remains unaffected with temperature.

**Exercise and Solutions**

Q1. Calculate the molar mass of the following: (i)H_{2}O (ii)CO_{2 }(iii)CH_{4}

**Solution**

(i) H_{2}O

Molecular weight of H_{2}O = (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)

= [2(1.0084) + 1(16.00 u)] = 2.016 u +16.00 u

= 18.016u

(ii) CO_{2 }

Molecular weight of CO2 = (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)

= [1(12.011 u) + 2(16.00 u)]

= 12.011 u +32.00 u

= 44.011 u

(iii) CH_{4}

Molecular weight of CH4 = (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)

= [1(12.011 u) +4 (1.008u)]

= 12.011u + 4.032 u

= 16.043 u

Q2. Calculate the mass per cent of different elements present in sodium sulphate (Na_{2}SO_{4}).

**Solution**

Molar mass of Na_{2}SO_{4}

= [(2 x 23.0) + (32.066) + 4(16.00)]

=142.066 g

Mass per cent of an element = (Mass of the element in the compound/ Mass of the compound) X 100

Hence Mass percent of the sodium =(46.0 g / 142.066 g) X 100 = 32.4 %

Mass percent of the sulphur = (36.066 g / 142.066 g) X 100 = 22.6 %

Mass percent of the oxygen = (64.0 g / 142.066 g) X 100 = 245.05 %

Q3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

**Solution**

Given

Percent of Fe by mass = 69.9 % and Percent of O_{2} by mass = 30.1 %

Relative moles of Fe in iron oxide = (percentage of iron by mass / atomic mass of iron)

= 69.9/55.85 = 1.25

Relative moles of O in iron oxide = (percentage of oxygen by mass / atomic mass of oxygen)

= 30.1/16.00 =1.88

molar ratio of Fe to O = 1.25:1.88 = 1:1.5

empirical formula of iron oxide is Fe_{2}O_{3}

Q4. Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

**Solution**

(i) 1 mole of carbon is burnt in air.

C + O_{2 }→ CO_{2}

1 mole of carbon reacts with 1 mole of O_{2} to form one mole of CO_{2}.

Amount of CO_{2} produced is 44 g

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

1 mole of carbon burnt in 32 grams of O_{2} it forms 44 grams of CO_{2}.

Therefore, 16 grams of O_{2} will form (44 X 16)/ 32 = 22 grams of CO_{2}

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Since oxygen is the limiting reactant here, the 16g (0.5 mol) of O_{2} will react with 6g of carbon (0.5 mol) to form 22 g of carbon dioxide. The remaining 18g of carbon (1.5 mol) will not undergo combustion.

Q5. Calculate the mass of sodium acetate (CH_{3}COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^{-1}

**Solution**

0.375 Maqueous solution of CH_{3}COONa

= 1000 mL of solution containing 0.375 moles of CH_{3}COONa

Therefore, no. of moles of CH_{3}COONa in 500 mL

= [ 0.375/1000 ] X 500

=0.1875 mole

Guven molar mass of sodium acetate = 82.0245 g mol^{-1}

Therefore, mass that is required of CH_{3}COONa

= [ 82.0245 g mol^{-1} ] X [ 0.1875 mole ]

= 15.38 gram

Q6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^{-1} and the mass per cent of nitric acid in it being 69%.

**solution**

Given mass percentage of nitric acid in sample = 69 %

thus 100g of niti acid contains 69 g of nitic acid by mass.

Molar mass of nitic acid (HNO_{3}) = 69g/ 63 g mol^{-1 }= 1.095 g mol^{-1}

volume of 100g of nitric acid solution = mass of solution/ density of solution

= 100g/ 1.41 g mL^{-1}

= 70.92mL = 70.92 X 10^{-3 }L

concentration of nitric acid = 15.44 mol/L

Q7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

**solution**

1 mole of CuSO_{4 }contains 1 mole of copper

Molar mass of CuSO_{4}

= (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 gram

159.5 gram of CuSO_{4} contains 63.5 gram of Cu.

Therefore, 100 gram of CuSO_{4} will contain (63.5×100g)/159.5 of Cu.

= 39.81 gram

Q8. Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

**solution**

Mass percent of Fe = 69.9%

Mass percent of O = 30.1%

No. of moles of Fe present in oxide = 69.90 / 55.85 = 1.25

No. of moles of O present in oxide = 30.1 / 16.0 = 1.88

Ratio of Fe to O in oxide = 1.25: 1.88 = 1: 1.5

= 2:3

Therefore empirical formula of oxide is Fe_{2}O_{3}

empirical formula mass of Fe_{2}O_{3}

[2(55.85) + 3(16.00)]gr = 159.69g

n = molar mass/empirical formula mass

= 159.69 g/159.7 g

= 0.999

= 1 (approx)

The molecular formula of a compound can be obtained by multiplying n and the empirical formula.

Thus, the empirical of the given oxide is Fe_{2}O_{3} and n is 1.

Q9. Calculate the atomic mass (average) of chlorine using the following data :

% Natural Abundance | Molar Mass | |

^{35}Cl |
75.77 | 34.9689 |

^{37}Cl |
24.23 | 36.9659 |

**Solution**

Average atomic mass of Chlorine

=[ (Fractional abundance of ^{35}Cl) (molar mass of ^{35}Cl) + (fractional abundance of ^{37}Cl ) (Molar mass of ^{37}Cl)]

= 77.77/100