Physical Chemistry

  Q. When 1.00 g of CaCO3 is heated, it produces 200 cm3 of CO2 (g). Calculate the % yeild of CO2.  Solution, CaCO 3 (s) CaO(s) + CO...

 

Q. When 1.00 g of CaCO3 is heated, it produces 200 cm3 of CO2 (g). Calculate the % yeild of CO2. 

Solution,

CaCO₃(s) CaO(s) + CO₂(g)

Mole of CO₂ 1/ Mole of CaCO₃ = 1/1

Or, (m/mr)/(vol/mv)  = 1/1

            Or, (1/100 ) / (vol/24000) = 1/1

Or, (1/100)    x (24000/vol)  = 1/1

Or, 24000 = 100 vol

Therefore, Vol = 240 cm3

Now,

                % yield = (200/240) x 100

                        = 83.33 %

Percentage Purity:

Formula:

% purity = (Pure mass / Total Mass) *100

 Q. When 1.00g of impure CaCO₃ is heated, it produced 200 cm³ of CO₂(g). Calculate the % purity of CaCO₃.

CaCO₃(s) CaO(s) + CO₂(g)

% purity = (Pure mass / Total Mass) *100

 CaCO₃ = 40+12+16x3 = 100

Mole of CaCO₃   / Mole of CO2 = 1/1

Or, (m/mr) / (vol/mv)  = 1/1

Or. (m/100) * (24000/200) = 1/1

Or, m = 0.12

Now,

% purity = (0.12/1) x 100 %

=12%

Therefore, % purity = 12%

When 10 cm³ of hydrocarbon , C_xH_y is burnt, it requires 65cm³ of oxygen and produced 40 cm³ of CO₂. Deduce the Hydrocarbon.

 C₃H₈(g) + 5O₂(g) 3 CO₂(g) + 4H₂0 (l)

1cm³ of C₃H₈ = x

10 cm³ of C₃H₈ = 40 cm³

X = 40/10

Therefore, X = 4

1cm³ of O₂ = x

1 cm³ of O₂ = x + y/4

10 cm³ of O₂ = 63 cm³

x + y/4 = 65/10

Or, y = 10

ii) 

Mole of C₃H_{3 /} Mole of O₂ = 1/5

(10/2400) / (2400/vol) = 1/5

or,  vol = 50 cm³

 Therefore, Volume of Oxygen reacted = 50 cm³

 iii)

Here,

10 cm³ of propane = 50 cm³ of oxygen,

Remained oxygen = 100 - 50

= 50 cm³

# Limiting : which completely reacts

In calculation use limiting reagent.

# excess : which remains.

 When 10 cm³ of propane, C₃H₈ is burnt in 100 cm³ of oxygen,Calculate :

i)Volume of CO₂ produced.

ii)Volume of O₂ reacted.

iii)Volume of O₂ remained.

 C₃H₈(g) + 5O₂(g) 3 CO₂(g) + 4H₂0

Mole of C₃H/ Mole of O₂ = 1/3

(10/2400) / (Vol/2400) = 1/3

 Or, vol = 30 cm³

Therefore, volume of CO₂ produced = 30 cm³

 ii) Solution,

  Mole of Mg(NO₃)₂/ Mole of O₂ = 1/ (1/2)

(1.48/148) = 2/1

 Or, 35.52= 296 Vol

Or, vol = 0.12 dm³

 g) When Magnesium nitrate is heated it produced Magnesium oxide, nitrogen dioxide and oxygen. Construct a balanced equation for it.

 i) mole = mass ( for a compound / element ) ; mole = mass (for atom)

ii) mole = Volume ( for gaseous substance )

 Mv = molar volume

= volume occupied by 1 mole of gaseous substance.

= 24.0 dm³ at RTP (Room temperature and pressure)

= 22.4 dm³ at STP (Standard temperature and pressure)

1 dm³ = 1000 cm³ = 1 liter

dm = decimeter

 iii) mole = concentration ( in mol/ dm³ ) x volume ( in dm³ )

   concentration = amount of solute / Volume of solvent

 Mole ratio formula :

C₃H₈ + 5O₂ 3 CO₂ + 4H₂0  

Mole of C₃H₃ / Mole of O₂  = 1/5

Mole of CO₂  / Mole of O₂ = 3/5

 Combustion Data :

0.1g of a compound containing carbon, hydrogen, and oxygen only as elements, on combustionproduced, 0.019g of CO₂ and 0.0117g of H₂0.Calculate the empirical formula of the compound.

Solution,

Mole of CO₂ = m/mr  = 0.0191/44 = 0.0043

Mole of C = 0.0043

Or, 0.0043 = m/12

 M = 0.052

 Now,

Mole of H₂0 = m/mr   = 0.0117/18 = 0.0065

 Mole of Hydrogen = 0.0065 x 2 = 0.0130

Mole = m/Ar

0.0180 = m/1

M = 0.0130

 Mass of Oxygen = 0.1 – 0.052 – 0.0130

                            = 0.035

 Q. The Compound has the following % composition, K 28.15%, Cl 25.63%, O 46.20% . Deduce the formula of the compound.

K = 28.15 = 28.15/39 = 0.72/0.72 = 1

Cl = 25.63 = 25.63/35.5 = 0.72/ 0.72 = 1

O = 46.20 = 46.20/16 = 2.88/0.72 = 4

 Therefore, the formula of the compound is KClO₄

 # In mass spectrum the ions are always positive.Eg: ³⁵Cl⁺

Empirical formula : The formula which contains least possible ratio of atom present in a molecule.

Molecular formula: The formula which contains actual ratio of atoms in a molecule.

 (E.F)_n = (M.F) where n is any positive integer.

 Q. The Compound has the following % composition, Na 26.43%, S 36.78%, O 27.58% . Deduce the formula of the compound.

Na = 26.43 = 26.43/23= 1.149 = 1.149/1.145 = 1.003 which is equivalent to 1, so, 1 * 2 = 2

S = 36.78 = 36.78/32.1 = 1/145 = 1/145/1.145 = 1. so, 1 * 2 = 2

O = 27.58 = 27.58/16 = 1.72 = 1.72 /1.145 = 1.5, so, 1.5 * 2 = 3

 Formula = Na₂S₂O₃

Therefore, the formula of the compound is Na₂S₂O₃

# Mass Spectrum: 1) To determine Ar (or Mr)

2) Particle is bombarded by electron so that particle dissolve into positive ions

with certain mass.

 3) A peak appears when % abundance Vs mass(m) is plotted.

 # Relative atomic mass (Ar) = ε isotopic abundance x isotopic mass

= isotopic abundance.1 x Isotopic mass.1 + isotopic abundance.2 x

Isotopic mass.2