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,

CaCO_{3}(s) CaO(s) + CO_{2}(g)

Mole of CO_{2} 1/ __Mole of CaCO___{3}_{ }= __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 CaCO_{3} is heated, it produced 200
cm^{3} of CO_{2}(g). Calculate the % purity of CaCO_{3}.

CaCO_{3}(s)
CaO(s) + CO_{2}(g)

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

CaCO_{3} = 40+12+16x3 = 100

__Mole of CaCO___{3 }_{ } __/ __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%

^{3}of hydrocarbon , C

_{x}H

_{y}is burnt, it requires 65cm

^{3}of oxygen and produced 40 cm

^{3}of CO

_{2}. Deduce the Hydrocarbon.

C_{3}H_{8}(g) + 5O_{2}(g)
3 CO_{2}(g) + 4H_{2}0 (l)

1cm^{3}
of C_{3}H_{8} = x

10 cm^{3}
of C_{3}H_{8} = 40 cm^{3}

X = 40/10

Therefore, X = 4

1cm^{3}
of O_{2} = x

1 cm^{3} of O_{2} = x + __y/4__

10 cm^{3} of O_{2} = 63 cm^{3}

x + __y/4__ = 65/10

Or, y = 10

ii)

__Mole of C___{3}__H___{3 / }Mole of O_{2 }= 1/5

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

or, vol = 50 cm^{3}

Therefore, Volume of Oxygen reacted = 50 cm^{3}

iii)

Here,

10 cm^{3}
of propane = 50 cm^{3 }of oxygen,

Remained oxygen = 100 - 50

= 50 cm^{3}

# Limiting : which completely reacts

In calculation use limiting reagent.

# excess : which remains.

When 10 cm^{3} of propane, C_{3}H_{8} is
burnt in 100 cm^{3} of oxygen,Calculate :

i)Volume of CO_{2}
produced.

ii)Volume of O_{2}
reacted.

iii)Volume of O_{2}
remained.

C_{3}H_{8}(g) + 5O_{2}(g)
3 CO_{2}(g) + 4H_{2}0

Mole of C_{3}H/ Mole of O_{2 }= 1/3

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

Or, vol = 30 cm^{3}

Therefore, volume
of CO_{2} produced = 30 cm^{3}

ii) Solution,

__Mole of Mg(NO___{3}__)___{2}/ Mole of O_{2 }= 1/ (1/2)

(1.48/148) = 2/1

Or, 35.52= 296 Vol

Or,
vol = 0.12 dm^{3}

^{}

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 dm^{3
}at RTP (Room temperature and pressure)

= 22.4 dm^{3
}at STP (Standard temperature and pressure)

1 dm^{3}
= 1000 cm^{3} = 1 liter

dm = decimeter

iii) mole = concentration ( in mol/ dm^{3} ) x volume ( in
dm^{3} )

concentration = amount of solute / Volume of solvent

Mole ratio formula :

C_{3}H_{8} + 5O_{2}
3 CO_{2} + 4H_{2}0

Mole of C_{3}H_{3 }/ Mole of O_{2} = 1/5

Mole of CO_{2} _{ }/ Mole of O_{2} = 3/5

Combustion Data :

0.1g of a
compound containing carbon, hydrogen, and oxygen only as elements, on
combustionproduced, 0.019g of CO_{2} and 0.0117g of
H_{2}0.Calculate the empirical formula of the compound.

Solution,

Mole of CO_{2} = m/mr = 0.0191/44 =
0.0043

Mole of C = 0.0043

Or, 0.0043 = __m/12__

M = 0.052

Now,

Mole of H_{2}0 = 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 KClO_{4}

_{}

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

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 = Na_{2}S_{2}O_{3}

Therefore, the
formula of the compound is Na_{2}S_{2}O_{3}

# 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) = Îµ isotopic abundance x isotopic mass

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

Isotopic mass.2

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