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%
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 C3H3 / 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
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