An organic compound consist of carbon, hydrogen and sulphur. The 3 volume of the organic compound undergoes combustion with excess of oxygen producing 3 volume of Co2, 3 volume of So2, 6 volume of vapour.
undergoes of Co2, 6 volume of vapour. Find out the molecular formula.
Answers
Answer:
By the end of this section, you will be able to:
Use the ideal gas law to compute gas densities and molar masses
Perform stoichiometric calculations involving gaseous substances
State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures
The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it.”[1]
As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.
The molecular formula of the compound is C
x
H
2y
O
y
C
x
H
2y
O
y
+xO
2
→xCO
2
+yH
2
O.
The amount of oxygen is twice the required amount i.e 2x.
The hot gases when cooled to 0
0
C and 1 atm pressure measured 2.24L.
This corresponds to 0.1 mole as 22.4 L of any gas at NTP corresponds to 1 mole.
x+x+y=0.1
2x+y=0.1 .....(1)
The water collected during cooling weighs 0.9 g. This correspnds to
18
0.9
=0.05 moles.
Thus y=0.05 .....(2)
Substitute equation (2) in equation (1)
2x+0.05=0.1
2x=0.05
x=0.025 ......(3)
The ration x:y is 0.025:0.05 or 1:2.
Hence, the empirical formula of the organic compound is CH
4
O
2
.
The empirical formula mass is 12+4+32=48g/mol.
p
0
p
0
−p
=
M
2
W
2
×
W
1
M
1
17.5
0.104
=
M
2
50
×
1000
0.018
17.5
0.104
=
M
2
50
×
1000
0.018
M
2
=0.151kg=151g
r=
empirical mass
Molecular mass
=
48
151
=3.1≃3
Therefore, the molecular formula of the organic gas is 3×CH
4
O
2
=C
3
H
12
O
6
.