ethylene glycol is used as an antifreeze combustion of 6.38 mg of Ethylene glycol give 9.06 mg of carbon dioxide and 5.58 mg of water ethylene glycol contains carbon hydrogen and oxygen and its molecular mass is 62 find empirical formula and molecular formula
Answers
Explanation:
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Answer:
Answer: The empirical and molecular formula for the given organic compound is CH_3OCH
3
O and C_4H_{12}O_4C
4
H
12
O
4
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
C_xH_yO_z+O_2\rightarrow CO_2+H_2OC
x
H
y
O
z
+O
2
→CO
2
+H
2
O
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of CO_2=9.06gCO
2
=9.06g
Mass of H_2O=5.58gH
2
O=5.58g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 9.06 g of carbon dioxide, \frac{12}{44}\times 9.06=2.47g
44
12
×9.06=2.47g of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 5.58 g of water, \frac{2}{18}\times 5.58=0.62g
18
2
×5.58=0.62g of hydrogen will be contained.
Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles
Molar mass of Carbon
Given mass of Carbon
=
12g/mole
2.47g
=0.206moles
Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles
Molar mass of Hydrogen
Given mass of Hydrogen
=
1g/mole
0.62g
=0.62moles
Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles
Molar mass of oxygen
Given mass of oxygen
=
16g/mole
3.29g
=0.206moles
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.
For Carbon = \frac{0.206}{0.206}=1
0.206
0.206
=1
For Hydrogen = \frac{0.62}{0.206}=3
0.206
0.62
=3
For Oxygen = \frac{0.206}{0.206}=1
0.206
0.206
=1
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 1 : 3 : 1
Hence, the empirical formula for the given compound is C_1H_{3}O_1=CH_3OC
1
H
3
O
1
=CH
3
O
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
n=\frac{\text{molecular mass}}{\text{empirical mass}}n=
empirical mass
molecular mass
We are given:
Mass of molecular formula = 124 amu = 124 g/mol
Mass of empirical formula = 31 g/mol
Putting values in above equation, we get:
n=\frac{124g/mol}{31g/mol}=4n=
31g/mol
124g/mol
=4
Multiplying this valency by the subscript of every element of empirical formula, we get:
C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4C
(1×4)
H
(3×4)
O
(1×4)
=C
4
H
12
O
4
Thus, the empirical and molecular formula for the given organic compound is CH_3OCH
3
O and C_4H_{12}O_4C
4
H
12
O
4
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