An organic acid contains C=70.59% and H=5.88%.Its vapour density was found 68.what is the molecular formulae of the compound.
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Answered by
3
It's formula is C8H8O2.
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evilgodfahim1:
can you post the procedure please
I did. Mark me as brainliest when you can!!
Answered by
66
Heya !
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◼Firstly , let's know what is molecular formula and empirical formula ?
→Empirical Formula - It gives a simple whole number ratio of the atoms and elements present in one molecule of compound .
→Molecular Formula - It gives an exact estimate of the atoms and elements present in one molecule of compound .
=> For example , the empirical formula of Benzene is CH and it's molecular formula is C6H6 .
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=> Let's move to the question.
Step. 1 ◼ Write down the % composition of Carbon ( C ) and Hydrogen ( H) as given . Here we see that the percentage doesn't add up to 100. So the remaining percentage is of Oxygen.
•°• % of Oxygen = 100 - (70.59 + 5.88 )
=> % Of Oxygen = 23.53%...............[ 1 ]
Step 2 ◼ Now , we shall calculate the number of moles (n) thus, divide the % composition by the atomic mass. We have -
◾For Carbon => 70.59 / 12 = 5.88 ( 2 )
◾For Hydrogen => 5.88/1 = 5.88 ( 3 )
◾For Oxygen => 23.53 / 16 = 1.47 ( 4 )
Step 3 ◼ Divide equations (2) (3) and (4) By the smallest value among the three.
The smallest value is 1.47 → So divide all three values by 1.47
▪For Carbon => 5.88/ 1.47 = 4
▪For Hydrogen => 5.88 / 1.47 = 4
▪For Oxygen => 1.47/1.47 = 1
°•° Empirical Formula = C4H4O ✔
Step 4 ◼ Now we need to calculate the molecular formula.
=> We have the relation -
→ n × empirical formula = molecular formula
[ n = molecular mass / empirical formula mass ]
→Molecular Formula = Vapour Density × 2
→Molecular Formula = 68 × 2 = > 136 ✔
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★ Empirical Formula mass = Mass of C4H4O
=> 12 × 4 + 1 × 4 + 16 = 48 + 4 + 16
=> 68 ✔
•°• n = 136 / 68 = 2 ✔
=> Molecular formula = 2 × ( C4H4O)
◼Molecular Formula is C8H8O ◼
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_____
______________________________________________________
◼Firstly , let's know what is molecular formula and empirical formula ?
→Empirical Formula - It gives a simple whole number ratio of the atoms and elements present in one molecule of compound .
→Molecular Formula - It gives an exact estimate of the atoms and elements present in one molecule of compound .
=> For example , the empirical formula of Benzene is CH and it's molecular formula is C6H6 .
____________________________________________
=> Let's move to the question.
Step. 1 ◼ Write down the % composition of Carbon ( C ) and Hydrogen ( H) as given . Here we see that the percentage doesn't add up to 100. So the remaining percentage is of Oxygen.
•°• % of Oxygen = 100 - (70.59 + 5.88 )
=> % Of Oxygen = 23.53%...............[ 1 ]
Step 2 ◼ Now , we shall calculate the number of moles (n) thus, divide the % composition by the atomic mass. We have -
◾For Carbon => 70.59 / 12 = 5.88 ( 2 )
◾For Hydrogen => 5.88/1 = 5.88 ( 3 )
◾For Oxygen => 23.53 / 16 = 1.47 ( 4 )
Step 3 ◼ Divide equations (2) (3) and (4) By the smallest value among the three.
The smallest value is 1.47 → So divide all three values by 1.47
▪For Carbon => 5.88/ 1.47 = 4
▪For Hydrogen => 5.88 / 1.47 = 4
▪For Oxygen => 1.47/1.47 = 1
°•° Empirical Formula = C4H4O ✔
Step 4 ◼ Now we need to calculate the molecular formula.
=> We have the relation -
→ n × empirical formula = molecular formula
[ n = molecular mass / empirical formula mass ]
→Molecular Formula = Vapour Density × 2
→Molecular Formula = 68 × 2 = > 136 ✔
===============================
★ Empirical Formula mass = Mass of C4H4O
=> 12 × 4 + 1 × 4 + 16 = 48 + 4 + 16
=> 68 ✔
•°• n = 136 / 68 = 2 ✔
=> Molecular formula = 2 × ( C4H4O)
◼Molecular Formula is C8H8O ◼
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