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Answers
Step-by-step explanation:
Given,
It contains 4.07 % H and 24.27 % C
Moles of H = Weight/Atomic wt.of H
= 4.07/1
= ~4.
Moles of C = Weight/Atomic wt. of C
= 24.27/12
= ~2
Moles of Cl = Weight/Atomic wt. of Cl
= 71.65/35.5
= ~2
On dividing the ratios by the least value we get,
(4/2) : (2/2) :( 2/2)
= 2 : 1 : 1
Therefore,
The empirical formula is : CH₂Cl.
(ii)
The empirical formula mass of CH₂Cl = 49.48 u
The molecular mass = 98.96 u
The molecular mass must be an integral multiple of the empirical formula mass.
= 98.96/49.48
= 2
It means,
The molecular formula must be twice the empirical formula.
Therefore,
Molecular Formula = C₂H₄Cl₂
Hope it helps!
4.07 % H and 24.27 % C
Moles of H = Weight/Atomic wt.of H
= 4.07/1
= ~4.
Moles of C = Weight/Atomic wt. of C
= 24.27/12
= ~2
Moles of Cl = Weight/Atomic wt. of Cl
= 71.65/35.5
= ~2
On dividing the ratios by the least value we get,
(4/2) : (2/2) :( 2/2)
= 2 : 1 : 1
Therefore,
The empirical formula is : CH₂Cl.
(ii)The empirical formula mass of CH₂Cl = 49.48 u
The molecular mass = 98.96 u
The molecular mass must be an integral multiple of the empirical formula mass.
= 98.96/49.48
= 2
It means,
The molecular formula must be twice the empirical formula.
Therefore,
Molecular Formula = C₂H₄Cl₂
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