A hydrocarbon contains 4.8 g of carbon per gram of hydrogen. Calculate: the g atom of each, find the emperical formula, find molecular formula, if its vapour density is 29
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Given:
Mass Ratio of Carbon and Hydrogen = 4.8g : 1g
Vapur Denisity (V.D): 29 ⇒ Molecular Wt. (M) = 2 x V.D = 2 x 29 = 58g
Let Wt of Carbon be 4.8K gms and Hydrogen be 1K gms.
Total Wt. 4.8K + 1K = 58gms
On Simplification K = 10
Wt of Carbon = 4.8 x 10 = 48 gms
Similarly Wt. Of Hydrogen = 10 gms
No. of Moles = Molar wt / Molcular Wt
No of Moles of Carbon = 48/12 = 4 moles
Similary No of moles of Hygrogen = 10/1 = 10 moles
Molecular Formula = C4H10
Empherical Formula = C2H5
Mass Ratio of Carbon and Hydrogen = 4.8g : 1g
Vapur Denisity (V.D): 29 ⇒ Molecular Wt. (M) = 2 x V.D = 2 x 29 = 58g
Let Wt of Carbon be 4.8K gms and Hydrogen be 1K gms.
Total Wt. 4.8K + 1K = 58gms
On Simplification K = 10
Wt of Carbon = 4.8 x 10 = 48 gms
Similarly Wt. Of Hydrogen = 10 gms
No. of Moles = Molar wt / Molcular Wt
No of Moles of Carbon = 48/12 = 4 moles
Similary No of moles of Hygrogen = 10/1 = 10 moles
Molecular Formula = C4H10
Empherical Formula = C2H5
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