The empirical and molecular formulas will be and
respectively.
The compound contains C, H, and O.
C = 61.15/12 = 5.0958
H = 5.3/1 = 5.3
O = 31.55/16 = 1.9719
Divide by the smallest
C = 2.6
H = 2.7
O = 1
Thus, the empirical formula is
Empirical formula mass = (12x5) + (1x5) + 16x2 = 97
n = 152.15/97 = 2
The molecular formula is
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Answer:
See attached picture.
Explanation:
Hello,
In this case, for the given name, you can verify the structure on the attached picture, wherein you can see verify the presence of both the ethyl and methyl radicals at the third carbon as well as the triple bond at the first carbon.
Best regards.
Answer : Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.
Explanation :
The given molecule is,
Three types of inter-molecular forces are present in this molecule which are Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.
Hydrogen-bonding is present between the oxygen and hydrogen molecule.
Dipole-dipole forces is present between the carbon and oxygen molecule.
London-dispersion forces is present between the carbon and carbon molecule.