Question

A gaseous compound has a composition by mass of 29.8% carbon, 2.1% hydrogen, and oxygen. The gas has a molar mass of about 50g/mol. Draw the most likely Lewis structure that fits these facts. Label each central atom with a hybridzation, bond angle, and Lewis notation. Indicate if the molecule has a dipole with an arrow, or else write that the molecule has no dipole.

Answer 1

Molecular mass of the compound is around 50 g/mol.

Mass percent of carbon = 29.8 %.

Hence, mass of carbon present is

molar mass of carbon = 12 g/mol.

number of moles of carbon present =

mass percent of Hydrogen = 2.1%

mass of carbon present =

Atomic mass of H = 1g/mol

hence, number of moles of H present is

The remaining mass must belong to O.

Hence, mass of oxygen is

Atomic mass of O is 16 g/mol.

hence, number of moles of O is

hence, the empirical formula of the compound is .

We can get the molecular formula by rounding up the fractions to nearest whole number,

Molecular formula is .

The central atom will be Carbon as it is less electronegative than O.

Valence electron, C =4 , O=6, and H=1.

We have to put an extra electron on one of the O atom to satisfy its octet.

Since there are 3 different group attached to the central Carbon atom, we need 3 hybrid orbital to put the bonding electrons. Hence, the hybridisation must be sp2.

In sp2 hybridisation, the bond angles will be 120.

Oxygen is more electronegative than Carbon. Hence, there will be bond dipoles on both C-O bonds. Also there will be resonance stabilisation of the C=O pi bond electrons.

the two bond vector dipoles along C-O bonds will be vectorically added to give a net dipole directed in between the two C-O bonds. Since, the electronegativity of C and H is almost similar, the C-H bond will be a non-polar bond and will contribute very small amount(almost zero) to the net dipole,