Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% c, 13.1% h, and 34.7% o by mass. write the lewis structures for the two molecules.

Answer : The Lewis structures for the two molecules are shown below.

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 52.2 g

Mass of H = 13.1 g

Mass of O = 34.7 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C =

Moles of H =

Moles of O =

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =

For H =

For O =

The ratio of C : H : O = 2 : 6 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =  =

The empirical formula weight = 12(2) + 6(1) + 1(16) = 46 gram/eq

Now we have to calculate the value of 'n'.

Formula used :

Molecular formula =

So, there are two possibilities for the arrangements of atoms. That means, it will be an ethanol or dimethyl ether .

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

As we know that carbon has '4' valence electrons, oxygen has '6' valence electrons and hydrogen has '1' valence electron.

Therefore, the total number of valence electrons in = 2(4) + 6(1) + 6 = 20

According to Lewis-dot structure, there are 16 number of bonding electrons and 4 number of non-bonding electrons.

Thus, the Lewis structures for the two molecules are shown below.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2