Question 41 E - Chemistry (OpenStax)

Question

A compound with a molar mass of about \({\rm{42 g/mol}}\) contains \({\rm{85}}{\rm{.7 \% }}\) carbon and \({\rm{14}}{\rm{.3 \% }}\) hydrogen by mass. Write the Lewis structure for a molecule of the compound.

Step-by-Step Solution

Answer

The formula is: \({{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}\).

The Lewis structure is:

Step 1: Define Chemical Bonding

A chemical bond is a long-term attraction between atoms, ions, or molecules that allows chemical compounds to form.

Step 2: Writing the Lewis symbol

In a \({\rm{100}}{\rm{.0 g}}\) sample, there are \({\rm{85}}{\rm{.7 g C}}\) and \({\rm{14}}{\rm{.3 g H}}\).

Now we must compute the mole of \({\rm{C}}\) and \({\rm{H}}\), which we shall accomplish by dividing their mass by their molecular weight, as follows:-

The moles of carbon is:

\(\dfrac{{{\rm{85}}{\rm{.7 g}}}}{{{\rm{12}}{\rm{.001 g mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{\rm{   =  7}}{\rm{.14 mol C}}\)

The moles of hydrogen is:

\(\dfrac{{{\rm{14}}{\rm{.3 g}}}}{{{\rm{1}}{\rm{.0079 g mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{\rm{   =  14}}{\rm{.19 mol H}}\)

To compute the formula, divide the mole by the smallest mole:-

\(\dfrac{{{\rm{7}}{\rm{.14 mol}}}}{{{\rm{7}}{\rm{.14 mol}}}}{\rm{  =  1 C}}\)

\(\dfrac{{{\rm{14}}{\rm{.19 mol}}}}{{{\rm{7}}{\rm{.14 mol}}}}{\rm{  =  2 H}}\)

The formula is \({\rm{C}}{{\rm{H}}_{\rm{2}}}\) having the molecular mass as \({\rm{14}}\).

But, here, the molecular mass of the compound is \({\rm{42}}\).

So, the correct formula will be \({\rm{3  \times   C}}{{\rm{H}}_{\rm{2}}}\) or \({{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}\) .

The Lewis structure is as follows:-

Therefore, the formula is \({{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}\) and the Lewis structure is: