Question 42 E - Chemistry (OpenStax)

Question

Two arrangements of atoms are possible for a compound with a molar mass of about \({\rm{45 g/mol}}\) that contains \({\rm{52}}{\rm{.2 \%  C, 13}}{\rm{.1 \%  H}}\), and \({\rm{34}}{\rm{.7 \%  O}}\) by mass. Write the Lewis structures for the two molecules.

Step-by-Step Solution

Answer

The formula is: \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}\).

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 \({\rm{52}}{\rm{.2 g C}}\), \({\rm{34}}{\rm{.7 g O}}\) and \({\rm{13}}{\rm{.1 g H}}\).

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 are:

\(\frac{{{\rm{52}}{\rm{.2 g}}}}{{{\rm{12}}{\rm{.001 g mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{\rm{   =  4}}{\rm{.346 mol C}}\)

The moles of hydrogen are:

\(\frac{{{\rm{13}}{\rm{.1 g}}}}{{{\rm{1}}{\rm{.0079 g mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{\rm{   =  12}}{\rm{.997 mol H}}\)

The moles of oxygen are:

\(\frac{{{\rm{34}}{\rm{.7 g}}}}{{{\rm{15}}{\rm{.9994 g mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{\rm{   =  2}}{\rm{.1668 mol O}}\)

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

\(\frac{{{\rm{4}}{\rm{.346 mol}}}}{{{\rm{2}}{\rm{.1688 mol}}}}{\rm{  =  2 C}}\)

\(\frac{{{\rm{12}}{\rm{.997 mol}}}}{{{\rm{2}}{\rm{.1688 mol}}}}{\rm{  =  6 H}}\)

\(\frac{{{\rm{2}}{\rm{.1688 mol}}}}{{{\rm{2}}{\rm{.1688 mol}}}}{\rm{  =  1 O}}\)

The formula is \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}\).

The Lewis structure is as follows:-

Therefore, the formula is \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}\) and the Lewis structure are: