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.

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

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: