Question

A compound with a molar mass of about\({\rm{28 g/mol}}\)contains\({\rm{85}}{\rm{.7 \% }}\)carbon and\({\rm{14}}{\rm{.3 \% }}\)hydrogen by mass. Write the Lewis structure for a molecule of the compound.

Short answer

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

The Lewis Structure is:

Step-by-step solution

Define Chemical Bonding

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

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 are:

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

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

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

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

Consequently, the formula is \({\rm{C}}{{\rm{H}}_{\rm{2}}}\).

The Lewis structure is as follows:-

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