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Chapter 7: Question 41 E (page 397)

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.

Short Answer

Expert verified

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

The Lewis structure is:

Step by step solution

01

Define Chemical Bonding

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

02

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:

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Most popular questions from this chapter

Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements: (a)\({\rm{CI}}\)(b)\({\rm{Na}}\)(c)\({\rm{Mg}}\)(d)\({\rm{Ca}}\)(e)\({\rm{K}}\)(f)\({\rm{Br}}\)(g)\({\rm{Sr}}\)(h)\({\rm{F}}\).

Determine the formal charge of each element in the following:

(a) \({\rm{HCl}}\)

(b) \({\rm{C}}{{\rm{F}}_{\rm{4}}}\)

(c) \({\rm{PC}}{{\rm{l}}_{\rm{3}}}\)

(d) \({\rm{P}}{{\rm{F}}_{\rm{5}}}\)

Question: Use principles of atomic structure to answer each of the following:

(a) The radius of the \({\rm{Ca}}\) atom is \({\rm{197 pm}}\); the radius of the \({\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}\) ion is \({\rm{99 pm}}\). Account for the difference.

(b) The lattice energy of \({\rm{CaO(s)}}\) is \({\rm{ - 3460 kJ/mol}}\); the lattice energy of \({{\rm{K}}_{\rm{2}}}{\rm{O}}\) is \({\rm{ - 2240 kJ/mol}}\). Account for the difference.

(c) Given these ionization values, explain the difference between \({\rm{Ca}}\) and \({\rm{K}}\) with regard to their first and second ionization energies.

(d) The first ionization energy of \({\rm{Mg}}\) is \({\rm{738 kJ/mol}}\) and that of \({\rm{Al}}\) is \({\rm{578 kJ/mol}}\). Account for this difference.

Why is the \({\rm{H - N - H}}\) angle in \({\rm{N}}{{\rm{H}}_{\rm{3}}}\) smaller than the \({\rm{H - C - H}}\) bond angle in \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)? Why is the \({\rm{H - N - H}}\) angle in \({\rm{NH}}_{\rm{4}}^{\rm{ + }}\) identical to the \({\rm{H - C - H}}\) bond angle in \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)?

Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?

  1. \({\rm{Cl}}{{\rm{F}}_{\rm{5}}}\)
  2. \({\rm{Cl}}{{\rm{O}}_{\rm{2}}}{\rm{ - }}\)
  3. \({\rm{TeC}}{{\rm{l}}_{\rm{4}}}^{{\rm{2 - }}}\)
  4. \({\rm{PC}}{{\rm{l}}_{\rm{3}}}\)
  5. \({\rm{Se}}{{\rm{F}}_{\rm{4}}}\)
  6. \({\rm{P}}{{\rm{H}}_{\rm{2}}}^{\rm{ - }}\)
  7. \({\rm{Xe}}{{\rm{F}}_{\rm{2}}}\)
See all solutions

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