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## Feb 2 Chapter 3 Notes

Compounds

• Types of chemical bonds
• Covalent Bonds
• Sharing of electrons between atoms
• Ionic Bonds
• Transfer of electrons from one atom to another
• Molecular Compound
• Made up of discrete units called molecules, which consist of a small number of atoms held together by covalent bonds.
• They are represented by chemical formulas, which indicate
• Elements
• Relative number of atoms
• Example:
• H2O
• The numbers, written in subscript, indicate the relative number of atoms inside the compound.
• Empirical Formula
• Simplest formula for a compound. Shows the types of atoms present and their relative numbers.
• So, it shows the elements and the subscripts are reduced to their simplest whole-number ratio.
• Example:
• Glucose is C6H12O6 (molecular formula), but in the empirical formula, it would be CH2O, just like Acetic Acid (C2H4O2)
• Molecular Formula
• This formula is based on an actual molecule. Sometimes, both formulas are identical, if the ratios cannot be reduced. Other times, they are multiples of the empirical formula.
• Example:
• Acetic Acid has two C atoms, four H atoms, and two O atoms, so the molecular formula is C2H4O2.
• They show nothing about how the molecule is attached, unlike the next formula.
• Structural Formula
• Shows the order in which atoms are bonded together and by what types of bonds. NOTE: The formula is a drawing of a molecule
• Covalent bonds are shown with dashes (-), and double bonds are shown with double lines (=)
• Double bond is tighter than a single bond
• Condensed Structural Formula
• Structural formula written out- shows order, just written on a line
• Example:
• Acetic Acid: CH3COOH or CH3CO2H, that formula shows all of the atoms in the molecule, and how the connect, while being on one line.
• Organic Compound
• Made up of primarily Carbon and Hydrogen, with Oxygen or Nitrogen
• Each Carbon atom has four covalent bonds.
• Sometimes, they're very complex, and they use lines to represent bonds, called the line-angle formula (shapes and lines are used, very common)
• Ionic Compound
• Combination of a metal and a nonmetal
• Made up of positive and negative ions joined together by electrostatic forces of attraction.
• Metallic elements lose electrons, and nonmetals gain electrons
• As a result, the metal becomes a positive ion, or a  cation, and the nonmetal atom becomes a nonmetal ion, or an anion.
• Example
• In the formation of salt (sodium chloride), each sodium atom gives up one electron to become a sodium ion, Na+, and each chlorine atom gains an electron, becoming Cl-
• Formula Unit
•  Smallest electrically neutral collection of ions. For example, in Magnesium Chloride, Magnesium gives up two electrons, and Chlorine only gets one, so it needs two Chlorine atoms, hence the molecular formula being MgCl2.

3-2

The Mole Concept and Chemical Compounds

• Formula mass
• Mass of a formula unit  (empirical formula) in atomic mass units.
• Molecular mass
• Mass of a molecule in atomic mass units
• Example:
• H20 = 2(atom. mass of H) + (atom mass of O)
• = 2(1.008) + 15.999 = 18.015
• Molar mass
• Mass of a mole (6.022 * 10^23 of molecules) of a compound.
• Diatomic Molecules
• Molecules composed of two atoms
• Diatomic Elements, Elements that occur made of two atoms
• H2 O2 N2 Cl2 I2 F2 Br2

3-3

Composition: Example: How many moles of F atoms are in a 75.0 mL sample of halothane (d = 1.871 g/mL)? (Halothane = C2HBrClF3. Its molar mass is 193.78 g/mol)

Solution:

First, convert the halothane from mL to grams.

75.0 mL of Halothane * 1.871 grams/(1 ml of Halothane) = 138.45 grams.

Now, we need the Halothane in moles.

Moles of Halothane = 138.45/(molar mass of Halothane) = 138.45/197.4 = .701 Moles of Halothane

Now, given a mole of halothane, we must now convert to moles of Fluorine. We know there are 3 Fluorine atoms in a Halothane molecule. So, the moles of Fluorine become:

Moles of Fluorine = .701 Moles * 3 Fluorine atoms = 2.103 mol F.

Percent Composition of Compounds

Example problem:

Dibutyl succinate is composed of 62.58% C, 9.63% H, and 27.79% O. Its molecular mass is 230 u. What are the empirical (simplest form) and molecular (actual) formulas of dibutyl succinate?

So, first, determine the mass of each in a 100.0-g sample.

62.58 g C, 9.63 g H, 27.79 O

The reason we do that is because we want to find the moles in each, then find them in their simplest forms. So we convert to moles.

62.58 g/12.011 g/mol = 5.21 mol C

9.63 g/1.00794 g/mol = 9.55 mol H

27.79/15.9994 = 1.736 mol O

Now, we divide by the smallest number of mols (1.736).

It becomes C3H5.5O. There cannot be 5.5 Hydrogen atoms in a molecule, so you have multiply by two to make the molecule possible. So, the empirical formula becomes C6H11O2

Now, the molar mass of that is (6 * 12.011) + (11 * 1.00794) + (2 * 15.9994) = around 115. So we multiply by two to get 230 u. The molecular formula is C12H22O4.

David Witten