Chapter 7: Chemical Formulas and Chemical Compounds

Section 1: Chemical Names and Formulas (Emily Brown)

Section 2: Oxidation Numbers (Anu Ninan)Section 3: Using Chemical Formulas (Johana Barxha)Section 4: Determining Chemical Formulas (Justine

Baird)

7.1: Chemical Names and FormulasVocabulary 

Hydrocarbons—molecular compounds composed solely of carbon and hydrogenMonatomic ions—ions formed from a single atomBinary compounds—compounds composed of two different elementsNomenclature—naming systemOxyanions—polyatomic ions that contain oxygenBinary acids—acids that consist of two elements, usually hydrogen and one of the halogens (fluorine, chlorine, bromine, iodine)Oxyacids—acids that contain hydrogen, oxygen, and a third element (usually a nonmetal)Hydrochloric acid—a water solution of the molecular compound hydrogen chlorideSalt—an ionic compound composed of a cation and the anion from an acid

7.1 The Naming System

These are the prefixes used in the stock system for naming binary molecular compounds:

1. The element with the smaller group number always goes first, except if both elements have the same group number (in which the greatest period number goes first)

2. The second element combines a prefix giving the number of atom contributed, the root name, and an -ide suffix

    

7.1 Naming Cont. 

The following is the naming system for binary ionic compounds:• In this method, the elements are

sorted according to cations (+) and anions (-). These can be determined by an atom's charge

• The positive element always goes first, followed by the anion.

• The cation retains its original name, but the anion drops its ending and adds an -ide suffix

7.2 Oxidation Numbers

VOCABULARY:•  Oxidation numbers: the general distribution of electrons

among the bonded atoms in a molecular compound or a polyatomic ion

• Oxidation states: another name for oxidation numbers

7.2 Steps Involved In Assigning Oxidation Numbers

The main key is that the sum of the oxidation numbers of the atoms equals 0• In a pure element, the oxidation numbers of all the atoms in

0. (ex. pure Na, each sodium atom's oxidation number is 0, and in P4 each phosphorus atom has an oxidation number of 0)

• In a binary molecular compound:o the more electronegative element is given the number

equal to its negative charge it would have as an aniono the less electronegative element is given the number

equal to its positive charge it would have as an anion• Fluorine, F, is ALWAYS given the oxidation number of -1,

because it is the most electronegative

7.2 Steps Involved In Assigning Oxidation Numbers

• Oxygen has an oxidation number of -2 in almost all compounds.  Exceptions: when it is in peroxides or in a compound with fluorine

• The algebraic sum of the oxidation numbers is always equal to 0, as long as the compound is neutral

• The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion

•  Oxidation numbers can also be assigned to ions• monoatomic ions have oxidation numbers equal to the

charge of the ion (ex. Ca2+ has an oxidation number of +2) • Hydrogen, H, has an oxidation number of +1 in compound

where the other atoms are more electronegative,but the oxidation number is -1 in compounds where there are metals

7.2 Oxidation numbers of certain nonmetals

7.2 Steps Involved In Assigning Oxidation Numbers

UF6

 

 

 

H2SO4

 

 

 

CS2

ClO3-

 

 

 

N2O5

 

 

 

P4O10  

7.2 Stock System and OxidationNumbers

Ionic Compound have names based on prefixes as well as names based on the Stock SystemExamplePbO2

can be written as: Lead dioxide  prefix systemLead (IV) oxide  Stock Systembecause oxygen has a charge of -2 and there are to oxygen so it become (-2)(2)=-4 and lead must make the chemical formular balance to 0, so-4+x=0x=4

7.2 Practice with naming using the Stock System

As2S3

  

 

SO3

Mo2O3

NCl3 

 

 

NO 

PCl5

7.3: Formula Masses

• The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula.

 For example, the mass of a water molecule is found by adding the masses of the three atoms in the molecule.         average mass of H: 1.01 amu          average mass of O: 16.00 amu   2 H atoms x  1.01 amu  = 02.02 amu    When added, the                          H atom                              sum equals the  1 O atom  x 16.00 amu  = 16.0 amu       average mass,                          O atom                              about 18.02 amu.               

7.3: Molar Masses

• The molar mass is calculated by summing the masses of the elements in a mole of the molecules that make up the compound.

 For example, the mass of a water molecule is found by adding the masses of the three atoms in the molecule.   2 mol H x  1.01 g H    = 02.02 g H      When added, the sum                     H atom                             equals the molar   1 mol O x 16.00 g O  = 16.0 g O        mass of H2O, about                      O atom                            18.02 g/mol.  

A compound's molar mass is numerically equal to its formula mass.       

       

 

Molar Mass as a Conversion Factor

7.3: Percentage Composition

• The percent composition is the percentage by mass of each element in a compound. It is calculated by dividing the mass of the element of the sample in a compound by the total mass of the sample. Then, multiply this value by 100.

  mass of element in sample of compound x 100 =   % element in         mass of sample of compound                             compound  mass of element in 1 mol of compound    x 100 = % element in        molar mass of compound                                     compound 

 

7.3: Percentage Composition

Find the percent composition of copper(I) sulfide, Cu2S.    

7.3: Percentage Composition

Find the percent composition of Ba(NO3)2.      

 

7.4: Determining Chemical Formulas Empirical Formula- Is the smallest is the smallest whole number ratio of the elements in a compound.    •     For ionic compounds the empirical formula is usually the

formula unit but...• For molecular compounds the empirical formula does not

always represent the number of atoms in the molecule. 

Practice: What is the empirical formula for

    N6H4                                            P2H5

 

    S8O4

 

 

7.4 Finding the Empirical Formula

Steps:1. If given the % composition (if given in grams skip to step 2) figure out the amount of each substance in the sample by putting the % into grams (remember 32% of a substance is 32 grams). 

2. Then divide the amount of each element by it's molar mass (found on the periodic table). 

Question: if a substance has 24% carbon and 76% oxygen what is the empirical formula?1. 24%= 24 grams    84%= 84 grams

2. 24g/12g = 2 mole    76g/16g = 4.75 mole

7.4 Finding the Empirical Formula Con.

3. Next divide the numbers from both elements ( which are in moles now) by the smallest number.

4.Finally round the results to a whole number and write them as a subscript of the appropriate element

3. 2 mol of carbon/ 2= 1    4.75 mol of oxygen/2 = 2.4

4. Co2

7.4 Finding the Molecular Formula 

The Molecular Formula, or the actual formula of a molecular compound can be found using the empirical formula ad the equation x(empirical formula)= the molecular formula. To solve for the molecular formula you...1. Dividing the molecular formula mass (given) by the empirical formula mass ( sum of the element's molar mass)

2. Then multiply the empirical formula by that number rounded to a whole  number.

What is the molecular formula for a  compound with the emperical formula HO and a formula mass of 34 amu?1. 34/ 17 = 2

2. 2(HO) = H2O2

7.4 Practice

What is the empirical formula for a compound that is 64% Sulfur,  20% Carbon and 16 % Oxygen?     If the molecular formula mass for the above empirical formula is 312 grams what is the molecular formula?  What is the difference between a molecular and an  empirical formula?