1. Chemical Formulas and Chemical Compounds
Chapter 7
Chemical Formulas and Chemical Compounds

2. 7.1 Chemical Names and Formulas
Chemical formulas show the relative #’s of atoms in a chemical compound
Examples
C12H22O11 C = 12 H = 22 O = 11
Pb(NO3)4 Pb = 1 N = O = 12
(NH4)2CrO4 N = 2 H = 8 Cr = 1 O = 4

3. Chemical Names and Formulas
Naming monatomic ions
End in -ide
Look up the charges on the periodic table
F vs F-1 F
F-1
S vs S-2 S
S-2
Fluorine
Fluoride
Sulfur
Sulfide

4. Writing Formulas for Ionic Compounds
Cations – positive ions (metal)
metal
Anions – negative ions
nonmetal
Charge on compound = 0

5. Writing Formulas for Ionic Compounds
Binary ionic compounds
Rules
if the charges on the ions are the same, drop ‘em
if the charges are different, criss-cross
Same charges –
Na+1 Cl-1 -
Mg+2 O-2+-
Different charges-
Na+1 S-2-
Mg+2 Cl-1-
NaCl Sodium Chloride
MgO Magnesium Oxide
Na2S Sodium Sulfide
MgCl Magnesium Chloride

6. Writing Formulas for Ionic Compounds
Ternary Ionic compounds
Metal + (Polyatomic ion)
When naming, do not use the ending –ide
Sodium Nitrate
Sodium Carbonate
Aluminum Nitrate
Na+1 NO3-1
Na(NO3)
Na+1 CO3-2
Na2(CO3)
Al+3 NO3-1
Al(NO3)3

7. Writing Formulas for Ionic Compounds
Aluminum Phosphate
Aluminum Bihypophosphite
Aluminum Carbonate
Al+3 PO4-3
Al(PO4)
Al+3 HPO2-2
Al2 (HPO2)3
Al+3 CO3-2
Al2 (CO3)3

8. Writing Names for Ionic Compounds
Front name – positive (cation – metal)
Back name – negative (anion – nonmetal)
Binary ionic compounds – composed of only 2 types of elements ( M + NM) – end in -ide
NaCl
MgCl2
Al2O3
NaH
Sodium Chloride
Magnesium Chloride
Aluminum Oxide
Sodium Hydride

9. The BIG Lie
Stock System – use Roman Numerals for naming compounds with metals that have multiple charges (the transitions!)
Normal Name
Symbol
Charge
Stock Name
Copper Cu +1
Cu (I) +2
Cu (II)
Mercury Hg
Hg (I)
Hg (II)
Lead Pb
Pb (II) +4
Pb (IV)
Tin Sn
Sn (II)
Sn (IV)
Iron Fe
Fe (II) +3
Fe (III)
Cory
Matthews
Loves
Topanga
Intensely

10. The BIG Lie More exceptions to the Lie! Ag is always = +1 charge
DON’T write Ag I
Zn always = +2 charge
DON’T write Zn II

11. Practice! Sn3N2 AgOH PbCO3 Zn(OH)2 Fe2(SO4)3 Tin (II) Nitride
Silver Hydroxide
Lead (II) Carbonate
Zinc Hydroxide
Iron (III) Sulfate

12. More Practice!! CuSO2 CuHSO2 Cu(HSO2)2 Copper (II) Bihyposulfite
Copper (II) Hyposulfite
Copper (I) Bihyposulfite

13. Practice! LiClO3 LiClO2 CaCO3 Ca(HCO2)2 Fe(NO3)3 Lithium Chlorate
Lithium Chlorite
Calcium Carbonate
Calcium Bicarbonite
Iron (III) Nitrate

14. Writing Names for Molecular Compounds
Molecular Compounds – covalent compounds
2 nonmetals
To name, we use prefixes #
Prefix 1
mono 6
hexa 2 di 7
hepta 3
tri 8
octa 4
tetra 9
nona 5
penta 10
deca
don’t use the prefix Mono on the first atom

15. Writing Names for Molecular Compounds
Prefix-name prefix-name-ide CO
CO2
PCl3
CBr4
N2O5
SF6
Carbon Monoxide
Carbon Dioxide
Phosphorous trichloride
Calcium tetrabromide
Dinitrogen pentoxide
Sulfur hexafluoride

16. Writing Formulas for Molecular Compounds
The prefixes = the subscripts.
Do NOT look at the charges.
Sulfur Dioxide
Disulfur Trioxide
Dinitrogen pentoxide
SO2
S2O3
N2O5

17. Naming Acids Acid - when a solution yields H+ ions in solution 2 types
Binary
H and one other type of atom
ternary (sometimes called oxy)
acids that have H with a polyatomic ion

18. Naming Binary Acids Rules Hydro__(begninng of name)__ic acid Ex. HCl
Hydrochloric acid
HBr HF
H2S
H3P
Hydrobromic acid
Hydrofluoric acid
Hydrosulfuric acid
Hydrophosphoric acid

19. Writing Formulas from Names for Acids
Do the criss-cross
Ex. Hydronitric acid
H+1 N H3N
Hydroiodic acid
H+1 I HI
Hydrosulfuric acid
H+1 S H2S

20. Naming Ternary Acids H + polyatomic Rules Do NOT start with hydro-
If the ending of polyatomic is –ate ic + acid
If the ending of polyatomic is –ite ous + acid
Ate/ite ic/ous
Example:
H2SO4
H+ and SO4-2 – sulfate
sulfuric acid

21. Naming Ternary Acids
H2SO3
HClO4
HClO3
HClO2
HClO

22. Formulas for Ternary Acids
Use the criss-cross method
Nitric acid
Phosphorous acid

23. 7.2 Oxidation Numbers
Since electrons are shared, there is no definite charge - we assign the more electronegative element the “apparent” negative charge - this is known as the oxidation #
oxidation numbers can also be positive.
oxidation # - a number assigned to an atom to show the distribution of elements

24. Oxidation Numbers Rules Free elements = 0 Ex. Mg = O Ions = charges
Ex. F = -1
S = -2
Oxygen (0) = -2
except in peroxides (H2O2) O = -1
H = +1
except in metal hydrides (MgH2, NaH) H = -1

25. Oxidation Numbers ….Rules More electronegative atom gets a (-) charge
Ox #’s add up to 0 in compounds
Ox #’s = the charge in polyatomic ions

26. Oxidation # Practice FeO (Iron II Oxide) O = -2 Fe = ?
Fe2O3 (Iron III oxide)
O = -2
Fe = ?
-2 + x = 0
x = 2
3(-2) + 2x = 0
x = 3

27. Oxidation # Practice H2SO4 (Hydrogen Sulfate or Sulfuric Acid) O = -2

28. Oxidation # Practice H2SO3 (Hydrogen sulfite or sulfurous acid) H =+1 +4 -2
SO3-2
X + 3(-2) = -2
X = 4

29. Oxidation # Practice H2Cr2O7 (Hydrogen Dichromate or Dichromic Acid
NO3-1 (Nitrate)
N =
O = -2
MgH2 (Magnesium hydride)
Mg =
H = -1 +6 +4 +2

30. 7.3 Using Chemical Formulas
Step 1 – be able to calculate molar mass (aka – formula mass, molecular weight, atomic weight, atomic mass, gram formula weight, etc.)
Add atomic weights from the periodic table
round to the nearest 10th place
Examples
CH4
MgSO4· 7H2O
(1.0) = 16.0 g/mol
(16.0) + 14(1.0) + 7(16.0) = g/mol

31. 7.3 Using Chemical Formulas
Step 2 – be able to convert between grams, moles, particles, and liters
Liters
Atoms, molecules, particles
Grams
Mole
22.4 L
6.022 x1023
Molar Mass

32. Using Chemical Formulas
Convert 32.0 g of CH4 to moles, liters, molecules, total atoms, atoms of H
Moles
Liters
Molecules
Atoms
Atoms H

33. Conversions Moles CH4 Liters 32.0 g 1 mole 1 16.0g 32.0g 1 mole 22.4L

34. Conversions Molecules Atoms Atoms H 32.0g 1 mole 6.022 x 1023molecules
32.0g 1 mole x 1023atoms 5
Atoms H
32.0g 1 mole x 1023atoms 4
1.20 x 1024molecules
6.00 x 1024molecules
4.80 x 1024molecules

35. Percent Composition
Percentage Composition - every compound has a certain percentage of each type of atom (we measure it by mass)
Formula
% composition = mass element
mass compound
X 100 =

36. Practice - % Composition
Calculate % composition if a compound contains 24 g of Carbon and 64 g of Oxygen
% composition = mass element
mass compound
Total mass compound = = 88 g
% Composition C = x 100 = 27 % C 88
% Composition O = x 100 = 73% O
X 100 =

37. Practice - % Composition
What is the % composition of Ba(OH)2?
Ba = g
O = 2 (16.0) = 32.0g
H = 2 (1.0) = 2.0g
Ba(OH)2 = g
%Ba = 137.3
171.3
= 80.2%
%O = 32.0
171.3
=18.7%
%H = 2.0
171.3
=1.1%

38. Practice - % Composition
What is the % composition of C6H12O6
C = 6 (12.0) = 72.0g
H = 12 (1.0) = 12.0g
O = 6 (16.0) = 96.0g
C6H12O6 = 180.0g
40.0%
6.7%
53.3%

39. 7.4 Determining a compound’s empirical and molecular formula
Empirical formula - the lowest whole number ratio of atoms in a compound (simplest formula)
4 rules to find empirical formula
Cross out the % → change to grams
Divide each by own molar mass
Divide by the smallest number
If needed, multiply by 2 or 3 ONLY if a whole number ratio isn’t the result of step 3

40. Determining a compound’s empirical formula
Ex1 Calculate the empirical formula if there is % C, 13.04% H, and % O
52.17 g C 1 mole = = g
13.04 g H 1 mole = = g
34.78 g O 1 mole = = g 2
2.17 6
2.17 1
2.17
C2H6O

41. Determining a compound’s empirical formula
Ex2 Calculate the empirical formula is there is % K, % Cr, and % O
26.56 g K mole =
39.1 g
35.41 g Cr 1 mole =
52.o g
38.03 g O mole =
16.0 g
.68
= 1 x2
.68
.68
= 1 x2
.68
2.38
= 3.5 x2
.68
K2Cr2O7

42. Determining a compound’s empirical formula
Ex3 Find the empirical formula if a sample contains 5.6 g N and 12.8 g O

43. Determining a compound’s molecular formula
Rules
Same steps as empirical formula + 3 more
Find the mass of the empirical compound
Divide this mass by the given molecular weight
Multiply the empirical formula by this number

44. Determining a compound’s molecular formula
Find the molecular formula of a compound (MW = g) with % C, % H, and % O
Empirical = C4H8O
Mass empirical = 72.0 g
144.0 g/72.0 g = 2
Molecular Formula = C8H16O2

45. Determining a compound’s molecular formula
Find the molecular formula of that compound that contains 19.80% C, 2.50% H, 66.10% O, 11.60% N and MW = 242.0
Empirical Formula = C2H3O5N
Mass Empirical = 121 g
242 g / 121 g = 2
C4H3O10N2