chapter 7 chemical formulas & chemical compounds

Chapter 7Chapter 7Chemical Formulas &Chemical Formulas &Chemical CompoundsChemical Compounds

7.1 Chemical Names & Formulas

IonsIons• Cation: A positive ion

• Mg2+, NH4+

• Anion: A negative ion

• Cl, SO42

• Ionic Bonding: Force of attraction between oppositely charged ions.

Predicting Ionic ChargesGroups 3 - 12Groups 3 - 12:: Many Many transitiontransition elements have elements have

more than one possible more than one possible oxidation state.oxidation state.

Iron (II) = Fe2+

Iron (III) = Fe3+

Predicting Ionic ChargesGroups 3 - 12Groups 3 - 12:: Some Some transitiontransition elements elements

have only one possible have only one possible oxidation state.oxidation state.

Zinc = Zn2+

Silver = Ag1+

Formula Writing for Binary Ionic Compounds

Magnesium Bromide

Mg+ 2 Br – 1

Mg1Br2

MgBr2

Calcium Sulfide

Ca + 2 S – 2

Ca2S2

CaS

criss-cross the oxidation numbers to balance criss-cross the oxidation numbers to balance out the charge.out the charge.

Writing Ionic Compound FormulasExample: Iron (III) chloride

1. Write the formulas for the cation and anion, including CHARGES!

FeFe3+3+ ClCl--2. Check to see if charges are balanced.

3. Balance charges, if necessary, using subscripts.

Not balanced!

33

Naming Ionic CompoundsNaming Ionic Compounds

1. Cation first, then anion

2. Monatomic cation = name of the element

• Ca2+ = calcium ion

3. Monatomic anion = root + -ide• Cl = chloride

• CaCl2 = calcium chloride

Naming Ionic CompoundsNaming Ionic Compounds

• some metal forms more than one cation

• use Roman numeral in name

• PbCl2

• Pb2+ is cation

• PbCl2 = lead (II) chloride

Metals with multiple oxidation statesMetals with multiple oxidation states

Elements with Multiple Oxidation Numbers

Copper ICopper I Cu Cu+1+1

Copper IICopper II Cu Cu+2+2

Iron IIIron II Fe Fe+2+2

Iron IIIIron III Fe Fe+3+3

Mercury IMercury I Hg Hg+1+1

Mercury IIMercury II Hg Hg+2+2

Lead IILead II Pb Pb+2+2

Lead IVLead IV Pb Pb+4+4

Tin IITin II Sn Sn+2+2

Tin IVTin IV Sn Sn+4+4

Chromium IIChromium II Cr Cr+2+2

Chromium III CrChromium III Cr+3+3

Chromium VI CrChromium VI Cr+6+6

Manganese IIManganese II MnMn+2+2

Manganese IIIManganese III MnMn+3+3

Manganese VIIManganese VII MnMn+7+7

Cobalt IICobalt II CoCo+2+2

Cobalt IIICobalt III CoCo+3+3

Gold IGold I AuAu+1+1

Gold IIIGold III AuAu+3+3

Nickel IINickel II NiNi+2+2

Nickel IIINickel III NiNi+3+3

Nickel IVNickel IV NiNi+4+4

**Silver**Silver Ag Ag+1+1

**Zinc**Zinc Zn Zn+2+2

**Cadmium Cd**Cadmium Cd+2+2

♥ Poly Atomic Ions to Know and Love ♥ Name Formula Name Formula

AcetateAcetateCC22HH33OO22

-1-1

(CH(CH33COO COO -1-1))

HypochloriteHypochlorite ClOClO-1-1

DichromateDichromate CrCr22OO77

-2-2 ChloriteChlorite ClOClO22

-1-1

AmmoniumAmmonium NHNH44

+1+1 ChlorateChlorate ClOClO33

-1-1

NitrateNitrate NONO33

-1-1 PerchloratePerchlorate ClOClO44

-1-1

NitriteNitrite NONO22

-1-1 CyanideCyanide CNCN-1-1

HydroxideHydroxide OHOH-1-1 CarbonateCarbonate COCO33

-2-2

PhosphatePhosphate POPO44

-3-3 ChromateChromate CrOCrO44

-2-2

SulfiteSulfite SOSO33

-2-2 Hydrogen Hydrogen CarbonateCarbonate

HCOHCO33

-1-1

SulfateSulfate SOSO44

-2-2 Hydrogen Hydrogen PhosphatePhosphate

HPOHPO44

-2-2

Hydrogen Hydrogen SulfiteSulfite

HSOHSO33

-1-1 Hydrogen Hydrogen SulfateSulfate

HSOHSO44

-1-1

PermanganatePermanganate MnOMnO44

-1-1 OxalateOxalate CC22OO44-2-2

HydroniumHydronium HH33OO++ SilicateSilicate SiOSiO

33-2-2

PeroxidePeroxide OO22-2-2 PhosphitePhosphite POPO

33-3-3

BromateBromate BrOBrO33

-1-1 ArsenateArsenate AsOAsO44

-2-2

♥ More Poly Atomic Ions to Know and Love ♥

Name Formula Name Formula

Naming Compounds with Polyatomic Ions

Formula Name

• (NH4)2SO4 ammonium sulfate

• ZnCO3 zinc carbonate

• NH4Br ammonium bromide

• Li2CO3 lithium carbonate

* Polyatomic & monatomic cation names remain the same, monatomic anions change their ending to –ide.

Writing Ionic Compound Formulas

Example: Barium nitrate


BaBa2+2+ NONO33--

2. Check to see if charges are balanced.

3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion.

Not balanced!

( )( )22


Example: Ammonium sulfate


NHNH44++ SOSO44

2-2-



Not balanced!

( )( )22


Example: Aluminum sulfide


AlAl3+3+ SS2-2-



Not balanced!

22 33


Example: Magnesium carbonate


MgMg2+2+ COCO332-2-


They are balanced!

3. Simplify to a formula unit.


Example: Zinc hydroxide


ZnZn2+2+ OHOH--



Not balanced!

( )( ) 22


Example: Aluminum phosphate


AlAl3+3+ POPO443-3-


They ARE balanced!

More Examples…

1. Cr2O3 2. Cr2O 3. CuSO4

4. Ni(OH)2

5. Cr2(C2O4)3

6. Cu2S7. CuS

1. chromium (III) oxide2. chromium (I) oxide3. copper (II) sulfate4. nickel (II) hydroxide5. chromium (III) oxalate6. copper (I) sulfide7. copper (II) sulfide

Chemical Formula Chemical Name

Hydrates• Hydrate – when a water molecule (s) are

chemically bonded to the ionic compound.

• Normal ionic naming protocol are used, then followed by the word “hydrate.”

• Prefixes are added to indicate the number of water molecules when naming hydrates.

Hydrate Prefixes

# of water molecules

prefix # of water molecules

prefix

1 mono- 6 hexa-

2 di- 7 hepta-

3 tri- 8 octa-

4 tetra- 9 nona-

5 penta- 10 deca-

Hydrates• Example: MgBr2 ∙ 6H2O

Magnesium bromide hexahydrate

• The “ ∙ ” means “loosely bonded”

• Hygroscopic - easily absorb water molecules from the air.

• Deliquescent- very hygroscopic; takes out water from the air to dissolve completely to form a liquid solution.

• Anhydrous – when all of the water has been removed.

Naming Binary Covalent CompoundsNaming Binary Covalent Compounds• Compounds between two nonmetals• First element in the formula is named first.• Second element is named as if it were an anion.• Use prefixes• Only use mono on second element

PP22OO55 = =COCO22 = =CO =CO =

NN22O =O =

didiphosphorus phosphorus pentpentoxideoxidecarbon carbon didioxideoxidecarbon carbon monmonoxideoxidedidinitrogen nitrogen monmonoxideoxide

Acids• Acids always begin with Hydrogen

Anion Formula Name

Cl-1 HCl Hydrochloric Acid

Br-1 HBr Hydrobromic Acid

SO4-2 H2SO4 Sulfuric Acid

SO3-2 H2SO3 Sulfurous Acid

NO3-1 HNO3 Nitric Acid

CN-1 HCN Hydrocyanic Acid

PO4-3 H3PO4 Phosphoric Acid

Bases

Cation Formula Name

Na+1 NaOH Sodium Hydroxide

K+1 KOH Potassium Hydroxide

NH4+1 NH3 Ammonia

Organic Compounds

• Organic compounds are named using a different set of rules.

• The simplest group is the hydrocarbons. These compounds are composed solely of the elements carbon and hydrogen.

• Carbon atoms can link to each other in chains and in rings.

Naming Hydrocarbons• The stem of the compound name is then

chosen from the following table:

# of carbon atoms

prefix # of carbon atoms

prefix

1 meth- 6 hexa-

2 eth- 7 hepta-

3 prop- 8 octa-

4 but- 9 nona-

5 penta- 10 deca-

Hydrocarbons: Alkanes

• These molecules have the generic formula: CnH2n+2

• They contain all single bonds.

CH4 methane

C2H6 ethane

C3H8 propane

C4H10 butane

C5H12 pentane

C6H14 hexane

Hydrocarbons: Alkenes

• These molecules have the generic formula:

CnH2n

• They contain double bonds between carbon atoms.

C2H4 ethene

C3H6 propene

C4H8 butene

C5H10 pentene

C6H12 hexene

Hydrocarbons: Alkynes

• These molecules have the generic formula:

CnHn

• They contain triple bonds between carbon atoms.

C2H2 ethyne

C3H3 propyne

C4H4 butyne

C5H5 pentyne

C6H6 hexyne


7.2 Oxidation Numbers

Oxidation Numbers

• Oxidation Number – numbers assigned to atoms composing a compound or ion that indicate the general distribution of electrons among bonded atoms


7.3 Using Chemical Formulas

Molar Mass• The mass of 1 mole of a pure substance is called

its Molar Mass.

• Ex: Molar mass of Iron is 55.847 g/mol

What is the molar mass of Platinum?

195.08 g/mol

Molar Mass• The molar mass depends on the particles that

compose the compound. If your element exists as a molecule, i.e. BrINClHOF, one mole of these particles contains 2 moles of the element as an atom.

• Determine the molar mass of oxygen molecules (O2)

(16.00 g/mol) x (2 atoms) = 32.00 g/mol

The molar mass of oxygen molecules (O2) is twice the molar mass of oxygen atoms!

Formula Mass

• The molar mass of a compound is the mass of the atomic mass units of one molecule.

• This takes into consideration the number of atoms of each element in a compound.

• Formula Mass is calculated the same way as molar mass except it is measured in amu, instead of g/mol.

Calculating Formula Mass

Calculate the formula mass of magnesium carbonate, MgCOCalculate the formula mass of magnesium carbonate, MgCO33..

24.31 + 12.01 + 3(16.00) =24.31 + 12.01 + 3(16.00) = 84.32 amu84.32 amu

Steps for Calculating Molar Mass for Compounds

1. List the elements

2. Determine how many atoms of each

3. Identify the atomic masses from the periodic table

4. Multiply how many atoms by the respective atomic mass

5. Add up the totals for the Molar Mass

Practice• H2O

H 2 x 1.008 = 2.016

O 1 x 15.99 = 15.99

18.006

• C6H12O6

C 6 x 12.01 = 72.06

H 12 x 1.008 = 12.096

O 6 x 15.99 = 95.94

180.096

• NaCl

Na 1 x 22.9 = 22.9

Cl 1 x 35.45 = 35.45

58.35

• K2O

K 2 x 39.1 = 78.2

O 1 x 15.99 = 15.99

94.19

g/mol

g/mol

g/molg/mol

Calculating Percentage Composition

Calculate the percentage composition of magnesium Calculate the percentage composition of magnesium carbonate, MgCOcarbonate, MgCO33..

24.31 + 12.01 + 3(16.00) = 84.32 amu24.31 + 12.01 + 3(16.00) = 84.32 amu

24.31100 28.83%

84.32Mg

12.01

100 14.24%84.32

C 48.00

100 56.93%84.32

O

100.00

• So…. In one mole of H2O, how many grams of Hydrogen are there?

2 mol H x 1.008g H = 2.016 g H in 1 mol H2O 1 mol H

• What % of Hydrogen, by mass, is in H2O?

2.016 g H x 100 = 11.2 % H 18 g H20

*Must also find molar mass of H2O

What % of Oxygen, by mass is in H2O?

Mass Percent

Formulas

molecular formula = (empirical formula)n

[n = integer] molecular formula = C6H6 = (CH)6

empirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.

Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas

Formulas for Formulas for ionic compoundsionic compounds are are ALWAYSALWAYS empirical (lowest whole number ratio). Often, empirical (lowest whole number ratio). Often, these are called these are called formula unitsformula units..

Examples:Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

Formulas

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be be empirical (lowest whole number ratio).empirical (lowest whole number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11


7.4 Determining Chemical Formulas

Empirical Formula Determination

1. Base calculation on assumption of 100 grams of compound.

2. Determine moles of each element in 100 grams of compound.

3. Divide each value of moles by the smallest of the values.

4. Multiply each number by an integer to obtain all whole numbers.

Empirical Formula DeterminationAdipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

49.32 14.107

12.01

g C mol Cmol C

g C

6.85 16.78

1.01

g H mol Hmol H

g H

43.84 12.74

16.00

g O mol Omol O

g O

Empirical Formula Determination(part 2)

4.1071.50

2.74

mol C

mol O

6.782.47

2.74

mol H

mol O

2.741.00

2.74

mol O

mol O

Divide each value of moles by the smallest of the values.Divide each value of moles by the smallest of the values.

Carbon:Carbon:

Hydrogen:Hydrogen:

Oxygen:Oxygen:

Empirical Formula Determination(part 3)

Multiply each number by an integer to obtain all whole Multiply each number by an integer to obtain all whole numbers.numbers.

Carbon: 1.50Carbon: 1.50 Hydrogen: 2.50Hydrogen: 2.50 Oxygen: 1.00Oxygen: 1.00x 2 x 2 x 2

33 55 22

Empirical formula: C3H5O2

Finding the Molecular FormulaThe empirical formula for adipic acid is The empirical formula for adipic acid is CC33HH55OO22. The molecular mass of adipic acid is . The molecular mass of adipic acid is

146 g/mol. What is the molecular formula of 146 g/mol. What is the molecular formula of adipic acid?adipic acid?

1. Find the formula mass of C1. Find the formula mass of C33HH55OO22

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g



3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

2. Divide the molecular mass by the 2. Divide the molecular mass by the mass given by the emipirical formula.mass given by the emipirical formula.

1462

73



3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

1462

73

3. Multiply the empirical formula by this 3. Multiply the empirical formula by this number to get the molecular formula.number to get the molecular formula.

(C(C33HH55OO22) x 2 =) x 2 = CC66HH1010OO44

Determining Chemical Formulas from Mass Percents

A sample has been analyzed, here are the results:

18.8 % Na

29.0 % Cl

52.2 % O

• How can you determine the chemical formula?

• Step 1: Assume a 100 g sample.

Then, your percent quantities become gram (mass) quantities.

18.8 g Na , 29.0 g Cl & 52.2 g O• Step 2: Convert those masses to moles.

18.8 g Na x 1 mol Na = 0.817 mol Na

23 g Na

29.0 g Cl x 1 mol Cl = 0.817 mol Cl

35.5 g Cl

52.2 g O x 1 mol O = 3.26 mol O

16 g O

• Step 3: Since your empirical formula is in small, whole number ratios, divide your mole amounts by the smallest mole quantity.

0.817 mol Na / 0.817 = 1.00 mol Na

0.817 mol Cl / 0.817 = 1.00 mol Cl

3.26 mol O / 0.817 = 3.99 ≈ 4.00 mol O

• Step 4: Use these values as subscripts in your formula

Na1Cl1O4 ≈ NaClO4

• Step 5: In the event the chemical formula is not the same as the empirical formula, you need the molar mass of the desired compound and you must compare it to the molar mass of the empirical formula.

• Step 6: Divide the given molar mass by the empirical molar mass to get the multiple quantity.

• Step 7: Multiply each subscript in the formula by that multiple quantity.

• Ex: MM of molecular formula = 180 g/mol

Using steps 1-4, you found that the empirical formula is CH2O.

Find the molar mass of the empirical formula: MM EF = 30 g/mol

Divide MM MF / MM EF to get a whole number.

Ex: 180 / 30 = 6

C1x6 H2x6 O1x6

C6H12O6

Practice Problems:• A sample has been analyzed to be 10.04 % C, 0.84

% H & 89.12% Cl. Find the Empirical Formula.

• A compound’s empirical formula has been determined to be HF. The compound’s molar mass is 40 g/mol. What is its chemical formula?

• A compound’s empirical formula has been determined to be CH2. The compound’s molar mass is 42 g/mol. What is its chemical formula?

chapter 7 chemical formulas & chemical compounds

Documents

balance charges

ionic compound formulasexample

ionic chargesgroups

so42ionic bonding

nh4 anion

naming ionic compounds1

possible oxidation state

fe2 iron