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Cartoon courtesy of NearingZero.net. Unit 4 – Conservation of Mass and Stoichiometry. Ions. Cation : A positive ion Mg 2+ , NH 4 + Anion : A negative ion Cl - , SO 4 2 - Ionic Bonding : Force of attraction between oppositely charged ions. Predicting Ionic Charges. Group 1 :.

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slide2
Ions
  • Cation: A positive ion
  • Mg2+, NH4+
  • Anion: A negative ion
  • Cl-, SO42-
  • Ionic Bonding: Force of attraction between oppositely charged ions.
predicting ionic charges
Predicting Ionic Charges

Group 1:

Lose 1 electron to form 1+ ions

H+

Li+

Na+

K+

predicting ionic charges1
Predicting Ionic Charges

Group 2:

Loses 2 electrons to form 2+ ions

Be2+

Mg2+

Ca2+

Ba2+

Sr2+

predicting ionic charges2
Predicting Ionic Charges

Group 13:

Loses 3

electrons to form

3+ ions

B3+

Al3+

Ga3+

predicting ionic charges3
Predicting Ionic Charges

Group 14:

Lose 4

electrons or gain

4 electrons?

Neither! Group 13 elements rarely form ions.

predicting ionic charges4
Predicting Ionic Charges

Nitride

N3-

Group 15:

Gains 3

electrons to form

3- ions

P3-

Phosphide

As3-

Arsenide

predicting ionic charges5
Predicting Ionic Charges

Oxide

O2-

Gains 2

electrons to form

2- ions

Group 16:

S2-

Sulfide

Se2-

Selenide

predicting ionic charges6
Predicting Ionic Charges

F1-

Fluoride

Br1-

Bromide

Group 17:

Gains 1

electron to form

1- ions

Cl1-

Chloride

I1-

Iodide

predicting ionic charges7
Predicting Ionic Charges

Group 18:

Stable Noble gases do not form ions!

predicting ionic charges8
Predicting Ionic Charges

Many transition elements

have more than one possible oxidation state.

Groups 3 - 12:

Iron(II) = Fe2+

Iron(III) = Fe3+

predicting ionic charges9
Predicting Ionic Charges

Some transition elements

have only one possible oxidation state.

Groups 3 - 12:

Zinc = Zn2+

Silver = Ag+

writing ionic compound formulas
Writing Ionic Compound Formulas

Example: Barium nitrate

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

( )

2. Check to see if charges are balanced.

Ba2+

NO3-

2

Not balanced!

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

writing ionic compound formulas1
Writing Ionic Compound Formulas

Example: Ammonium sulfate

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

( )

NH4+

SO42-

2. Check to see if charges are balanced.

2

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

Not balanced!

writing ionic compound formulas2
Writing Ionic Compound Formulas

Example: Iron(III) chloride

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

Fe3+

Cl-

2. Check to see if charges are balanced.

3

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

Not balanced!

writing ionic compound formulas3
Writing Ionic Compound Formulas

Example: Aluminum sulfide

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

2. Check to see if charges are balanced.

Al3+

S2-

2

3

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

Not balanced!

writing ionic compound formulas4
Writing Ionic Compound Formulas

Example: Magnesium carbonate

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

Mg2+

CO32-

2. Check to see if charges are balanced.

They are balanced!

writing ionic compound formulas5
Writing Ionic Compound Formulas

Example: Zinc hydroxide

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

( )

2. Check to see if charges are balanced.

Zn2+

OH-

2

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

Not balanced!

writing ionic compound formulas6
Writing Ionic Compound Formulas

Example: Aluminum phosphate

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

2. Check to see if charges are balanced.

Al3+

PO43-

They ARE balanced!

naming ionic compounds
Naming Ionic Compounds
  • 1. Cation first, then anion
  • 2. Monatomic cation = name of the element
  • Ca2+ = calciumion
  • 3. Monatomic anion = root + -ide
  • Cl- = chloride
  • CaCl2= calcium chloride
naming ionic compounds continued
Naming Ionic Compounds(continued)

Metals with multiple oxidation states

  • - some metal forms more than one cation
  • - use Roman numeralin name
  • PbCl2
  • Pb2+is cation
  • PbCl2 = lead(II) chloride
naming binary compounds
Naming Binary 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 -

P2O5 =

diphosphorus pentoxide

CO2 =

carbon dioxide

CO =

carbon monoxide

N2O =

dinitrogen monoxide

calculating formula mass
Calculating Formula Mass

Calculate the formula mass of magnesium carbonate, MgCO3.

24.31 g + 12.01 g + 3(16.00 g) =

84.32 g

calculating percentage composition
Calculating Percentage Composition

Calculate the percentage composition of magnesium carbonate, MgCO3.

From previous slide:

24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

100.00

formulas
Formulas

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.

  • molecular formula = (empirical formula)n [n = integer]
  • molecular formula = C6H6 = (CH)6
  • empirical formula = CH
formulas continued
Formulas(continued)

Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).

Examples:

NaCl

MgCl2

Al2(SO4)3

K2CO3

formulas continued1
Formulas(continued)

Formulas for molecular compoundsMIGHT be empirical (lowest whole number ratio).

Molecular:

C6H12O6

H2O

C12H22O11

Empirical:

H2O

CH2O

C12H22O11

empirical formula determination
Empirical Formula Determination
  • Base calculation on 100 grams of compound.
  • Determine moles of each element in 100 grams of compound.
  • Divide each value of moles by the smallest of the values.
  • Multiply each number by an integer to obtain all whole numbers.
empirical formula determination1
Empirical Formula Determination

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

empirical formula determination part 2
Empirical Formula Determination(part 2)

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

Carbon:

Hydrogen:

Oxygen:

empirical formula determination part 3
Empirical Formula Determination(part 3)

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

Carbon: 1.50

Hydrogen: 2.50

Oxygen: 1.00

x 2

x 2

x 2

3

5

2

C3H5O2

Empirical formula:

finding the molecular formula
Finding the Molecular Formula

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

1. Find the formula mass of C3H5O2

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

finding the molecular formula1
Finding the Molecular Formula

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

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

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

finding the molecular formula2
Finding the Molecular Formula

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

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

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

(C3H5O2) x 2 =

C6H10O4

combination synthesis reactions
Combination (Synthesis) Reactions

Two or more substances combine to form a

new compound.

A + X  AX

  • Reaction of elements with oxygen and sulfur
  • Reactions of metals with Halogens
  • Synthesis Reactions with Oxides
  • There are others not covered here!
decomposition reactions
Decomposition Reactions

A single compound undergoes a reaction that

produces two or more simpler substances

AX  A + X

  • Decomposition of:
  • Binary compounds H2O(l )  2H2(g) + O2(g)
  • Metal carbonates CaCO3(s)  CaO(s) + CO2(g)
  • Metal hydroxides Ca(OH)2(s)  CaO(s) + H2O(g)
  • Metal chlorates 2KClO3(s)  2KCl(s) + 3O2(g)
  • Oxyacids H2CO3(aq)  CO2(g) + H2O(l )
single replacement reactions
Single Replacement Reactions

A + BX  AX + B

BX + Y  BY + X

Replacement of:

  • Metals by another metal
  • Hydrogen in water by a metal
  • Hydrogen in an acid by a metal
  • Halogens by more active halogens
the activity series of the metals
The Activity Series of the Metals
  • Lithium
  • Potassium
  • Calcium
  • Sodium
  • Magnesium
  • Aluminum
  • Zinc
  • Chromium
  • Iron
  • Nickel
  • Lead
  • Hydrogen
  • Bismuth
  • Copper
  • Mercury
  • Silver
  • Platinum
  • Gold

Metals can replace other metals

provided that they are above the

metal that they are trying to

replace.

Metals above hydrogen can

replace hydrogen in acids.

Metals from sodium upward can

replace hydrogen in water

the activity series of the halogens
The Activity Series of the Halogens
  • Fluorine
  • Chlorine
  • Bromine
  • Iodine

Halogens can replace other

halogens in compounds, provided

that they are above the halogen

that they are trying to replace.

2NaCl(s) + F2(g) 

???

2NaF(s) + Cl2(g)

MgCl2(s) + Br2(g) 

???

No Reaction

double replacement reactions
Double Replacement Reactions

The ions of two compounds exchange places in an

aqueous solution to form two new compounds.

AX + BY  AY + BX

One of the compounds formed is usually a

precipitate, an insoluble gas that bubbles out of

solution, or a molecular compound, usually water.

combustion reactions
Combustion Reactions

A substance combines with oxygen, releasing a large

amount of energy in the form of light and heat.

  • Reactive elements combine with oxygen

P4(s) + 5O2(g)  P4O10(s)

(This is also a synthesis reaction)

  • The burning of natural gas, wood, gasoline

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

stoichiometry
Stoichiometry

“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”

Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

review atomic masses
Review: Atomic Masses
  • Elements occur in nature as mixtures of isotopes
  • Carbon = 98.89% 12C
  • 1.11% 13C
  • <0.01% 14C
  • Carbon’s atomic mass = 12.01 amu
review the mole
Review: The Mole
  • The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.
  • 1 moleof anything = 6.022 ´ 1023units of that thing
review molar mass
Review: Molar Mass

A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.

CO2 = 44.01 grams per mole

H2O = 18.02 grams per mole

Ca(OH)2 = 74.10 grams per mole

review chemical equations

Review: Chemical Equations

C2H5OH + 3O2®2CO2 + 3H2O

reactants

products

When the equation is balanced it has quantitative significance:

Chemical change involves a reorganization of

the atoms in one or more substances.

1 mole of ethanol reacts with 3 moles of oxygen

to produce 2 moles of carbon dioxideand 3 moles of water

calculating masses of reactants and products
Calculating Masses of Reactants and Products
  • Balance the equation.
  • Convert mass to moles.
  • Set up mole ratios.
  • Use mole ratios to calculate moles of desired substituent.
  • Convert moles to grams, if necessary.
working a stoichiometry problem
Working a Stoichiometry Problem

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.

1. Identify reactants and products and write the balanced equation.

4

Al

+

3

O2

2

Al2O3

a. Every reaction needs a yield sign!

b. What are the reactants?

c. What are the products?

d. What are the balanced coefficients?

working a stoichiometry problem1
Working a Stoichiometry Problem

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

4 Al + 3 O2 2Al2O3

6.50 g Al

1 mol Al

2 mol Al2O3

101.96 g Al2O3

=

? g Al2O3

4 mol Al

1 mol Al2O3

26.98 g Al

6.50 x 2 x 101.96÷ 26.98 ÷ 4 =

12.3 g Al2O3

standard molar volume
Standard Molar Volume

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

- Amedeo Avogadro

At STP (Standard Temperature and Pressure):

1 mole of a gas occupies 22.4 liters of volume

gas stoichiometry 1
Gas Stoichiometry #1

If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios.

3 H2(g) + N2(g)  2NH3(g)

3molesH2+1moleN2 2molesNH3

3 liters H2 + 1 liter N2 2 liters NH3

gas stoichiometry 2
Gas Stoichiometry #2

How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen?

3 H2(g) + N2(g)  2NH3(g)

12 L H2

2

L NH3

= L NH3

8.0

3

L H2

gas stoichiometry 3
Gas Stoichiometry #3

How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?

2 KClO3(s)  2 KCl(s) + 3 O2(g)

50.0 g KClO3

1 mol KClO3

3 mol O2

22.4 L O2

122.55 g KClO3

2 mol KClO3

1 mol O2

= L O2

13.7

gas stoichiometry 4
Gas Stoichiometry #4

How many liters of oxygen gas, at 37.0C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?

2 KClO3(s)  2 KCl(s) + 3 O2(g)

50.0 g KClO3

1 mol KClO3

3 mol O2

= “n” mol O2

0.612

mol O2

122.55 g KClO3

2 mol KClO3

= 16.7 L

limiting reactant

Limiting Reactant

The limiting reactantis the reactant

that is consumed first,limiting the amounts of products formed.

solving a stoichiometry problem
Solving a Stoichiometry Problem
  • Balance the equation.
  • Convert masses to moles.
  • Determine which reactant is limiting.
  • Use moles of limiting reactant and mole ratios to find moles of desired product.
  • Convert from moles to grams.