Chapter 3

Equations for Chemical Reactions

Chemical ReactionsReactions involve rearrangement and exchange

of atoms to produce new pure substances.

Reactants Products

Chemical EquationsShorthand way of describing a reaction

Provides information about the reaction

1. formulas of reactants and products2. states of reactants and products3. relative numbers of reactant and product molecules

Combustion of MethaneMethane gas reacts with oxygen gas to produce

carbon dioxide gas and gaseous water.

CH4(g) + O2(g) ➜ CO2(g) + H2O(g)

This equation reads “1 molecule of CH4 gas combines with 1 molecule of

O2 gas to make 1 molecule of CO2 gas and 1 molecule of H2O gas.”

H

H C

H

H O O +

O

O

C + O

H H

H

H C

H

H O O +

O

O

C + O

H H

1 C + 4 H + 2 O 1 C + 2 O + 2 H + O

1 C + 2 H + 3 O

What about conservation of mass ??

Combustion of Methane, BalancedTo show the reaction obeys the Law of Conservation of Mass,

the equation must be balanced.

CH4(g) + O2(g) ➜ CO2(g) + H2O(g)

CH4(g) + O2(g) ➜ CO2(g) + 2 H2O(g)

CH4(g) + 2 O2(g) ➜ CO2(g) + 2 H2O(g)

“1 molecule of CH4 gas combines with 2 molecules of O2 gas to make 1 molecule of CO2 gas and 2

molecules of H2O gas.”

Symbols used to indicate state after chemical:

(g) = gas; (l) = liquid; (s) = solid(aq) = aqueous = dissolved in water

Energy symbols used above the arrow for conditions for reactions:

Δ = heat hν = lightshock = mechanicalelec = electrical

Symbols Used in Equations

Steps in Balancing Equations1. In compounds balance elements other than H and O.

a. Balance elements which occur only once on each side of the equation.b. Start with the elements which occur the most.c. Balance polyatomic ions which do not change in the reaction.

2. Be prepared to rebalance if something changes!!!

3. Balance H.

4. Balance O.

5. Balance elements which appear in their “elemental” forms.

When aluminum metal reacts with air, it produces a white, powdery compound, aluminum oxide.

aluminum(s) + oxygen(g) ➜ aluminum oxide(s)

Al(s) + O2(g) ➜ Al2O3(s)

2 Al(s) + O2(g) ➜ Al2O3(s)

2 Al(s) + 3 O2(g) ➜ 2 Al2O3(s)

4 Al(s) + 3 O2(g) ➜ 2 Al2O3(s)

Solid phosphorous (P4) reacts with hydrogen gas to produce phosphorous trihydride.

P4 (s) + H2 (g) ----------> PH3 (g)

P4 (s) + H2 (g) ----------> 4 PH3 (g)

P4 (s) + 6 H2 (g) ----------> 4 PH3 (g)

Solid potassium chlorate decomposes to produce oxygen gas and potassium chloride.

KClO3 (s) -------------------> O2 (g) + KCl (s) 2 KClO3 (s) -------------------> 3 O2 (g) + KCl (s)

2 KClO3 (s) -------------------> 3 O2 (g) + 2 KCl (s)

H2SO4 (aq) + NaCN (s) ------> Na2SO4 (aq) + HCN(g)

H2SO4 (aq) + 2 NaCN (s) ------> Na2SO4 (aq) + HCN(g)

H2SO4 (aq) + 2 NaCN (s) ------> Na2SO4 (aq) + 2 HCN(g)

Aqueous sulfuric acid reacts with solid sodium cyanide to produce aqueous sodium sulfate and hydrogen cyanide gas.

K3PO4 (aq) + Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + KNO3 (aq)

K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + KNO3 (aq)

K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + 6 KNO3 (aq)

2 K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + 6 KNO3 (aq)

Aqueous potassium phosphate reacts with aqueous calcium nitrate to produce solid calcium phosphate and aqueous potassium nitrate.

Write a balanced equation for the combustion of butane, C4H10.

C4H10 (g) + O2 (g) ➝ CO2 (g) + H2O (g)

C4H10 (g) + O2 (g) ➝ 4 CO2 (g) + H2O (g)

C4H10 (g) + O2 (g) ➝ 4 CO2 (g) + 5 H2O (g)

C4H10 (g) + 13/2 O2 (g) ➝ 4 CO2 (g) + 5 H2O (g)

2 C4H10 (g) + 13 O2 (g) ➝ 8 CO2 (g) + 10 H2O (g)

Organic Chemistry

Classifying CompoundsOrganic vs. Inorganic

In the18th century, compounds from living things were called organic; compounds from the nonliving environment were called inorganic.

Organic compounds easily decomposed and could not be made in the 18th-century lab.

Inorganic compounds were very difficult to decompose, but could be synthesized.

Modern Classification of CompoundsOrganic vs. Inorganic

Today we commonly make organic compounds in the lab and find them all around us.

Organic compounds are mainly made of C and H, sometimes with O, N, P, S, halogens, and trace amounts of other elements.

The main element that is the focus of organic chemistry is carbon.

Carbon Bonding in Organic Compounds

Carbon atoms bond almost exclusively covalently in organic compounds.

When C bonds, it forms four covalent bonds.

Carbon is unique in that it can form limitless chains of C atoms, both straight and branched, and rings of C atoms.

Carbon Bonding in Organic Compounds

Classifying Organic Compounds

There are two main categories of organic compounds, hydrocarbons and functionalized hydrocarbons.

Most fuels are mixtures of hydrocarbons

Hydrocarbons

Alkanes Aldehydes

Alkenes Ketones

Alkynes Carboxylic Acids

Aromatics Esters

Alcohols Amines

Ethers Amides

Families of Organic Compounds

C C

C C

C C

CC

CC

C

C

C O

H

C OC

CH

O

CC

O

C

CO

H

O

C

O

OC

N

CN

O

Chapter 4

Chemical Quantitiesand

Aqueous Reactions

StoichiometryThe study of the numerical relationship between

chemical quantities in a chemical reaction

Making Pizza

The number of pizzas you can make depends on the amount of the ingredients you use.

1 crust + 5 oz. tomato sauce + 2 cu cheese ➜ 1 pizza

This relationship can be expressed mathematically1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza

If you want to make more than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make.

Predicting Amounts from Stoichiometry

According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose?

C6H12O6 + 6 O2 ➜ 6 CO2 + 6 H2O

mol H2O mol C

6H

12O

6

1 mol C6H12O6 : 6 mol H2O

Moles of A

Moles of B

Grams of A

Grams of B

Particles of A

Particles of B

Avogadro’s Number Avogadro’s Number

Molar MassMolar Mass

Coefficients

The amounts of any other substance produced or consumed in a chemical reaction can be

determined from the amount of just one substance.

Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g of octane (C8H18).

C8H18(l) + O2(g) ➜ CO2(g) + H2O(g)

2 C8H18(l) + 25 O2(g) ➜ 16 CO2(g) + 18 H2O(g)

g C8H18 mol CO2 g CO2 mol C8H18

How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis?

6 CO2 + 6 H2O➜C6H12O6+ 6 O2

g C6H12O6 mol CO2 g CO2 mol C6H12O6

6 12 6 6 12 6

6 12 6 2

6 12 6 2

2 2

O H C g 5.8 2 O H C mol 1 O H C g 180.2

CO mol 6 O H C mol 1

CO g 44.01 CO mol 1

CO g 7.8 3

=

× × ×

g 44.01 mol 1

2 6 12 6

CO mol 6 O H C mol 1

mol 1 g 180.2

How many grams of O2 can be made from the decomposition of 100.0 g of PbO2?

2 PbO2(s) → 2 PbO(s) + O2(g)(PbO2 = 239.2, O2 = 32.00)

g O2 mol PbO2 g PbO2 mol O2

Stoichiometry Road Map

Moles of A

Moles of B

Grams of A

Grams of B

Particles of A

Particles of B

Avogadro’s Number Avogadro’s Number

Molar MassMolar Mass

Coefficients

More Making Pizzas1 crust + 5 oz. tomato sauce + 2 cu cheese ➜1 pizza

What would happen if we only had 4 crusts,15 oz. tomato sauce, and 10 cu cheese?

Limitingreagent Theoretical

yield

The Limiting Reactant

For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others.

When this reactant is used up, the reaction stops and no more product is made.

Limiting and Excess Reactants in the Combustion of Methane

CH4(g) + O2(g) ➜ CO2(g) + H2O(g)

CH4(g) + 2 O2(g) ➜ CO2(g) + 2 H2O(g)

➜

➜

If we have five molecules of CH4 and eight molecules of O2, which is the limiting reactant?

➜

If we have five molecules of CH4 and eight molecules of O2, which is the limiting reactant?

How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the

reaction 3 Si + 2 N2 ➜ Si3N4 ?

Theoretical yield

Limitingreactant

More Making PizzasLet’s now assume that as we are making pizzas, we

burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield.

We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield.

When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield,

and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

kg

TiO2

kg

C

smallest

mol Ti

} smallestamount is

from limitingreactant

Collect needed relationships:

1000 g = 1 kg Molar Mass Ti = 47.87 g/mol

Molar Mass C = 12.01 g/mol Molar Mass TiO2 = 79.87 g/mol

1 mole TiO2 : 1 mol Ti 2 mole C : 1 mol Ti

When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield,

and percent yield.

TiO2(s) + 2 C(s)➜ Ti(s) + 2 CO(g)

limiting reactant Theoretical yield

When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield,

and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

1.19 x 103 mol Ti

1.10 x 103 mol Ti

theoretical yield

percent yield

When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield,

and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(II) oxide?

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

If 4.61 g of N2 are made, what is the percent yield?

g NH3 mol N2 mol NH3

g CuO mol N2 mol CuO

g N2 ?

Theoretical yield

How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(II) oxide?

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

If 4.61 g of N2 are made, what is the percent yield?