the study of chemical change is the heart of chemistry

The study of chemical change is the heart of chemistry.

LAW OF CONSERVATION OF MASS“We may lay it down as an

incontestable axiom that, in all the operations of art and nature, nothing is

created; an equal amount of matter exists both before and after the

experiment. Upon this principle, the whole art of performing chemical

experiments depends.”--Antoine Lavoisier, 1789

“Atoms are neither created nor destroyed during any chemical reaction.”

3.1 CHEMICAL EQUATIONS Consider a simple chemical equation:

2 H2 + O2 → 2 H2O + : “reacts with”, → : “produces”

•reactants and products•coefficients•balanced

3.1 CHEMICAL EQUATIONS

The difference between a subscript and a coefficient Subscripts should not be change when balancing

equation

BALANCING EQUATIONS

3.1 CHEMICAL EQUATIONS

Consider a chemical reaction:

BALANCING EQUATIONS

3.1 CHEMICAL EQUATIONS

Express the chemical reaction

Balance carbon and hydrogen

Complete the chemical equation by balancing oxygen

BALANCING EQUATIONS

3.1 CHEMICAL EQUATIONS

3.1 CHEMICAL EQUATIONS INDICATING THE STATES OF

REACTANTS AND PRODUCTS

3.2 SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITYCOMBINATION AND DECOMPOSITION REACTIONS

3.2 SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITYCOMBINATION AND DECOMPOSITION REACTIONS

Figure 3.7

3.2 SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITY

3.2 SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITYCOMBUSTION IN AIR

When hydrocarbons are combusted in air, they react with O2 to form CO2 and H2O.

Combustion of hydrocarbon derivatives•CH3OH, C2H5OH•Glucose (C6H12O6) – oxidation reaction

3.3 FORMULA WEIGHTS How do we relate the numbers of atoms or molecules to

the amounts we measure in the laboratory?

Although we can not directly count atoms or molecules, we can indirectly determine their numbers if we know their masses.

3.3 FORMULA WEIGHTSFORMULA AND MOLECULAR WEIGHTS

3.3 FORMULA WEIGHTSPERCENTAGE COMPOSITION FROM FORMULAS

Needed for identifying unknown samples

3.4 AVOGADRO’S NUMBER & THE MOLE The definition of a mole

•The amount of matter that contains as many objects (atoms, molecules, etc) as the number of atoms in exactly 12 g of isotopically pure 12C.

Avogadro’s number•The number of objects in 1 mole of matter.

•6.0221421 × 1023

3.4 AVOGADRO’S NUMBER & THE MOLE

3.4 AVOGADRO’S NUMBER & THE MOLEMOLAR MASS

The mass of a single atom of an element (in amu) is numerically equal to the mass (in grams) of 1mol of that element.

3.4 AVOGADRO’S NUMBER & THE MOLEMOLAR MASS

The molar mass of a substance is the mass in grams of one mole of the substance.

Figure 3.10

3.4 AVOGADRO’S NUMBER & THE MOLE

0.02989 mol

6.05 mol

180.0 g/mol

84.0 g/mol

3.4 AVOGADRO’S NUMBER & THE MOLE

71.1 g

(a) 532 g, (b) 0.0029 g

164.1 g/mol

84.0 g/mol98.1 g/mol

3.4 AVOGADRO’S NUMBER & THE MOLE

Molecules C6H12O6

3.5 EMPIRICAL FORMULAS FROM ANALYSES The ratio of the number of moles of each element in a

compound gives the subscript in a compound’s empirical formula.

Consider a compound containing Hg & Cl (MWHg 200.6, MWCl 35.5) (73.9% Hg, 26.1% Cl by mass) How to get the empirical formula for the compound?

HgCl2

Procedure for calculating an empirical formula from percentage composition.

3.5 EMPIRICAL FORMULAS FROM ANALYSES

40.92 g C, 4.58 g H, and 54.50 g O.

C:H:O = 3(1:1.33:1) = 3:4:3C3H4O3

3.5 EMPIRICAL FORMULAS FROM ANALYSES

The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula.

MOLECULAR FORMULA FROM EMPIRICAL FORMULA

The formula weight of the empirical formula C3H4 is 3(12.0 amu) + 4(1.0 amu) = 40.0 amu

the molecular formula: C9H12

3.5 EMPIRICAL FORMULAS FROM ANALYSES

A common technique used for the determination of the

empirical formula for compounds containing principally

carbon and hydrogen.

COMBUSTION ANALYSIS

Fig 3.14 Apparatus for combustion analysis.

3.5 EMPIRICAL FORMULAS FROM ANALYSES COMBUSTION ANALYSIS

Mass of O = mass of sample - (mass of C + mass of H) = 0.255 g - (0.153 g + 0.0343 g) = 0.068 g O

C3H8O

3.6 QUANTITATIVE INFORMATION FROM BALANCED EQUATIONS

3.6 QUANTITATIVE INFORMATION FROM BALANCED EQUATIONS

3.6 QUANTITATIVE INFORMATION FROM BALANCED EQUATIONS Procedure for calculating amounts of reactants consumed or

products formed in a reaction

KClO3 122.55, KCl 74.5, O2 32.00 Answer: 1.76 g

MWLiOH 23.95

2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)

Answer: 3.64 g

3.7 LIMITING REACTANTSLet’s consider a sandwich-making process:

If you have 10 slices of bread and 7 slices of cheese,

You will have 5 sandwiches and 2 slices of cheese leftover. In this case,

•Limiting reactant (limiting reagent) : Bd•Excess reactant (excess reagent) : Ch

3.7 LIMITING REACTANTS

Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting Reactant

The most important commercial process for converting N2 from the air into nitrogen-

containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3):

N2(g) + 3 H2(g) → 2 NH3(g)

How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?

Answer: (a) AgNO3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn

Zn: 65.39 g/molAgNO3: 169.87 g/molAg: 107.9 g/molZn(NO3)2: 189.39 g/mol

3.7 LIMITING REACTANTS THEORETICAL YIELD The quantity of product that is calculated to form when all of

the limiting reactant reacts

(a) The theoretical yield is

146.14 g/mol84.16 g/mol

3.7 LIMITING REACTANTS THEORETICAL YIELD The quantity of product that is calculated to form when all of

the limiting reactant reacts

146.14 g/mol84.16 g/mol

EXERCISES

EXERCISES

EXERCISES3.94

3.94

EXERCISES

Atomic number 57, Lanthanum

3.96

EXERCISES

3.101

3.101