Section 9.1

Using Chemical Equations Chapter 9 Chemical Quantities:

• Georgia Performance Standards:

• SC2 (c, d & e): Students will relate how the Law of Conservation of Matter is used to determine chemical composition in compounds and chemical reactions.– c. Apply concepts of the mole

and Avogadro’s number to conceptualize and calculate the empirical & molecular formulas, Mass, moles and molecules relationships, and molar volumes of gases.

• Essential Questions:– -How is the molar mass

calculated for various compounds?

– How do you convert between moles, mass, and number of atoms?

– How is the mass % of an element in a compound calculated?

Section 9.1

Using Chemical Equations

Mass, Moles, Number of

atoms relationships

Section 9.1

Using Chemical Equations

Performance Tasks:

• Calculate the molar mass for various compounds.

• Use the mole concept, Avogadro’s number, & molar mass to convert among moles, mass, and number of atoms

• Calculate the mass percent of an element in a compound

Section 9.1

Using Chemical Equations

Law of Conservation of Mass

• We balance chemical equations to satisfy the law of conservations of mass– Mass is neither created nor destroyed, only

transformed

Section 9.1

Using Chemical Equations

Chemical Equations

Chemical equations are concise representations of chemical reactions.

Section 9.1

Using Chemical Equations Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Section 9.1

Using Chemical Equations Anatomy of a Chemical Equation

Reactants appear on the left side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Section 9.1

Using Chemical Equations Anatomy of a Chemical Equation

Products appear on the right side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Section 9.1

Using Chemical Equations Anatomy of a Chemical Equation

The states of the reactants and products are written in parentheses to the right of each compound.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Section 9.1

Using Chemical Equations Anatomy of a Chemical Equation

Coefficients are inserted to balance the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Section 9.1

Using Chemical Equations

Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule.

Section 9.1

Using Chemical Equations

Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule

• Coefficients tell the number of molecules.

Section 9.1

Using Chemical Equations

Notice• The number of atoms of each type of element must be the same on both

sides of a balanced equation.

• Subscripts must not be changed to balance an equation.

• A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction.

• Coefficients can be fractions, although they are usually given as lowest integer multiples.

• Trial and error is a valid method to balance a chemical equation.

Section 9.1

Using Chemical Equations

Balancing Chemical Equations

Section 9.1

Using Chemical Equations

Chemical Equation

• C2H5OH + 3O2 2CO2 + 3H2O• The equation is balanced.• 1 mole of ethanol reacts with 3 moles of oxygen to

produce 2 moles of carbon dioxide and 3 moles of water

Section 9.1

Using Chemical Equations

Formula Weights

Section 9.1

Using Chemical Equations

Formula Weight (FW)

• A formula weight is the sum of the atomic weights for the atoms in a chemical formula.

• So, the formula weight of calcium chloride, CaCl2, would be

Ca: 1(40.1 amu) + Cl: 2(35.5 amu)

111.1 amu• Formula weights are generally reported for ionic

compounds.

Section 9.1

Using Chemical Equations

Molecular Weight (MW)

• A molecular weight is the sum of the atomic weights of the atoms in a molecule.

• For the molecule ethane, C2H6, the molecular weight would be

C: 2(12.0 amu)

30.0 amu+ H: 6(1.0 amu)

Section 9.1

Using Chemical Equations

Atomic Masses

• Elements occur in nature as mixtures of isotopes• Carbon = 98.89% 12C

1.11% 13C <0.01% 14C Carbon atomic mass =

12.01 amu

Section 9.1

Using Chemical Equations

Percent Composition

One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

% element =(number of atoms)(atomic weight)

(FW of the compound)x 100

Section 9.1

Using Chemical Equations

Percent Composition

So the percentage of carbon in ethane is…

%C =(2)(12.0 amu)

(30.0 amu)24.0 amu30.0 amu

= x 100

= 80.0%

Section 9.1

Using Chemical Equations

Percent Composition

• Mass percent of an element:

• For iron in iron (III) oxide, (Fe2O3)

massmass of e lem en t in com pound

mass o f com pound% 10 0 %

mass Fe%..

. 11 1 6 915 9 6 9

10 0 % 6 9 9 4%

Section 9.1

Using Chemical Equations

Moles

Section 9.1

Using Chemical Equations

The Mole

• The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.

• 1 mole of anything = 6.022 1023 units of that thing

Section 9.1

Using Chemical Equations

Avogadro’s Number

• 6.02 x 1023

• 1 mole of 12C has a mass of 12 g.

Section 9.1

Using Chemical Equations

Molar Mass

• By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).– The molar mass of an element is the mass number for

the element that we find on the periodic table.– The formula weight (in amu’s) will be the same

number as the molar mass (in g/mol).

Section 9.1

Using Chemical Equations

Using Moles

Moles provide a bridge from the molecular scale to the real-world scale.

Section 9.1

Using Chemical Equations

Mole Relationships

• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.

• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.