The Mole and Molar Mass

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• Molar mass

is the mass of one mole of a

substance.

• Molar mass is numerically equal to atomic mass,

molecular mass, or formula mass. However …

• … the units of molar mass are g/mol.

• Examples:

1 atom

Na = 22.99

u

1 mol

Na = 22.99

g

1 formula unit

KCl = 74.56

u

The Mole and Molar Mass

1 mol

CO

₂

= 44.01

g

1 molecule

CO

₂

= 44.01

u

1 mol

KCl = 74.56

g

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We can read formulas

in terms of moles of atoms or ions.

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• Molecular mass

: sum of the masses of the

atoms represented in a molecular formula.

– Simply put: the mass of a molecule.

– Molecular mass is specifically for molecules.

– Ionic compounds don’t exist as molecules; for

them we use …

• Formula mass

: sum of the masses of the

atoms or ions present in a formula unit.

Molecular Masses and

Formula Masses

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Determining the Formula/Molar

Mass of Ammonium Sulfate

The molar mass of ammonium sulfate is _______ g mol-1

The molar mass of acetic acid (CH₃COOH) is ______ g mol-1

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We can use these equalities to construct conversion factors:

Conversion Factors involving Mass, Moles, and

Number of Atoms/Molecules

1 mol CH₃COOH = 6.022 ×1023CH₃COOH molecules ≡ 60.05 g CH₃COOH

60.05 g mol CH3COOH 1 60.05 mol g CH3COOH 6.022 × 1023 molec. mol (CH3COOH) 6.022 × 1023 _molec. mol (CH₃COOH) 1

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The

mass percent composition

of a compound

refers to the mass proportion of the

constituent elements:

Mass Percent Composition

from Chemical Formulas

g element

% element = –––––––––––

×

100 g compound

X

g element

X

% element = ––––––––––––––

100 g compound

OR …

_{#grams of the element}

per 100 grams of the compound.

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Percentage Composition of Butane

g/mol g/mol

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• … is one method of determining empirical

formulas in the laboratory.

• This method is used primarily for simple organic

compounds (that contain carbon, hydrogen,

oxygen).

– The organic compound is burned in oxygen.

– The products of combustion (usually CO₂and H₂O) are weighed.

– The amount of each element is determined from the mass of products.

Elemental Analysis

תודוסי תזילנא

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Elemental Analysis Setup

The sample is burned in a stream of oxygen gas, producing … … H₂O, which is absorbed by MgClO₄, and … … CO2, which is absorbed by NaOH.

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Elemental Analysis

(cont’d) If our sample were CH₃OH, every two molecules of CH₃OH …

… would give two molecules of CO₂…

… and four molecules of H₂O.

12 Elemental analysis of butane (cooking gas) gave the following mass

composition:

C – 82.66%, H – 17.34%.

a) Determine the empirical formula of butane.

b) The molar mass of butane is 58.12 g/mol; what is the molecular formula?

Determine the empirical formula of phenol (a general disinfectant) from its elemental analysis:

C – 76.57% H – 6.43% H.

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• A chemical equation is a shorthand description of a chemical reaction, using symbols and formulas to represent the elements and compounds involved.

Writing Chemical Equations

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• Sometimes

additional

information

about the

reaction is

conveyed in

the equation.

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Balancing chemical equations

תואוושמ ןוזיא

• A balanced equation must conform to the law of conservation of mass! הסמ ה רומ יש קוח

• To balance an equation we need to adjust the

stoichiometric coefficients םיירטמ ויכיוטס ם י מדק מ

Balance the following equations:

H₂+ O₂→ H₂O C₂H₆+ O₂→ CO₂+ H₂O

Fe + O₂→ Fe₂O₃

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• If an element is present in just one compound on each side of the equation, try balancing that element first.

• Balance any reactants or products that exist as the free element

last.

• In some reactions, certain groupings of atoms (such as

polyatomic ions) remain unchanged. In such cases, treat these groupings as a unit.

• At times, an equation can be balanced by first using a fractional coefficient(s). The fraction is then cleared by

Guidelines for Balancing

Chemical Equations

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• A

stoichiometric factor

or mole ratio is a conversion

factor obtained from the stoichiometric coefficients in

a chemical equation.

ירטמויכיוטס

סחי

• In the equation: CO

(g)

+ 2 H

_2(g)

→ CH

₃

OH

(l)

Stoichiometric Equivalence

and Reaction Stoichiometry

1 mol CO

–––––––––

2 mol H

₂

1 mol CO

–––––––––––––

1 mol CH

₃

OH

1 mol CH

₃

OH

–––––––––––––

2 mol H

₂

2 mol H

₂

–––––––––

1 mol CO

1 mol CO reacted per 2 mol H₂ 2 mol H2 reacted per 1 mol CO 1 mol CO was consumed per 1 mol CH₃OH that evolved

1 mol CH₃OH was formed for each 2 mol

of H₂that reacted

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Outline of Simple Reaction Stoichiometry

Stoichiometric factor (or: mole ratio)

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When 0.105 mol propane is burned in an

excess of oxygen, how many moles of oxygen

are consumed? The reaction is

C

₃

H

₈

+ 5 O

₂

→ 3 CO

₂

+ 4 H

₂

O

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Outline of Stoichiometry Involving Mass

To our simple stoichiometry scheme … … we’ve added a conversion from mass at the beginning … … and a conversion to mass at the end.

Substances A and B may be two reactants,

two products, or reactant and product.

Think: If we are given moles of substance A initially, do we need to convert

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dioxide with water; nitrogen monoxide is also produced.

How many grams of nitric acid are produced for every 5 kg of nitrogen dioxide that reacts?

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Summary of Concepts

• Mass percent composition can be used to calculate the empirical formula.

– Additional information on the molecular weight of the material allows the calculation of the molecular formula

• A balanced equation must conform to the law of conservation of mass.

• Calculations involving reactions use stoichiometric factors based on stoichiometric coefficients in the balanced equation.