Chemical QuantitiesChemical QuantitiesChapter 10:The MoleChapter 10:The Mole

Measuring Matter:If one marble has a mass of 4.5g, how manykilograms will 230 marbles have?

ans: 1.035 KgHow many grass seeds are in a 40 kg bag of seeds if

one grass seed has a mass of 0.125g?

ans: 320 000 seeds

The MoleThe Mole::• An expression of the quantity of matter. The

quantity of matter is referred to a representative particles.

• One mole = 6.023 x 1023 quantity (Avogadro’s #)• The representative particles are:

1. Atoms: One mole of atoms = 6.023 x 1023

atoms.2. Molecules: One mole of molecules =

6.023 x 1023 molecules.3. Ions: One mole of ions = 6.02 x 1023 ions4. Compounds: One mole of compound =

6.023 x 1023 compounds.The Mole Concept

To find Molar Mass:To find Molar Mass:• Gram atomic mass (gam) is the mass of the atom

Reference off the periodic table.ClNiBaAgN

To find Molar Mass:To find Molar Mass:Gram molecular mass (gmm) is the sum of mass of

atoms in a molecule.

CO2

C2H8

NO2

Cl2CCl4 N2

To find Molar Mass:To find Molar Mass:• Gram formula mass (gfm) is the sum of the mass

of atoms in an ioinc compound.

NaClNa2SO4

Fe2O3

Ba3(PO4)2

Determine the Molar Determine the Molar MassMass

Find the gam for Ca ______ and Pb ______Find the gfm for K2CrO4

Find the gmm for N2O5

TLS: Complete Molar Mass ReviewHW: #34-36 all page 335

The Mole Concept The Mole Concept SamplerSampler

• Text page 323, Practice #1-4.• Text page 324, Practice #5-6• Text page 335, Problems #34-36

To determineTo determine::

The mass in moles of an The mass in moles of an

atom:atom:• reference the periodic table.• Note: 1 mole of Na atoms = 6.02 x 1023 Na

atoms = 23g (the atomic mass of Na)• *the atomic mass of any element is the molar

mass (in g/mol) of that element.• The Mole Concept flowchart.

The Mole Concept

The mass to mole The mass to mole

relationship.relationship.Find the gfm, gmm, or gam for the given

compoundConvert given # if grams to moles of

compound using 1 mole = gfm, gmm or gam

1. How many moles in 2.5 g of NaCl?

2. How many moles of N2O5 are in 35 g of N2O5?

**the mole to mass the mole to mass

relationship.relationship.3. How many grams of CO2 gas are in 2.5 moles of

CO2 gas?

4. How many grams of chlorine gas are in 40 moles of chlorine gas?

*The Mole to Mass and *The Mole to Mass and

Mass to Mole Relationship Mass to Mole Relationship

for Elements :for Elements :Text page 328-329, Practice #15-

#18#15b

#17b

The # of molecules in a The # of molecules in a

given # of molesgiven # of moles..Write the # of moles givenUse the molecules to mole relationship.

1 mole of molecule = 6.02 x 1023 molecules.1. How many molecules are in 2 moles of H2O?

2. How many moles of hydrogen are in 2 moles of H2O?

• TLS: Practice #29-#36 pp335.

Apply the Mole Concept Apply the Mole Concept Given mass converts to mole using (1 mole = gfm,

gmm,or gam) Mole converts to molecule using (1 mole = 6.023 x 1023 ) Number of Molecules.

How many molecules of CO2 are there in 23 g of CO2?

Ans: 3.15 x 1023 molecules of CO2

HW: Problem Set ThreeTextbook: page 331 #19-20 and page 339 #42-46

Percent Composition Percent Composition • refers to what part by mass of a compound

does each element contribute. NOTE page 342.Formula: MolarMass of element X 100 = %

Molar Mass of compound

SampleSample• Calcium Chloride : CaCl2

Ca: 1 x 40 = 40Cl: 2 x 35 = 70

110 is the total mass• % Comp Ca = 40 x 100 = 36.36 %

110• % Comp Cl = 70 x 100 = 63.6 %

110 100 %• TLS: Practice #54-57 pp 344.

Empirical Formula Empirical Formula Expression of the smallest whole

number ratio of elements in a molecule.

1. May or may not be the same as the molecular formula.

2. Total percentage of any compound is 100%. The % = massa. Given 36.63% Ca will = 36.63 g Ca

Procedures: Empirical Procedures: Empirical FormulaFormula

1. Convert % to mass (total percentage is 100%).2. Convert mass to moles via the gfm.3. Divide the moles by the smallest mole value to

determine the mole ratio within the compound.4. If not a whole number, multiply the mole ratio

from #3 by the smallest whole #.5. The answers are the subscript in the empirical

formula.

Sampler #1Sampler #1The % composition of an unknown compound is

found to be 38.43% Mn, 16.80% C and 44.77% O. What is the empirical formula for this compound?

Sampler #2Sampler #2Calculate the empirical formula for a compound

whose % composition is 74.97% Al and 25.03% C ?

Sampler #3Sampler #3What is the empirical formula of compound

composed of 37.01 % Carbon, 2.22 % Hydrogen, 18.50 % Nitrogen, and 42.27% Oxygen?

HW: #58-61 page 346

Molecular Formula Molecular Formula Actual number of elements expressed in a

molecule.a. molecular formula = empirical formula x nb. Formula:

molar mass of the molecule = MW = nmass of the empirical formula EW

n = the number of moles

Sampler Molecular Sampler Molecular FormulaFormula

A compound has a molecular mass of 118 g/mol and is composed of 40.68% carbon, 5.08% H and 54.24% O. What is the molecular formula?

Page 350, #65TLS: Practice #62-64 pp 350.

Hydrates Hydrates Compounds with water molecules attached.

a. Review the prefixes. (Table 10-1, pp 351)b. The mass of the water is included in the mass of the total molecule.c. Calculation of % composition, empirical formula and molecular formula is the same with hydrate. EXCEPT the mass of the water molecules are included.

TLS: Practice #74 and #75 p 353

Hydrated SampleHydrated SampleA hydrate of aluminum bromide is composed of

71.16% AlBr3 and 28.84% H2O. What is the formula for this hydrate?

Hydrated Sampler #2Hydrated Sampler #2A 17.44 g sample of hydrated zinc (II) sulfate is

heated to drive off the water. After heating, 9.79 g of anhydrous zinc(II) sulfate remain. Find the hydrated formula.

Hydrated Sampler #3Hydrated Sampler #3What is the % composition of a hydrated Lead (II)

acetate if the molecule is composed of 14.25% water by mass?

What is the hydrated formula?

Page 361; #184 and #185