The Mole & FormulasThe Mole & Formulas

Dr. Ron RusayDr. Ron Rusay

Mole - Mass Relationships Chemical Reactions

Stoichiometry

The Mole

•% Composition: Determining the Formula of an Unknown Compound•Writing and Balancing Chemical Equations•Calculating the amounts of Reactant and Product•Limiting Reactant

The MoleThe Mole

• The The numbernumber of carbon atoms in exactly 12 of carbon atoms in exactly 12 grams of pure grams of pure 1212C. The C. The numbernumber equals equals

6.02 6.02 10 10 2323 1 mole of anything = 1 mole of anything = 6.02 6.02 10 10 2323

unitsunits

• 6.02 6.02 10 10 2323 “units” of anything: atoms, “units” of anything: atoms,

people, stars, $s, etc., etc. = people, stars, $s, etc., etc. = 1 mole

Avogadro’s NumberAvogadro’s Number

Avogadro’s number equals Avogadro’s number equals 1 mole1 mole….which equals….which equals

6.022 6.022 10 102323 “units” “units”

Counting by Weighing

12 red marbles @ 7g each = 84g12 yellow marbles @4g each=48g

55.85g Fe = 6.022 x 1023 atoms Fe32.07g S = 6.022 x 1023 atoms S

CaCO3

100.09 gOxygen 32.00 gCopper 63.55 gWater 18.02 g

Relative Masses of 1 Mole

Atomic and Molecular WeightsAtomic and Molecular WeightsMass MeasurementsMass Measurements

• 11H weighs 1.6735 x 10H weighs 1.6735 x 10-24-24 g and g and 1616O O 2.6560 x 102.6560 x 10-23-23 g. g.

•DEFINITION: mass of DEFINITION: mass of 1212C = C = exactly 12 amu.exactly 12 amu.

•Using atomic mass units:Using atomic mass units:

• 1 amu = 1.66054 x 101 amu = 1.66054 x 10-24-24 g g

• 1 g = 6.02214 x 101 g = 6.02214 x 102323 amu amu

QUESTIONQUESTION

ANSWERANSWER

–22

23 –22

E) 1.06 10 g

A mole of copper atoms has a mass of 63.55 g. The mass of 1 copper atom is

63.55 g/(6.022 10 ) = 1.06 10 g.

Atomic and Molecular WeightsAtomic and Molecular Weights

• Formula Weight Formula Weight a.k.a.a.k.a. Molecular Weight Molecular Weight

• Formula weights (FW): sum of Atomic Formula weights (FW): sum of Atomic Weights (AW) for atoms in formula.Weights (AW) for atoms in formula.

• FW (HFW (H22SOSO44) = ) = 22AW(H) + AW(S) + AW(H) + AW(S) +

44AW(O)AW(O)

• = = 22(1.0 amu) + (32.0 amu) + (1.0 amu) + (32.0 amu) + 44(16.0)(16.0)

• = 98.0 amu= 98.0 amu

Atomic and Molecular WeightsAtomic and Molecular Weights

• Molecular weight (MW) is the weight of Molecular weight (MW) is the weight of the molecular formula in amu.the molecular formula in amu.

• MW of sugar (CMW of sugar (C66HH1212OO6 6 ) = ?) = ?

• MW =MW = 6 6(12.0 amu) + (12.0 amu) + 1212(1.0 amu) + (1.0 amu) + 66(16.0 (16.0 amu)amu)

• = 180 amu= 180 amu

Molar MassMolar Mass

• A substance’s A substance’s molar mass molar mass (equal to the (equal to the formula weight: atomic or molecular weight in formula weight: atomic or molecular weight in grams) is the mass in grams of one mole of the grams) is the mass in grams of one mole of the element or compound. element or compound.

C = 12.01 grams per mole (g/mol)C = 12.01 grams per mole (g/mol)

COCO22 = ?? = ??

44.01 grams per mole (g/mol)44.01 grams per mole (g/mol)12.01 + 2(16.00) = 44.0112.01 + 2(16.00) = 44.01

QUESTIONQUESTION

ANSWERANSWER

C) 46.07

The molar mass is the sum of masses of all the atoms in the molecule.

2 12.01 + 6 1.008 + 1 16.00 = 46.07

QUESTIONQUESTION

ANSWERANSWER

3

C) CH Cl

The molar mass has units of g/mol.(12.8 g/0.256 mol) = 50.0 g/mol. The molecule with the closest molar mass is CH3Cl.

QUESTIONQUESTION

ANSWERANSWER

B) 278 g

The molar mass of sodium hydroxide, NaOH, is 22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 40.00 g/mol. Convert to grams: 6.94 mol (40.00 g/mol) = 278 g.

Percent CompositionPercent Composition

• Mass percent of an element:Mass percent of an element:

• For iron in (FeFor iron in (Fe22OO33), iron (III) oxide = ? ), iron (III) oxide = ?

mass Fe%..

.= × =11169159 69

100% 69 94%

massmass of element in compound

mass of compound% = ×100%

QUESTIONQUESTION

ANSWERANSWER

8 8

E) All of these

The ratio of C to H in C H is 1:1. This is the same ratio found for each of the compounds, so all have the same percent composition by mass.

QUESTIONQUESTIONMorphine, derived from opium plants, has the potential for use and abuse. It’s formula is C17H19NO3. What percent, by mass, is the carbon in this compound?

A. 42.5%B. 27.9%C. 71.6%D. This cannot be solved until the mass of the sample is given.

ANSWERANSWERMorphine, derived from opium plants, has the potential for use and abuse. It’s formula is C17H19NO3. What percent, by mass, is the carbon in this compound?

A. 42.5%B. 27.9%C. 71.6%D. This cannot be solved until the mass of the sample is given.

(17 12) / ((17 12) + (19 1) + (1 14) + 3 16)) = 0.716

0.716 100 = 71.6 %

QUESTIONQUESTION

ANSWERANSWER

2 7

C) 4.04 g

The molar mass of K CrO is 2 39.10 + 52.00 + 7 16.00 = 242.2. The mass fraction of potassium is (2 39.10)/242.2 = 0.3229.0.3229 12.5 g = 4.04 g.

Formulas: Dalton’s LawFormulas: Dalton’s Law

• Dalton’s law of multiple proportions: Dalton’s law of multiple proportions:

When two elements form different compounds, When two elements form different compounds, the the mass ratiomass ratio of the elements in one of the elements in one compound compound is relatedis related to the mass ratio in the to the mass ratio in the other other by a small whole numberby a small whole number..

Formulas: Multiple ProportionsFormulas: Multiple Proportions

Formulas & Multiple ProportionsFormulas & Multiple Proportions Components of acid rain, SOComponents of acid rain, SO22(g) and SO(g) and SO33(g)(g)

• Compound A contains: Compound A contains: 1.000 g Sulfur & 1.500 g Oxygen1.000 g Sulfur & 1.500 g Oxygen

• Compound B contains: Compound B contains: 1.000 g Sulfur & 1.000 g Oxygen1.000 g Sulfur & 1.000 g Oxygen

• Mass ratio A: 2 to 3; Mass ratio B: 1 to 1Mass ratio A: 2 to 3; Mass ratio B: 1 to 1

• Adjusting for atomic mass differences: AW Adjusting for atomic mass differences: AW sulfur is 2x the AW oxygen; the atom ratios sulfur is 2x the AW oxygen; the atom ratios therefore are Stherefore are S11OO33 and S and S11OO22 respectively respectively

Formulas &Formulas &Molecular RepresentationsMolecular Representations

molecular formula = Cmolecular formula = C66HH66 BenzeneBenzene

empirical formula = CH = Cempirical formula = CH = C6/66/6HH6/66/6

molecular formula = (empirical formula)molecular formula = (empirical formula)nn

[[nn = integer] = integer] (CH)(CH)66

• Other representations: Lewis Dot Other representations: Lewis Dot formulas, structural formulas, 2-D, 3-Dformulas, structural formulas, 2-D, 3-D

Formulas &Formulas &Molecular RepresentationsMolecular Representations

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Empirical Formula DeterminationEmpirical Formula Determination

• 1. Use percent analysis. 1. Use percent analysis.

Let 100 % = 100 grams of compound.Let 100 % = 100 grams of compound.

• 2. Determine the moles of each element.2. Determine the moles of each element.

(Element % = grams of element.)(Element % = grams of element.)

• 3. 3. Divide each value of moles by the smallest of Divide each value of moles by the smallest of the mole values.the mole values.

• 4.4. Multiply each number by an integer to Multiply each number by an integer to

obtain all whole numbersobtain all whole numbers..

QUESTIONQUESTIONThe dye indigo is a compound with tremendous economic importance (blue jeans wouldn’t be blue without it.) Indigo’s percent composition is: 73.27% C; 3.84% H; 10.68%N and 12.21% O. What is the empirical formula of indigo?

A.C6H4NOB.C8H3NOC.C8H5NOD. I know this should be whole numbers for each atom, but I do not

know how to accomplish that.

ANSWERANSWERThe dye indigo is a compound with tremendous economic importance (blue jeans wouldn’t be blue without it.) Indigo’s percent composition is: 73.27% C; 3.84% H; 10.68%N and 12.21% O. What is the empirical formula of indigo?

A.C6H4NOB.C8H3NOC.C8H5NOD. I know this should be whole numbers for each atom, but I do not

know how to accomplish that.

Empirical & Molecular Formula Empirical & Molecular Formula DeterminationDetermination

The Molecular Formula is the important objective.The Molecular Formula is the important objective.The Molar Mass (molecular weight) must be The Molar Mass (molecular weight) must be

determined in order to reach this objective since determined in order to reach this objective since the Molar Mass may not be equal to the the Molar Mass may not be equal to the Empirical formula’s Mass.Empirical formula’s Mass.

The experimental process involves different The experimental process involves different processes as does the calculations.processes as does the calculations.

Using mass percent data and molar mass is the Using mass percent data and molar mass is the most straightforward.most straightforward.

Combustion analysis is more involved.Combustion analysis is more involved.COMPARISON of CALCULATIONS

Empirical & Molecular Formula Empirical & Molecular Formula DeterminationDetermination

Quinine:Quinine:C 74.05%, H 7.46%, N 8.63%, O 9.86%C 74.05%, H 7.46%, N 8.63%, O 9.86%• 74.05/74.05/12.0112.01, 7.46/, 7.46/1.0081.008, 8.63/, 8.63/14.0114.01, ,

9.86/9.86/16.0016.00

CC6.1666.166 H H7.407.40 N N0.6160.616 O O0.616 0.616

• Empirical Formula: CEmpirical Formula: C1010 H H1212 N N11 O O11

Empirical Formula Weight = ?Empirical Formula Weight = ?• Molecular Weight = Molecular Weight = 324.42324.42Molecular Formula = 2x empirical Molecular Formula = 2x empirical

formulaformula

• Molecular Formula = CMolecular Formula = C2020 H H2424 N N22 O O22

A Mass SpectrometerRecords a mass spectrum

A mass spectrum records only positively charged fragments

m/z = mass to charge ratio of the fragment

http://chemconnections.org/pdb/Quinine.html

N

HON

H

HH2C

H3CO

A Carbon atom is at each angle.Each C has 4 bonds (lines + Hs).Hs are not always drawn in &must be added.

From the structures, determine the molecular formula of quinine.

A)C18H24NO2 B)C20H20NO3 C)C20H24N2O2 D)C20H26N2O2

QUESTION

http://chemconnections.org/pdb/Quinine.html

N

HON

H

HH2C

H3CO

A Carbon atom is at each angle.Each C has 4 bonds (lines + Hs).Hs are not always drawn in &must be added.

C = 20H = 24N = 2O = 2

From the structures, determine the molecular formula of quinine.

C20H24N2O2

ANSWER

A)C18H24NO2 B)C20H20NO3 C)C20H24N2O2 D)C20H26N2O2

QUESTIONQUESTION

ANSWERANSWER

B) C8H8

The mass of CH is 13.018. 13.018 divides into 104.1 about 8 times. Therefore there are 8 CH groups in this compound. C8H8 is the molecular formula.

Examine the condensed structural formulas shown below:

I) Acetic acid (main ingredient in vinegar), CH3COOH

II) Formaldehyde (used to preserve biological specimens), HCHO

III) Ethanol (alcohol in beer and wine), CH3CH2OH

For which molecule(s) are the empirical formula(s) and the molecular formula(s) the same?

A) II B) I and II C) II and III D) I, II, and III

QUESTIONQUESTION

Examine the condensed structural formulas shown below:

I) Acetic acid (main ingredient in vinegar), CH3COOH

II) Formaldehyde (used to preserve biological specimens), HCHO

III) Ethanol (alcohol in beer and wine), CH3CH2OH

For which molecule(s) are the empirical formula(s) and the molecular formula(s) the same?

A) II B) I and II C) II and III D) I, II, and III

ANSWERANSWER