Mass Relationships of Atoms - Dr. Marten’s AP Chemistryteachermarten.com/APChem/APLectureNotes_files/08Ch03Zum7...Calculate the number of grams of lead (Pb) In 12.4 moles of lead.

Mass Relationships of Atoms

Atomic number and mass number

Atomic number (Z) = the number of protons in the nucleus.

Mass number (A) = the sum of the number of protons + neutrons in the nucleus.

ZA X

Some isotopes

11

H 126

C 168

O 23592

U

21H 13

6C 17

8O 238

92U

31H 14

6C

188O

Atomic masses- synonymous with atomic weight

- is a relative scale- the isotope of carbon with 6 p+ and 6

no(carbon-12) is the reference atom and assigned an atomic mass of exactly 12

- one atomic mass unit (amu) is defined as a mass exactly equal to 1/12th the mass of one

carbon-12 atom- so 1 amu is about the mass of 1 ___ or 1 ___p+ no

Average Atomic Mass and the Periodic Table

Average Atomic Mass: The Story of Boron

• Any chunk of Boron you dig up is a mixture of 2 isotopes: on average ~20% will be 10B, ~80% will be 11B

• Say we get a small chunk of B.

• 100 atoms.

• 20 are 10B with a mass of 10 amu each

• 80 are 11B with a mass of 11 amu each

• What is the average mass of Boron?

The Story of Boron (cont.)• Average mass of a B atom

• = (total mass)/(total # of B atoms)

• Total mass for these 100 B atoms

• = 20 x 10 amu + 80 x 11 amu

• Average mass of a B atom

• = (20 x 10 amu + 80 x 11 amu)/100

• = 20 x 10 amu + 80 x 11 amu

• = 0.20 x 10 amu + 0.80 x 11 amu = 10.8 amu100 100

So to get the average atomic mass, multiply % abundance times each isotope mass and add up!

20% 80%

The Mole- the fundamental SI measure of “amount of

substance”- the amount of substance that contains as

many elementary entities as there are atoms in exactly 12 g of carbon-12

- this number of atoms is 6.022045 x 1023

Avogadro’s number

The Mole vs. The Dozen

The Dozen - the amount of substance that contains 12 entities.

The Mole - the amount of substance that contains Avogadro’s number (6.022x1023) of

entities.

Dozen Apples = 10 Lbs.

Mole of Helium atoms = 4.0026g

Dozen Apples = 12 Apples

Mole of Helium atoms = 6.022x1023 atoms

Example

1.3 Lbs

How many dozens of apples are represented by 1.3 Lbs. of apples.

Converting to Dozens

10 Lbsx

1 dozen.13 dozen=

Example

How many moles of He are in 6.46 g of He?

Converting to Moles

1.61 mol=6.46 gHe4.003 g1 molx

Mass Relationships of Atoms - Extended

ExampleCalculate the number of grams of lead (Pb)

In 12.4 moles of lead.

12.4 mol 2569 g=207.2 g

1molx

ExampleWhat is the mass in grams of one silver atom?

107.9 g

1 mol

= 17.91 x 10–23 g 1 atom

6.022 x 1023 atoms x

1 molSolution: Convert grams per mole to grams per atom.

Experimental DeterminationOf Atomic & Molecular Masses

Atomic mass is measured by mass spectrometry

An atom is ionized in the instrument.

+

e–

Ion is deflected by magnetic field

+

amount of deflection depends on mass to charge

ratio

highest m/z ratiodeflected ________

lowest m/z ratiodeflected ________

N

Sleast

most

Ions are detected after passagethrough magnetic field

+ ++ ++

+

N

S

target

Ions are detected after passagethrough magnetic field

++++ ++mixture of ions of

different mass givesseparate peak for each m/zintensity of peak proportionalto percentage of each atom of

different mass in mixtureseparation of peaks

depends on relative mass

N

S

target

19 20 21 22 23

90.92%

0.26%8.82%

The mass spectrum ofthe three isotopes of

neon.

Percent Composition of Compounds

Percent composition is the percent by mass of each element the compound contains.

Obtained by dividing the mass of each element in one mole of the compound by the

molar mass of the compound and multiplying by 100%

Example

Calculate the percent composition by mass of H,P and O for phosphoric acid (H3PO4)

Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99

Solution: Assume you have 1 mole of the compound.

3.086%

Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99

31.61%

65.31%

x 100% =%H = 97.99g

3(1.008g)

100% =%P = x30.97g

97.99g

100% =%O = x4(16.00)

97.99g

Determining Formula

Determining the Formula with a CHN Analyzer

Determining the Formula with a CHN Analyzer

0.1156 g compound containing only

C,H,&N H2O absorber CO2 absorber

O2

O2compound + H2O + CO2 + other gasses

Burning the sample completely

CO2 , H2O, O2 , and N2 O2, and other gases

Reacting the C,H,N sample with O2 , we assume all of the hydrogen and carbon present in the 0.1156g

sample is converted to H2O and CO2 respectively

Determining the Formula

0.1638g CO2

0.1676g H2O

grams collected

x12.01g C

44.009 g CO2

x2.016g H

18.015 g H2O

= 0.04470 g C

0.01876 g H=

(27.3% C in CO2)

(11.1% H in H2O)

Remembering the original mass of the sample the percentages of the components can be determined

Determining the Formula

38.67% C + 16.23% H + % N = 100%

x 100%0.1156 g sample

0.04470 g C

0.01876 g H

0.1156 g samplex 100%

=

= 16.23% H

38.67% C

% N = 45.10%

Elemental Composition

Levels of Structure

Empirical Formula

Molecular Formula

Constitution

Configuration

Conformation

Examples:

Elemental Composition

Formaldehyde Glucose

C: 40.00% C: 40.00%

H: 6.73% H: 6.73%O: 53.27% O: 53.27%

Elemental Composition

Empirical Formula

Molecular Formula

Constitution

Configuration

Conformation

Levels of Structure

√

The empirical formula tells us which elements are present and the simplest

whole-number ratio of their atoms.

Empirical Formula

Examples: Formaldehyde and Glucose

Empirical Formula

Elemental Composition

C: 40.00%H: 6.73%O: 53.27%

assume a 100g samplecalculate atom ratios by dividing by atomic

weight

40.00 g6.73 g

53.27 g

40.00 g x 1 mol

12.01 g= 3.33 mol

6.73 g x 1 mol

1.00 g= 6.73 mol

53.27 g x 1 mol

16.0 g= 3.33 mol

100 g

Calculating Empirical Formula

C:

H:

O:

Examples: Formaldehyde and Glucose

Empirical Formula

Elemental CompositionC: 40.00%H: 6.73%O: 53.27%

assume a 100g sample

calculate atom ratios by dividing by atomic weight

40.00 g6.73 g

53.27 g

3.33 mol6.73 mol3.33 mol

determine the smallest whole number ratio by dividing by the smallest molar value

40.00 g x 1 mol

12.01 g= 3.33 mol

6.73 g x 1 mol

1.00 g= 6.73 mol

53.27 g x 1 mol

16.0 g= 3.33 mol

100 g

Calculating Empirical Formula

3.33 mol

3.33 mol

3.33 mol

= 1.00

= 2.02

= 1.00

C:

H:

O:

Examples: Formaldehyde and Glucose

Empirical Formula

Elemental CompositionC: 40.00%H: 6.73%O: 53.27%

40.00 g6.73 g

53.27 g

3.33 mol6.73 mol3.33 mol

12

1

Empirical Formula: CH2O

A 1.723 g sample of aluminum oxide (which consists of aluminum and oxygen only) contains 0.912g of Al. Determine the empirical formula of the compound.

Example

0.912 g Al = 0.811 g O 1.723 g sample -

0.912 g Al x 1 mol Al26.98 g Al

0.811 g O x 1 mol O16.0 g O

= 0.0338 mol

= 0.0507 mol0.0338 mol

0.0338 mol

= 1.0

1.5=

x 2 =

x 2 =

Al2O3

2

3

Elemental Composition

Empirical Formula

Molecular Formula

Constitution

Configuration

Conformation

Levels of Structure

√√

determined from empirical formula and experimentally determined molecular mass

Molecular Formula (the REAL formula)

Compound EmpiricalFormula

True molar mass

formaldehyde CH2Oglucose CH2O

30180

Calculation of empirical mass1 mol C = 12.01 g

2 mol H x 2 = 2.016 g1 mol O = 16.00g

30.026g

30g 30g

= 1formaldehyde

180g 30g

= 6glucoseEmpirical mass

Molecular mass

determined from empirical formula and experimentally determined molecular mass

Molecular Formula

Compound EmpiricalFormula

Molar mass

Molecularformula

formaldehyde CH2Oglucose CH2O

30180

CH2OC6H12O6

Mass Relationships

Stoichiometry

mass relationships

how much reactant is needed to yield a certain amount of product

Amounts of Reactantsand Products

The mole method

1. Write and balance the equation.2. Convert to moles.

3. Use the coefficients in the balanced equation to relate the number of moles of known

substances to the unknown one.

4. Convert to desired units (g, L, etc.).5. Check your answer.

Stoichiometry

X YMass of X Mass of Y

Moles of X

(using molar

ratio of Y to X)

nynx

Moles of Y

Example

How many grams of nitrogen dioxide can be formed by reaction of 1.44 g of nitrogen

monoxide with oxygen?

NO + O2 NO22 2

Stoichiometry

1.44 g of NO Mass of NO2

Moles of X

Molar ratio of Y to X nynx

Moles of Y

NO + O2 NO22 2

Stoichiometry

1.44 g of NO Mass of NO2

0.048 Moles NO

Molar ratio of Y to X nynx

Moles of Y

NO + O2 NO22 2

Stoichiometry

1.44 g of NO Mass of NO2

0.048 Moles NO

Molar ratio NO2 to NO 22

Moles of Y

NO + O2 NO22 2

Stoichiometry

1.44 g of NO Mass of NO2

0.048 Moles NO

Molar ratio NO2 to NO 22

0.048 Moles NO2

NO + O2 NO22 2

Stoichiometry

1.44 g of NO 2.21 g of NO2

0.048 Moles NO

Molar ratio NO2 to NO 22

0.048 Moles NO2

NO + O2 NO22 2

Example

1.44g NO

NO + O2 NO22 2

= 2.21g NO2

x 2mol NO2 2mol NO

x46g NO2

1mol NO2

1mol NO

30g NO x

Limiting Reagents

Limiting Reagent

Reactants are not always present (or available) in “stoichiometric” quantities.

One reactant may be present in quantities such that it is completely consumed while excess

amounts of other reactants remain.

- called “limiting reactant” or “limiting reagent”

The limiting reagent will limit the amount of product produced.

Example

How many moles of MgCl2 will be produced?

Mg + Cl2 MgCl2

Start 1 mol 1 mol 0

Finish 0 0 1 mol

Example

How many moles of MgCl2 will be produced?

Mg + Cl2 MgCl2

Start 1 mol 2 mol 0

Finish 0 1 mol 1 mol

magnesium is the limiting reagent1 mol of chlorine will be left unchanged

W + X YMass of X Mass of Y

Moles of X

Molar ratio Y to X nynx

Moles of Y

Mass of W

Moles of W

Limiting Reagent

W + X YMass of X

Moles of X

Mass of W

Moles of W

Compare molar ratio W to X to theircoefficients in balancd equation; identify LR

Molar ratio Y to LR

Mass of Y

Moles of Y

ExampleDetermine the limiting reagent and the amount of PI3

produced when 6.00g P4 reacts with 25.0g of I2.

P4 + 6I2 4PI3

6.00g P4 = .0484mol P41mol P4

124g P4 x

25.0g I2 = .0984mol I21mol I2

254g I2 x

Example cont...Determine how much I2 would be needed to react

completely with the available amount of P4.

P4 + 6I2 4PI3

.0484mol P4 x 6mol I2

1mol P4 0.290mol I2 =

but only .0984mol I2 is availableI2 is the limiting reagent

amount of I2

needed

Example cont...…the amount of PI3 produced from the limiting

reagent...

.0984mol I2 x 4mol PI3

6mol I2 x 412g PI3

1mol PI3

=

27.0g PI3

another methodDetermine the limiting reagent and the amount of PI3

produced when 6.00g P4 reacts with 25.0g of I2.

P4 + 6I2 4PI3

Compute the moles of product formed 2 ways (assuming P4 is completely consumed and then assuming I2 is

completely consumed. See which reactant, if used up completely, gives the least amount of product. It’s limiting.

Reaction Yield

Theoretical yield the amount of product that would result if

all the limiting reagent reacted

Actual yield the amount of product actually obtained

from the reaction

Almost always less than the theoretical yield

Percent Yield

%yield = x 100%Theoretical yield

Actual yield

Determines how efficient a reaction is

Example

In a certain industrial operation 3.54 x 107g of TiCl4 is reacted with 1.13 x 107g of Mg. (a) Calculate the

theoretical yield of Ti in grams. (b) Calculate the percent yield if 7.91 x 106g are actually obtained.

TiCl4 + 2Mg Ti + 2MgCl2

= 1.87 x 105mol TiCl4

= 4.65 x 105 mol Mg

1.13 x 107g Mg1mol Mg

24.31g Mg x

1mol TiCl4

187.7g TiCl4

x3.54 x 107g TiCl4

1.87 x 105mol TiCl4

2mol Mg

1mol TiCl4

x = 3.74 x 105 mol Mg

TiCl4 is limiting

Calculate theoretical yield

there is more than enough Mg

1mol TiCl4

187.7g TiCl4

x3.54 x 107g TiCl4

8.93 x 106g Ti=

x1mol Ti

1mol TiCl4

x47.88g Ti

1mol Ti

%yield = x 100%

Theoretical yield

Actual yield

8.93 x 106g Ti

7.91 x 106g Ti100% x =

= 88.6%