Stoichiometry Dr. Ron Rusay. Chemical Stoichiometry ð Stoichiometry is the study of mass in...

StoichiometryStoichiometry

Dr. Ron RusayDr. Ron Rusay

Chemical StoichiometryChemical Stoichiometry

Stoichiometry Stoichiometry is the study of mass in chemical reactions. It is the study of mass in chemical reactions. It deals with both reactants and products.deals with both reactants and products.

It quantitatively and empirically relates the behavior of It quantitatively and empirically relates the behavior of atoms and molecules in a balanced chemical equation to atoms and molecules in a balanced chemical equation to observable chemical change and measurable mass effects.observable chemical change and measurable mass effects.

It accounts for mass and the conservation of mass, just as the It accounts for mass and the conservation of mass, just as the

conservation of atoms in a balanced chemical equation.conservation of atoms in a balanced chemical equation.

Chemical ReactionsChemical ReactionsAtoms, Mass & Balance: eg. Zn(s) + S(s) Atoms, Mass & Balance: eg. Zn(s) + S(s)

Chemical EquationChemical Equation

_ C_ C22HH55OH + _ OOH + _ O22 _ CO_ CO22 + _ H + _ H22OO

Reactants Reactants ProductsProducts• C=2; H =5+1=6; O=2+1C=2; H =5+1=6; O=2+1 C=1; H=2; O=2+1C=1; H=2; O=2+1

• 11 C C22HH55OH + OH + 33 O O22 22 CO CO22 + + 33 H H22OO

Representation of a chemical reaction:


The balanced equation can be completely stated as:The balanced equation can be completely stated as:

1 mole of ethanol1 mole of ethanol reacts with reacts with 3 moles of 3 moles of oxygenoxygen to produce to produce 2 moles of carbon 2 moles of carbon dioxide dioxide and and 3 moles of water.3 moles of water.

CC22HH55OH OH ++ 3 O3 O22 2 CO2 CO22 ++ 3 H3 H22OO

All Balanced Equations relate on a molar mass basis. For All Balanced Equations relate on a molar mass basis. For the combustion of octane:the combustion of octane:

2 2 CC88HH18(l)18(l)+ + 25 25 OO2(g)2(g) 16 16 COCO2(g) 2(g) ++18 18 HH22OO(l)(l)

• moles moles of octane react with of octane react with 25 moles25 moles of oxygen of oxygen to to produce produce 16 moles16 moles of carbon dioxide and of carbon dioxide and 18 moles18 moles of of water. water.

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QUESTIONQUESTIONThe fuel in small portable lighters is butane (C4H10). After using a lighter for a few minutes, 1.0 gram of fuel was used. How many moles of carbon dioxide would it produce?

• 58 moles• 0.017 moles• 1.7 10–24 moles• 0.068 moles

ANSWERANSWERThe fuel in small portable lighters is butane (C4H10). After using a lighter for a few minutes, 1.0 gram of fuel was used. How many moles of carbon dioxide would it produce?

A. 58 molesB. 0.017 molesC. 1.7 10–24 molesD. 0.068 moles

2 C4H

10(g) + 13 O

2(g) → 8CO2(g)+10H2O( )g

The molar mass of butane: 4 12g/mol = 48 for carbon + 10 1g/mol = 10.0 for hydrogen. Total = 48.0 + 10.0 = 58.0 g/mol. Next; 1.0 gram of butane 1 mol/58.0 g = 0.017 mol

The Chemical Equation: The Chemical Equation: Mole & MassesMole & Masses

46g (1 mole) of ethanol46g (1 mole) of ethanol reacts with reacts with 3 moles 3 moles of oxygen (96g)of oxygen (96g) to produce to produce 2 moles of 2 moles of carbon dioxide carbon dioxide and and 3 moles of water.3 moles of water.

How many grams of How many grams of carbon dioxide carbon dioxide and and waterwater are respectively produced from 46g are respectively produced from 46g (1 mole) of ethanol(1 mole) of ethanol ??

2 mol x 44 g/mol = 88g2 mol x 44 g/mol = 88g 3 mol x 18 g/mol = 54 g3 mol x 18 g/mol = 54 g

CC22HH55OH OH ++ 3 O3 O22 2 CO2 CO22 ++ 3 H3 H22OO

The Chemical Equation: The Chemical Equation: Moles & MassesMoles & Masses

• CC22HH55OH OH ++ 3 O3 O22 2 CO2 CO22 ++ 3 H3 H22OO

How many grams of How many grams of oxygen oxygen are needed to are needed to react with 15.3g of ethanol in a 12oz. beerreact with 15.3g of ethanol in a 12oz. beer ??

15.3g15.3gethanol ethanol x mol x molethanol ethanol /46.0g/46.0gethanol ethanol = 0.333mol = 0.333molethanol ethanol

0.333mol0.333molethanol ethanol x x 3mol3moloxygenoxygen / / 1mol1molethanol ethanol = = 0.999mol0.999moloxygenoxygen

32.0g32.0goxygenoxygen /mol /moloxygenoxygen x 0.999mol x 0.999moloxygen oxygen = 32.0g= 32.0goxygenoxygen

NOTE: It takes approximately 1 hour for the biologically equivalent amount of oxygen available from cytochrome p450 to consume the

alcohol in a human in 1 beer to a level below the legal limit of 0.08%.

Chemical StoichiometryChemical Stoichiometry

Epsom salt (magnesium sulfate heptahydrate) is one Epsom salt (magnesium sulfate heptahydrate) is one of five possible hydrates: mono-, di-, tri-, hexa-, or of five possible hydrates: mono-, di-, tri-, hexa-, or heptahydrate.heptahydrate.

How can stoichiometry be used to determine, which How can stoichiometry be used to determine, which hydrate is present in a pure unknown sample, by hydrate is present in a pure unknown sample, by heating the sample in a kitchen oven at 400 heating the sample in a kitchen oven at 400 oo C for C for 45 minutes? 45 minutes?

MgSO4 . x H2O(s) MgSO4 (s) + x H2O (g)

Mass CalculationsMass Calculations

All Balanced Equations relate on a molar and mass basis. All Balanced Equations relate on a molar and mass basis. For the combustion of octane:For the combustion of octane:

2 2 CC88HH18(l)18(l)+ + 25 25 OO2(g)2(g) 16 16 COCO2(g) 2(g) ++18 18 HH22OO(l)(l)

228 g of octane (228 g of octane (2 moles)* 2 moles)* will react with 800 g of will react with 800 g of oxygen (oxygen (25 moles) 25 moles) to produce (to produce (16 moles) 16 moles) 704 g of 704 g of carbon dioxide and (carbon dioxide and (18 moles) 18 moles) 324 g of water. 324 g of water.

*(*(2 moles octane x 114 g/mol = 228 g )2 moles octane x 114 g/mol = 228 g )

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Mass Calculations: Mass Calculations: Reactants ProductsReactants Products

Something (S) Another Thing (AT)

Mass (S) Mass (AT)

grams (S) grams (AT)

Chemically Relate:

1. Balance the chemical equation.2. Convert mass of reactant or product

to moles.3. Identify mole ratios in balanced equation:

They serve as the “Gatekeeper”.4. Calculate moles of desired product or

reactant.5. Convert moles to grams.



grams (S)

grams (S)

(Molecular Weight)

1 mol (S)

grams (AT)

grams (AT)

1 mol (AT)

(Molecular Weight)

Avogadro's NumberAtomsMoleculesStoichiometry

?

?"Gatekeeper"


• How many grams of salicylic acid are needed to produce How many grams of salicylic acid are needed to produce 1.80 kg of aspirin?1.80 kg of aspirin?

• Balanced Equation:Balanced Equation:

C7H6O3

MW = 138.12C2H3OClMW = 78.49

C9H8O4

MW = 180.15HClMW= 36.45

O

H

C

O

O

H

H

H

H

H

+ C

H

3

C

O

C

l

O

C

C

H

3

C

O

O

H

H

H

H

H

O

+ Cl

H

Salicylic Acid Aspirin


grams (Aspirin) grams (Salicylic Acid)

1800 grams (A)

grams (A)

(Molecular Weight A)

1 mol (A)

grams (SA)

? (SA)

1 mol (SA)

(Molecular Weight SA)


1 mol SA

1 mol A

"Gatekeeper"


Mass Calculations: Mass Calculations: How many grams of How many grams of salicylic acid are needed to produce salicylic acid are needed to produce 1.80 kg1.80 kg of aspirin? of aspirin?

• Balanced Equation:Balanced Equation:

C7H6O3

MW = 138.12C2H3OClMW = 78.49

C9H8O4

MW = 180.15HClMW= 36.45

O

H

C

O

O

H

H

H

H

H

+ C

H

3

C

O

C

l

O

C

C

H

3

C

O

O

H

H

H

H

H

O

+ Cl

H

Salicylic Acid Aspirin

?g C7 = 1.80 x 103 gC9 x [molC9 /180.15gC9 ] x [1 molC7/ 1 molC9] x 138.12 gC7/molC7

11

?g C7 = 1380 grams

QUESTIONQUESTIONThe fuel in small portable lighters is butane (C4H10). After using a lighter for a few minutes, 1.0 gram (0.017 moles) of fuel was used. How many grams of carbon dioxide would it produce?

2 C4H

10(g) + 13 O

2(g) → 8CO2(g)+10H2O( )g

How many grams of carbon dioxide would this produce?

A.) 750 mg B.) 6.0 g

C) 1.5 g D.) 3.0 g

ANSWERANSWERD.) 3.0 g


grams (S)

grams (S)

(Molecular Weight)

1 mol (S)

grams (AT)

grams (AT)

1 mol (AT)

(Molecular Weight)


?

?"Gatekeeper"

1.0 g C4H10 ?g CO2

C4H10

mol C4H10

1.0 g

58 g C4H10

mol CO2

44 g CO2

?g CO2

2 C4H10(g) + 13 O2(g) → 8CO2( )g +10H2O( )g

2 molC4H10

8 molCO2

3.0 g CO2

Combustion AnalysisCombustion Analysis

SEE: COMPARISON to wt % CALCULATIONS

Molecules with oxygen in their formula are more difficult

to solve for Oz knowing only the respective masses of

CxHyOz sample, CO2 and H2O.

• CxHyOz + (x + y/4 - z) O2 x CO2 + y/2 H2O

• CxHy + (x + y/4) O2 x CO2 + y/2 H2O

Combustion Analysis CalculationCombustion Analysis CalculationAscorbic Acid ( Vitamin C )Ascorbic Acid ( Vitamin C )

• Combustion of a 6.49 mg sample in excess oxygen, Combustion of a 6.49 mg sample in excess oxygen, yielded 9.74 mg COyielded 9.74 mg CO2 2 and 2.64 mg H and 2.64 mg H22OO

• Calculate it’s Empirical formula!Calculate it’s Empirical formula!

• C: 9.74 x10C: 9.74 x10-3-3g COg CO22 x(12.01 g C/44.01 g CO x(12.01 g C/44.01 g CO22))

= = 2.65 x 102.65 x 10-3-3 g C g C

• H: 2.64 x10H: 2.64 x10-3-3g Hg H22O x (2.016 g HO x (2.016 g H22/18.02 gH/18.02 gH22O)O)

= = 2.92 x 102.92 x 10-4-4 g H g H

• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg

= = 3.54 mg O 3.54 mg O

Vitamin C: CalculationVitamin C: Calculation(continued)(continued)

• C = 2.65 x 10C = 2.65 x 10-3-3 g C / ( 12.01 g C / mol C ) = g C / ( 12.01 g C / mol C ) =

= = 2.21 x 102.21 x 10-4-4 mol C mol C

• H = 0.295 x 10H = 0.295 x 10-3-3 g H / ( 1.008 g H / mol H ) = g H / ( 1.008 g H / mol H ) =

= = 2.92 x 102.92 x 10-4-4 mol H mol H

• O = 3.54 x 10O = 3.54 x 10-3-3 g O / ( 16.00 g O / mol O ) = g O / ( 16.00 g O / mol O ) =

= = 2.21 x 102.21 x 10-4-4 mol O mol O

• Divide each by 2.21 x 10-4

• C = 1.00 Multiply each by 3 = 3.00 = 3.0C = 1.00 Multiply each by 3 = 3.00 = 3.0

• H = 1.32 = 3.96 = 4.0H = 1.32 = 3.96 = 4.0

• O = 1.00O = 1.00 = 3.00 = 3.0= 3.00 = 3.0C3H4O3

QUESTIONQUESTION

Erythrose contains carbon, hydrogen and oxygen (MM = 120.0 g/mol). It is an important sugar that is used in many chemical syntheses.

Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and 0.4194 g H2O. Mass Spectrometry produced a molecular ion @ 120 mass units (m/z). What is the molecular formula of erythrose?

A) CH2OB) C6H12O6

C) C3H6O3

D) C4H8O4

Alternative Solution Method

Mass ratio of C in CO2 = =

= = 0.2729 g C / 1 g CO2

Mass ratio of H in H2O = =

= = 0.1119 g H / 1 g H2O

Calculating masses of C and H:

Mass of Element = mass of compound x mass ratio of element

mol C x M of Cmass of 1 mol CO2

1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2

mol H x M of Hmass of 1 mol H2O

2 mol H x 1.008 g H / 1 mol H 18.02 g H2O

ANSWERANSWER D) C4H8O4

Alternative Method

Mass (g) of C = 1.027 g CO2 x = 0.2803 g C

Mass (g) of H = 0.4194 g H2O x = 0.04693 g HCalculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g OCalculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O

C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula120 g /mol / 30 g / formula = 4 formula units / cmpd = C4H8O4

0.2729 g C 1 g CO2

0.1119 g H 1 g H2O

Stoichiometry Dr. Ron Rusay. Chemical Stoichiometry ð Stoichiometry is the study of mass in...

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Transcript of Stoichiometry Dr. Ron Rusay. Chemical Stoichiometry ð Stoichiometry is the study of mass in...