stoichiometry chapter 3 -mw. what is stoich? stoichiometry is the study of reactions: why do...

Stoichiometry

Chapter 3 -MW

What Is Stoich?

Stoichiometry is the study of reactions: Why do reactions occur? How fast do they proceed? What intermediary products if any are used? How much of the reactants react?

Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Mass SpectrometerCompares mass of atomsAtomic mass is defined by Carbon 12 = 12amuParts of a mass spectrometer:

VaporizerElectron beamsIon-accelerating electric fieldMagnetic fieldDetector Plate

Copyright©2000 by Houghton Mifflin Company. All rights

reserved.

4

Mass Spectrometer Process

1. A heater vaporizes a sample2. A beam of high speed electrons knocks electrons

off test atoms/molecules.3. An electric field accelerates the sample ions4. The accelerating ions have a magnetic field. They

interacts with an applied magnetic field deflecting their path. The ions separate.

5. A detector plate measures the deflections-comparison of deflections gives

ions’ masses-less massive particles deflect more

. 6

Atomic Masses

Elements occur in nature as mixtures of isotopes

Carbon = 98.89% 12C

1.11% 13C

<0.01% 14C

Carbon atomic mass = 12.01 amu

Change % abundance into decimals & multiply by respective isotopic weights.

Add together.

If want % abundance; use “x” & “1 – x” to represent abundance.

7

Examples1) There are two isotopes of carbon 12C with a mass

of 12.00000 amu(98.892%), and 13C with a mass of 13.00335 amu (1.108%).

2) There are two isotopes of nitrogen , one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundance of each if the weighted average is 14.01amu?

Answers

1) 0.98892(12.00000amu) + 0.01108(13.00335amu)

= 11.86704amu + .1440771amu = 12.011117amu

Matches P.table

14.0031(x) + 15.0001(1-x) = 14.01

14.0031x + 15.0001amu - 15.001x = 14.01 amu

-0.997x = -0.9901

X = .9930192 or 99.3%, 1-x = .692%

9

10

The Mole

The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.

1 mole of anything = 6.022 x 1023 units of that

thing

Avogadro’s number equals 6.022 x 1023 units

11

Molar Mass

A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.

CO2 = 44.01 grams per mole

Find the molar mass of

CH4

Mg3P2

Ca(NO3)3

Al2(Cr2O7)3

CaSO4 · 2H2O

CH4 =12.0 + (4) 1.01 = 16.0g/mol

Mg3P2 = (3) 24.3 + (2) 30.974 = 135g/mol

Ca(NO3)3 = 40.1 + (3) 14.0 + (9) 16.0 = 226g/mol

Al2(Cr2O7)3 = (2) 27.0 + (6) 52.0 + (21) 16.0 =

= 702g/mol

CaSO4 · 2H2O = 40.1 + 32.1 + (4) 16.0 + (2) 18.0

= 156g/mol13

Percent Composition

Mass percent of an element:

For iron in iron (III) oxide, (Fe2O3)

mass Fe%..

. 111 69159 69

100% 69 94%

massmass of element in compound

mass of compound% 100%

Working backwards

From percent composition, you can determine the empirical formula.

Empirical Formula the lowest ratio of atoms in a molecule.

Based on mole ratios.

16

Empirical Formula Determination

1. Base calculation on 100 grams of compound.

2. Determine moles of each element in 100 grams of compound.

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

4. Multiply each number by an integer to obtain all whole numbers.

A sample is 59.53% C, 5.38% H, 10.68% N, and 24.40% O. What is its empirical formula?

17

C H N O59.53 5.38 10.68 24.4012.0 1.01 14.0 16.0

4.96 5.33 .7628 1.525.7628 .7628 .7628 .7628

6.5 7 1 2 (mult. By 2)

C13H14N2O4

A 0.2000 gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O2 .

0.2998 g of CO2 and 0.0819 g of H2O are

produced. What is the empirical formula?

Get C from CO2, H from H2O and O from subtracting C + O from original amount.

C 0.2998g x 12g = 0.0817636g C

CO2 44g

H0.0819g x 2.02g = 0.00919g H

H20 18.0g 0.0909546g

0.2000g - 0.0909546g = 0.1090454g O19

0.0817636g C 0.00919g H 0.1090454g O

12.0 1.01 16.0

1 1.33 1(x 3)

C3 H4 O3

20

21

Molecular Formula

Molar mass = (empirical formula)n [n = integer]

empirical formula = CH, Molar mass = 78.0g

(CH)x = 78.0g, (12 + 1.01)x= 78.0g, x = 6

molecular formula = (CH)6 = C6H6

Example

A compound is made of only sulfur and nitrogen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its formula?

69.6/32.1 = 2.168 30.4/14.0 = 2.171

46.1x = 184, x = 3.99

S4N4

23

Chemical Equations

Chemical change involves a reorganization of the atoms in one or

more substances.

24

Chemical Equation

A representation of a chemical reaction:

C2H5OH + 3O2 2CO2 + 3H2O

reactants products

Chemical Equations

Are sentences.

Describe what happens in a chemical reaction.

Reactants Products

Equations should be balanced.

Have the same number of each kind of atoms on both sides because ...

Abbreviations

(s) (g) (aq)

heat

catalyst

27

Chemical Equation

C2H5OH + 3O2 2CO2 + 3H2O

The equation is balanced.

1 mole of ethanol reacts with 3 moles of oxygen

to produce

2 moles of carbon dioxide and 3 moles of water

Practice

Ca(OH)2 + H3PO4 H2O + Ca3(PO4)2

Cr + S8 Cr2S3

KClO3(s) Cl2(g) + O2(g)

Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.

Fe2O3(s) + Al(s) Fe(s) + Al2O3(s)

3Ca(OH)2 + 2H3PO4 6H2O + Ca3(PO4)2

16Cr + 3S8 8Cr2S3

2KClO3(s) Cl(s) + 3O2(g)

Fe2S3(s) + 6HCl(g) FeCl3(s) + 3H2S(g)

Fe2O3(s) + 2Al(s) 2Fe(s) + Al2O3(s)

29

All chemical reactions can be placed into one of six categories.  Here they are, in no particular order:

1) Combustion: A combustion reaction is when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An example of this kind of reaction is the burning of napthalene:

C10H8 + 12 O2 ---> 10 CO2 + 4 H2O

 . 30

2) Synthesis: A synthesis reaction is when two or more simple compounds combine to form a more complicated one. These reactions come in the general form of:

A + B ---> AB

One example of a synthesis reaction is the combination of iron and sulfur to form iron (II) sulfide:

8 Fe + S8 ---> 8 FeS

 31

3) Decomposition: A decomposition reaction is the opposite of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions come in the general form:

AB ---> A + B

One example of a decomposition reaction is the electrolysis of water to make oxygen and hydrogen gas:

2 H2O ---> 2 H2 + O2

 32

 4) Single displacement: This is when one element trades places with another element in a compound. These reactions come in the general form of:

A + BC ---> AC + B

One example of a single displacement reaction is when magnesium replaces hydrogen in water to make magnesium hydroxide and hydrogen gas:

Mg + 2 H 2O ---> Mg(OH)2 + H2

33

5) Double displacement: This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These reactions are in the general form:

AB + CD ---> AD + CB

One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium iodide to form lead (II) iodide and potassium nitrate:

Pb(NO3) 2 + 2 KI ---> PbI 2 + 2 KNO3

6) Acid-base: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water:

HA + BOH ---> H2O + BA

One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium hydroxide:

HBr + NaOH ---> NaBr + H2O

34

6) Acid-base: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water:

HA + BOH -> H2O + BA

One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium hydroxide:

HBr + NaOH -> NaBr + H2O35

Copyright©2000 by Houghton Mifflin Company. All rights

reserved.

36

Calculating Masses of Reactants and Products

1. Balance the equation.

2. Convert mass to moles.

3. Set up mole ratios.

4. Use mole ratios to calculate moles of desired substituent.

5. Convert moles to grams, if necessary.

ExamplesOne way of producing O2(g) involves the

decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed.

1) How many moles of O2(g) are produced?

2) How many grams of potassium chloride?

3) How many grams of oxygen?

2KClO3(s) Cl(s) + 3O2(g)

25.5g KClO3 x 1 mol x 3 mol = 0.3109756 mol O2

123g 2 mol

25.5g KClO3 x 1 mol x 2 mol x 74.6g = 15.5g Cl

123g 2 mol 1 mol

25.5g KClO3 x 1 mol x 3 mol x 32.0g = 9.95g O2

123g 2 mol 1 mol38

Examples1) A piece of aluminum foil 5.11 in x 3.23 in x

0.0381 in is dissolved in excess HCl(aq). How many grams of H2(g) are produced?

2) How many grams of each reactant are needed to produce 15 grams of iron from the following reaction?

Fe2O3(s) + Al(s) Fe(s) + Al2O3(s)

5.11in=12.98, 3.23in=8.20cm, 0.0381in=0.0968cm

vol = 10.3cm3 D = 2.7g/cm3

27.8g Al x 1mol x 3 mol H2 x 27.0g = 3.12g H2

27g 2 mol Al 1 mol

------------------------------------------------------------------

15g Fe x 1 mol x 2 mol Al x 27.0g = 7.25g Al

55.85g 2 mol Fe 1 mol

15g Fe x 1 mol x 1 mol Fe2O3 x 159.7g = 21.5g Al

55.85g 2 mol Fe 1 mol 40

ExamplesK2PtCl4(aq) + NH3(aq)

Pt(NH3)2Cl2 (s)+ KCl(aq)

What mass of Pt(NH3)2Cl2 can be produced

from 65 g of K2PtCl4 ?

How much KCl will be produced?

How much from 65 grams of NH3?

K2PtCl4(aq) + 2NH3(aq) Pt(NH3)2Cl2 (s)+ 2KCl(aq)

65 g K2PtCl4x 1mol x 1mol x 300g = 47g Pt(NH3)2Cl2

415g 1mol 1mol

How much KCl will be produced?

65 g K2PtCl4x 1mol x 2mol x 74.6g = 23g KCl

415g 1mol 1mol

How much from 65 grams of NH3?

65g NH3 x 1mol x 1mol x 300g = 574g Pt(NH3)2Cl2

17.0g 2mol 1mol

65g NH3x 1mol x 2mol x 74.6g = 285g KCl

17.0g 2mol 1mol 42

Limiting Reagent

Reactant that determines the amount of product formed.

The one you run out of first.

Makes the least product.

Book shows you a ratio method.

It works.

So does mine

Example

Ammonia is produced by the following reaction

N2 + H2 NH3

What mass of ammonia can be produced from a mixture of 100. g N2 and 500. g H2 ?

How much unreacted material remains?

100g N2 x 1mol x 2mol x 17.0g = 121g NH3

28.0g 1mol 1mol

500g H2 x 1mol x 2mol x 17.0g = 2805g NH3

2.02g 3mol 1mol

121g NH3 x 1mol x 3mol x 2.02g = 21.6g H2

17.0g 2mol 1mol

500g – 21.6g = 478g H2 unreacted 45

Excess Reagent

The reactant you don’t run out of.

The amount of stuff you make is the yield.

The theoretical yield is the amount you would make if everything went perfect.

The actual yield is what you make in the lab.

Percent Yield

% yield = Actual x 100% Theoretical

% yield = what you got x 100% what you could have got

Examples

Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.

2Al + 3Br2 2AlBr3

6.0g Al x 1mol x 2mol x 267g = 59.3g AlBr3

27.0g 2mol 1mol

50.3 x 100 = 84.8%

59.3

Copyright©2000 by Houghton Mifflin Company. All rights

reserved.

49

Examples

Years of experience have proven that the percent yield for the following reaction is 74.3%

Hg + Br2

HgBr2 If 10.0 g of Hg and

9.00 g of Br2 are reacted, how much HgBr2 will

be produced?

If the reaction did go to completion, how much excess reagent would be left?

Hg + Br2 HgBr2 If 10.0 g of Hg and 9.00 g of Br2 are reacted, how much HgBr2

will be produced?

10.0 g Hg x 1mol x 1mol x 361g = 17.96g

201g 1mol 1mol

9.00 g Br2 x 1mol x 1mol x 361g = 20.31g

160g 1mol 1mol

how much excess reagent would be left?

.743 x 17.96g = 13.34 g HgBr2

13.34 g HgBr2 x 1mol x 1mol x 160g = 5.91gBr2

361g 1mol 1mol 9.00-5.91=3.09g excess

Examples

Commercial brass is an alloy of Cu and Zn. It reacts with HCl by the following reaction Zn(s) + 2HCl (aq) ZnCl2 (aq) + H2(g)

Cu does not react.

When 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnCl2 are eventually

isolated. What is the composition of the brass?

Zn(s) + 2HCl(aq) ZnCl2 (aq) + H2(g)

0.5065 g of brass is reacted with excess HCl, 0.0985g ZnCl2 x 1mol x 1mol x 65.39g

= .0473597g

136g 1mol 1mol

0.5065 - 0.0473597g = 0.4591 x 100 = 90.6% Cu

9.35% zn

53