Chapter 3Chapter 3

StoichiometryStoichiometry

Chapter 3 - Stoichiometry

3.1 Atomic Masses3.2 The Mole3.3 Molar Mass3.4 Percent Composition of Compounds3.5 Determining the Formula of a Compound3.6 Chemical Equations3.7 Balancing Chemical equations3.8 Stoichiometric Calculations: Amounts of Reactants and Products3.9 Calculations Involving a Limiting Reactant

Reaction of zinc and sulfur.

Figure 3.1: Mass spectrometer

Chemists using a mass spectrometer to analyze for copper in blood plasma.

Source: USDA Agricultural Research Service

A herd of savanna-dwelling elephants

Source: Corbis

Figure 3.2: Relative intensities of the signals recorded when natural neon is

injected into a mass spectrometer.

Figure 3.3: Mass spectrum of natural copper

Atomic Definitions II: AMU, Dalton, 12C Std.

Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom on this scale Hydrogen has a mass of 1.008 AMU.

Dalton (D) = The new name for the Atomic Mass Unit, one dalton = one Atomic Mass Unit on this scale, 12C has a mass of 12.00 daltons.

Isotopic Mass = The relative mass of an Isotope relative to the Isotope 12C the chosen standard.

Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances.

Isotopes of Hydrogen

• 11H 1 Proton 0 Neutrons 99.985 % 1.00782503 amu

• 21H (D) 1 Proton 1 Neutron 0.015 % 2.01410178 amu

• 31H (T) 1 Proton 2 Neutrons -------- ----------

The average mass of Hydrogen is 1.008 amu • 3H is Radioactive with a half life of 12 years.• H2O Normal water “light water “ • mass = 18.0 g/mole , BP = 100.000000C• D2O Heavy water• mass = 20.0 g/mole , BP = 101.42 0C

Element #8 : Oxygen, Isotopes

168O 8 Protons 8 Neutrons

99.759% 15.99491462 amu

• 178O 8 Protons 9 Neutrons

0.037% 16.9997341 amu

• 188O 8 Protons 10 Neutrons

0.204 % 17.999160 amu

Calculating the “Average” Atomic Mass of an Element

24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu

___________ amu

With Significant Digits = __________ amu

Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%).

Calculate the Average Atomic Mass of Zirconium, Element #40

Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr.

Isotope (% abd.) Mass (amu) (%) Fractional Mass

90Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu91Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu92Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu94Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu96Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu

____________ amu

With Significant Digits = ___________ amu

Problem: Calculate the abundance of the two Bromine isotopes: 79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that the average mass of Bromine is 79.904 g/mol.

Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0

Solution: X(78.918336) + Y(80.91629) = 79.904

X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904

78.918336 - 78.918336 Y + 80.91629 Y = 79.904

1.997954 Y = 0.985664 or Y = 0.4933

X = 1.00 - Y = 1.00 - 0.4933 = 0.5067

%X = % 79Br = 0.5067 x 100% = 50.67% = 79Br %Y = % 81Br = 0.4933 x 100% = 49.33% = 81Br

LIKE SAMPLE PROBLEM 3.2

During a perplexing dream one evening you come across 200 atoms of Einsteinium. What would be the total mass of this substance in grams?

SOLUTION:

200 Atoms of Es X 252 AMU/Atom = 5.04 x 104 AMU

5.04 x 104 AMU x (1g / 6.022 x 1023 AMU) =

______________ g of Einsteinium

MOLE

• The Mole is based upon the following definition:• The amount of substance that contains as many

elementary parts (atoms, molecules, or other?) as there are atoms in exactly

• 12 grams of carbon -12.

• 1 Mole = 6.022045 x 1023 particles

Figure 3.4: One-mole samples of copper, sulfur, mercury, and carbon

One moleof commonSubstances

CaCO3

100.09 gOxygen 32.00 gCopper 63.55 gWater 18.02 g

MolesMoles

Molecules

Avogadro’s Number

Molecular Formula

Atoms

6.022 x 1023

Mole - Mass Relationships of Elements

Element Atom/Molecule Mass Mole Mass Number of Atoms

1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms

1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms

1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 1023 atoms

1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 1023 atoms

1 molecule of O2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule

1 molecule of S8 = 256.56 amu 1 mole of S8 = _______ g = 6.022 x 1023 molecules

Molecular Mass - Molar Mass ( M )

The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu

Molar mass = ( 2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 g ) + 16.00 g = __________ g 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

LIKE SAMPLE PROBLEM 3.4

How many carbon atoms are present in a 2.0 g tablet of Sildenafil citrate (C28H38N6O11S) ?

SOLUTION:

MW of Sildenafil citrate = 28 X 12 amu (C) +

38 X 1 amu (H) +

6 X 14 amu (N) +

11 X 16 amu (O) +

1 X 32 amu (S) =

666 AMU

LIKE SAMPLE PROBLEM 3.4 (cont…)

2.0 g (C28H38N6O11S) X 1 mol/666g =

3.0 X 10-3 mol (C28H38N6O11S)

3.0 X 10-3 mol (C28H38N6O11S) X

6.022 X 1023 molecules / 1 mol (C28H38N6O11S) =

1.8 X 1021 molecules of C28H38N6O11S

1.8 X 1021 molecules of C28H38N6O11S X

28 atoms of C / 1 molecules of C28H38N6O11S =

Carbon Atoms

Calculating the Number of Moles and Atoms in a Given Mass of Element

Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal?Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number!Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol

NO. of W atoms = 1.90 x 10 - 4 mol W x = = __________________ atoms of Tungsten

1 mol W183.9 g W

6.022 x 1023 atoms 1 mole of W

Calculating the Moles and Number of Formula Units in a given Mass of Cpd.

Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample?Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients.Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol

Converting mass to moles:Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4

= 0.23545 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4= ______________ formula units

Flow Chart of Mass Percentage Calculation

Moles of X in one mole of Compound

Mass % of X

Mass fraction of X

Mass (g) of X in onemole of compound

M (g / mol) of X

Divide by mass (g) of one mole of compound

Multiply by 100

Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I

Problem: Sucrose (C12H22O11) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose?

(a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd

= 0.421046 To find mass % of C = 0.421046 x 100% = ______%

Mass Fraction of C = =

Calculating Mass Percents and Masses of Elements in a Sample of Compound - II

(a) continued Mass % of H = x 100% = x 100%

= 6.479% H

Mass % of O = x 100% = x 100%

= 51.417% O

(b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose)

Mass (g) of C = 24.35 g sucrose X = C

mol H x M of H 22 x 1.008 g Hmass of 1 mol sucrose 342.30 g

mol O x M of O 11 x 16.00 g Omass of 1 mol sucrose 342.30 g

0.421046 g C 1 g sucrose

Mol wt and % composition of NH4NO3

• 2 mol N x 14.01 g/mol = 28.02 g N

• 4 mol H x 1.008 g/mol = 4.032 g H

• 3 mol O x 15.999 g/mol = 48.00 g O

80.05 g/mol

%N = x 100% = 35.00% N28.02g N2

80.05g

%H = x 100% = 5.037% H 4.032g H2

80.05g

%O = x 100% = 59.96% O48.00g O2

80.05g99.997%

Calculate the Percent Composition of Sulfuric Acid H2SO4

Molar Mass of Sulfuric acid = 2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol

%H = x 100% = 2.06% H2(1.008g H2) 98.09g

%S = x 100% = 32.69% S1(32.07g S) 98.09g

%O = x 100% = 65.25% O4(16.00g O) 98.09 g

Check = 100.00%

Penicillin is isolated from a mold

Source: Getty Images

Empirical and Molecular Formulas

Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms.

Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.

Figure 3.6: Examples of substances whose empirical and molecular formulas differ.

Steps to Determine Empirical Formulas

Mass (g) of Element

Moles of Element

Preliminary Formula

Empirical Formula

M (g/mol )

use no. of moles as subscripts

change to integer subscripts

Some Examples of Compounds with the same Elemental Ratio’s

Empirical Formula Molecular Formula

CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8

OH or HO H2O2

S S8

P P4

Cl Cl2

CH2O (carbohydrates) C6H12O6

Figure 3.7: Structural Formula of P4O10

Determining Empirical Formulas from Masses of Elements - I

Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is theempirical formula and name of the compound?Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.Solution: Finding the moles of the elements:

Moles of Na = 5.678 g Na x = _________ mol Na

Moles of Cr = 6.420 g Cr x = ___________ mol Cr

Moles of O = 7.902 g O x = ____________ mol O

1 mol Na22.99 g Na 1 mol Cr52.00 g Cr

1 mol O16.00 g O

Determining Empirical Formulas from Masses of Elements - II

Constructing the preliminary formula:

Na0.2469 Cr0.1235 O0.4939

Converting to integer subscripts (dividing all by smallest subscript):

Na1.99 Cr1.00 O4.02

Rounding off to whole numbers:

Na2CrO4 Sodium Chromate

Determining the Molecular Formula from Elemental Composition and Molar Mass - I

Problem: The sugar burned for energy in cells of the body is Glucose(M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the Molecular formula.Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements.Solution: Mass Carbon = 40.00% x 100g/100% = 40.00 g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = 53.27 g O

99.989 g Cpd

Determining the Molecular Formula from Elemental Composition and Molar Mass - II

Converting from Grams of Elements to moles:

Moles of C = Mass of C x = 3.3306 moles C

Moles of H = Mass of H x = 6.6657 moles H

Moles of O = Mass of O x = 3.3294 moles O

Constructing the preliminary formula C 3.33 H 6.67 O 3.33

Converting to integer subscripts, divide all subscripts by the smallest:

C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O

1 mole C 12.01 g C

1 mol H1.008 g H 1 mol O16.00 g O

Determining the Molecular Formula fromElemental Composition and Molar Mass - III(b) Determining the Molecular Formula:

The formula weight of the empirical formula is: 1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03

Whole-number multiple = =

= = 6.00 = 6

M of Glucoseempirical formula mass

180.16 30.03

Therefore the Molecular Formula is:

C 1 x 6 H 2 x 6 O 1 x 6 = C6H12O6

Adrenaline is a very Important Compound in the Body - I

• Analysis gives :

• C = 56.8 %

• H = 6.50 %

• O = 28.4 %

• N = 8.28 %

• Calculate the Empirical Formula !

Adrenaline - II

• Assume 100g!• C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C• H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H• O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O• N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N• Divide by 0.591 = • C = 8.00 mol C = 8.0 mol C or• H = 10.9 mol H = 11.0 mol H

• O = 3.01 mol O = 3.0 mol O C8H11O3N

• N = 1.00 mol N = 1.0 mol N

Figure 3.5: Combustion device

Ascorbic acid ( Vitamin C ) - I contains C , H , and O

• Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O

• Calculate it’s Empirical formula!

• C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)

= 2.65 x 10-3 g C

• H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)

= 2.92 x 10-4 g H

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

= 3.54 mg O

Vitamin C combustion - II

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

= 2.21 x 10-4 mol C

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

= 2.92 x 10-4 mol H

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

= 2.21 x 10-4 mol O

• Divide each by 2.21 x 10-4

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

• H = 1.32 = 3.96 = 4.0

• O = 1.00 = 3.00 = 3.0

C3H4O3

Computer generated molecule:Caffeine, C8H10N4O2

Determining a Chemical Formula from Combustion Analysis - I

Problem: Erthrose (M = 120 g/mol) is an important chemical compound as a starting material in chemical synthesis, and contains Carbon Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and 0.4194 g H2O.Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula.

Determining a Chemical Formula from Combustion Analysis - II

Calculating the mass fractions of the elements:

Mass fraction of C in CO2 = =

= = 0.2729 g C / 1 g CO2

Mass fraction 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 fraction 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

Determining a Chemical Formula from Combustion Analysis - III

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 H

Calculating 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 / cpd = C4H8O4

0.2729 g C 1 g CO2

0.1119 g H 1 g H2O

Like Example 3.7 (P 65) - ISucrose is the common sugar used in all homes, and chemical analysis tells us that the chemical composition is 42.14% carbon,6.48% hydrogen and 51.46% oxygen. What is the molecular formula of sucrose if its molecular mass is approximately 340 g/mol?

First determine the mass of each element in 1 mole (342.3g) of thecompound, sucrose. 42.14g C 342.3g 144.24g C 100.0g sucrose mol mol

6.48g H 342.3g 22.18g H 100.0g sucrose mol mol

51.46g O 342.3g 176.15g O 100.0g sucrose mol mol

=X

=

=

X

X

Like Example 3.7 (P 65) - II

Now we convert to moles:

C: X = 144.24g C 1 mol C 12.01g Cmol sucrose 12.011g C mol sucrose2 22.18g H 1 mol Hg 22.00g Hmol sucrose 1.008g H mol sucrose

176.15g O 1 mol O 11.01g O Mol sucrose 16.00g O mol sucrose

H: X =

O: X =

Divide by the smallest number:

C: 1.09 Therefore Empirical Formula = CH2O since the molecular mass = 340g/mol,H: 2.00 we must divide the formula mass into the molecular mass or 340/30 = 11.3 = 11O: 1.00 therefore the molecular formula is C11H22O11 !!!

Chemical Equations

Reactants Products

Qualitative Information:

Phases (States of Matter): (s) solid (l) liquid (g) gaseous (aq) aqueous

2 H2 (g) + O2 (g) 2 H2O (g)

Balanced Equations

1 CH4 (g) + O2 (g) 1 CO2 (g) + H2O (g)

• mass balance (atom balance)- same number of each element

(1) start with simplest element(2) progress to other elements(3) make all whole numbers(4) re-check atom balance

•charge balance (no “spectator” ions)

1 CH4 (g) + 2 O2 (g) 1 CO2 (g) + 2 H2O (g)

1 CH4 (g) + O2 (g) 1 CO2 (g) + 2 H2O (g)

Ca2+ (aq) + 2 OH- (aq) Ca(OH)2 (s)+ Na+ + Na+

Information Contained in a Balanced Equation

Viewed in Reactants Productsterms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy

Molecules 2 molecules of C2H6 + 7 molecules of O2 = 4 molecules of CO2 + 6 molecules of H2O

Amount (mol) 2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O

Mass (amu) 60.14 amu C2H6 + 224.00 amu O2 = 176.04 amu CO2 + 108.10 amu H2O

Mass (g) 60.14 g C2H6 + 224.00 g O2 = 176.04 g CO2 + 108.10 g H2O

Total Mass (g) 284.14g = 284.14g

Flare in a natural gas field

Source: Stock Boston

Figure 3.8: Methane/oxygen reaction

Table 3.2 (P 66) Information Conveyed by the Balanced Equation for the Combustion of Methane

Reactants Products

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

1 molecule CH4 + 1 molecule CO2 + 2 molecules of O2 2 molecules H2O

1 mol CH4 molecules + 1 mol CO2 molecules + 2 mol O2 molecules 2 mol H2O molecules

6.022 x 1023 CH4 molecules + 6.022 x 1023 CO2 molecules + 2 x (6.022 x 1023) O2 molecules 2 x (6.022 x 1023) H2O molecules

16g CH4 + 2 (32g) O2 44g CO2 + 2 (18g) H2O

80g reactants 80g products

Molecular model: Balanced equation

C2H6O(aq) + 3 O2(g) 2 CO2 (g) + 3 H2O(g) + Energy

Decomposition of ammonium dichromate

Balancing Chemical Equations - I

Problem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C6H14).Plan: Write the skeleton equation from the words into chemical compounds with blanks before each compound. begin the balance with the most complex compound first, and save oxygen until last!Solution:

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

Begin with one Hexane molecule which says that we will get 6 CO2’s!

1 6

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

The H atoms in the hexane will end up as H2O, and we have 14 H atoms, and since each water molecule has two H atoms, we will geta total of 7 water molecules.

Balancing Chemical Equations - II

1 6 7

Since oxygen atoms only come as diatomic molecules (two O atoms, O2),we must have even numbers of oxygen atoms on the product side. We do not since we have 7 water molecules! Therefore multiply the hexane by 2, giving a total of 12 CO2 molecules, and 14 H2O molecules.

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy2 12 14

This now gives 12 O2 from the carbon dioxide, and 14 O atoms from thewater, which will be another 7 O2 molecules for a total of 19 O2 !

C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy2 12 1419

Chemical Equation Calc - I

Reactants ProductsMolecules

Atoms (Molecules)

Avogadro’sNumber

6.02 x 1023

Chemical Equation Calc - II

Reactants ProductsMolecules

Moles

MassMolecularWeight g/mol

Atoms (Molecules)

Avogadro’sNumber

6.02 x 1023

Sulfuric Acid Plant

Source: Southern States Chemical

S(s) + O2 (g) SO2 (g)

SO2 (g) + O2 (g) SO3 (g)

SO3 (g) + H2O(l) H2SO4 (aq)

The process for finding the mass of carbon dioxide produced from 96.1 grams of propane

Sample Problem: Calculating Reactants and Products in a Chemical Reaction - I

Problem: Given the following chemical reaction between AluminumSulfide and water, if we are given 65.80 g of Al2S3: a) How many molesof water are required for the reaction? b) What mass of H2S & Al(OH)3

would be formed? Al2S3 (s) + 6 H2O(l) 2 Al(OH)3 (s) + 3 H2S(g)

Plan: Calculate moles of Aluminum Sulfide using it’s molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it’s molecular weight.Solution: a) molar mass of Aluminum Sulfide = 150.17 g / mol

moles Al2S3 = = ____________ Al2S3

65.80 g Al2S3

150.17 g Al2S3/ mol Al2S3

Calculating Reactants and Products in a Chemical Reaction - II

a) cont. 0.4382 moles Al2S3 x = 2.629 moles H2O

b) 0.4382 moles Al2S3 x = 1.314 moles H2S

molar mass of H2S = 34.09 g / mol

mass H2S = 1.314 moles H2S x = 44.81 g H2S

0.4382 moles Al2S3 x = 0.4764 moles Al(OH)3

molar mass of Al(OH)3 = 78.00 g / mol mass Al(OH)3 = 0.4764 moles Al(OH)3 x = = ________________ g Al(OH)3

6 moles H2O1 mole Al2S3

3 moles H2S1 mole Al2S3

34.09 g H2S1 mole H2S

2 moles Al(OH)3

1 mole Al2S3

78.00 g Al(OH)3

1 mole Al(OH)3

Calculating the Amounts of Reactants and Products in a Reaction Sequence - I

Problem: Calcium Phosphate could be prepared in the following reaction sequence:

4 P4 (s) + 10 KClO3 (s) 4 P4O10 (s) + 10 KCl (s)

P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq)

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

Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What massof Calcium Phosphate could be formed?

Plan: (1) Calculate moles of P4. (2) Use molar ratios to get moles of Ca3(PO4)2. (3) Convert the moles of product back into mass by using the molar mass of Calcium Phosphate.

Calculating the Amounts of Reactants and Products in a Reaction Sequence - II

Solution: moles of Phosphorous = 15.50 g P4 x = 0.1251 mol P4

For Reaction #1 [ 4 P4 (s) + 10 KClO4 (s) 4 P4O10 (s) + 10 KCl (s) ]

For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq) ]

For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2 1 Ca3(PO4)2 + 6 H2O]

0.1251 moles P4 x x x

= _______ moles Ca3(PO4)2

1 mole P4

123.88 g P4

4 moles H3PO4

1 mole P4O10

4 moles P4O10

4 moles P4

1 mole Ca3(PO4)2

2 moles H3PO4

Molar mass of Ca3(PO4)2 = 310.18 g mole

mass of product = 0.2502 moles Ca3(PO4)2 x =

= g Ca3(PO4)2

Calculating the Amounts of Reactants and Products in a Reaction Sequence - III

310.18 g Ca3(PO4)2

1 mole Ca3(PO4)2

Hydrochloric acid reacts with solid sodium hydrogen carbonate

Two antacid tablets

Molecular model: N2 molecules require 3H2 molecules for the reaction

N2 (g) + 3 H2 (g) 2 NH3 (g)

Figure 3.9: Hydrogen and Nitrogen reacting to form Ammonia, the Haber process

Limiting Reactant Problems

a A + b B + c C d D + e E + f F

Steps to solve1) Identify it as a limiting Reactant problem - Information on the: mass, number of moles, number of molecules, volume and molarity of a solution is given for more than one reactant!2) Calculate moles of each reactant!3) Divide the moles of each reactant by the coefficient (a,b,c etc....)!4) Which ever is smallest, that reactant is the limiting reactant!5) Use the limiting reactant to calculate the moles of product desired then convert to the units needed (moles, mass, volume, number of atoms etc....)!

Limiting Reactant Problem: A Sample Problem

Problem: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4). They ignite on contact ( hypergolic!) to form nitrogen gas and water vapor. How many grams of nitrogen gas form when exactly 1.00 x 102 g N2H4

and 2.00 x 102 g N2O4 are mixed?Plan: First write the balanced equation. Since amounts of both reactants are given, it is a limiting reactant problem. Calculate the moles of each reactant, and then divide by the equation coefficient to find which is limiting and use that one to calculate the moles of nitrogen gas, then calculate mass using the molecular weight of nitrogen gas.Solution:

2 N2H4 (l) + N2O4 (l) 3 N2 (g) + 4 H2O (g) + Energy

Sample Problem cont.

Moles N2H4 = = 3.12 moles N2H4

Moles N2O4 = = 2.17 moles N2O4

dividing by coefficients 3.12 mol / 2 = 1.56 mol N2H4

2.17 mol / 1 = 2.17 mol N2O4

Nitrogen yielded = 3.12 mol N2H4 = = 4.68 moles N2

Mass of Nitrogen = 4.68 moles N2 x 28.02 g N2 / mol = g N2

molar mass N2H4 = ( 2 x 14.01 + 4 x 1.008 ) = 32.05 g/molmolar mass N2O4 = ( 2 x 14.01 + 4 x 16.00 ) = 92.02 g/mol

1.00 x 102 g 32.05 g/mol

2.00 x 102 g92.02 g/mol

Limiting !

3 mol N2

2 mol N2H4

• 2Al(s) + 6HCl(g) 2AlCl3(s) + 3H2(g)

• Given 30.0g Al and 20.0g HCl, how many moles of Aluminum Chloride will be formed?

• 30.0g Al / 26.98g Al/mol Al = 1.11 mol Al

• 1.11 mol Al / 2 = 0.555

• 20.0g HCl / 36.5gHCl/mol HCl = 0.548 mol HCl

• O.548 mol HCl / 6 = 0.0913

• HCl is smaller therefore the Limiting reactant!

Acid - Metal Limiting Reactant - I

• since 6 moles of HCl yield 2 moles of AlCl3

• 0.548 moles of HCl will yield:

• 0.548 mol HCl / 6 mol HCl x 2 moles of

• AlCl3 = ______________ mol of AlCl3

Acid - Metal Limiting Reactant - II

• What mass of NO could be formed by the reaction 30.0g of Ammonia gas and 40.0g of Oxygen gas?

• 4NH3 (g) + 5 O2 (g) 4NO(g) + 6 H2O(g)

• 30.0g NH3 / 17.0g NH3/mol NH3 = 1.76 mol NH3

1.76 mol NH3 / 4 = 0.44 mol NH3

• 40.0g O2 / 32.0g O2 /mol O2 = 1.25 mol O2

1.25 mol O2 / 5 = 0.25 mol O2

• Therefore Oxygen is the Limiting Reagent!

• 1.25 mol O2 x = 1.00 mol NO

• mass NO = 1.00 mol NO x = g NO4 mol NO5 mol O2

30.0 g NO1 mol NO

Ostwald Process Limiting Reactant Problem

Chemical Reactions in Practice: Theoretical, Actual, and Percent Yields

Theoretical yield: The amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation. Side Reactions: These form smaller amounts of different products that take away from the theoretical yield of the main product.

Actual yield: The actual amount of product that is obtained.

Percent yield: (%Yield)

% Yield = x 100 Actual YieldTheoretical Yield

Percent Yield Problem:

Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe3O4 and Hydrogen gas given below. If 4.55g of Iron is reacted with sufficient water to react all of the Iron to form rust, what is the percent yield if only 6.02g of the oxide are formed?Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield.Solution: 3 Fe(s) + 4 H2O(l) Fe3O4 (s) + 4 H2 (g)

4.55 g Fe55.85 g Fe mol Fe

= 0.081468 mol = 0.0815 mol

0.0815 mol Fe x = 0.0272 mol Fe3O41 mol Fe3O4

3 mol Fe

0.0272 mol Fe3O4 x = 6.30 g Fe3O4231.55 g Fe3O4

1 mol Fe3O4

Percent Yield = x 100% = x 100% = ____ %

Actual YieldTheoretical Yield

6.02 g Fe3O4

6.30 g Fe3O4

Percent Yield / Limiting Reactant Problem - IProblem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction.

N2 (g) + 3 H2 (g) 2 NH3 (g)

Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield.Solution: Moles of Nitrogen and Hydrogen:

moles N2 = = 3.066 mol N285.90 g N2

28.02 g N2

1 mole N2

moles H2 = = 10.74 mol H221.66 g H2

2.016 g H2

1 mole H2

Divide by coefficientto get limiting: 3.066 g N2

1

10.74 g H2

3

= 3.066

= 3.582

Percent Yield/Limiting Reactant Problem - II

Solution Cont. N2 (g) + 3 H2 (g) 2 NH3 (g)

We have 3.066 moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is:

3.066 mol N2 x = 6.132 mol NH3

(Theoretical Yield)

6.132 mol NH3 x = 104.427 g NH3 (Theoretical Yield)

2 mol NH3

1 mol N2

17.03 g NH3

1 mol NH3

Percent Yield = x 100%

Percent Yield = x 100% = %

Actual YieldTheoretical Yield

98.67 g NH3

104.427 g NH3

Flowchart : Solving a stoichiometry problem involving masses of reactants

Mass Percent Composition of Na2SO4

Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen

Elemental masses

2 x Na = 2 x 22.99 = 45.981 x S = 1 x 32.07 = 32.07

4 x O = 4 x 16.00 = 64.00

142.05

Percent of each Element

% Na = Mass Na / Total mass x 100%% Na = (45.98 / 142.05) x 100% =32.37%

% S = Mass S / Total mass x 100%% S = (32.07 / 142.05) x 100% = 22.58%

% O = Mass O / Total mass x 100%% O = (64.00 / 142.05) x 100% = 45.05%

Check

% Na + % S + % O = 100%

32.37% + 22.58% + 45.05% = 100.00%

Calculating the Mass of an Element in a Compound Ammonium Nitrate

Ammonium Nitrate = NH4NO3

How much Nitrogen is in 455 kg of Ammonium Nitrate?

The Formula Mass of Cpd is:

4 x H = 4 x 1.008 = 4.032 g2 x N = 2 X 14.01 = 28.02 g3 x O = 3 x 16.00 = 48.00 g

80.052 g

Therefore gm Nitrogen/ gm Cpd

28.02 g Nitrogen

80.052 g Cpd = 0.35002249 g N / g Cpd

455 kg x 1000g / kg = 455,000 g NH4NO3

455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 105 g Nitrogen

28.02 kg Nitrogen

80.052 kg NH4NO4

= 159 kg Nitrogen 455 kg NH4NO3 Xor: