Stoichiometry of Formulas and Equations

CHEM 101 – CHAPTER 3

STOICHIOMETRY OF FORMULAS AND EQUATIONS

Stoichiometry – the study of the quantitative aspects of formulas and reactions

Defining the Mole

Mole (mol) – the SI unit for the amount of substance

Defined as the amount of a substance that contains the same number of entities as the number of atoms in 12 g of carbon-12
This number is called Avogadro’s number
One mole (1 mol) contains 6.022 x 1023 entities
1 mol of carbon-12 contains 6.022 x 1023 atoms and has a mass of 12 g

Examples (elements)…(The mass in atomic mass units (amu) of one atom of an element is the same numerically as the mass in grams (g) of 1 mole of atoms of the element.)

Sulfur
1 atom has a mass of 32.07 amu
1 mol [6.022 x 1023 atoms] has a mass of 32.07 g
Iron
1 atom has a mass of 55.85 amu
1 mol [6.022 x 1023 atom] has a mass of 55.85 g

Examples (compounds)…(The mass in atomic mass units (amu) of one molecule of a compound is the same numerically as the mass in grams (g) of 1 mole of molecules of the compound.

Water (H2O)
1 molecule of water has a mass of 18.02 amu
1 mol [6.022 x 1023 molecules] has a mass of 18.02 g

Sodium Chloride (NaCl)
1 molecule of NaCl has a mass of 58.44 amu
1 mol [6.022 x 1023 molecules] has a mass of 58.44 g

Two key points about the mole unit…

(1)The mole lets us relate the number of entities to the mass of a sample of those entities.

(2)The mole maintains the same numerical relationship between mass on the atomic scale (atomic mass units, amu) and mass on the macroscopic scale (grams, g)

Determining Molar Mass

Molar mass – the mass per mole of the entities (atoms, molecules, or formula units) of a substance and has the units of grams per mole (g/mol)

For elements… look up the atomic mass on the periodic table
For monatomic elements the molar mass is the periodic table value in grams per mole
Molar mass of neon is 20.18 g/mol
Molar mass of gold is 197.0 g/mol
For molecular elements you would need to know the formula to determine its molar mass
Molar mass for O2 (2 x 16.00 g/mol = 32.00 g/mol)
Molar mass for S8 (8 x 32.07 g/mol) = 256.6 g/mol)
For compounds… the molar mass is the sum of the molar masses of the atoms in the formula
Example… SO2
Molar mass SO2 = (1 x molar mass of S) + (2 x molar mass of O)
Molar mass SO2 = (1 x 32.07 g/mol) + (2 x 16.00 g/mol) = 64.07 g/mol
Example… K2S
Molar mass K2S = (2 x molar mass of K) + (1 x molar mass of S)
Molar mass K2S = (2 x 39.10 g/mol) + (1 x 32.07 g/mol) = 110.27 g/mol

Converting Between Amount, Mass, and Number of Chemical Entities

Converting from amount (mol) and mass (g)

1.25 mol of carbon / 12.01 g carbon / = 15.0 g C
1 mole of carbon

Converting from mass (g) to amount (mol)

34.6 g carbon / 1 mol carbon / = 2.88 mol C
12.01 g carbon

Converting from amount (mol) to number of entities

1.25 mol of carbon / 6.022 x 1023 entities / = 7.53 x 1023 entities
1 mol carbon

Converting from number of entities to amount (mol)

8.13 x 1025 entities / 1 mol carbon / = 1.35 x102 mol carbon
6.022 x 1023 entities

Sample Problem 3.1 p. 86

Calculating the Mass of a Given Amount of an Element

0.0342 mol Ag / 107.9 g Ag / = 3.69 g Ag
1 mol Ag

Follow-Up Problem 3.1

315 mg C / 1 g C / 1 mol C / = 0.0262 mol C
1000 mg C / 12.01 g C

= 2.62 x 10-2 mol C

Sample Problem 3.2 p. 86

Calculating the Number of Entities in a Given Amount of an Element

2.85 x 10-3 mol Ga / 6.022 x 1023Ga atoms / = 1.72 x 1021 Ga atoms
1 mol Ga

Follow-Up Problem 3.2 p. 87

9.72 x 1021 nitrogen molecules / 1 mol nitrogen molecules / 2 mol nitrogen atoms / = 3.23 x 10-2 mol of nitrogen atoms
6.022 x 1023 nitrogen molecules / 1 mol nitrogen molecules

Sample Problem 3.3 p. 87

Calculating the Number of Entities in a Given Mass of an Element

95.8 g Fe / 1 mol Fe / 6.022 x 1023 Fe atoms / = 1.03 x 1024 Fe atoms
55.85 g Fe / 1 mol Fe

Follow-Up Problem 3.3 p. 87

3.22 x 1020 Mn atoms / 1 mol Mn atoms / 54.94 g Mn / = 2.94 x 10-2 g Mn
6.022 x 1023 Mn atoms / 1 mol Mn atoms

Sample Problem 3.4 p. 88

Calculating the Number of Chemical Entities in a Given Mass of a Compound

8.92 g NO2 / 1 mol NO2 / 6.022 x 1023 NO2 molecules / = 1.17 x 1023 NO2 molecules
46.01 g NO2 / 1 mol NO2

Follow-Up Problem 3.4 p. 88

1.19 x 1019 formula units NaF / 1 mol NaF / 41.99 g NaF / = 8.30 x 10-4 g NaF
6.022 x 1023 formula units NaF / 1 mol NaF

Sample Problem 3.5 p. 89

Calculating the Number of Chemical Entities in a Given Mass of a Compound (Part 2)

(a)

41.6 g (NH4)CO3 / 1 mol (NH4)CO3 / 6.022 x 1023 formula units (NH4)CO3 / = 2.61 x 1023 formula units (NH4)CO3
96.09 g (NH4)CO3 / 1 mol (NH4)CO3

(b)

2.61 x 1023 formula units (NH4)CO3 / 3 O atoms / = 7.83 x 1023 O atoms
1 formula unit (NH4)CO3

Determining Mass Percent from a Chemical Formula

For a molecule…

Mass % of element X =

[(atoms of X in formula) x (atomic massX (g/mol)] x 100%

molecular or formula mass of compound (amu)

For a mole of compound

Mass % of element X =

[(moles of X in formula) x (molar mass of X (g/mol)] x 100%

mass (g) of 1 mol of compound

Sample Problem 3.6 p. 90

Calculating the Mass Percent of Each Element in a Compound from the Formula

Mass percent carbon (C)

(a)Calculate mass of carbon in 1 mol C6H12O6

6 mol carbon / 12.01 g carbon / = 72.06 g carbon
1 mol carbon

(b)Calculate mass percent of carbon in C6H12O6

72.06 g carbon / = .4000 x 100% / = 40.00% carbon
180.16 g C6H12O6

Mass percent of hydrogen (H)

(a)Calculate the mass of hydrogen in 1 mol of C6H12O6

12 mol of hydrogen / 1.008 g hydrogen / = 12.096 g hydrogen
1 mol hydrogen

(b)Calculate the mass percent hydrogen in C6H12O6

12.096 g hydrogen / = 0.06714 x 100% / = 6.714% hydrogen
180.16 g C6H12O6

Mass percent of oxygen (O)

(a)Calculate the mass of oxygen in 1 mol of C6H12O6

6 mol of oxygen / 16.00 g oxygen / = 96.00 g oxygen
1 mol oxygen

(b)Calculate the mass percent of oxygen in C6H12O6

96.00 g oxygen / = 0.5329 x 100% / = 53.29% oxygen
180.16 g C6H12O6

Follow-Up Problem 3.6 p. 91

Ammonium nitrate – NH4NO3

Formula mass = 80.05 g/mol

(a)Calculate the mass of oxygen in one mole of NH4NO3

2 mol of nitrogen / 14.01 g nitrogen / = 28.02 g nitrogen
1 mol nitrogen

(b)Calculate the mass percent of nitrogen in NH4NO3

28.02 g nitrogen / = 0.3500 x 100% / = 35.00% nitrogen
80.05 g NH4NO3

Determining the Mass of an Element from its Mass Percent

What if we wanted to know how much oxygen was in 15.5 g of NO2?

15.5 g NO2 / 32.00 g oxygen / = 10.8 g oxygen
46.01 g NO2

Sample Problem 3.7 p. 91

Calculating the Mass of an Element in a Compound

16.55 g C6H12O6 / 72.06 g carbon / = 6.620 g carbon
180.16 g C6H12O6

Follow-Up Problem 3.7 p. 91

35.8 kg NH4NO3 / 1000 g NH4NO3 / 28.02 g nitrogen / = 1.25 x 104 g nitrogen
1 kg NH4NO3 / 80.05 g NH4NO3

Determining the Formula of an Unknown Compound

Empirical Formula – shows the lowest whole number of moles, and thus the relative number of atoms, of each element in the compound.

Molecular Formula – shows the actual number of atoms of each element in a molecule.

Structural Formula – shows the relative placement and connections of atoms in the molecule

Sample Problem 3.8 p. 92

Determining an Empirical Formula from Amounts of Elements

(1)Zn0.21P0.14O0.56

(2)Zn0.21P0.14O0.56

0.14 0.14 0.14

(3) Zn1.5P1.0O4.0

(4) Zn3P2O8

Follow-Up Problem 3.8 p. 93

(1)H0.255B0.170

(2)H0.255B0.170

0.170 0.170

(3) H1.5B1.0

(4) H3B2

Sample Problem 3.9 p. 93

Determining an Empirical Formula from Masses of Elements

Calculate moles of Na

2.82 g Na / 1 mol Na / = 0.123 mol Na
22.99 g Na

Calculate moles of Cl

4.35 g Cl / 1 mol Cl / = 0.123 mol Cl
35.45 g Cl

Calculate moles of O

7.83 g O / 1 mol O / = 0.489 mol O
16.00 g O

(1)Na0.123Cl0.123O0.489

(2)Na0.123Cl0.123O0.489

0.123 0.123 0.123

(3)Na1.0Cl1.0O4.0

(4)NaClO4

Follow-Up Problem 3.9 p. 93

2.88 g S / 1 mole S / = 0.0898 mol S
32.07 g S
0.0898 mol S / 2 mol M / = 0.0599 mol M
3 mol S
3.12 g M / = 52.09 g/mol (Chromium)
0.0599 mol M

Cr2S3 Chromium (III) Sulfide

Molecular Formulas

Sample Problem 3.10 p. 94

Determining a Molecular Formula from Elemental Analysis and Molar Mass

Step 1 – Assume 100.0 g of compound to express each mass percent directly as mass (g)

We have 40.0 g C, 6.71 g H, and 53.3 g O

Step 2 – Convert each mass (g) to amount (mol)

40.0 g C / 1 mol C / = 3.33 mol C
12.01 g C
6.71 g H / 1 mol H / = 6.66 mol H
1.008 g H
53.3 g O / 1 mol O / = 3.33 mol O
16.00 g O

Step 3 – Derive the empirical formula

(a)C3.33H6.66O3.33

(b)C3.33H6.66O3.33

3.33 3.33 3.33

Step 4 – Determine the compound’s empirical formula mass

Formula mass = (1 x 12.01) + (2 x 1.008) + (1 x 16.00)

Formula mass = 30.026 g/mol

Step 5 – Divide the compound’s molar mass by the empirical formula mass to find the whole number multiple

90.08 g/mol / 30.026 g/mol= 3.00

Step 6 – Multiply each subscript in the empirical formula by the multiple found in step 5

C3H6O3

Follow-Up Problem 3.10 p. 95

Step 1 – Assume 100.0 g of compound to express each mass percent directly as mass (g)

95.21 g C, 4.79 g

Step 2 – Convert each mass (g) to amount (mol)

95.21 g carbon / 1 mol carbon / = 7.93 mol carbon
12.01 g carbon
4.79 g hydrogen / 1 mol hydrogen / = 4.75 mol hydrogen
1.008 g hydrogen

Step 3 – Derive the empirical formula

(a)C7.93H4.75

(b)C7.93H4.75

4.75 4.75

(c)C1.67H1.00

(d)C5H3

Step 4 – Determine the compound’s empirical formula mass

Formula mass = (5 x 12.01) + (3 x 1.008)

Formula mass = 63.074 g/mol

Step 5 – Divide the compound’s molar mass by the empirical formula mass to find the whole number multiple

252.30 g/mol / 63.074 = 4.00

Step 6 – Multiply each subscript in the empirical formula by the multiple found in step 5

C20H12

Sample Problem 3.11 p. 95

Determining a Molecular Formula from Combustion Data

Step 1 – Determine mass of CO2 collected

Mass of CO2 absorber after combustion / – Mass of CO2 absorber before combustion
85.35 g / - 83.85 g / = 1.50 g CO2 collected

Step 2 – Determine the mass of H2O collected

Mass of H2O absorber after combustion / – Mass of H2O absorber before combustion
37.96 g / - 37.55 g / = 0.41 g H2O collected

Step 3 – Determine mass of carbon collected

1.50 g CO2 / 12.01 g C / = 0.409 g C
44.01 g CO2

Step 4 – Determine the mass of hydrogen collected

0.41 g H2O / 2.016 g H / = 0.046 g H
18.02 g H2O

Step 5 – Determine the mass of oxygen collected

Mass of oxygen collected / = 1.000 g vitamin C / - mass of carbon collected / - mass of hydrogen collected

Mass of oxygen collected = 1.000 g – 0.409 g – 0.046 g

Mass of oxygen collected = 0.545 g O

Step 6 – Calculate the amount (mol) of each element

0.409 g C / 1 mol C / = .0341 mol C
12.01 g C
0.046 g H / 1 mol H / = .0456 mol H
1.008 g H
0.545 g O / 1 mol O / = .0341 mol O
16.00 g O

Step 7 – Determine the empirical formula of vitamin-C

(a)C.0341H.0456O.0341

(b)C.0341H.0456O.0341

.0341 .0341 .0341

(c)C1.0H1.33O1.0

(d)C3H4O3

Step 8 – Determine empirical formula mass

Formula mass = (3 x 12.01) + (4 x 1.008) + (3 x 16.00)

Formula mass = 88.06 g/mol

Step 9 - Divide the compound’s molar mass by the empirical formula mass to find the whole number multiple

176.12 g/mol / 88.06 g/mol = 2.00

Step 10 - Multiply each subscript in the empirical formula by the multiple found in step 9

C6H8O6

Follow-Up Problem 3.11 p. 96

Step 1 – Determine mass of carbon collected

0.451 g CO2 / 12.01 g C / = 0.123 g C
44.01 g CO2

Step 2 – Determine mass of hydrogen collected

0.0617 g H2O / 2.016 g H / = 0.00690 g H
18.02 g H2O

Step 3 – Determine mass of chlorine collected

0.250 g sample / - 0.123 g C / - 0.00690 g H / = 0.120 g Cl

Step 4 – Calculate mol of carbon

0.123 g C / 1 mol C / = 0.0102 mol C
12.01 g C

Step 5 – Calculate mol of hydrogen

0.00690 g H / 1 mol H / = 0.00685 mol H
1.008 g H

Step 6 – Calculate mol of chlorine

0.120 g Cl / 1 mol Cl / = 0.00339 mol Cl
35.45 g Cl

Step 7 – Determine empirical formula

(a)C.0102H.00685Cl.00339

.00339 .00339 .00339

(b)C3H2Cl

Step 8 – Calculate the empirical formula mass

Formula mass = (3 x 12.01) + (2 x 1.008) + (1 x 35.45)

Formula mass = 73.50 g/mol

Step 9 – Determine the Multiplier

146.99 g/mol / 73.50 g/mol = 2.00

Step 10 – Calculate the molecular formula

C6H4Cl2

Chemical Formulas and Molecular Structure; Isomers

Different compounds can have the same empirical formula as we can observe in table 3.3 (the empirical formula tells us nothing about molecular structure because it is based solely on mass analysis)
Isomers – compounds with the same molecular formula and molar mass, but different properties. They have different structural formulas
Constitutional (structural) isomers –isomers where the atoms link together in different arrangements

Writing & Balancing Chemical Equations

Macroscopic level…2.016 g of H2 and 38.00 g F2 react to form 40.02 g HF
Macroscopic level… 1 mol of H2 and 1 mol of F2 react to form 2 mol of HF
Molecular level… 1 molecule of H2 and 1 molecule of F2 react to form 2 molecules of HF

Steps for Balancing Equations

Translating the statement
Reactants – substance(s) on the left side of the arrow
Products – substance(s) on the right side of the arrow
Arrow – means “yields” or “produces”
Balancing the atoms
Start with any metals that you have in the equation and balance them first (if there is more than one, balance them one at a time)
Inventory the number of atoms on reactant and product side of the equation
Place a balancing (stoichiometric) coefficient in front of the metal in the equation (or the formula containing the metal) to balance the number of metal atoms on both sides of the equation.
Proceed to any other metals (if there are other metals) in the equation and balance them in the same manner.
After the metals are balanced, balance any nonmetals other than hydrogen or oxygen by placing balancing (stoichiometric) coefficients in front of their formulas.
Once you have balanced all other elements, balance hydrogen atoms and finally oxygen atoms.
Re-inventory the number of atoms of each element on both the reactant and product side of the equation and check to make sure that the coefficients are the smallest possible whole numbers

Sample Problem 3.12 p. 101

Balancing Chemical Equations

Step 1 – Translate the statement

C8H18 + O2 CO2 + H2O

Step 2 – Balance the atoms

Balance the carbons (there is 8 on the left and only 1 on the right)… place a balancing coefficient of 8 in front of the compound with carbon in it on the left (CO2)

C8H18 + O2 8 CO2 + H2O

Balance the hydrogens (there are 18 on the left and only 2on the right)… place a balancing coefficient of 9in front of the compound that contains hydrogen on the right side of the equation(H2O)

C8H18 + O2 8 CO2 + 9 H2O

At this point in the process, the carbons and hydrogens are balanced (8 carbons and 18 hydrogens on both sides of the equation). When you atom inventory the oxygens, there are now 2 oxygens on the left and 25 on the right (remember the balancing coefficients multiply every element in the formula)
Balancing coefficients have to be whole numbers…so you cannot do the following…

C8H18 + 12 ½O2 8 CO2 + 9 H2O

To get rid of the ½ in the balancing coefficient, you need to multiply all of the coefficients in the equation by 2. This gives the following result…

2C8H18 + 25O2 16 CO2 + 18 H2O

Specify states of matter

2C8H18 (l) + 25O2 (g) 16 CO2 (g) + 18 H2 (g)

Follow-up Problem 3.12 p. 101

(a)

Na (s) + H2O (l)  H2 (g) + NaOH (aq)

Na (s) + H2O (l)  H2 (g) + 2NaOH (aq)

Na (s) + 2H2O (l)  H2 (g) + 2NaOH (aq)

2Na (s) + 2H2O (l)  H2 (g) + 2NaOH (aq)

(b)

CaCO3 (s) + HNO3 (aq)  CO2 (g) + H2O (l) + Ca(NO3)2 (aq)

CaCO3 (s) + 2HNO3 (aq)  CO2 (g) + H2O (l) + Ca(NO3)2 (aq)

(c)

PCl3 (g) + HF (g)  PF3 (g) + HCl (g)

PCl3 (g) + 3HF (g)  PF3 (g) + 3HCl (g)

(d)

C3H5N3O9 (l)  CO2 (g) + H2O (g) + N2 (g) + O2 (g)

2C3H5N3O9 (l)  CO2 (g) + H2O (g) + 3N2 (g) + O2 (g)

2C3H5N3O9 (l) 6CO2 (g) + H2O (g) + 3N2 (g) + O2 (g)

2C3H5N3O9 (l) 6CO2 (g) + 5H2O (g) + 3N2 (g) + O2 (g)

4C3H5N3O9 (l) 12CO2 (g) + 10H2O (g) + 6N2 (g) + O2 (g)

Sample Problem 3.13 p. 102

Balancing an Equation from a Molecular Scene

4N2O5 (g)  8NO2 (g) + 2O2 (g)

2N2O5 (g) 4NO2 (g) + O2 (g)

Follow-Up Problem 3.13 p. 102

6CO (g) + 3O2 (g)  6CO2 (g)

2CO (g) + O2 (g) 2CO2 (g)

Calculating Quantities or Reactants and Products

Stoichiometrically Equivalent Molar Ratios from the Balanced Equation

In a balanced equation, the amounts (mol) of substances are stoichiometrically equivalent to each other…

A specific amount of one substance is formed from, produces, or reacts with a specific amount of the other.
The quantitative relationships are expressed as stoichiometrically equivalent molar ratios
Use the molar ratios as conversion factors to calculate amounts

Example… the combustion of propane

C3H8 (g) + 5O2 (g)  3CO2 (g) + 4 H2O

From the balanced equation we know that…

1 mol C3H8 reacts with 5 mol of O2
1 mol C3H8 produces 3 mol of CO2
1mol C3H8 produces 4 mol of H2O

Therefore, in this reaction…

1 mol C3H8 is stoichiometrically equivalent to 5 mol O2
1 mol C3H8 is stoichiometrically equivalent to 3 mol CO2
1 mol C3H8 is stoichiometrically equivalent to 4 mol H2O

This also means that…

3 mol CO2 is stoichiometrically equivalent to 4 mol H2O
5 mol O2 is stoichiometrically equivalent to 3 mol of CO2
5 mol O2 is stoichiometrically equivalent to 4 mol CO2

If we want to convert between mol of O2 and mol of H2O, we could construct to conversion factors depending on which quantity we are solving for…

Mol H2O / 5 mol O2 / Mol of O2
4 mol H2O
Mol O2 / 4 mol H2O / Mol H2O
5 mol O2

If we wanted to know how many mol of O2 is consumed when 10.0 mol of H2O is produced…

10 mol H2O / 5 mol O2 / = 12.5 mol O2
4 mol H2O

Sample Problem 3.14 p. 104

Calculating Quantities of Reactants and Products: Amount (mol) to Amount (mol)

Step 1 – Make sure the equation is balanced

Copper (I) sulfide + oxygen Copper (I) oxide + Sulfur dioxide

Cu2S (s) + O2 (g) Cu2O (s) + SO2 (g)

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g)

Step 2 – Convert from mol Cu2S to mol O2

10.0 mol Cu2S / 3 mol O2 / = 15.0 mol O2
2 mol Cu2S

Follow-Up Problem 3.14 p. 105

Step 1 – Make sure the equation is balanced

Iron (III) oxide + Aluminum  Aluminum oxide + Iron

Fe2O3 + Al  Al2O3 + Fe

Fe2O3 + 2Al  Al2O3 + 2Fe

Step 2 – Convert from mol Fe to mol Fe2O3

3.60 x 103 mol Fe / 1 mol Fe2O3 / = 1.80 x 103 mol Fe
2 mol Fe

Sample Problem 3.15 p. 105

Calculating Quantities of Reactants and Products: Amount (mol) to Mass (g)

Step 1 – Make sure the equation is balanced

Same equation as sample problem 3.14

Copper (I) sulfide + oxygen Copper (I) oxide + Sulfur dioxide

Cu2S (s) + O2 (g)  Cu2O (s) + SO2 (g)

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g)

Step 2 – Convert from mol Cu2S to grams SO2

10.0 mol Cu2S / 2 mol SO2 / 64.07 g SO2 / = 641 g SO2
2 mol Cu2S / 1 mol SO2

Follow-Up Problem 3.15 p. 105

Step 1 – Make sure the equation is balanced

*Same equation as follow-up problem 3.14 on page 105

Iron (III) oxide + Aluminum  Aluminum oxide + Iron

Fe2O3 + Al  Al2O3 + Fe

Fe2O3 + 2Al  Al2O3 + 2Fe

Step 2 – Convert from formula units of Al2O3 to mol Fe

1.85 x 1025 formula units Al2O3 / 1 mol Al2O3 / 2 mol Fe / = 61.4 mol Fe
6.022 x 1023 formula units Al2O3 / 1 mol Al2O3

Sample Problem 3.16 p. 105

Calculating Quantities of Reactants and Products: Mass to Mass

Step 1 – Make sure the equation is balanced

Same equation as sample problem 3.14 and 3.15

Copper (I) sulfide + oxygen Copper (I) oxide + Sulfur dioxide

Cu2S (s) + O2 (g)  Cu2O (s) + SO2 (g)

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g)

Step 2 – Convert from kg Cu2O to kg O2

2.86 kg Cu2O / 1000 g Cu2O / 1 mol Cu2O / 3 mol O2 / 32.00 g O2 / 1 kg O2 =
1 kg Cu2O / 143.10 g Cu2O / 2 mol Cu2O / 1 mol O2 / 1000 g O2

= .960 kg O2

Follow-Up Problem 3.16 p. 106

Step 1 – Make sure the equation is balanced

*Same equation as follow-up problem 3.14 on page 105

Iron (III) oxide + Aluminum  Aluminum oxide + Iron

Fe2O3 + Al  Al2O3 + Fe

Fe2O3 + 2Al  Al2O3 + 2Fe

Step 2 – Convert from grams Al2O3 to atoms of aluminum

1 g Al2O3 / 1 mol Al2O3 / 2 mol Al / 6.022 x 1023 Al atoms / = 1.18 x 1022 Al atoms
101.96 g Al2O3 / 1 mol Al2O3 / 1 mol Al atoms

Reactions That Occur in a Sequence

In many situations, a product of one reaction becomes a reactant for the next in a sequence of reactions.

For stoichiometric purposes, when the same (common) substance forms in one reaction and reacts in the next, we eliminate it in an overall (net) equation.

Sample Problem 3.17 p. 106

Writing an Overall Equation for a Reaction Sequence

Step 1 – Write the sequence of balanced equations.

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g) [equation 1]

Cu2O (s) + C (s)  2 Cu (s) + CO (g) [equation 2]

Step 2 – Adjust the equations arithmetically to cancel the common substance.

2Cu2S (s) + 3O2 (g) 2Cu2O (s) + 2SO2 (g) [equation 1]

(2x) 2 Cu2O(s) + 2C (s)  4Cu (s) + 2CO (g) [equation 2]

We multiplied equation 2 through by 2 so that we would have equal coefficients in front of the common substance (Cu2O) in both equations

Step 3 – Add the adjusted equations together to obtain the overall balanced equation

2Cu2S (s) + 3O2 (g) + 2C (s)  2SO2 + 4Cu (s) + 2CO (g)

Follow-Up Problem 3.17 p. 107

Step 1 – Write the sequence of balanced equations

2SO2 (g) + O2 (g)  2SO3 (g) [equation 1]

SO3 (g) + H2O (l)  H2SO4 (aq) [equation 2]

Step 2 – Adjust the equations arithmetically to cancel the common substance.

2SO2 (g) + O2 (g) 2SO3 (g) [equation 1]

(2x)2SO3 (g) + 2H2O (l) 2H2SO4 (aq) [equation 2]

Step 3 – Add the adjusted equations together to obtain the overall balanced equation

2SO2 (g) + O2 (g) + 2H2O (l)  2H2SO4 (aq)

Example of Reaction Sequence in Biology…

C6H12O6 (aq) + 6O2 (aq)  6CO2 (g) + 6H2O (l)

Cellular respiration (net reaction)… this is the net equation of a 30 reaction process!

Reactions That Involve a Limiting Reactant

2 scoops ice cream + 1 cherry + 50 mL syrup  1 sundae

We can only make 2 sundaes because we run out of syrup
The syrup is our “limiting reactant”
We have excess ice cream and cherries

Sample Problem 3.18 p. 109

Using Molecular Depictions in a Limiting Reactant Problem

Step 1 – Make sure you have a balanced equation

Cl2 (g) + 3F2 (g)  2ClF3 (g)

Step 2 – Determine the limiting reactant

3 molecules Cl2 / 2 ClF3 molecules / = 6 molecules ClF3
1 molecule Cl2
6 molecules F2 / 2 ClF3 molecules / = 4 molecules of ClF3
3 molecules F2

F2 is the limiting reactantbecause it limited the number of ClF3 molecules that can be produced

Follow-Up Problem 3.18 p. 110

Step 1 – Make sure you have a balanced equation

B2 (g) + 2AB (g)  2AB2 (g)

Step 2 – Determine the limiting reactant

3 molecules B2 / 2 molecules AB2 / = 6 molecules AB2
1 molecule B2
4 molecules AB / 2 molecules AB2 / = 4 molecules of AB2
2 molecules AB

AB is the limiting reactantbecause it produces the least amount of AB2

Sample Problem 3.19 p. 110

Calculating Quantities in a Limiting Reactant Problem: Amount to Amount

Step 1 – Make sure you have a balanced equation

Cl2 (g) + 3F2 (g)  2ClF3 (g)

Step 2 – Determine the limiting reactant

0.750 mol Cl2 / 2 mol ClF3 / = 1.50 mol ClF3
1 mol Cl2
3.00 mol F2 / 2 mol ClF3 / = 2.00 mol ClF3
3 mol F2

Cl2 is the limiting reactantbecause it produces the least amount of ClF3

Follow-Up Problem 3.19 p. 110

Step 1 – Make sure you have a balanced equation

B2 (g) + 2AB (g)  2AB2 (g)

Step 2 – Determine the limiting reactant

1.50 mol B2 / 2 mol AB2 / = 3.00 mol AB2
1 mol B2
1.50 mol AB / 2 mol AB2 / = 1.50 mol AB2
2 mol AB

AB is the limiting reactantbecause it can only form 1.50 mol of AB2

Sample Problem 3.20 p. 111

Calculating Quantities in a Limiting Reactant Problem: Mass to Mass

Step 1 – Make sure you have a balanced equation

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

Step 2 – Determine the limiting reactant

100. g N2H4 / 1 mol N2H4 / 3 mol N2 / 28.02 g N2 / = 131 g N2
32.05 g N2H4 / 2 mol N2H4 / 1 mol N2
200. g N2O4 / 1 mol N2O4 / 3 mol N2 / 28.02 g N2 / = 183 g N2
92.02 g N2O4 / 1 mol N2O4 / 1 mol N2

N2H4 is the limiting reactantbecause it would produce less N2

Follow-Up Problem 3.20 p. 112

Step 1 – Make sure you have a balanced equation

2Al (s) + 3S (s)  Al2S3(s)

Step 2 – Determine the limiting reactant

10.0 g Al / 1 mol Al / 1 mol Al2S3 / 150.15 g Al2S3 / = 27.8 g Al2S3
26.98 g Al / 2 mol Al / 1 mol Al2S3
15.0 g S / 1 mol S / 1 mol Al2S3 / 150.15 g Al2S3 / = 23.4 g Al2S3
32.07 g S / 3 mol S / 1 mol Al2S3

Sulfur (S) is the limiting reactantbecause it produces the least mass of product

Step 3 – Calculate the mass of excess reactant

23.4 g Al2S3 / 1 mol Al2S3 / 2 mol Al / 26.98 g Al / = 8.40 g Al (used)
150.15 g Al2S3 / 1 mol Al2S3 / 1 mol Al

10.0 g Al – 8.40 g Al = 1.60 g Al remaining

Theoretical, Actual, and Percent Yields

Up until now we have assumed that…

100% of the limiting reactant becomes product
Ideal methods exist for isolating the product
We have perfect lab techniques and collect all the product

In reality, these three assumptions are rarely observed, so chemist recognize three types of reaction yields…

Theoretical yield – The amount of product calculated from the molar ratio in the balanced equation.

Side reactionsform different products than the main product.