Stoichiometry Objectives1. Identify the quantitative relationships in a balanced chemical

chemical equation.2. Determine the mole ratios from from a balanced chemical

equation.3. Explain the sequence of steps used in solving stoichiometric

problems.4. Use the steps to solve stoichiometric problems.5. Identify the limiting reactant in a chemical equation.6. Identify the excess reactant and calculate the amount remaining

after the reaction is complete.7. Calculate the mass of a product when the amounts of more than

one reactant are given.8. Calculate the theoretical yield of a chemical reaction from data.9. Determine the percent yield for a chemical reaction.

The Mole (Ch 11)Chemists need a convenient method for counting accurately

the number of atoms, molecules, or formula units in a sample of a substance.

-atoms and molecules are extremely small -even in the smallest sample it’s impossible to actually

count each individual atom.

To fix this problem chemists created their own counting unit called the mole.

-mole: commonly abbreviated mol, is the SI base unit used to measure the amount of a substance

The Mole

Through experimentation, it has been established 1 mole = 6.022 136 7 x 1023 representative particles.

-representative particle: any particle such as atoms, molecules, formula units, electrons, or ions.

-called Avogadro’s number in honor of the Italian physicist and lawyer Amedeo Avogadro who, in 1811, determined the volume of one mole of a gas.

-we round Avogadro’s number to three significant figures— 6.02 x 1023.

- If you write out Avogadro’s number, it looks like this: - 602 000 000 000 000 000 000 000

One Mole Quantities

Other Mole & Stoichiometry Vocabulary 1. What is stoichiometry?

Study of quantitative relationships among amounts of reactants and products.

2. What is a mole ratio?Ratio between the moles of any two substances in a balanced chemical equation

3. What is molar mass? Mass, in grams, of one mole of pure substance.

-numerically equal to its atomic mass

-has the units g/mol

Converting Moles to Particles (11.1)Determine how many particles of sucrose are in 3.50mol

of sucrose. - write a conversion factor using Avogadro’s number

that relates representative particles to moles of a substance.

Converting Particles to Moles (11.1)

Now, suppose you want to find out how many moles are represented by a certain number of representative particles, such as 4.50 x 1024 atoms of zinc.

-You can use the inverse of Avogadro’s number as a conversion factor.

Mole Practice 1Identify and calculate the number of representative particles

in each of the following quantities.1. 2.15 moles of gold2. 0.151 mole of nitrogen oxide3. 11.5 moles of potassium bromide

Calculate the number of moles of the substance that contains the following number of representative particles.4. 8.92 x 1023 atoms of barium5. 5.50 x 1023 molecules of carbon monoxide6. 2.66 x 1023 formula units of potassium iodide

Practice-Homeworkp 311 # 1-3; p 312 # 4a-c

1. Determine the number of atoms in 2.50 mol Zn2. Determine the number of formula units in 3.25 mol

AgNO3

3. Determine the number of molecules in 11.5 mol H2O4. Determine the number of moles in a. 5.75 x 1024 atoms Al b. 3.75 x 1024 molecules CO2

c. 3.58 x 1023 formula units ZnCl2

Moles to Mass (11.2)

To convert between moles and mass, you need to use the atomic mass found on the periodic table.

Calculate the mass of 0.625 moles of calcium. -According to the periodic table, the atomic mass of

calcium is 40.078 amu, so the molar mass of calcium is 40.078 g/mol.

Mass to Moles (11.2)

How many moles of copper are in a roll of copper that has a mass of 848g?

Practice-Homeworkp 316 # 11ab-12ab

11. Determine the mass in grams of a. 3.57 mol Al b. 42.6 mol Si12. Determine the number of moles of a. 25.5 g Ag b. 300.0 g S

Mass to Atoms (11.2)

To find the number of atoms in a sample, you must first determine the number of moles.

Calculate the number of atoms in 4.77 g lead.1. Determine moles

Mass to Atoms (cont.)

2. Determine atoms

You can also convert from number of particles to mass by reversing the procedure above and dividing the

number of particles by Avogadro’s number to determine the number of moles present.

Atoms to Mass

Example problem 11-5, p 318A party balloon has 5.50x1022 atoms of helium.

What is the mass in grams of the helium?5.50x1022 atoms He x 1 mol He = 0.0914 mol He

6.02 x1023 atoms He0.0914 mol He x 4.00 g He = 0.366 g He 1 mol He

Mole Practice 2

How many atoms are in the following samples?1. 1.24 g cobalt2. 0.575 g cesium

How many grams are in the following samples?3. 4.16 x1023 atoms of radium4. 1.50 x 1020 atoms of cadmium

Practice-Homworkp 316 # 11ab-12ab, p 318 # 13ab-14ab

Moles of Compounds (11.3)

A mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula.

-For example, a mole of ammonia (NH3) consists of one mole of nitrogen atoms and three moles of hydrogen atoms.

-the molar mass of the compound is found by adding the molar masses of all of the atoms in the representative particle.

Molar mass of NH3 = 1(molar mass of N) + 3(molar mass of H)

Molar mass of NH3 = 1(14.007 g) + 3(1.008 g) = 17.031 g/mol

Practicep 322 # 25

P 318 # 13ab-14abp32225. Determine the molar mass of each of the following: NaOH, CaCl2, Sr(NO3)2

p31813. How many atoms are in a. 55.2 g Li b. 0.230 g Pb14. What is the mass of a. 6.02x1023 atoms Bi b. 1.00x1024 atoms Mn

-Mole relationships from a formula (p 321)Determine the number of moles of aluminum ions

in 1.25 moles of aluminum oxide (Al2O3).

First we need the ratio of Al ions to Al2O3.

2 mol Al ions1 mole Al2O3

1.25 mol Al2O3 x 2 mol Al ions = 2.50 mol Al ions 1 mole Al2O3

-Mole relationships from a formula (p 321)

P321# 20-2120. Determine the number of moles of chloride ions

in 2.50 mol ZnCl2.

21. Calculate the number of moles of each element in 1.25 mole glucose (C6H12O6).

-Mole to Mass for compounds ( p 323)What is the mass of 2.50 moles of allyl sulfide,

(C3H5)2S?

1.Calculate the mass of allyl sulfide.6(12.01 g/mol) = 72.06 g/mol

10(1.01 g/mol) = 10.10 g/mol 1(32.07 g/mol) = 32.07 g/mol 114.23 g/mol

-Mole to Mass for compounds ( p 323)What is the mass of 2.50 moles of allyl sulfide,

(C3H5)2S?

2. Convert the moles to mass. 2.50mol (C3H5)2S x 114.23g (C3H5)2S

1 mol (C3H5)2S

= 286g (C3H5)2S

Mass of Compound to Moles

Calculate the number of moles of water that are in 1.000 kg of water?

1. Before you can calculate moles, you must determine the molar mass of water (H2O).

molar mass H2O = 2(molar mass H) + molar mass O

2. Now you can use the molar mass of water as a conversion factor to determine moles of water.

-Notice 1.000 kg is converted to 1.000 x 103 g

Practicep 323 # 27-28, p 324 # 30a&b

27. What is the mass of 3.25 moles H2SO4?

28. What is the mass of 4.35x10-2 moles of ZnCl2?

30. Determine the number of moles present in each of the following:

a. 22.6 g AgNO3

b. 6.50 g ZnSO4

Mole Practice 3Calculate the molar mass of the following:1. C2H5OH

2. HCNWhat is the mass of the following:3. 2.25 moles of KMnO4

4. 1.56 moles of H2O

Determine the number of moles in the following:5. 35.0 g HCl6. 254 g PbCl4

What is the mass in grams of one molecule of the following:7. H2SO4

Percent Composition, Molecular & Empirical Formulas

Recall that every chemical compound has a definite composition—a composition that is always the same wherever that compound is found.

The composition of a compound is usually stated as the percent by mass of each element in the compound, using the following process.

Percent Composition

Example: Determine the percent composition of calcium chloride (CaCl2).

1. Determine mass of each ion in CaCl2.

-1mol CaCl2 consists of 1mol Ca+2 ions and 2mol Cl- ions.

1mol Ca+2 ions x 40.08g Ca+2 ions = 40.08g Ca+2 ions

1mol Ca+2 ions 2mol Cl- ions x 35.45g Cl- ions = 70.90g Cl- ions 1mol Cl- ions

Percent Composition

Example: Determine the percent composition of calcium chloride (CaCl2).

2. Calculate molar mass of CaCl2.

- 40.08g Ca+2 ions + 70.90g Cl- ions = 110.98 g CaCl2

1 mole CaCl2 1 mole CaCl2

3. Determine percent by mass of each element.

Percent Composition

Example: Determine the percent composition of calcium chloride (CaCl2).

3. Determine percent by mass of each element.

% Ca = 40.08 g Ca+2 x 100 = 36.11 % Ca+2

110.98 g CaCl2

% Cl = 70.90 g Cl- x 100 = 63.89 % Cl-

110.98 g CaCl2

4. Make sure your percent compositions equal 100%. 36.11% Ca+2 + 63.98% Cl- = 100%

Practicep 331 # 43, 45

43. Calculate the percent composition of sodium sulfate (Na2SO4).

45. What is the percent composition of phosphoric acid (H3PO4).

Empirical Formula

You can use percent composition data to help identify an unknown compound by determining its empirical formula.

-empirical formula-simplest whole-number ratio of atoms of elements in the compound.

~In many cases, the empirical formula is the actual formula for the compound. the empirical formula of sodium chloride is Na1Cl1,

or NaCl, which is the true formula ~sometimes, the empirical formula is not the actual formula of the compound. the empirical formula for N2O4 (the actual) is NO2.

Empirical Formula

Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.

- Because percent means “parts per hundred parts,” assume that you have 100 g of the compound.

1. Calculate the number of moles of each element in the 100 g of compound.

Empirical Formula

Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.

1. Calculate the number of moles of each element in the 100 g of compound.

Empirical Formula

Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.

2. The results show the following relationship

3. Obtain the simplest whole-number ratio of moles: -divide each number of moles by the smallest number of

moles. 0.6995 mol Mn : 1.339 mol C : 2.798 mol O

0.6995 mol 0.6995 mol 0.6995 mol

1 : 2 : 4

Empirical Formula

Example: The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.

3. Obtain the simplest whole-number ratio of moles: Mn : C : O

1 : 2 : 4 4. Determine the empirical formula.

MnC2O4

Practice p 333 # 46-4746. A blue solid is found to contain 36.84% N

and 63.16% O. What is the empirical formula?

47. Determine the empirical formula for a compound that contains 35.98% Al and 64.02% S.

Molecular Formula

For many compounds, the empirical formula is not the true formula.

-Chemists have learned, though, that acetic acid is a molecule with the formula C2H4O2, which is the molecular formula for acetic acid.

-molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound.

Notice the molecular formula for acetic acid (C2H4O2) has exactly twice as many atoms of each element as the empirical formula (CH2O).

-The molecular formula is always a whole-number multiple of the empirical formula.

Molecular Formula

Example: Determine the molecular formula for maleic acid, which has a molar mass of 116.1g/mol.

1. empirical formula of the compound

-composition of maleic acid is 41.39% C, 3.47% H, and 55.14% O (change the % to g)

Molecular Formula

Example: Determine the molecular formula for maleic acid.

1. empirical formula of the compound -the ratio of C:H:O is 1:1:1, making the empirical formula CHO2. calculate the molar mass of CHO (empirical formula). -29.01g/mol

3. Determine the molecular formula for maleic acid,

Molecular Formula

Example: Determine the molecular formula for maleic acid.

3. Determine the molecular formula for maleic acid,

-shows the molar mass of maleic acid is 4x that of CHO. 4. Multiply CHO by 4 to get C4H4O4

Practice p 335 # 51, 5351. A substance has a chemical composition of 65.45% C,

5.45% H and 29.09% O. The molar mass of the molecular formula is 110.0 g/mol. Determine the molecular formula.

53. A compound contains 46.68 % N and 53.32 % O. It has a molar mass of 60.01 g/mol. What is the molecular formula?

Empirical Formula from Mass

You can also calculate the empirical formula of a compound from mass of individual elements.

Example: Determine the empirical formula for ilmenite, which contains 5.41g Fe, 4.64g Ti and 4.65g O.

1. Multiply the mass by molar mass to get moles 5.41g Fe x 1 mol Fe = 0.0969 mol Fe 55.85 g Fe 4.64g Ti x 1 mol Ti = 0.0969 mol Ti 47.88g Ti 4.65g O x 1 mol O = 0.291 mol O 16.00g O

Empirical Formula from Mass

Example: Determine the empirical formula for ilmenite, which contains 5.41g Fe, 4.64g Ti and 4.65g O.

2. Multiply by the smallest number to get the mole ratio. 0.0969 mol Fe : 0.0969 mol Ti : 0.291 mol O 0.0969 mol 0.0969 mol 0.0969 mol 1 : 1 : 3 3. Calculate the empirical formula. FeTiO3

Practice

p 337 # 54-55

TEST!!!!

Stoichiometry Practice – Conservation of Mass

For the following balanced chemical equations, determine all possible mole ratios.

1. HCl(aq) + KOH(aq) KCl(aq) + H2O(l)

2. 2Mg(s) + O2(g) 2MgO(s)

Stoichiometric CalculationsMany times we need to determine a certain amount of product

from a reaction or want to know how much product will form from a given amount of reactant.

To do this you need:

1. balanced chemical equations

2. mole ratios

3. molar mass

Stoichiometric Calculations: mole-moleExample: If you put 0.0400 mol of K into water, how much hydrogen gas will be produced? 2K(s) + 2H2O(l) → 2KOH(aq) +H2(g)

Use ‘modified’ RICE table:R : Reaction (must be balanced) I : Initial (amount in moles)C : Change (also in moles)E : “End” (really means equilibrium, but….)

Stoichiometric Calculations: mole-moleExample: If you put 0.0400 mol of K into water, how many

moles of hydrogen gas will be produced?R : 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

(0.400 mol) (_____ mol) I : 0.400 0C : -0.400 +1/2(0.400) E : 0 0.200

~subtract from reactants, add to products~products start at ‘0’ mol~what about water? it is in ‘excess’.~why multiply by ½? b/c K:H2 = 2:1 mole ratio (1/2 as much

H2 as K)

Stoichiometric Calculations Practice

1. How many moles of carbon dioxide are produced when 10.0 moles of propane (C3H8) are burned in excess oxygen in a gas grill. Water is also a product.

2. Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write the balanced chemical equation for the reaction. If 12.5 mol SO2 reacts, how many moles H2SO4 can be produced? How many mole O2 is needed?

Practice

p 359 #10

10. CH4(g) + S8(s) CS2(l) + H2S(g)

a. Balance the equation. b. Calculate moles of CS2 produced when 1.50 mol S8 is

used. c. How many mol H2S is produced?

Stoichiometric Calculations: mole-massExample: If you put 0.0400 mol of K into water, how many

grams of hydrogen gas will be produced?R : 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

(0.0400 mol) (_____ g) I : 0.0400 0C : -0.0400

+1/2(0.0400) E : 0 0.0200

0.0404 g~must change moles into grams if asked for mass!!

Practicep 360 #11-12

11. If you begin with 1.25 mol TiO2, what mass of Cl2 is

needed? TiO2 + C + Cl2 TiCl4 + CO2

12. Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy. How many grams of chlorine gas are produced from 2.50 mol sodium chloride?

Stoichiometric Calculations: mass-massExample: If you put 15.0g of K into water, how many

grams of hydrogen gas will be produced?R : 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

15.0g (_____ g) I : 0.384 0C : -0.384 +1/2(0.384)E : 0 0.192

0.388 g~now we must change initial mass to moles!!

Practicep 361 #13-14

13. Determine the mass of N2 produced if 100.0g NaN3 is

decomposed. NaN3(s) Na(s) + N2(g)

14. If 2.50 g sulfur dioxide reacts with excess oxygen and water, how many grams of sulfuric acid are produced?

Extra Practice

p 379-380 #61, 64, 70

QUIZ!!!!

Limiting Reactants

Rarely are the reactants in a chemical reaction present in the exact mole ratios specified in the balanced equation.

-usually, one or more of the reactants are present in excess, and the reaction proceeds until all of one reactant is used up.

-the reactant that is used up is called the limiting reactant

The limiting reactant limits the reaction and, thus, determines how much of the product forms.

- The left-over reactants are called excess reactants

Limiting Reactants

In the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4).

a. Calculate the limiting reactant b. Determine the reactant in excess c. Calculate the mass of water produced d. Calculate the mass of reactant in excess.

Limiting ReactantsR : 40.0g 60.0 g (_____ g) I : 1.00 0.612 0C : -1.00 -1/2(1.00) +1.00E : 0 0.112 1.00~change both initial masses to moles~compare ratio, I need a NaOH : H2SO4 of 2 : 1, so ask:

“Do I have 2x as much NaOH as H2SO4?”

No, b/c 2(0.612) = 1.22 and I only have 1.00, thereforeNaOH is my limiting reactant and H2SO4 is excess.

Limiting ReactantsR : 40.0g 60.0 g (_____ g) I : 1.00 0.612 0C : -1.00 -1/2(1.00) +1.00E : 0 0.112 1.00~change mol product to mass show work:~to determine amount in excess, take moles excess and change

to mass show work:

Practice

p 368 # 20-21

Percent Yield

Most reactions never succeed in producing the predicted amount of product.

-not every reaction goes cleanly or completely ●liquids may stick to surfaces of containers ●liquids may vaporize/evaporate ●solids may be left behind on filter paper ●solids may be lost in the purification process ●sometimes unintended products formThe amount you have been calculating so far has been the

theoretical yield, the maximum amount of product that can be produced from a given amount of reactant.

Percent Yield

A chemical reaction rarely produces the theoretical yield. -actual yield is the amount of product produced in the

chemical reaction

We can measure the efficiency of the reaction by calculating the percent yield.

-percent yield (% yield) of the product is the ratio of actual yield to the theoretical yield, expressed as a percent.

% yield = actual yield (from the experiment) x 100 theoretical yield (from calculations)

Percent Yield

When potassium chromate is added to a solution containing 0.500 g of silver nitrate, solid silver chromate is formed.

a. Determine the theoretical yield of silver chromate b. If 0.455 g of silver chromate is actually obtained, calculate the percent yield.