# Stoichiometry Objectives

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Stoichiometry Objectives
Identify the quantitative relationships in a balanced chemical chemical equation. Determine the mole ratios from from a balanced chemical equation. Explain the sequence of steps used in solving stoichiometric problems. Use the steps to solve stoichiometric problems. Identify the limiting reactant in a chemical equation. Identify the excess reactant and calculate the amount remaining after the reaction is complete. Calculate the mass of a product when the amounts of more than one reactant are given. Calculate the theoretical yield of a chemical reaction from data. 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 = 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:

One Mole Quantities

Other Mole & Stoichiometry Vocabulary
1. What is stoichiometry? Study of quantitative relationships among amounts of reactants an products. 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 1 Identify and calculate the number of representative particles in each of the following quantities. moles of gold mole of nitrogen oxide moles of potassium bromide Calculate the number of moles of the substance that contains the following number of representative particles. x 1023 atoms of barium x 1023 molecules of carbon monoxide x 1023 formula units of potassium iodide

Practice-Homework p 311 # 1-3; p 312 # 4

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 moles of calcium. -According to the periodic table, the atomic mass of calcium is amu, so the molar mass of calcium is 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-Homework p 316 # 11ab-12ab

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. 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 318

Mole Practice 2 Determine the mass in grams of the following quantities. moles of beryllium moles of calcium Calculate the number of moles in each of the following. g lithium g zinc How many atoms are in the following samples? g cobalt g cesium How many grams are in the following samples? x1023 atoms of radium x 1020 atoms of cadmium

Practice-Homwork p 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( g) + 3(1.008 g) = g/mol

Practice p 322 # 25

Book Practice P 312 # 4

-Mole relationships from a formula (p 321)
-Mole to Mass for compounds ( p 323)

Mass of Compound to Moles
Calculate the number of moles of water that are in 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 kg is converted to x 103 g

Practice p 323 # 27-28 p 324 # 30a&b

Mole Practice 3 Calculate the molar mass of the following: C2H5OH HCN
What is the mass of the following: 2.25 moles of KMnO4 1.56 moles of H2O Determine the number of moles in the following: 35.0 g HCl 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. g Ca+2 ions g Cl- ions = g CaCl2 1 mole CaCl 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 = g Ca+2 x 100 = % Ca+2 g CaCl2 % Cl = g Cl x 100 = % Cl- 4. Make sure your percent compositions equal 100%. 36.11% Ca % Cl- = 100%

Practice p 331 # 42-43

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 0.6995 mol 0.6995 mol 0.6995 mol
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. mol Mn : mol C : mol O mol mol mol : :

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 : : 4. Determine the empirical formula. MnC2O4

Practice p 333 # 46-47

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.

-composition of maleic acid is 41.39% C, 3.47% H, and
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 CHO 2. 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 # 51-52

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 = mol Fe 55.85 g Fe 4.64g Ti x 1 mol Ti = mol Ti 47.88g Ti 4.65g O x 1 mol O = 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. mol Fe : mol Ti : mol O mol mol mol : : 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. HCl(aq) + KOH(aq)  KCl(aq) + H2O(l) 2. 2Mg(s) + O2(g)  2MgO(s) 2HgO(s)  2Hg(l) + O2(g)

Stoichiometric Calculations
Many 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 Moles of known x moles of unknown = moles of known

Stoichiometric Calculations: mole-mole
Example: If you put mol of K into water, how much hydrogen gas will be produced? 2K(s) + 2H2O(l) → 2KOH(aq) +H2(g) Mole ratio between K and KOH is 2 mol K or 1 mol H2 1 mol H mol K moles of known x moles of unknown moles of known mol K x 1 mol H2 = 2 mol K mol H2

Stoichiometric Calculations Practice
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 357 #2a, 3a p 359 #10

Stoichiometric Calculations: mole-mass
Example: If you put mol of K into water, how many grams of hydrogen gas will be produced? 2K(s) + 2H2O(l) → 2KOH(aq) +H2(g) Mole ratio between K and KOH is 2 mol K or 1 mol H2 1 mol H mol K moles of known x moles of unknown moles of known mol K x 1 mol H2 = 2 mol K mol H2 mol H2 x 2.02 g H2 = 1 mol H2 g H2

Practice p 360 #11-12

Stoichiometric Calculations: mass-mass
Example: If you put 15g of K into water, how many grams of hydrogen gas will be produced? 2K(s) + 2H2O(l) → 2KOH(aq) +H2(g) 15 g K x 1 mol K = mol K g K Mole ratio between K and KOH is 2 mol K or 1 mol H2 1 mol H mol K 0.384 mol K x 1 mol H2 = 2 mol K 0.192 mol H2 0.192 mol H2 x 2.02 g H2 = 1 mol H2 0.388 g H2

Practice p 361 #13-14

Stoichiometric Review
Why is a balanced chemical equation needed in solving stoichiometric calculations? When solving stoichiometric problems, how is the correct mole ratio expressed? List the steps in solving stoichiometric equations. How many grams of zinc (II) chloride can be obtained in 50.0 g of zinc reacts completely in the following reaction: Zn + 2HCl → ZnCl2 + H2 How many grams of KCl is produced when mol of KClO3 decomposes in the following reaction: 2KClO3 + heat → 2KCl + 3O2

Stoichiometric Calculations Practice 1
What mass 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 grams of H2SO4 can be produced?

Stoichiometric Calculations Practice 2
What mass of carbon dioxide are produced when 20.0 grams 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 10.0 grams SO2 reacts, how many grams of H2SO4 can be produced?

QUIZ!!!!

- The left-over reactants are called excess reactants
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 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 1. calculate the actual number of moles of reactants

Limiting Reactants In the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4). a. determine the limiting reactant 2. look at the actual ratio to the available ratio. ~actual ratio = 1.00 mol NaOH 0.612 mol H2SO4 ~available ratio = 2.00 mol NaOH = mol NaOH 1.00 mol H2SO mol H2SO4 You can see that when mol H2SO4 has reacted, all of the 1.00 mol of NaOH would be used up. -NaOH is the limiting reactant

Limiting Reactants In the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4). b. determine the reactant in excess -since NaOH is the limiting reactant, H2SO4 is the reactant in excess c. calculate the mass of sodium sulfate produced -use the limiting reactant 1.00mol NaOH x 1.00mol Na2SO4 x 142g Na2SO4 = 71.0g Na2SO4 2.00mol NaOH mol Na2SO4

Limiting Reactants In the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4). d. calculate the mass of reactant in excess -use NaOH, the limiting reactant to determine moles and mass of H2SO4 used ~1.00mol NaOH x 1.00mol H2SO4 x 98.09g H2SO = 49.05g H2SO4 2.00mol NaOH mol H2SO4 -subtract the mass needed from the mass available. 60.0g H2SO4 – 49.05g H2SO4 = 11.0g H2SO4 in excess

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 form The 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 g of silver nitrate, solid silver chromate is formed. a. Determine the theoretical yield of silver chromate b. If g of silver chromate is actually obtained, calculate the percent yield.