Stoichiometry
I. Atomic Masses.
* Atomic Masses.
1. Actual masses of subatomic particle and atoms in general are small and impractical to use.
2. Scientists chose the carbon-12 isotope to be the foundation for a new, more effective and efficient way of manipulating atomic masses.
a. This modern system was instituted in 1961.
b. The carbon-12 isotope is assigned a value of exactly 12.00 amu.
3. An atomic mass unit (amu) is defined as one-twelfth the mass of a carbon-12 isotope.
4. All other isotopic masses are determined relative to this isotope and are written in “amu’s.”
5. In nature, most elements occur as a mixture of two or more isotopes.
6. Each isotope of an element has a fixed mass and a natural percent abundance.
7. The atomic mass of an element is a weighted average of the atoms in a naturally occurring sample of an element.
a. It is a “weighted average,” also known as the average atomic mass.
b. These values appear on the periodic table.
c. Even though natural carbon does not contain a single atom with mass 12.01 (its periodic table value), for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01.
8. To determine the atomic mass, three values must be known.
a. The number of stable isotope of the element.
b. The mass of each isotope.
c. The natural percent abundance of each isotope.
9. The Average Mass of an Element.
* When a sample of natural copper is vaporized and injected into a mass spectrometer, the result is that 69.09% of Cu has a mass of 62.93 amu and 30.91% has a mass of 64.93 amu. Use this information to compute the average mass of natural copper.
II. The Mole.
A. Defining the mole.
- The SI definition of the mole is the amount of a substance that contains as many entities as there are in exactly 12 g of carbon-12.
- Avogadro’s number is 6.022 x 1023. One mole of anything is 6.022 x 1023 units of that substance.
- The mass of 1 mole of an element is equal to its atomic mass in grams.
- The relationship between the atomic mass unit and the gram:
B. Problems.
- Determining the Mass of a Sample of Atoms.
* Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of americium containing six atoms.
- Determining Moles of Atoms.
* Aluminum is a metal with a high strength-to-mass ration and a high resistance to corrosion; thus it is often used for structural purposes. Compute both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum.
- Calculating Number of Atoms.
* A silicon chip is an integrated circuit of a microcomputer has a mass of 5.68 mg. How many silicon atoms are present in the chip?
- Calculating the Number of Moles and Mass.
* Cobalt is a metal that is added to steel to improve its resistance to corrosion. Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample.
III. Molar Mass.
A. Definition.
* The molar mass of any substance is the mass (in grams) of one mole of the substance.
B. Problems.
1. Calculating Molar Mass I.
* Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree does not affect grass and other noncompetitive plants. The formula for juglone is C10H6O3.
a. Calculate the molar mass of juglone.
b. A sample of 1.56 x 10-2 g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent?
2. Calculating Molar Mass II.
* Calcium carbonate (CaCO3), also known as calcite, is the principal mineral found in limestone, marble, chalk, pearls, and the shells of marine animals such as clams.
a. Calculate the molar mass of calcium carbonate.
b. A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample? What is the mass of the CO32- ions present?
3. Molar Mass and Numbers of Molecules.
* Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bees release about 1 g if this compound when they sting. The resulting scent attracts other bees to join the attack. How many molecules of isopentyl acetate are released in a typical bee sting? How many atoms of carbon are present?
V. Percent Composition of Compounds.
* Calculating the Percent Composition of a Compound.
1. The percent composition (mass percent) is the percent by mass each element in a compound.
2. % mass of element E = grams of element E x 100%
grams of compound
3. The percentages should total 100%.
4. To calculate the percent composition of a known compound, use the chemical formula to calculate the molar mass.
a. This gives the mass of one mole of the compound.
b. Then for each element, calculate the percent by mass in one
mole of the compound.
c. % mass = grams of element in 1 mol compound x 100%
molar mass of compound
5. Calculating Mass Percent I.
* Carvone is a substance that occurs in two form having different arrangements of the atoms but the same molecular formula (C10H14O) and mass. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone.
6. Calculating Mass Percent II.
* The Scottish bacteriologist Alexander Fleming discovered penicillin, the first of a now large number of antibiotics, accidentally in 1928, but he was never able to isolate it as a pure compound. This and similar antibiotics have saved millions of lives that might have been lost to infections. Penicillin F has the formula C14H20N2SO4. Compute the mass percent of each element.
V. Determining the Formula of a Compound.
A. Calculating Empirical Formulas.
1. Percent composition data is important in that it can be used to determine the empirical formula of a compound.
2. The empirical formula gives the lowest whole-number ratio of the atoms of the elements in a compound.
a. The empirical formula provides valuable information about the kinds and relative amounts of atoms or moles of atoms in molecules or formula units of a compound.
b. An empirical formula may or may not be the same as the molecular formula. If the formulas are different, the molecular formula is a simple multiple of the empirical formula.
3. Empirical Formula Determination.
a. Since the mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound. Each percent will then represent the mass in grams of that element.
b. Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present.
c. Divide each value of the number of moles by the smallest of the values. If the resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the element in the empirical formula.
d. If the number obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers.
B. Calculating Molecular Formulas.
1. Many substances have the same empirical formula. This formulas alone would make it impossible to classify compounds.
2. The molecular formula, a multiple of the empirical formula must be determined to further identify and classify a compound
* Some moleculs have the same empirical and molecular formulas! Therefore, structural formulas must be determined (more later).
3. Molecular Formula Determination
a. Method 1
* Obtain the empirical formula.
* Compute the mass corresponding to the empirical formula.
* Calculate the ratio Molar Mass .
Empirical formula mass
* The integer from the previous step represents the number of empirical formula units in one molecule. When the empirical formula subscripts are multiplied by this integer, the molecular formula results. This procedure is summarized by the equation:
molecular formula = (empirical formula) x molar mass .
empirical formula mass
b. Method 2
* Using the mass percentages and the molar mass, determine the mass of each element present in one moles of compound.
* Determine the number of moles of each element present in one mole of compound.
* The integers from the previous step represent the subscripts in the molecular formula.
C. Determining Empirical and Molecular Formulas I.
* Determine the empirical and molecular formulas for a compound that gives the following analysis (in mass percents):
71.65% Cl24.27% C4.07% H
The molar mass is known to be 98.96 g/mol.
D. Determining Empirical and Molecular Formulas II.
* A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas?
E. Caffeine, a stimulant food in coffee, tea, and chocolate, contains 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula of caffeine.
F. Combustion Analysis I.
* Maleic acid is an organic compound composed of 41.39% C, 3.47% H, and the rest oxygen. If 0.129 mol of maleic acid has a mass of 15.0 g, what are the empirical and molecular formulas of maleic acid?
G. Combustion Analysis II.
* A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg CO2 and 41.1 mg H2O. What is the empirical formula of the compound?
VI. Stoichiometric Calculations: Amounts of Reactants and Products.
A. Guidelines.
1. The coefficients in chemical equations represent numbers of molecules, not masses of molecules.
2. However, in a laboratory, amounts of substances needed cannot be determined by counting molecules directly.
3. Counting is done using masses.
4. The mole ratio is the “bridge” which allows you to determine a quantity of one substance, given information about another.
B. Rules.
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant or product to moles of that
substance.
3. Use the balanced equation to set up the appropriate mole ratios.
4. Use the appropriate mole ratios to calculate the number of moles of
the desired reactant or product.
5. Convert the mole ratio back to grams if required by the problem.
C. Chemical Stoichiometry I.
* Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?
D. Chemical Stoichiometry II.
* Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach:
NaHCO3(s) + HCl(aq) -----> NaCl(aq) + H2O(l) + CO2(g)
Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also used as an antacid:
Mg(OH)2(s) + 2HCl(aq) -----> 2H2O(l) + MgCl2(aq)
Which is the most effective antacid per gram, NaHCO3 or Mg(OH)2?
VII. Calculations Involving a Limiting Reactant.
A. What Is a Limiting Reagent?
1. When experimenting with chemical reactions, chemists are faced with the task of producing a specific amount of a product without leaving any reactants unreacted (waste).
2. In most reactions, some reactant is used completely while another is left unreacted.
a. A limiting reagent is used up completely in a reaction.
* When the limiting reactant runs out, the reaction ceases as one of the “ingredients” required to produce a product is no longer present.
* The limiting reactant, therefore, limits or determines the amount of product that can be formed in a reaction.
b. An excess reactant is not completely used up in a reaction.
3. When working problems, all quantities must be converted to moles first before the amount of product, based on the moles of the limiting reactant, can be calculated.
4. Rules.
a. Write and balance the equation for the reaction.
b. Convert the known masses of substances to moles.
c. Determine which reactant is limiting.
d. Using the amount of limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.
e. Convert from moles to grams, using the molar mass.
5. Problems.
a. Stoichiometry: Limiting Reactant I.
* Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1 g of NH3 is reacted with 90.4 g CuO, which is the limiting reactant? How many grams of N2 will be formed?
b. Stoichiometry: Limiting Reactant II.
* Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane:
2NH3(g) + 3O2(g) + 2CH4(g) 2HCN(g) + 6H2O(g)
If 5.00 x 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?
B. Calculating Percent Yield.
1. In theory, every chemist hopes for 100% efficiency and return of
products for the amount of reactants used.
2. In actuality, 100% yield in chemical reactions in virtually impossible.
3. The theoretical yield is the maximum amount of a product that could
be formed from given amounts of reactants.
4. The actual yield is the amount of product that actually forms when
the reaction is carried out in a laboratory.
* The actual yield is less than the theoretical yield.
5. The percent yield is the ratio of the actual yield to the theoretical
yield expressed as a percent.
Percent yield = actual yield x 100%
theoretical yield
6. Reasons for percent yields less than 100%.
a. Reactions do not always go to completion (limiting reagent).
b. Impure reactants and competing side reactions may cause unwanted product formation.
c. Loss of product during filtration, transferring, or evaporation.
d. Poor human measuring.
7. Problems.
a. Calculating Percent Yield I.
* Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in racecars and is a potential replacement for gasoline. Methanol can be manufactures by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO(g) is reacted with 8.60 kg H2(g). Calculate the theoretical yield of methanol. If 3.57 x 104 g CH3OH is actually produced, what is the percent yield of methanol?
VIII. Water and the Nature of Aqueous Solutions
A. The Water Molecule.
1. Water is the foundation of all life on Earth.
2. Water is a triatomic molecule.
a. Each O--H covalent bond is highly polar because the electronegativity of O is so much greater than that of hydrogen.
b. The EN creates a slightly negative O and a slightly positive H as the electrons are pulled strongly toward O.
3. Water is a bent molecule with a bond angle of 105o.
a. This creates a highly polar molecule.
b. Diagram:
4. Polar molecules are attracted to one another by dipole attractions.
a. In water, however, the attraction is the result of hydrogen bonding.
b. Instead of being able to easily slide past one another, water temperature to escape the surface of the liquid.
c. The strength of the hydrogen bonds accounts for water’s high surface tension, its low vapor pressure, its high specific heat, its high heat of vaporization, and its high boiling point.
B. Solvents and Solutes.
1. Chemically pure water never exists in nature because water dissolves so many substances.
* An aqueous solution is one in which water samples contain dissolved substances.
2. Components of a solution.
a. A solvent is the dissolving medium.
b. A solute is the dissolved particles.
c. Examples:
d. Solutions are homogeneous mixtures in which solute particles are usually less than 1 nm in diameter. The solute and solvent are not capable of being separated by filtration.
3. Substances that dissolve most readily in water include ionic compounds and polar covalent molecules.
a. Nonpolar molecules do not dissolve in water (a polar molecule).
b. Rule: Like dissolves like.
C. The Solution Process.
1. Because water molecules are in constant motion as a results of their kinetic energy, the molecules collide with solute particles.
a. The solvent molecules attract the solute ions.
b. The ionic crystal breaks apart by the action of the solvent.
c. Solvation is the process that occurs when a solute dissolves.
d. Diagram:
2. In some ionic compounds, the attractions between the ions in the crystal are stronger than the attractions exerted by water.
* These compounds are insoluble.
3. Nonpolar substances form a solution because there are no repulsive forces between them, not because the solute and solvent are attracted.
D. Electrolytes and Nonelectrolytes.
1. Compounds that conduct an electric current in aqueous solution or the molten state are called electrolytes.
a. All ionic compounds are electrolytes.
b. Soluble ionic compounds conduct electricity in both a solution and in the molten state.
c. Insoluble ionic compounds only conduct electricity in the molten state.
2. Compounds that do not conduct an electric current in either aqueous solution or the molten state are called nonelectrolytes.
a. Many molecular compounds are nonelectrolytes because they do not contain ions.
b. Most compounds of carbon are nonelectrolytes.
3. Some very polar molecular compounds are nonelectrolytes in the pure state, but are electrolytes when they dissolve in water.
a. This occurs because such compounds ionize in solution.
b. Examples:
4. Not all electrolytes conduct electricity to the same degree.
a. A weak electrolyte conducts electricity poorly because only a fraction of the solute exists as ions.
* Most of the compound is in the original form.
* The most common weak electrolytes are weak acids and weak bases.
* Examples:
b. A strong electrolyte conducts electricity very well because almost all of the solute exists as separated ions.
* Very little of the original compound remains intact.
* Classes of electrolytes include soluble salts, strong acids and strong bases.
* Examples:
IX. The Composition of Solutions.
A. Introduction.
* To perform stoichiometric calculations when two solutions are mixed, two things must be known.
- The nature of the reaction, which depends on the exact forms the chemical takes when dissolved.
- The amounts of chemical present in solution, usually expressed as concentrations.
B. Molarity.
1. Molarity (M) is defined as the moles of solute per volume of solution in liters.
M = molarity = moles of solute
liters of solution
2. Calculation of Molarity I.
* Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.
3. Calculation of Molarity II.
* Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.
4. Concentrations of Ions I.
* Give the concentration of each type of ion in the following solutions:
a. 0.50 M Co(NO3)2
b. 1 M Fe(ClO4)3
5. Concentrations of Ions II.
* Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3M ZnCl2.