Stoichiometry Notes

Counting by Weighing for Carbon Atoms

• Conversion factor with a more common mass unit (grams) is preferred.
• Describing the number of atoms in terms of a collective unit (mole).
• A mole is the collective unit.
• Step 1: Choose an easily measurable property, such as mass for carbon atoms.
• Step 2: Choose a convenient unit for measurement, such as atomic mass units (u) for carbon atoms.
  • Atomic mass unit (u) = 1/12 the mass of a carbon-12 atom (with 6 protons, 6 neutrons, and 6 electrons).
• Step 3: If the measurable property is mass, determine the mass of the individual objects being measured.
  • For carbon: 98.90% are 12 u and 1.10% are 13.003355 u.
• Step 4: If the objects do not all have the same mass, determine the weighted average mass of the objects.
  • $0.9890(12 \text{ u}) + 0.0110(13.003355 \text{ u}) = 12.011 \text{ u}$

Mole

• A mole (mol) is an amount of substance that contains the same number of particles as there are atoms in 12 g of carbon-12.
• To four significant figures, there are $6.022 \times 10^{23}$ atoms in 12 g of carbon-12.
• A mole of natural carbon is the amount of carbon that contains $6.022 \times 10^{23}$ carbon atoms.
• The number $6.022 \times 10^{23}$ is often called Avogadro's number.

Avogadro's Number

• If the extremely tiny atoms in just 12 grams of carbon are arranged in a line, the line would extend over 500 times the distance between Earth and the sun.

Molar Mass Development

• From the definition of mole:
  • 12 g C-12 = 1 mol C-12
• From relative atomic masses:
  • 12.011 g C = 1 mol C
  • 15.9994 g O = 1 mol O
  • 24.3050 g Mg = 1 mol Mg
  • 1.00794 g H = 1 mol H

Molar Mass of Elements

• The atomic masses found on the periodic table can be used to get molar masses, which can be used to convert between grams and moles of any element.
• $( \frac{\text{(atomic mass) g element}}{1 \text{ mol element}} )$

Example Calculations

• The masses of diamonds and other gemstones are measured in carats. There are exactly 5 carats per gram. How many moles of carbon atoms are in a 0.55 carat diamond? (Assume that the diamond is pure carbon.)

Our Calculation: Maximum Mass of P4O10 from 1.09 x 10^4 kg P

• The general steps for our calculation:
  • Mass P → moles P → moles P4O10 → mass P4O10
• We'll see how to do the first two steps in this lesson, and I'll tell you how to do the last step in another lesson.
• We can convert grams of P to moles of P using the molar mass of P, which comes from its atomic mass that is found on the periodic table.
• Before we can convert grams P to moles P, we need to convert kg to g.
• The chemical formula provides a conversion factor for converting from moles of phosphorus atoms to moles of tetraphosphorus decoxide molecules in the second step of our calculation.
• If we have the reaction: $4P + 5O<em>2 \rightarrow P</em>4O_{10}$

Making Phosphoric Acid: Furnace Process

• Used to make fertilizers, detergents, and pharmaceuticals.
  • React phosphate rock with sand and coke at 2000 ºC: $2Ca<em>3(PO</em>4)<em>2 + 6SiO</em>2 + 10C \rightarrow 4P + 10CO + 6CaSiO_3$
  • React phosphorus with oxygen to get tetraphosphorus decoxide: $4P + 5O<em>2 \rightarrow P</em>4O_{10}$
  • React tetraphosphorus decoxide with water to make phosphoric acid: $P<em>4O</em>{10} + 6H<em>2O \rightarrow 4H</em>3PO_4$

Sample Calculation: Maximum Mass of P4O10 from 1.09 x 10^4 kg P

• The formula for P4O10 provides us with a conversion factor that converts from units of P to units of P4O10.

Goal: Develop Conversion Factors

• To convert between a measurable property (mass) and number of particles:
  • Measurable Property 1 → Number of Particles 1 → Number of Particles 2 → Measurable Property 2
  • Mass 1 → Number of Particles 1 → Number of Particles 2 → Mass 2

Molar Conversions

• Mass 1 → Number of Particles 1 → Number of Particles 2 → Mass 2
• Mass 1 → Moles 1 → Moles 2 → Mass 2

Molecular Mass

• Whole = sum of parts
• Mass of a molecule = sum of the masses of the atoms in the molecule
• Molecular mass = the sum of the atomic masses of the atoms in the molecule

Molar Mass For Molecular Compounds

• Molecular Mass = Sum of the atomic masses of the atoms in one molecule
• $( \frac{\text{(molecular mass) g molecular compound}}{1 \text{ mol molecular compound}} )$

Our Calculation: Molar Mass of P4O10

*What is the maximum mass of P4O10 that can be formed from 1.09×104 kg P?

Mass P → moles P → moles P4O10 → mass P4O10

• We can now take the next step in our calculation using the molar mass of P4O10 that comes from its molecular mass to convert from mol P4O10 to g P4O10.
• $4(30.9738) + 10(15.9994) = 283.889$

Formula Units

• A formula unit of a substance is the group represented by the substance’s chemical formula, that is, a group containing the kinds and numbers of atoms or ions listed in the chemical formula.
• Formula unit is a general term that can be used in reference to elements, molecular compounds, or ionic compounds.

Formula Unit Examples

• Neon gas (element):
  • A formula unit of neon contains one Ne atom.
• Liquid water (molecular compound):
  • Liquid water is composed of discrete H2O molecules.
  • A formula unit of water contains one oxygen atom and two hydrogen atoms.
• Ammonium chloride (ionic compound):
  • There are no separate ammonium chloride, NH4Cl, molecules. Each ion is equally attracted to eight others.
  • A formula unit of ammonium chloride contains one ammonium ion, NH4*, and one chloride ion, Cl-, (or one nitrogen atom, four hydrogen atoms, and one chloride ion).

Formula Mass for Ionic Compounds

• Whole = sum of parts
• Mass of a formula unit = sum of the masses of the atoms in the formula unit
• Formula mass = the sum of the atomic masses of the atoms in the formula

Molar Mass For Ionic Compounds

• Formula Mass = Sum of the atomic masses of the atoms in a formula unit
• $( \frac{\text{(formula mass) g ionic compound}}{1 \text{ mol ionic compound}} )$

Important Skills from Previous Sections

• Determination of atomic mass, molecular mass, and formula mass
• Using molar mass to convert between mass and the number of particles expressed in moles
• How chemical formulas can be used to convert between moles of element and moles of compound containing that element.

General Conversions

• Measurable property of substance 1 → Moles of substance 1 → Moles of substance 2 → Measurable property of substance 2

Units of One Substance to Units of Another

• Given units of substance 1 → Grams of substance 1 → Moles of substance 1 → Moles of substance 2 → Grams of substance 2 → Desired units of substance 2

Units of Element to Units of Compound

• Any unit of an element
  • Unit analysis conversion factors
• Grams of element
  • Using molar mass derived from atomic mass: $(\frac{1 \text{ mol element}}{\text{(atomic mass) g element}})$
• Moles of element
  • Using the mole ratio from the compound's formula: $(\frac{1 \text{ mol compound}}{\text{(number of atoms in formula) mol element}})$
• Moles of compound containing the element
• Grams of compound
  • Using molar mass derived from formula mass: $(\frac{\text{(formula mass) g compound}}{1 \text{ mol compound}})$
• Any unit of a compound

Study Sheets

• Write a description of the “tip-off” that helps you to recognize the type of problem the calculation represents.
• Write a description of the general procedure involved in the particular type of problem.
• Write an example of the type of calculation.

Sample Study Sheet: Converting Between Mass of Element and Mass of Compound Containing the Element

• Tip-off: When you analyze the type of unit you have and the type of unit you want, you recognize that you are converting between a unit associated with an element and a unit associated with a compound containing that element.

Sample Study Sheet - General Steps

• If necessary, convert from the given unit to grams.
• Convert grams to moles of the first substance using its molar mass.
• Convert moles of the first substance to moles of the second substance using the molar ratio derived from the formula for the compound.
• Convert moles of the second substance to grams of the second substance using its molar mass.
• If necessary, convert from grams to the desired unit.

Exercise 1 – First Steps

• Disulfur dichloride, S2Cl2, is used in vulcanizing rubber and hardening soft woods. It can be made from the reaction of pure sulfur with chlorine gas. What is the mass of S2Cl2 that contains 123.8 g S?
• First steps:
  • Grams to moles 1, using the molar mass of sulfur that comes from its atomic mass on the periodic table.
  • Moles 1 to moles 2, using the molar ratio that comes from the formula.

Exercise 1: Moles of 2 to Grams 2

• Moles of 2 to grams 2.
• $? \text{ g } S<em>2Cl</em>2 = 123.8 \text{ g S} \times \frac{1 \text{ mol S}}{32.066 \text{ g S}} \times \frac{1 \text{ mol } S<em>2Cl</em>2}{2 \text{ mol S}} \times \frac{135.037 \text{ g } S<em>2Cl</em>2}{1 \text{ mol } S<em>2Cl</em>2} = 260.7 \text{ g } S<em>2Cl</em>2$\newline

Exercise 2

• Vanadium metal, used as a component of steel and to catalyze various industrial reactions, is produced from the reaction of vanadium(V) oxide, V2O5, and calcium metal. What is the mass in kilograms of vanadium in 2.3 kilograms of V2O5?
  • Given unit to grams 1.
  • First steps:
    • Grams of 1 to moles 1, using the molar mass of V2O5 that comes from the sum of the atomic masses of 2 vanadium atoms and 5 oxygen atoms.
    • Moles 1 to moles 2.
    • Moles 2 to grams 2.

Exercise 2: Grams to the Desired Unit

• Grams to the desired unit.
• $? \text{ kg V} = 2.3 \text{ kg } V<em>2O</em>5 \times \frac{10^3 \text{ g}}{1 \text{ kg}} \times \frac{1 \text{ mol } V<em>2O</em>5}{181.880 \text{ g } V<em>2O</em>5} \times \frac{2 \text{ mol V}}{1 \text{ mol }V<em>2O</em>5} \times \frac{50.9415 \text{ g V}}{1 \text{ mol V}} \times \frac{1 \text{ kg}}{10^3 \text{ g}} = 1.3 \text{ kg V}$

Empirical and Molecular Formulas

• When the subscripts in a chemical formula represent the simplest ratio of the kinds of atoms in the compound, the formula is called an empirical formula.
  • Most ionic compounds are described with empirical formulas.
• A molecular formula describes the actual numbers of atoms of each element in a molecule.

Examples of Empirical and Molecular Formulas

• Hydrogen peroxide
  • Molecular formula: H2O2
  • Empirical formula: HO
• Glucose
  • Molecular formula: C6H12O6
  • Empirical formula: CH2O

Calculating Empirical Formulas

• Step 1: If you are not given mass in grams for each element, convert the data you are given to grams of each element.
  • This may involve simple unit conversions. For example, you may be given pounds or milligrams, which you convert to grams using unit analysis.
  • Sometimes you are given the percentage of each element in the compound. Assume that you have 100 g of compound, and change the numbers for the percentages to grams.
• Step 2: Convert grams of each element to moles by dividing by the atomic mass of the element.
• Step 3: Divide each mole value by the smallest and round your answers to whole numbers or common mixed fractions.
• Step 4: If you have a fraction after the last step, multiply all the mole values by the denominator of the fraction.
• Step 5: The resulting mole values correspond to the subscripts in the empirical formula.

Calculating Empirical Formulas: Flowchart

• Start with:
  • Any mass unit for each element in the compound
    • Unit analysis
    • OR Percentage of each element in the compound
      • Assume 100 g compound, and convert % to g.
  • ↓ Gram ratio of elements
  • Divide each value by its atomic mass.
  • ↓ Mole ratio of elements
  • (1) Divide each mole value by the smallest, and round to positive integers or common mixed fractions.
  • (2) If you have a fraction after step 1, multiply each mole value by the denominator of the fraction.
  • ↓ Simplest molar ratio (empirical formula)

Example Empirical Formula Calculation

• An ionic compound used in the brewing industry to clean casks and vats and in the wine industry to kill undesirable yeasts and bacteria is composed of 35.172% potassium, 28.846% sulfur, and 35.982% oxygen. What is the empirical formula for this compound?
  • Step 1: Convert percentages to a gram ratio of the elements by assuming 100 g.
    • 35.172 g K : 28.846 g S : 35.982 g O
  • Step 2: Convert grams of each element to moles by dividing by the atomic mass of the element.
  • Step 3: Divide each mole value by the smallest and round your answers to whole numbers or common mixed fractions.
  • Step 4: If you have a fraction after the last step, multiply all the mole values by the denominator of the fraction.
  • Step 5: The resulting mole values correspond to the subscripts in the empirical formula.
    • K2S2O5

Calculating Molecular Formulas

• Step 1: If necessary, calculate the empirical formula of the compound from the data given.
• Step 2: Divide the given molecular mass by the empirical formula mass.
• Step 3: Multiply each of the subscripts in the empirical formula by n to get the molecular formula.

Calculating Molecular Formulas: Flowchart

• Simplest molar ratio (empirical formula)
  • Divide the molecular mass by the empirical formula mass.
  • Multiply the subscripts in the empirical formula by the value from the preceding step.
  • ↓ Molecular formula

Example Molecular Formulas

• Compounds called polychlorinated biphenyls (PCBs) have structures similar to chlorinated insecticides, such as DDT. They have been used in the past for a variety of purposes, but because they have been identified as serious pollutants, their use today is limited to insulating fluids in electrical transformers. They have been banned for even this use in the U.S., but because they and the transformers last a long time, they are still in many transformers, even in the United States. One PCB is 39.94% carbon, 1.12% hydrogen, and 58.94% chlorine and has a molecular mass of 360.88. What is its molecular formula?

  • Step 1: If necessary, calculate the empirical formula of the compound from the data given.

• Step 2: Divide the given molecular mass by the empirical formula mass.

• Step 3: Multiply each of the subscripts in the empirical formula by n to get the molecular formula.