Unit 4: Formula Stoichiometry

What is stoichiometry? Deals with the quantitative information in chemical formula or chemical reaction. Using our reference tables we will learn how to calculate the following: Mass of an atom Mass of a compound Moles of an atom or compound (molar mass) Percent Composition in a compound

Key Terms Review Subscripts - tell you the amount of atoms in a formula Coefficients - tell you the amount of molecules of that substance

How many oxygen atoms in each? NH4NO3 C8H8O4 O3 C3H5(NO3)3 (3) (4) (3) (9)

Counting Atoms Practice Example #1: (NH4)2CO3 * First list the types of atoms and then count each. N: H: C: O: 2 8 1 3

Counting Atoms Practice Example #2: CaCrO4 * First list the types of atoms and then count each. Ca: Cr: O: 1 4

Counting Atoms Practice Example #3: Ca3(PO4)2 * First list the types of atoms and then count each. Ca: P: O: 3 2 8

Mass Ex) Ne Ex) CHCl3 Ex) CaCl2 - All measured in amu (atomic mass units) Atomic Mass Molecular Mass Formula Mass Mass of one atom of one element from periodic table. Mass of all of the atoms in a “Covalent” or “Molecular” compound (non-metal atoms only) Mass of all of the atoms in an ionic compound (metal + non-metal ions) Ex) Ne Ex) CHCl3 Ex) CaCl2

Gram Formula Mass & Gram Molecular mass The same as before, but instead of “amu” Mass expressed in grams (g) H2O O + H + H = + + Molecular Mass = 15.9994 amu + 1.00794 amu +1.00794 amu Gram Molecular Mass = 18.0 g

How to calculate molecular or formula mass… First: Identify and count atoms in the compound. Second: Locate atomic mass of each element. Third: Multiply mass by number of atoms, then get the total of all elements.

Gram Molecular Mass Example (Covalent Compound) Example: H2O H- 2 x 1 O- 1 x 16 Molecular Mass = 18 g

Gram Formula Mass Example (ionic compound) Example: NaCl Na- 1 x 23 Cl- 1 x 35 Formula Mass = 58 g

Percent Composition by Mass Used to calculate how much of a substance’s mass is made up of a particular element… Formula is given on Table T in reference table Mass of Part Mass of Whole = % Composition “by Mass” X 100

Example 1: In a sample of calcium carbonate, what is the percent composition by mass of calcium? Formula = CaCO3 Ca: 1 C: 1 O: 3 x 40 = 40 g x 12 = 12 g x 16 = 48 g } total = 100 g 40 g 100 g % Composition Calcium = X 100 = 40 % Ca

Example 2: A compound containing carbon and hydrogen has a mass of 16 grams. When decomposed, 12.0 grams are found to be carbon. What is the percent by mass of carbon in the compound? Formula = C?H? } total = 16 g C = 12 g O = __ g 12 g 16 g % Composition Carbon = X 100 = 75 % Carbon

Hydrated Crystals: Some ionic compounds have surrounding water molecules. These are called Hydrated Crystals. Percent water hydration can be found using the formula for % composition… Example: CuSO4 . 5H2O 5 H2O molecules are attached to 1 CuSO4

Percent Water Hydration (cont.) % Hydration = Total mass of Water x 100 Formula Mass Formula = CuSO4 . 5H2O Cu = 1 x 64 = 64 g S = 1 x 32 = 32 g O = 4 x 16 = 64 g Separately H2O: O = 1 x 16 = 16 g H = 2 x 1 = 2 g _________________ H2O = 18 x 5 = 90 g g ✔

The Mole Pt. 1 A mole is essentially one complete unit of any particular substance Think of it like the term “dozen” Meaning 12 items of any substance make up a dozen of that substance 1 mole is always equal to 6.02 x 1023 particles of a substance

The Mole Pt. 2 Avagadro’s Number is 6.02 x 1023 Avagadro’s Number is the number of: - Atoms in 1 mole of any element - Molecules in 1 mole of any compound - Particles in 1 mole of any substance

The Mole Pt. 3 The mass of 1 mole will be different for all substances. Just like a dozen eggs being different from a dozen tires (size and weight) Gram Atomic / Formula / Molecular Mass = Molar Mass - because it is the mass of one Units (g/mol) mole of any substance. Example: 1 mole of H2O = 18g Molar Mass of Water = 18 g/mol

Mole Relationships or Equalities 1 mole = formula mass in grams 1 mole = 6.02 x 1023 particles 1 mole = 22.4L (for gases) So… Formula mass = 6.02x1023 particles 22.4L of a gas = 6.02x1023 particles Formula mass = 22.4L of a gas

Conversion Problems Some problems involve mass-mole conversions you can use the “Mole Calculation” formula on Table T. You will usually have to calculate the gram formula mass of the compound… # of Moles = given mass (g) gram formula mass

Example 1 - What is the mass of 1.75 moles of oxygen gas (Hint: oxygen is diatomic)? 1st find the molecular mass of oxygen remember oxygen is a diatomic element (O2) O: 2 x 16 g= 32 g Then use formula on table T: Moles = Given Mass (g) Gram Formula Mass 1.75 = ? 32 g ✔ ? = 56 g

Example 2 - How many moles are in 42 g of water? ✔ 1st find the molecular mass of H2O: H: 2 x 1 g = 2 g O: 1 x 16 g = 16 g 2 + 16 = 18 g Convert from grams to moles… ? Moles = 42 g 18 g ✔ ? =2.3 mol

Example 3 - How many moles of potassium chromate are in a 500 g sample? K2CrO4 1) Find the formula of Potassium Chromate: 2) Calculate formula Mass: K:2 x 39 g = 78 g Cr: 1 x 52 g = 52 g O: 4 x 16 g = 64 g 78+52+64 = 194 g 3) Calculate the Moles using the formula on table T: ? Moles = 500 g 194 g ? = 2.6 mol

Empirical Formula Review Empirical Formula is when the subscripts in a chemical formula are simplified to their lowest common multiples. Ex) C4H8 vs. CH2 N2O4 vs. NO2 NaCl vs. ? Molecular Empirical Molecular and Empirical

Empirical Mass - Is the mass of the empirical formula of a compound In a question you are given the molecular mass and empirical formula… 1st you must find the empirical mass Then divide molecular mass by empirical mass Multiply all subscripts in the empirical formula by your answer to get molecular formula

Empirical Mass Example Molecular Mass = 180 Empirical Formula = CH2O Step 1: CH2O Empirical Mass C – 1 x 12 = 12 H – 2 x 1 = 2 O – 1 x 16 = 16 30 amu Step 2: Divide (Molecular mass/Empirical mass) 180/30 = 6 Step 3: Multiply Empirical Formula by Step 2 6(CH2O) = C6H12O6

Moles in Chemical Equations In problems involving chemical reactions we balance chemical reactions using coefficients These coefficients also represent the mole ratios in a reaction Ex) 2C2H6 + 7O2  4CO2 + 6H2O Moles C2H6 Moles O2 Moles CO2 Moles H2O 2 7 4 6 1 3.5 3

Moles in Chemical Equations Pt. 2 After balancing we can determine the amount of a reactant or product involved in the reaction As long as your given the amount of one ingredient… Ex) 4Al + 3O2  2Al2O3 What is the Minimum # of moles of Oxygen gas needed to make 1 mole of aluminum oxide?

Moles in Chemical Equations Pt. 3 The mole ratio in a balanced reaction can also be used to calculate the mass of reaction ingredients Ex) Based on the reaction below a student calculates 20g of H2 will react with 20g Cl2 to form 40g HCl, is this true? H2 + Cl2  2HCl

Moles in Chemical Equations Pt. 4 We can use these mole ratios to calculate the volume of reaction ingredients also (only in gas reactions) Ex) Mole ratio = 2 : 7 : 4 : 6 2C2H6 (g)+ 7O2 (g) 4CO2 (g)+ 6H2O (g) How many Liters of CO2 (g) will be produced from the combustion of 30 Liters of C2H6 (g) ? 30 L 2 = 15 L Then, 15 L x 4 = 60 L

A pound of meth is worth 40,000 $ 1 lb = ~454 g Balance the reaction and figure out the amount of reactants needed to synthesize 454 g of meth… _C9H10O + _CNH5  _C10H13N + _H2O ? g + ? g  454 g