Unit 5: Stoichiometry and Chemical Reactions

I. What is a mole? 6.02 x 10 23 6.02 x 10 23 atoms of carbon A mole (mol) represents a certain amount. Just like a dozen bagels represent 12 bagels. 1 mole = ________________________ particles (Avogadro’s Number) A particle can mean molecule, an atom, an ionic compound. 12 amu = 1 carbon atom 12 g of carbon = 1 mole of carbon atoms = __________________________ 6.02 x 10 23 6.02 x 10 23 atoms of carbon

II. Gram Formula Mass 1 mole Cl = Gram Atomic Mass (GAM) – mass of 1 mole of atoms in an element Example: Cl: Subscripts – represent the number of moles of each atom Examples: Al2O3 – _______ moles of Al ions, ______ moles of O ions   Ca(NO3)2 - ______ mole of Ca ions, ______ moles of N atoms, ______ moles of O atoms 1 mole Cl = 35 g of Cl (round to the nearest whole number) 2 3 1 2 6

II. Gram Formula Mass Mass of 1 mole of a compound Gram Formula Mass (GFM) – To find the GFM you add the masses of all of the elements in the compound or molecule. Examples: H2O 3. Al2(SO4)3 2. K2CO3 4. Zn3(PO4)2 • 4H2O Mass of 1 mole of a compound 2 (1g) + 1 (16g) = 18 g 2 (27g) + 3 (32g) + 12(16 g) = 342 g 2 (39 g) + 1 (12 g) + 3(16 g) = 138 g 3 (65g) + 2 (31g) + 8(16 g) + 8 (1g) + 4 (16 g) = 457 g

III. Grams   Moles (1 Step Mole Problems) Moles  Grams:   Example: How many grams are present in 40.5 mol of sulfuric acid (H2SO4)? Example: What is the total mass of 0.75 mole of SO2? STEP 1: H2SO4 = 2 (1g) + 1 (32g) + 4 (16 g) = 98 g STEP 2: 40.5 mol H2SO4 x 98 g H2SO4 = 3970 g 1 mol H2SO4 48 g SO2

III. Grams   Moles (1 Step Mole Problems) Example: How many moles are equivalent to 4.75 g of sodium hydroxide (NaOH)? Example: How many moles are equivalent to 39 g of LiF? STEP 1: NaOH = 1 (23g) + 1 (16g) + 1(1g) = 40 g NaOH STEP 2: 4.75 g NaOH x 1mol NaOH = 0.119 mol NaOH 40 g NaOH 1.5 mol LiF

IV. Mole  Volume(1 Step Mole Problems) 1 mole of any gas is equal to ___________ (at standard pressure and temperature - STP) 1 mole = _____________________________ = _______________ of any gas at STP Example: Find the number of moles of 2.35 L of chlorine gas. Example: Find the amount of liters of 3.5 moles of hydrogen gas. 22. 4 L 22.4 L 6.02 x 10 23 particles 2.35 L Cl2 x 1 mol Cl2= 0.105 mol Cl2 22. 4 L Cl2 3.5 mol H2 x 22.4 L H2= 78 L H2 1 mol H2

V. Mole  Particles (1 Step Mole Problems) 1 mole of any substance is equal to _____________________ particles (Avogadro’s Number) Example: How many molecules of water are there in 3.5 moles of NH3? Example: How many moles of H2O are there if there are 1.204 x 10 34 molecules? 6.02 x 10 23 6.02 x 10 23 molecules NH3 = 2.1 x 10 24 molecules NH3 3.5 mol NH3 x 1 mol NH3 1.204 x 10 34 H2O x 1 mol H2O__________= 2.0 x 10 10 mol H2O 6.02 x 10 23 molecules H2O

VI. Avogadro’s Hypothesis Any 2 samples of gas at the same temperature, pressure, and volume have the same number of particles Examples: Which of the following would occupy the same volume as 2.5 mol of O2 at STP? (a) 1.5 mol of CO2 at STP (b) 3.5 mol of CH4 at STP (c) 3.0 mol of H2 at STP (d) 2.5 mol of Ne at STP (a) 33.6L of CO2 at STP (b) 78.4L of CH4 at STP (c) 67.2L of H2 at STP (d) 56.0L of Ne at STP

VII. Stoichiometry Applying math to chemical equations Reasons: 1. To find the perfect “recipe” of reactants, producing the most products 2. To predict the mass or volume of every reactant and product Coefficients: Tell you how many moles of the compound or element there are

VII. Stoichiometry Think about this: The “recipe” for bread is: 3 cups of flour + 1 eggs  4 loafs of bread 1) How many loafs of bread is produced if you use 6 cups of flour? 2) How many eggs are needed to produce 12 loafs of bread? 8 loafs 3 eggs

VII. Stoichiometry 2 2 1 2 1 2 1 mol O2 : 2 mol H2 2 mol H2 5 mol O2 x Mole – Mole Problems: Using moles means using coefficients Example 1: _____ H2 + ____ O2  ____ H2O How many moles of each of the substances? ____ H2 ____ O2 ____ H2O What is the mole ratio of O2 to H2? If I want to burn 5 moles of O2, how many moles of H2 gas is needed? 2 2 1 2 1 2 1 mol O2 : 2 mol H2 2 mol H2 5 mol O2 x = 10 mol H2 1 mol O2

VII. Stoichiometry 1 3 2 0.60 mol N2 x 2 mol NH3= 1.2 mol NH3 1 mol N2 Example 2: How many moles of ammonia (NH3) are produced when 0.60 mole of nitrogen reacts with hydrogen? (first balance) _____ N2 + ____ H2  ____ NH3 Example 3: How many moles of aluminum are needed to form 3.7 mol of Al2O3? (first balance) _____ Al (s) + ____ O2 (g)  ____ Al2O3 (s) 1 3 2 0.60 mol N2 x 2 mol NH3= 1.2 mol NH3 1 mol N2 4 3 2 3.7 mol Al2O3 x 4 mol Al = 7.4 mol Al 2 mol Al2O3

VIII. Percent Composition Think about this: How would you find the percent of boys in our class? # of boys (part)____ x 100% Total # of students (whole)

VIII. Percent Composition Equation: (See Ref Tabs) Table ________ Example 1: A 14.80 g sample contains 3.83 g of iron and 10.97 g of bromine. What is the percent composition of bromine? Represents the composition as a percentage of each element compared with the total mass of the compound T % Composition = mass of part * 100 by mass mass of whole % Composition = mass of part * 100 = 10.97 g Br____ x 100 = 74.12 % by mass mass of whole 14.80 g Sample

VIII. Percent Composition Example 2: Find the percent by mass of nitrogen in NH4NO3. Example 3: Which species contains the greatest percent by mass of hydrogen? A) OH– B) H2O C) H3O- D) H2O2 Step 1: Find the GFM: 2 (14 g) + 4 (1 g) + 3 (16 g) = 80 g Step 2: % Composition = mass of part * 100 = 28 g N x 100 = 35 % by mass mass of whole 80 g A) 1/17 = 6 % B) 2/18 = 11 % C) 3/19 = 16 % D) 2/66 = 3 %

IX. Percent Hydrate Think about this: A furry mole gets caught in a rain storm. His fur becomes wet and raises his mass to 125 g. He dries off and has a mass of 120 g. How would you find the percent of water that was on the wet mole? Difference in Masses of the Mole x 100% Total Mass of Wet Mole 5 g x 100% = 4% 125 g

IX. Percent Hydrate Hydrate: Anhydrate: The water is NOT bonded to the compound. It is attracted by a molecule-ion force of attraction. Anhydrates can be used to absorb moisture in packaged electronic equipment and clothing. You have probably seen packages of sodium silicate (an anhydrate) used when buying a pair of shoes. A compound that contains a specific amount of water molecules bound to its atoms A hydrate that has had the water driven off (evaporated)

IX. Percent Hydrate Example 1: What is the percentage, by mass, of water in sodium carbonate crystals, Na2CO3 • 5H2O? Step 1: Find the total GFM: 2(23g) + 1(12 g) + 3(16 g) + 10 (1g) + 5 (16 g) = 196 g Step 2: % Composition = Total mass of H2O* 100 = 5 (18 g) x 100 = 46% by mass GFM 196 g

IX. Percent Hydrate Example 2: What is the percentage, by mass, of water in barium chloride crystals, BaCl2 • 2H2O? Example 3: Find the percent of water of hydration for CuSO4 • 5H2O. % Composition = Total mass of H2O* 100 = 2 (18 g) x 100 = 15% by mass GFM 243 g % Composition = Total mass of H2O* 100 = 5 (18 g) x 100 = 36% by mass GFM 250 g

IX. Percent Hydrate Example 4: A hydrate has a mass of 45.3g. After being heated to a constant mass, the substance has a mass of 39.8g. What is the percent water of hydration?   Example 5: Given the lab data below, find the percent water of hydration. crucible 5.13 g crucible and hydrate 7.14 g crucible and anhydrate 6.29 g Step 1: Find mass of H2O = 45.3 g – 39.8 g = 5.5 g of H2O Step 2: % Composition = mass of H2O* 100 = 5.5 g x 100 = 12% by mass Total mass 45.3 g Step 1: Find mass of hydrate = 7.14 g – 5.13 g = 2.01 g of hydrate Step 2: Find mass of water = 7.14 g – 6.29 g = 0.85 g of water Step 3: % Composition = mass of H2O* 100 = 0.85 g x 100 = 42% by mass Total mass 2.01 g

IX. Percent Hydrate (Practice) Complete problems 4 and 6 from pg 15 (if you finish early, move on to problem 2 and 7)

X. Writing Empirical Formulas   – Examples: 1) BaCl2 2) CH4 What is the empirical formula for C8H18? Molecular Formula: Only molecular compounds have molecular formulas – Example: C6H12O6 the simplest ratio in which atoms (moles) combine to form a compound Ionic compounds are ALWAYS written as empirical formulas (most reduced) C4H9 the actual ratio of the atoms (moles) in a molecule (NO IONIC COMPOUNDS) Molecular compounds have only COVALENT BONDS

XI. Finding an Empirical Formula Given % Composition Given: 11.1% H and 88.9% O, find the empirical formula Steps 1. Convert % to grams (assuming 100 g sample) 2. Convert grams to moles (Divide by gam) 3. Divide the moles by the SMALLEST # of moles to get ratio 4. Write formula 11.1 % H  11.1 g H 88.8% O  88.9 g O 11.1 g H x 1 mol H = 11.1 mol H 1 g H 88.9 g O x 1 mol O = 5.5 mol O 16 g O 11.1 mol H = 2 mol H 5.56 5.56 mol O = 1 mol O 5.56 H2O

XI. Finding an Empirical Formula Given % Composition Examples: Find the empirical formulas for the following problems. 69.6 % barium, 6.1 % carbon, 24.3% oxygen 40.5% zinc, 19.9 % sulfur, 39.6% oxygen 0.508 mol Ba = 1 mol Ba 0.508 69.9 g Ba x 1 mol Ba = 0.508 mol Ba 137 g Ba 6.1 g C x 1 mol C = 0.508 mol C 12g C 0.508 mol C = 1 mol C 0.508 BaCO3 24.3 g O x 1 mol O = 1.52 mol O 16 g O 1.52 mol O = 3 mol O 0.508 0.623 mol Zn = 1 mol Zn 0.623 40.5 g Zn x 1 mol Zn = 0.623 mol Zn 65 g Zn 19.9 g S x 1 mol S = 0.622 mol S 65 g S 0.622 mol S = 1 mol S 0.623 ZnSO4 39.6 g O x 1 mol O = 2.48 mol O 16 g O 2.48 mol O = 4 mol O 0.623

XI. Finding an Empirical Formula Given % Composition If you don’t end up within a tenth of a whole number, multiply both numbers by 2 to get whole number ratio Example - A compound of zinc and phosphorous, when analyzed, showed 76.0% Zn and 24.0% P by mass. Calculate the simplest formula for the compound. 76.0 g Zn x 1 mol Zn = 1.17 mol Zn 65 g Zn 24.0 g P x 1 mol P = 0.774 mol P 31 g P Zn3P2 1.17 mol Zn= 1.5 mol Zn 0.774 * 2 = 3 Zn 0.774 mol P = 1 mol P 0.774 * 2 = 2 P

XII. Finding a Molecular Formula What is the molecular formula of a compound that has a GFM of 92 g and an empirical formula of NO2? Steps 1. Find the empirical formula Find the mass of the empirical formula   Find how many times larger the GFM is than the empirical formula and multiply the subscripts by that number. NO2 1 (14 g) + 2 (16 g) = 46 g 92 g/ 46 g = 2 times larger than the empirical formulas N2O4

XII. Finding a Molecular Formula Example: 1. A compound of phosphorus and oxygen, when analyzed, showed 39.24% P and 60.76% O by mass. Calculate the simplest formula for the compound. 2. Find the molecular formula if the GFM of this compound is 158g. 1.27 mol P = 1 mol P 1.27 39.24 g P x 1 mol P = 1.27 mol P 31 g P 60.76 g O x 1 mol O = 3.80 mol O 16 g O 3.80 mol O = 3 mol O 1.27 PO3 1 (31 g) + 3 (16 g) = 79 g 158 g/ 79 g = 2 times larger than the empirical formulas P2O6

Comparing Molecular vs. Empirical Formulas Empirical Formula: Summarize what an empirical formula is. Molecular Formula: Summarize what a molecular formula is. Simplest ratio which atoms combine to form a compound Actual ratio of atoms (moles) in a molecule

Molecular Formula vs. Covalent Formulas C2H2 Ca3N2 SCl3 C8H8 1. Identify the empirical formulas: ______________________________________________ 2. A) Identify the molecular formulas: ___________________________________________ B) What is the empirical formula of these substances? _______________________   C) Why would these substances have different chemical properties? Empirical formulas involve what types of bonding? _____________________________ 4. Molecular formulas involve what type of bonding? ______________________________ Ca3N2 SCl3 C2H2 C8H8 CH The molecular formula indicates how many atoms are bonded. The way and number of bonds affect the properties. Ionic and covalent Covalent only

Practice Problems: Show your work on the whiteboard and call Mr Practice Problems: Show your work on the whiteboard and call Mr. Mellon over once your group has completed the entire problem. Be sure to show all work for each problem. 1. A compound has an empirical formula Al2O3. A)Why does this compound not have a molecular formula? B) Calculate the percent composition of each element.   2. A substance has a formula of C8H18. A) What is the empirical formula of this substance? B) Using the empirical formula, calculate the percent composition of each element? 3. A substance contains 30.4% N and 69.6% O. A) Calculate the empirical formula of the compound (show all work). B) If the compound has a gram formula mass of 92 g, what is the molecular formula of the substance? 4. A hydrate consists of 19.17 % Na, 13.33 % S, and 67.50 % H2O. A) Calculate the empirical formula of the hydrate (show all work). B) Why does this hydrate not have a molecular formula?