By, Rachel Smith and Matt Rose

1. Chapter 3: Stoichiometry By, Rachel Smith and Matt Rose

2. Basics • It is used to find the quantities of materials consumed and produced in chemical reactions • Average mass: total mass/amount • Once you have the average mass you can use it to find the estimate of how much of something you will need to meet a certain weight or amount • Example: The average weight of jelly beans is 5.0 grams so how many beans will you have to weigh to have 1000?

3. Atomic Masses • The most accurate method for comparing masses of atoms involves the use of the mass spectrometer • Atoms are passed into a beam of high-speed electrons which knock electrons off the atoms being analyzed and change them into positive ions • Example: When 12C and 13C are analyzed in a mass spectrometer the ratio of their masses are mass 13C/mass 12C=1.0836129 • Mass of 13C=(1.0836129)(12 amu)= 13.003355 amu • The mass of 12C is exactly 12 atomic mass units

4. Average Atomic Mass • It is known that natural carbon is composed of 98.89% 12C atoms and 1.11% 13C atoms • Average atomic mass= 98.89% of 12 amu+ 1.11% of 13.003355 amu • (0.9889)(12amu) + (.0111)(13.003355amu)= 12.01amu • 12 amu is the atomic mass for carbon

5. The Mole • Avogradro’s number- one mole of something consists of 6.022x10^23 units of that substance • In exactly 12 grams of 12C mean that 12 grams of 12C contains 6.022x10^23 atoms • Example: How many moles are there in 43.26 grams of carbon? • (43.26g)x(1mol/12.01g)= 3.60 moles of carbon in 43.26 grams

6. Molar Mass • Mass of 1 mole of methane: CH4 • Mass of 1 mol C= 12.01g • Mass of 4 mol H= 4x1.008g • Mass of 1 mol CH4= 16.04g • 16.04g represents 1 mole of methane molecules which is the molar mass • Molar Mass- the mass in grams of one mole of the compound

7. Percent Composition of Compounds • Can be found by comparing the mass of each element present in 1 mole of the compound to the total mass of 1 mole compound • Example: Mass percent of carbon in ethanol • (mass of C in 1 mol C2H5OH/mas of 1 mol C2H5OH) x 100% • (24.02g/46.07g) x 100% = 52.14%

8. Section 3.7 Determining the Formula of a Compound Solving for Empirical Formula Determine mass percentage, and base that on 100 grams From the amount of each type of element in the 100 grams determine moles of a given element by dividing by the atomic mass Divide each amount of moles by the smallest mole number present If each of the resulting numbers is whole, then that is your empirical compound If even a single whole number is not whole, you must multiply by an integer which will make all the number whole, the resulting numbers then make you empirical formula

9. Section 3.7 Determining the Formula of a Compound Determining Molecular Formula from Empirical Formula • Obtain Empirical Formula • Calculate mass of Empirical Formula • Calculate Ration of Molar Mass over Empirical Formula

10. Section 3.7 Determining the Formula of a Compound Empirical Formula Mass 1 x 12.01 g C= 12.01 g C 5 x 1.008 g H= 5.040 g H 1 x 14.01 g N= 14.01 g N ----------------------------------------------------- Empirical Formula Mass = 36.06 g 72.12/36.06 = 2 2 x C₁H₅N₁  Molecular Formula C₂H₁₀N₂ • Ex. A compound is found to be 38.67% Carbon, 16.22% Hydrogen and 45.11% Nitrogen. The molecular mass is found to be 72.12 g, determine Molecular formula 100g x .3876 = 38.67g C 38.67g C x (1 mole C/12.01 g C) = 3.220 mol C 3.220/3.220 = 1 100g x .1622 = 16.22g H 16.22g H x (1 mole H/1.008g H) = 16.09 mol H 16.09/3.220 = 4.997 5 100g x .4511 =45.11g N 45.11g N x (1 mole N/14.01g N) = 3.220 mol N • 3.220/3.220 = 1 • Empirical Formula: C₁H₅N₁

11. Section 3.8 Chemical Equations Chemical Equations • Chemical Equations have 2 side • Left side are the reactants, right side are the products • Important to remember that in a reaction atoms are never created or destroyed so the same amount of atoms should be found on each side of the equation • Chemical Equations tell us 2 important things • The amount of each given atom • The state of each atom • (g)-gas • (l)-liquid • (s)-solid • (aq)-aqueous solution Ex. 2C₂H₆(l) + 7O₂(g)  4CO₂(g) + 6H₂O(g) 4 Carbon 4 Carbon 12 Hydrogen 12 Hydrogen 14 Oxygen 14 Oxygen

12. Section 3.9 Chemical Balancing Equations Balancing an Equation • Done mostly through trial and error • The main objective is to always have the same number of atoms on both side • In a chemical equation the formulas may no be changed in any manner, only amount may be changed Writing and Balancing Equations • Determine what type of reaction is occurring, reactants, products and physical states • Create a summarizing unbalanced equation • Balance equation by inspection and trial and error • Determine coefficients • Don’t change the identity of any atoms or formulas

13. Section 3.10 Stoichiometric Equations • Chemical equations determine the number of atoms and compounds by moles • However in application settings all amounts of materials are determined by mass • Stoichiometry • The branch of chemistry concerned with the proportions in which elements are combined in compounds and the quantitative relationships between reactants and products in chemical reactions¹ Solving for Stoichiometric Equations • Balance the equation • Convert masses into moles • Set up the appropriate mole rations • Convert back to grams

14. Section 3.10 Stoichiometric Equations Ex. What amount of oxygen would it take to react fully with 96.1 grams of Propane? 91.6g C₃H₈(g) + 5O₂(g)3CO₂(g) + 4H₂O(g) 96.1 grams C₃H₈ · · · = 349 g

15. Section 3.11 The Concept of Limiting Reagent Stoichiometric mixture-contain relative amounts of that reactants that match the numbers in the equation • When doing a stoichiometric equation you must consider the possibility that not all the reactants will be used because the amount of one may be limited Limiting Reagent-The reactant which runs out first and thus limits the amount of product that can be formed • How to fund limiting Reagent • Find moles • Find mole ratios • Find the reactants which does not satisfy the equation

16. Section 3.11 The Concept of Limiting Reagent Ex. In a reaction with between methane and oxygen, 67.02 grams of Methane and 103.45 grams of oxygen gas are present. Which is the limiting reagent CH₄(g) + 2O₂(g)  CO₂(g) + 2H₂O(g) 103.45g · = 3.233 molO₂ 3.222 mol · = 1.616 mol of CH₄ to react 67.02g · = 4.176 mol CH₄ 4.176 mol CH₄ · = 8.352 mol O₂ to react O₂ is the limiting reagent