1-Stoichiometry of Chemical Reactions-I.pdf

Stoichiometry of Chemical Reactions I Dr. Alya A. Arabi College of Medicine and Health Sciences UAEU Stoichiometry Example of stoichiometry in Medicine: Drug Dosage Calculation How much of the active drug in the tablet do I need for an effective treatment? What is the appropriate dosage of a medi...

Stoichiometry of Chemical Reactions I Dr. Alya A. Arabi College of Medicine and Health Sciences UAEU Stoichiometry Example of stoichiometry in Medicine: Drug Dosage Calculation How much of the active drug in the tablet do I need for an effective treatment? What is the appropriate dosage of a medication given a certain weight for the patient? We need the proper amount to achieve the desired therapeutic effect while minimizing potential side effects or toxicity. What ratio of reactants is crucial to produce a consistent and effective drug formulations? Stoichiometry Stoichiometry is… • Greek for “measuring elements” • Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. • Stoichiometry is based on the principle of exchange of matter. Stoichiometry Molar Mass or molecular weight • By definition, these are the mass of 1 mol of a substance (i.e., g/mol). – The molar mass of an element is the mass number for the element that we find on the periodic table – The formula weight (in amu’s) will be the same number as the molar mass (in g/mol). Water Ammonium chromate Stoichiometry Barium nitrate Molecular Weight (MW) • Round the number you find to the nearest tenth – Hydrogen is 1.00797 → 1.0 g/mol Weight/Mole is an INTENSIVE property doesn’t depend on amount Stoichiometry Practice • CaCl2 Ca: 1 x (40.1 amu) + Cl: 2 x (35.5 amu) 111.1 amu or g/mol amu stands for atomic mass unit Stoichiometry Try these MW’s? • Ca 1 mole of Ca weighs 40.1 grams g 40.1 mol • H2 2 (1.0) = 2.0 g mol 1 mole of H2 weighs 2.0 grams • BaF2 [137.3 + 2(19.0)] = 175.3 g mol Practice Problem: (NH4)3PO4 ??? Stoichiometry Balanced reactions A chemical equation of a chemical reaction includes: ✓ the molecular formulae of reactants (starting materials) and products (resulting substance), ✓ the phases (solid, liquid, gas, aqueous) (sometimes it is omitted), and ✓ the proportion (stoichiometry, coefficients). Stoichiometry Jynto Robert A. Rohde Jacek FH Jynto and one more author - Methane-3D-space-filling.svg Oxygen Molecule VdW.png Carbon-dioxide-3D-vdW.svg Water molecule 3D.svg Examples 2H2 + O2 → 2H2O 2Al2O3 → coefficients 4Al + 3O2 2Na + 2H2O → 2NaOH + H2 Reactants → Products Stoichiometry • The coefficients tell us the proportions of each substance 2Al2O3 4Al + 3O2 2Na + 2H2O 2NaOH + H2 Stoichiometry Rules in a balanced equation: 1. Conservation of mass, mass cannot be created or destroyed. Therefore, The number of elements should be the same in both sides. 2. Net charges must be equal in both sides 3. The ratio of moles/coefficient (not individual masses, not volumes, not moles alone!) are equal Stoichiometry Rules in a balanced equation: 1. Conservation of mass, mass cannot be created or destroyed. Therefore, The number of elements should be the same on both sides. Stoichiometry https://in.pinterest.com/pin/444871269413992687/ Rules in a balanced equation: 1. Conservation of mass, mass cannot be created or destroyed. Therefore, The number of elements should be the same on both sides. 2H2 + 1O2 → 2H2O (2 mol X 2.02 g/mol H2) + (1 mol X 32.00 g/mol O2 ) = (2 mol X 18.02 g/mol H2O) 36.04 g TOTAL mass of reactans = 36.04 g TOTAL mass of products Note: in conservation of mass, we refer to the TOTAL masses (and NOT the mass of individual chemicals) e.g. mass of H2 alone (4.04g) ≠ mass of O2 alone (32.00 g) ≠ massStoichiometry of H2O alone (36.04 g) Rules in a balanced equation: 2. Net charges must be equal on both sides H3PO4 (aq) + H2O(l) --> H3O+ + H2PO4- Stoichiometry Rules in a balanced equation: 1. Conservation of mass, mass cannot be created or destroyed. Therefore, The number of elements should be the same on both sides. 2. Net charges must be equal on both sides 3. The ratio of moles/coefficient (not individual masses, not volumes!) are equal Stoichiometry Using Moles Moles provide a bridge from the molecular scale to the real-world scale Stoichiometry Steps to Calculate Stoichiometric Problems ✓ Correctly balance the equation. ✓ Convert the given amount to moles using n=m/M (to convert from mass in g) or PV=nRT (to convert from volume in L). ✓ Set up mole ratios. ✓ Use mole ratios to calculate moles of desired chemical. ✓ Convert moles back into final unit using n=m/M (to convert to mass in g) or pv=nRT (to Stoichiometry convert to volume in L). Practice How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2Al2O3 4Al + 3O2 𝑚𝑜𝑙 𝐴𝑙2𝑂3 𝑚𝑜𝑙 𝑂2 = 2 3 3.34 𝑚𝑜𝑙 𝑂2 = 2 3 mol O2 = 5.01 mol Stoichiometry Practice How many moles of Al are produced when 430 g of Al2O3 decompose? 430 g of Al2O3 means 4.22 moles of Al2O3 2Al2O3 4Al + 3O2 𝑚𝑜𝑙 𝐴𝑙2𝑂3 𝑚𝑜𝑙 𝐴𝑙 = 2 4 4.22 𝑚𝑜𝑙 𝐴𝑙 = 2 4 mol Al = 8.44 mol Stoichiometry Practice How many moles of Al are produced when 6.11 moles of O2 are produced? 2Al2O3 4Al + 3O2 𝑚𝑜𝑙 𝐴𝑙 𝑚𝑜𝑙 𝑂2 = 4 3 𝑚𝑜𝑙 𝐴𝑙 6.11 = 4 3 mol Al = 8.15 mol Stoichiometry General plan for stoichiometry calculations Mass reactant Mass product Stoichiometric factor Moles Reactant/coefficient Moles Product/coefficient Stoichiometry Practice 2C2H2 + 5 O2 4CO2 + 2 H2O • How many moles of C2H2 are needed to produce 8.95 mole of H2O? (8.95 mol) • If 2.47 moles of C2H2 are oxidized, how many moles of CO2 are formed? Mol C2H2/2 = mol CO2/4 2.47/2=mol CO2/4 (4.94 mol) Stoichiometry Practice 2C2H2 + 5 O2 4CO2 + 2 H2O • If 3.84 moles of C2H2 are oxidized, how many moles of O2 are needed for oxidation of C2H2? What is the volume of O2 required? First solve for moles of O2 (9.6 mol) Then use PV =nRT to convert moles to L R=8.21×10−5 m3⋅atm⋅K−1⋅mol−1 p = 1 atm T = 25 oC = 298.15 K n = 9.6 mol Solve for V: V = nRT/P = 2.35 x 10-1 m3 UNITS!! Stoichiometry PV = nRT n= PV RT If P and T are constant, and R is already a constant Then n = constant x V In other words, n and V are directly proportional Stoichiometry Shortcut for Volume-Volume? • How many liters of CH4 at STP are required to completely react with 17.5 L of O2? CH4 + 2O2 CO2 + 2H2O 17.5 L O2 1 L CH4 2 L O2 = 8.75 L CH4 Note: This shortcut only works for VolumeVolume problems when Temperature and Stoichiometry Pressure are kept constant. Why?