Unit 4 Lecture 1- Balancing Eqns and the Mole Stoichiometry Balancing Equations The Mole
Chemical Equations Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) special information goes here: Δ (heat), catalyst, spark, etc. Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) reactants products the arrow in the equation means: “produce”, “yield”, “react to form” Letters in parentheses show the physical state: (s) solid, (l) liquid, (g) gas, (aq) aqueous (in water) Correctly written formulas for all reactants and products are a “must.” Coefficients give the relative amounts of reactants used and products formed. Do NOT balance equations by changing the subscripts!!!
Balancing Chemical Equations Gaseous nitrogen and hydrogen react to form ammonia: Write the correct formulas for the reactants and products. N2(g) + H2(g) NH3(g) Coefficients (numbers in front of the formulas) are used to balance the numbers of atoms of each element. 2 atoms N 2 atoms H 1 atom N, 3 atoms H N2(g) + H2(g) 2NH3(g) 2 atoms N 2 atoms H 2 atoms N, 6 atoms H N2(g) + 3H2(g) 2NH3(g) 2 atoms N 6 atoms H 2 atoms N, 6 atoms H The equation is now balanced. The Law of Conservation of Mass has been satisfied.
Balancing Chemical Equations Tip: Balance first the elements which appear least often. NaOH(aq) + H3PO4(aq) Na3PO4(aq) + H2O(l) sodium hydroxide phosphoric acid sodium phosphate Balance Na or P first. 3NaOH(aq) + H3PO4(aq) Na3PO4(aq) + H2O(l) 3NaOH(aq) + (1)H3PO4(aq) (1)Na3PO4(aq) + H2O(l) 6 atoms of H, 7 atoms of O 2 atoms of H, 5 atoms of O 3NaOH(aq) + (1)H3PO4(aq) (1)Na3PO4(aq) + 3H2O(l) Balanced equation: 3NaOH(aq) + H3PO4(aq) Na3PO4(aq) + 3H2O(l)
Balancing Chemical Equations CaC2 (s) + H2O(l) Ca(OH)2 (aq) + C2H2 (g) calcium carbide calcium hydroxide acetylene CaC2 (s) + 2 H2O(l) Ca(OH)2 (aq) + C2H2 (g) PCl3(g) + H2O(l) H3PO3 (aq) + HCl (aq) phosphorus trichloride phosphorous acid hydrochloric acid PCl3 (g) + 3 H2O(l) H3PO3 (aq) + 3 HCl (aq) H2S(g) + Fe(OH)3 (s) Fe2S3 (s) + H2O(l) hydrogen iron(III) iron(III) sulfide hydroxide sulfide 3 H2S(g) + 2 Fe(OH)3(s) Fe2S3(s) + 6 H2O(l)
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) Stoichiometry The application of the laws of definite proportions and the conservation of mass to chemical reactions is called stoichiometry. Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) Applications: Computing the formula mass or molar mass of a compound from its chemical formula Computing the % composition of a compound from its chemical formula Computing the amounts of reactants required and products formed (the yield) for a specific chemical reaction
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) Stoichiometry and the Mole Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) 2 molecules 2 molecules 1 formula unit 1 formula unit In the laboratory, it is virtually impossible to work with single molecules or formula units. Instead, we generally work with amounts of chemicals large enough to see and manipulate. A MOLE is a number (Avogadro’s number) large enough to give appreciable amounts of material. A MOLE is 6.022 x 1023 of anything…anything!!! The mass of a mole of any element is its atomic mass in grams. 1 mole of magnesium has a mass of 24.305 g (see periodic table). 1 atom of magnesium has a mass of 24.305 amu.
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) Stoichiometry and the Mole Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) 1 mole 2 moles 1 mole 2 moles One way to tell if we have a mole of a chemical is to weigh it. The mass of a mole (aka the molar mass) of any element is its atomic mass in grams. The mass of a mole (aka the molar mass) of any compound is its formula mass in grams.
Stoichiometry and the Mole Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) 1 mole 2 moles 1 mole 2 moles 1 formula unit of Mg(OH)2 has: 1 ion Mg2+ 24.3050 amu 2 atoms O 2 x 15.9994 = 31.9988 amu 2 atoms H 2 x 1.00794 = 2.01588 amu formula mass 58.3197 amu
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) Stoichiometry and the Mole Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) 1 mole 2 moles 1 mole 2 moles 1 mole of Mg(OH)2 has: 1 mole of Mg2+ ions 24.3050 g 2 moles of O atoms 2 x 15.9994 = 31.9988 g 2 moles of H atoms 2 x 1.00794 = 2.01588 g molar mass 58.3197 g
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) Stoichiometry and the Mole Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) 1 mole 2 moles 1 mole 2 moles 58.320 g 2 x 36.461 g = 72.922 g 95.211 g 2 x 18.015 g = 36.030 g Balanced chemical equations tell us the proportions in which chemicals react. These are generally in MOLES. One way to measure moles in the lab is to weigh the chemical. Molar masses: Mg(OH)2 58.320 g HCl 36.461 g MgCl2 95.211 g H2O 18.015 g
Converting from Mass to Moles Once we know the molar mass of a compound, we can calculate the number of moles present in ANY MASS of that compound. Simply treat the molar mass as a conversion factor. 1 mol Mg(OH)2 = 58.320 g How many moles of Mg(OH)2 are in 29.16 g of the compound? 29.16 g x 1 mol Mg(OH)2 = 0.5000 mol Mg(OH)2 58.320 g How many moles of Mg(OH)2 are in 4.07 x 10-3 g of the compound? 4.07 x 10-3 g x 1 mol Mg(OH)2 = 6.98 x 10-5 mol Mg(OH)2
Converting from Moles to Mass If we can convert from mass to moles, we can also convert from moles to mass. What is the mass in grams of 1.5 moles of Mg(OH)2 ? 1.5 mol Mg(OH)2 x 58.320 g = 87 g 1 mol Mg(OH)2
Unit 4 Lecture 1- Balancing Eqns and the Mole Examples 1. What is the mass in grams of 0.250 mol of MgCl2? 2. How many moles of NH4Cl are in 76.5 g? 3. How many moles of nitrate ions are present in 2.5 mol of Al(NO3)3? 4. How many grams of nitrate ions are present in 10.0 g of Al(NO3)3?
Converting from Moles to Numbers of Formula Units, Molecules, or Atoms Moles can be converted into numbers of formula units (for ionic compounds) OR numbers of molecules OR numbers of atoms. How many formula units of Mg(OH)2 are in 1.5 moles of the compound? 1.5 mol Mg(OH)2 x 6.022 x 1023 formula units = 9.0 x 1023 formula units 1 mol Mg(OH)2 How many hydroxide ions (OH-) are in 1.5 moles of Mg(OH)2? 1.5 mol Mg(OH)2 x 2 mol OH- ions x 6.022 x 1023 ions = 1.8 x 1024 ions 1 mol Mg(OH)2 1 mol OH- ions
The Map Number of things Moles Mass conversion factor: 1 mol = molar mass conversion factor: 1 mol = 6.022 x 1023
Examples 1. How many molecules of HCOOH are present in 0.0772 mol of the compound? 2. Calculate the mass of 6.626 x 1026 formula units of NH4Cl.