Chapter 3 The Mole The Mole: The amount of a substance that contains the same number of “entities” as 12 grams of Carbon-12 (C-12) 1 mol C-12 = 6.022 x 1023 Carbon atoms 1 mol H20 = 6.022 x 1023 molecules H20 1 mol H20 = 2 mols H and 1 mol O (ratio never changes)

Chapter 3 The Mole Definitions The mole: the SI unit for amount of substance Specifies the # of objects in a fixed mass of substance Avogadro's #: 6.022 x 1023 All isotopes are considered to have the same atomic mass

The mass on the periodic table is an average of the The mass on the periodic table is an average of the masses of the naturally occurring isotopes The atomic mass of an element expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams

Example: 1 Fe atom has a mass of 55.85 amu and 1 mole of Fe has a mass of 55.85 g therefore 55.85 g of Fe contains 6.022 x 1023 atoms Same thing with compounds: 1 molecule of H2O has a mass of 18.02 amu and 1 mol of H2O (6.022 x 1023 molecules) has a mass of 18.02g so 1 mol = 6.022  1023 “things” = molar mass (g/mol)

Conversions 1. How many moles of Calcium phosphate are in 43.67 g of the compound? 3 Ca  40.08 = 120.24 2 P  30.97 = 61.94 8 O  16.00 = 128.00 310.18 g/mol 43.67 g Ca3(PO4)2  =.1408 mol Ca3(PO4)2

Conversions 2. How many molecules (formula units) of Calcium phosphate are in 43.67 g of the compound? 43.67 g Ca3(PO4)2 x 3. 33.64 g of ethanol would contain how many moles of Carbon in the molecule? 33.64 C2H5OH x 1 mol x 2 mol C 46.00 g C2H5OH 1 molC2H5OH

Percent Composition The percentage (by mass) an element contributes to the overall mass of the compound. (Mass of element X) x (number of atoms of X) x 100 Formula Mass of the compound *** The percent that an element contributes to the compound is fixed. This percentage can be used to find the number of grams of the element no matter the amount of the compund present.

Example: How many grams of carbon are in 1.75 kg of ethanol?

Empirical Formula The simplest whole number ratio of the atoms in a compound This ratio is found using moles Steps: 1. Convert grams (or %) to mole (dividing each element by their molar mass) 2. Divide all moles by the smallest amount of moles 3. Round to nearest whole #(if less than .2) 4. If mole value ends with....Multiply by what’s shown .5 (x2); .25 (x4); .75 (x4); .333 (x3)

Molecular Formulas Actual ratio of the atoms in a molecules The M.F. is a whole # multiple (n) of the empirical formula E.F. The multiple is found by this formula Multiple= Mass of Molecule Mass of EF

Molecular Formulas Samples What is molecular formula for lactic acid (ℳ= 90.08 g/mol) if by analysis the compound contains 40.00% C, 6.71% H, 52.29% O by mass? 40%= 40.00 g x 1 mol C = 3.331 mols C 3.331 = 1 C 12.01 g C 6.71%= 6.71g x 1 mol H = 6.64 mols H  3.331  2 H 1.01 g H 52.29%= 52.29 g x 1 mol 0 =3.331 mols O3.331= 1 O 16.00 g O

In the formula carbon goes first. If carbon is not in the compound, the first in the formula is the most metallic element. Therefore the empirical formula for this analysis is CH2O To get Molecular Formula (M.F.) you divide the molar mass of the analyzed one (lactic acid) by the molar mass of the empirical formula In this case the M.F. of lactic acid found by doing MF = n=3, EF where n is the whole # multiple mentioned earlier So multiply n, which is 3, by the subscripts of the empirical formula to get molecular: 3(CH2O) = C3H6O3

Example 2) Find molecular formula when a 1.00 g sample of Vitamin C (Mol. Mass= 176.17 g/mol) is combusted, the following data is obtained [remember that these are products after burning a hydrocarbon!] Mass CO2 before combustion= 83.85 g Mass CO2 after combustion= 85.35 g Mass H2O before combustion= 37.55 g Mass H2O after combustion= 37.96 g *find molecular formula, you need CxHyO z For CO2 = 85.35-83.85= 1.50 g CO2 X = .409 g C For H2O = 37.96-37.55= .41 g H2O x = .046 g H For O, since you don’t have before and after, just subtract the total of the other two from 1 gram, since it is total of the whole sample O = 1.00 g – .409 g C – .046 g H = .545 g O

.409 g C x 1 mol C = .0341 mol C ÷ .0341 → 1 x 3 = 3 12.01 g C .046 g H x 1 mol H = .0455 mol H ÷ .0341 → 1.333 x 3 ≈ 4 1.01 g H .545 g O x 1 mol O = .0341 mol O ÷ .0341 → 1 x 3 = 3 16.00g O Empirical form. is C3H4O3, mol. Mass= 88.07 g/mol N= MF= n=2; 2(C3H4O3)= C6H8O6 EF

Chemical Equations A symbol representation of the substances involved in a chemical reaction Shows microscopic (atomic level) and macroscopic (mole) ratios aA(aq) + bB(aq)  cC(s) + dD(aq) Requirement Atoms cannot be created or destroyed, only rearranged The # and types of atoms must be the same on both sides of the reaction

Chemical Equations Sample Write the balanced equation for the combustion of octane to produce carbon dioxide and water 2 C8H18 +25 O2  16 CO2 + 18 H2O

Write the balanced equation to show the reaction between Write the balanced equation to show the reaction between solutions of Barium Nitrate and Sodium Sulfate to produce a precipitate of Barium Sulfate and aqueous Sodium Nitrate Ba(NO3)2 + Na2SO4  BaSO4 + 2NaNO3

Stoichiometry: the study of the quantitative aspects of Stoichiometry: the study of the quantitative aspects of chemical formulas and reactions In a balanced chemical equation, all of the substances involved show a ratio of moles/molecules that is unique to that reaction The ratio comes from the coefficients of the materials involved (it’s not a mass ratio, it’s a particle ratio!!) 2H2 (g) + O2 (g) → 2H2O (g) 2 H2 : 1 O2 2 H2O : 2 H2 (reduces to a 1 : 1 ratio!)

Molarity: a measure of concentration Example: 2.50 M → 2.50 mol/L or 1 L/2.5 mols The process IN × Switch × OUT g A x = grams B (Coefficients) mL A x = mL B

The process IN × Switch × OUT kg A x = kg B (Coefficients) Sample: Copper (I) sulfide is heated in the presence of oxygen gas to produce solid copper (I) oxide and sulfur dioxide gas. How many Kg of O2 are required to form 2.8 Kg copper (I) oxide? 2Cu2S + 3O2  2Cu2O + 2SO2

Limiting Reactant Problems How many grams of nitrogen gas should be produced (H2O is also produced) if 100.0g of N2H4 and 200.0g of N2O4 are ignited? What is the % yield of N2 gas if 125g are actually produced? How much excess reactant remains? Equation 2N2H4 + N2O4  4H2O + 3N2 Which runs out first? 131.1 g N2 is the theoretical yield because it is the smallest amount from each reactant. N2H4 is limiting reaction. % yield = actual / theoretical x 100 = 95.3 % How much excess reactant is left? 100.0 g N2H4 = ? g N2O4 used 200.0 gN2O4 – 143.6 g N2O4 = 56.4 gN2O4 have – used = excess

Molar Dilutions Solute dissolved in solvent Concentration: the amount of solute dissolved in a given amount of solution – independent of the volume of solution Concentration = Moles Solute Amt. Solvent Molarity expresses concentrations in units of moles of solute per liter of solution A solution contains a fixed number of moles. When a portion of that solution is removed, a fraction of the original moles is removed. M = mol  mol1 = MxL L

Whan a new solution is prepared, the # of moles does not Whan a new solution is prepared, the # of moles does not change, but the molarity does. M2 x L2 = mol1 Since the moles are constant, we can set the two molarity equations equal to each other M1 x L1 = mol1 = M2 x L2 M1V1 = M2V2 Samples 1. How would you prepare 250.0 mL of 1.50 M NaCl from a stock solution of 13.0 M NaCl? (250.0 mL) x (1.50) = (13.0)(X) = 28.8 mL of 13.0 NaCl 250.0 mL – 28.8 mL = 221.2 mL H2O needed. Take 28.8 mL of 13.0 M NaCl and add 221.2 mL H2O 2. If 100.0 mL of 1.50 M Ca(OH)2 is diluted to 750.0 mL, how many mg/mL of OH- ions are present?