Chapter 3

1. Formulas

   A. Types of Formulas

   1. Simplest formula: gives simplest atom ratio –mole ratio

      K₂CrO₄ 2 atom K:1 atom Cr: 4 atom O

      2 mol K: 1 mol Cr: 4 mol O

      CH₂O 1 atom C: 2 atom H: 1 atom O

      1 mol C: 2 mol H: 1 mol O

   2. Molecular Formula: gives composition of molecule. May be identical with the simplest formula of an integral multiple of it.

   Formaldehyde: molecular formula CH₂O

   Glucose: molecular formula C₆H₁₂O₆

   See p. 64

   1. Mass percent from formula

      Percent composition of K₂CrO₄

      Molar mass = (78.20 + 52.00 + 64.00g/mol)

      = 194.20 g/mol

      %K = 78.20/194.20 x 100 = 40.27

      %Cr = 52.00/194.20 x 100 = 26.78

      %O = 64.00/194.20 x 100 = 32. 96

      Percents must add to 100.

   2. Simplest Formula from percent composition

      Find mass of each element in sample of compound. Then find numbers of moles of each element and finally the mole ratio.

      Simplest formula of compound containing 26.6%K, 35.4%Cr, and 38.0% 0?

      Work with 100. g sample: 26.6g K, 35.4g Cr, 38.0 g O

       

      No. moles K = 26.6g*mol/39.10g = 0.680 mol K

      No. moles Cr = 35.4g*mol/52.00g = 0.681 mol Cr

      No. moles O = 38.0g*mol/16.00g = 2.38 mol O

      Note that 2.38/0.680 = 3.50 = 7/2. Simplest formula K₂Cr₂O₇

   3. Simplest Formula From Analytical Data

      A sample of acetic acid (C,H, and O atoms weighing 1.000g burns to give 1.466 g CO₂ and 0.6001g H₂O). Simplest formula?

      Find mass of C in sample (from mass CO₂), then mass of H (from H₂O), and finally mass of oxygen by difference.

      Mass C = 1.466g CO₂*12.01gC/44.01gCO₂

      = 0.4001g C

      Mass H = 0.6001g H₂O*2.02gH/18.02gH₂O

      = 0.0673g H

      Mass O = 1.000g – 0.400g – 0.067g = 0.533g O

      No. moles C = .4001g*mol/12.01g = 0.0333mol C

      No. moles H = .0673g*mol/1.01g = 0.0666mol H

      No. moles O = 0.533g*mol/16.00g = 0.0333mol O

      Simplest formula is CH₂O, as with formaldehyde and glucose.

   4. Molecular Formula from simplest formula

Must know molar mass. For acetic acid,

MM = 60 g/mol.

Formula mass CH₂O = 30 amu; 60/30 =2

Molecular formula: C₂H₄O₂

III. Chemical Equations

1. Balancing Must have same number of atoms of each type on both sides. Achieve this by adjusting coefficients in front of formulas. Example: combustion of propane in air to give carbon dioxide and water.

   C₃H₈(g) + O₂(g) à CO₂(g)+ H₂O(g)

   Balance C: C₃H₈(g) + O₂(g) à 3CO₂(g)+ H₂O(g)

   Balance H: C₃H₈(g) + O₂(g) à 3CO₂(g)+ 4H₂O(g)

   Balance O: C₃H₈(g) + 5O₂(g) à 3CO₂(g)+ 4H₂O(g)

   Meaning: 1 mol C₃H₈ reacts with 5 mol O₂ to form 3 mol CO₂ and 4 mol H₂O

2. Mass relations in reactions
   1. Moles of CO₂ produced when 1.65 mol C₃H₈ burns?

      1.65 mol C₃H₈ * 3 mol CO₂/ mol C₃H₈ = 4.95 mol

   2. Moles H₂O form combustion of 20.0 g C₃H₈?

      1 mol C₃H₈ = 44.09 g C₃H₈

      20.0g C₃H₈* 1 mol C₃H₈/44.09g C₃H₈* 4 mol H₂O/1 mol C₃H₈

      =1.81 mol H₂O

   3. Mass of O₂ required to react with 12.0 g of C₃H₈?

12.0 g C₃H₈ * 1 mol C₃H₈ * 5 mol O₂ * 32.00g O₂ = 43.6 g O₂

44.09g C₃H₈  1 mol C₃H₈    1 mol O₂

1. Yield of Product in Reaction
   1. Limiting Reactant, theoretical yield, ordinarily, reactants are not present in the exact ratio required for reaction. Instead, one reactant I sin excess; some of it is left when the reaction is over. The other, limiting reactant, is completely converted to give the theoretical yield of product.

      To calculate the theoretical yield and identify the limiting reactant:

      1. Calculate the yield expected if the first reactant is limiting.
      2. Repeat this calculation for the second reactant.
      3. The theoretical yield is the smaller of these two quantities. The reactant that gives the smaller calculated yield is the limiting reactant.

2Ag(s) +I₂(s) à 2 AgI (s)

Calculate the theoretical yield of AgI and determine the limiting reactant starting with:

1. 1.00 mol Ag, 1.00 mol I₂

theor. Yield of AgI if Ag is limiting:

1.00 mol Ag*2 mol AgI/2 mol Ag = 1.00 mol AgI

theor. Yield of AgI if I₂ is limiting:

1. mol I₂*2 mol AgI/1 mol I₂ = 2.00 mol AgI

Theoretical yield = 1.00 mol AgI; Ag is limiting

b. Given: 1.00g Ag and 1.00g I₂

Find: What is the theoretical yield in grams?

1.00 g Ag  * 1 mol Ag *  2 mol AgI * 234.77g AgI = 2.18 g Ag

107.9 g Ag    2 mol Ag      1 mol AgI

1.00 g I₂  * 1 mol I₂ * 2 mol AgI * 234.77g AgI = 1.85 g AgI

253.8 g I₂   1 mol I₂    1 mol AgI

So I₂ is the limiting reagent.

c. Calculate the number of grams of excess reagents remaining at the end of the reaction [Assume 100% conversion].

2Ag(s) + I₂(s) à 2AgI(s)

use this one—1.00g 1.85g

1.00g I₂    * 1 mol I₂  * 2 mol Ag * 107.9g Ag = 0.850g Ag

253.8g I₂     1 mol I₂     1 mol Ag

1.00-0.850g = .150 g in excess

d. If 1.51 g of AgI is actually obtained, what is the % yield?

% yield = act. Yield *100% = 1.51 * 100% = 81.6% yield

theor. Yield = 1.85

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