4.1 Percent Composition

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1 Chapter 4 Page Percent Composition Thursday, September 22, :33 PM Percent composition of mass (percent composition) = the portion of the total mass contributed by each element. This is consistent no matter how much of a sample you have. However, this does not account for the ratio of atoms between the different elements. EXAMPLES If you have a kg Emerald, a precious variant of the mineral Beryl, and percent is Beryllium, what mass of the emerald is Beryllium? kg = 1000 g 1000 g % = 1000 g = g If you have g of Emerald and g is determined to be Oxygen, what is the percent composition of Oxygen for Emerald? g % = % g PROBLEMS Given our Emerald, what is the mass of Aluminum in a 400 g Emerald if the mass percentage of Aluminum is 10.04%?

2 Chapter 4 Page 2 If there is a 31.35% mass percentage of Silicon in Emeralds, what is the mass of Silicon in a 45 g sample? Given the information above, is there anything other than Aluminum, Beryllium, Oxygen and Silicon in Emeralds? Why?

3 4.2 Mole Quantities Thursday, September 22, :34 PM If we look at a simple compound like water we can determine that the mass percentages are as follows: Oxygen = 88% Hydrogen = 12% Total = 100% While, this tells us the complete composition of the compound, it does not tell us the formula of the compound, why? We are not accounting for differing relative atomic masses of the elements! Consider the following Element Percentage Rel. AM H O Do we have scales that measure in 'amu?' NO We need to express the mass of a collection of atoms, therefore we need a standard number of atoms to count. Atoms are so small that this number will have to be very large to equal to a significant mass. The solution is a unit called a mole. Mole = a collection of atoms, molecules, ions, or formula units equal to units. Why !? This number, called Avagadro's Number, is the number of atoms of 12 C needed to mass at 12 g (remember how we standardized relative atomic mass to this isotope). You use this to calculate the number of atoms in a sample. EXAMPLE In a mole of Water, H 2 O, how many atoms of hydrogen are present? Of oxygen? Chapter 4 Page 3

4 1 mole water = 1 mole of Oxygen and 2 moles of Hydrogen Therefore 1 mole oxygen = atoms of oxygen 2 moles hydrogen = 2 X = atoms of hydrogen Molar Mass = Since the definition of a mole is based off the basis of the relative atomic mass, a mole can also be extended to be used to define mass in grams. A mole of a substance has a mass in grams equal to its relative atomic mass in atomic mass units. EXAMPLE 1 mole of 12 C = 12 g (exact) 1 mole of hydrogen = g We can use this to determine ratios in formula units or chemical formula. EXAMPLE Take our example of Water. Say we have a 1 g of water this means we have 0.12 g Hydrogen and 0.88 g Oxygen. Let us calculate the moles of each. Moles of H = 0.12g H X (1 mole H/1.008g H) = 0.11 moles H Moles of O = 0.88g O X (1 mole O/16.00 g O) = moles O Moles H 0.11 moles H = = 2:1 hydrogen:oxygen = H 2 O Moles O moles O PROBLEM Given what we determined about the mass percentages of Emeralds last section, can you calculate the formula ratios? Beryllium = 5.04% Chapter 4 Page 4

5 section, can you calculate the formula ratios? Beryllium = 5.04% Aluminum = 10.04% Silicon = 31.35% Oxygen = 53.56% Chapter 4 Page 5

6 4.3 Determining Empirical and Molecular Formula Thursday, September 22, :34 PM Empirical Formula : The simplest ratio of atoms in a compound EXAMPLE Benzene (C 6 H 6 ) and Acetylene (C 2 H 2 ) have the same empirical formula: CH Molecular formula: The actual number of atoms for each element in a molecule. EXAMPLE In the above examples C 6 H 6 and C 2 H 2 are the molecular formulas. Sometimes the Molecular Formula and the Empirical Formula are the same (such as with H 2 0 or SO 3 ), but not always. Remember that Ionic Compounds DO NOT have Molecular Formulas, the Empirical Formula will be equal to the Formula Unit (see Chapter 3) PROBLEMS Determine the Empirical Formula for each of the following. CaCl = C 2 H 4 (OH) 2 = CH 2 Cl 2 = H 2 C 2 O 4 = N 2 H 4 = Chapter 4 Page 6

7 N 2 H 4 = This allows us to go back to our problem from last time and see if you remember how to calculate an the Empirical Formula from the Percent Composition. PROBLEM Substance: Jadeite Percent Composition Na = 11.37% Al = 13.35% Si = 27.79% O = 47.49% Substance: Potassium Persulfate Percent Composition K = 28.93% S = 23.72% O = 47.35% Determine the Empirical Formula? If a molar mass is g/mol, what is the molecular formula? Chapter 4 Page 7

8 Chapter 4 Page Chemical Composition of Solutions Thursday, September 22, :34 PM A Solution is a solute dissolved in a solvent. The Concentration of a solution is the relative amount of solute in solvent. Solutions can be Dilute (low solute to solvent) or Concentrated (high solute to solvent) Ways to Quantify Concentration 1. Percent Mass (accurate by mass but lacks in specificity of number of molecules) Mass of Solute % Mass = % Mass of Solution 2. Molarity (reflects the molar amount of solute present) Moles of Solute Molarity (M) = Liters of Solution (Note: NOT liters of Solvent) EXAMPLES 1. If 34g of NaCl is dissolved into water, and the total solution weighs 245 g, what is the Percent Mass of the solution for NaCl? % Mass of NaCl = (34 g NaCl / 245 g Solution) X 100% = 14%

9 Chapter 4 Page 9 2. If aspirin ( C 9 H 8 O 4 ) is dissolved to a M solution, how much aspirin must be used per liter of solution? Aspirin = g/mol moles aspirin g Aspirin g Aspirin = M aspirin = = 22.5 g 1 liter solution mole aspirin PROBLEM If we dissolved 34 g BaCl 2 into water and the total solution volume is 1.45 l, what is the molarity of the solution for Ba + and Cl -? Dilution Moles conc = M conc X V conc of solution Relation of moles to molarity and volume Dilution is adding solvent so it does not change the number of moles of solute.

10 Chapter 4 Page 10 Therefore, Moles conc = Moles dil Since Moles dil = M dil X V dil ==> M conc X V conc = M dil X V dil Simple algebra rearranges this equation to: M conc X V conc M dil = V dil PROBLEMS 1. If a 0.23M solution is dissolved from 1.0 l to 1.5 l, what will be the new concentration? 2. If I want a 1.5 liter 0.95M solution, but want to add 745ml of water to do this, what must the concentration of the original solution be?

11 Chapter 4 Page 11 Review Problems Friday, September 30, :53 AM Here are a set of review problems for the next exam. I am going to post these first and then later tonight I will post the answers for your use. Name these Compounds 1. NH4IO4 : 2. K2CrO4 : 3. B2O5 : 4. H2Te : 5. Rb2SeO3 : Write the Empirical Formula (for Ionic) or Molecular Formula (for Molecular) 1. Cobalt (III) Cyanide : 2. Silver Chlorate : 3. Beryllium Permanganate : 4. Lead (IV) Oxalate : 5. Dinitrogen Hexaphospide : Determine the Empirical Formulas 1. C4H3OH :

12 Chapter 4 Page N2H4O2 : 3. K2SO4: 4. N2(OH)2 : Determine the Empirical Formula from Percent Composition 1. Substance: Caffeine Carbon : 49.48% Hydrogen : 5.18% Nitrogen: 28.86% Oxygen: 16.48% Given a molar mass of g/mol, determine the Molecular formula. Determine Concentrations 1. What is the Molarity of a solution where you add g of Potassium Permanganate to a 2.8 l total volume?

13 2. Given an original volume of 1.3 l at 1.3 M, what volume would be needed for a 0.8 M solution? Chapter 4 Page 13

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