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Unit 4 – The Mole Honors Chemistry Part 1

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WHAT IS A MOLE? 6.02 x 1023

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Mole Facts 6.02 X 1023 Pennies: Would make at least 7 stacks that would reach the moon. 6.02 X 1023 Watermelon Seeds: Would be found inside a melon slightly larger than the moon. 6.02 X 1023 Blood Cells: Would be more than the total number of blood cells found in every human on earth. 1 Liter bottle of Water contains 55.5 moles H20

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Definition of Mole The amount of atoms in 12.0 grams of Carbon 12 (6.02 x 1023 atoms known as Avogadro’s number). A sample of any element with a mass equal to that element's atomic weight (in grams) will contain precisely one mole of atoms (6.02 x atoms).

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**The sum of the masses of all the atoms in a molecule of a substance**

Molecular Mass The sum of the masses of all the atoms in a molecule of a substance CaCO3 1 atom of Ca = amu 1 atom of C = amu 3 atoms of O = amu amu } Add these

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**Formula Mass (Molar Mass)**

The mass of 1 mole (in grams) Equal to average atomic mass but the unit is grams 1 mole of C atoms = g 1 mole of Na atoms = g 1 mole of Cu atoms = g

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**Example Problem Find the mass of 1 mole of KAl(SO4)2 ● 12H2O**

2 (SO4) 2( ((16.00 x 4))=192.12 12 H2O 12( ) =216.24 Mass of 1 mole = g/mol

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**Try These: Find the molecular mass for these :**

HNO3 CO2 Find the molar mass for these compounds: C6H10O5 H2SO4

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The Mole 1 mole of gas always contains 6.02 x molecules of that gas 1 mole Cl2 gas = 6.02 x 1023 molecules of Cl2 1 mole NO2 gas = 6.02 x 1023molecules of NO2 1 mole CO gas = 6.02 x 1023 molecules of CO 1 mole CO2 gas = 6.02 x 1023molecules of CO2

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The Mole Also applies to other particles! (not only molecules in a gas) 1 mole C = x 1023 C atoms 1 mole H2O = x 1023 H2O molecules 1 mole NaCl = x 1023 NaCl formula units 1 mole of Na+ = 6.02 x 1023 Na+ ions 1 mole of Cl- = x 1023 Cl– ions

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**Avogadro’s Number We can use Avogadro’s # as a conversion factor:**

1 mole 6.02 x 1023 particles Or Note that a particle could be an atom, molecule, formula unit, or ion !

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**Example Problems How many molecules are in 3.5 moles of H2O?**

How many moles are present in molecules of NO2?

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**Mass and Mole Relationship**

1 mole of any substance = the molar mass of that substance (in grams) Find the number of moles present in 56.7 g of HNO3. 56.7 g HNO mole HNO3 63.01 g HNO3

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Example Problems Find the number of grams present in 4.5 moles of C6H10O5. Find the number of moles present in g of H2SO4. How many molecules are in 4.5 grams of NaCl?

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**Gas Volumes and Molar Mass**

Avogadro’s Law Equal volumes of gases under the same conditions of temperature and pressure contain equal numbers of molecules 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters Standard temperature: 0ºC or 273K Standard pressure: 1 atm or kPa

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**Gas Volumes and Molar Mass**

32.00 g O2 = 1 mole = 22.4 L 2.02 g H2 = 1 mole = 22.4 L 44.01 g CO2 = 1 mole = 22.4 L

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**Example Problems How many liters are present in 5.9 moles of O2?**

How many liters are present in 3.67 moles of CO2? How many atoms of O are present in 78.1 g of O2?

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**Total mass of the element in the compound**

Percent Composition Finding what percent of the total weight of a compound is made up of a particular element Formula for calculating % composition: Total mass of the element in the compound Total formula mass X 100

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Example Problem: Calculate the % composition of calcium in Ca(OH)2.

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Example Problems Find the percentage composition of a compound that contains 1.45 g of carbon, g of sulfur, and 1.00 g of hydrogen in a g sample. 21.7% carbon 63.3% sulfur 15.0% hydrogen

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Example Problem A sample of an unknown compound with a mass of 5.00 grams is made up of 75% carbon and 25% hydrogen. What is the mass of each element? 3.75 g of carbon 1.25 g of hydrogen

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Formulas Empirical Formula - expresses the smallest whole number ratio of atoms present E.g. CH2O Ionic formulas are always empirical formulas Molecular Formula - states the actual number of each kind of atom found in one molecule of the compound. E.g. C6H12O6

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**Empirical Formula Determine mass in grams of each element**

Calculate the number of moles of each Divide each by the smallest number of moles to obtain the simplest whole number ratio If whole numbers are not obtained in step 3, multiply all by the smallest number that will give whole numbers

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**Empirical Formula Remember this: Percent to mass Mass to mole**

Divide by small Multiply ‘till whole

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Example Problem Given that a compound is composed of 60.0% Mg and 40.0% O, find the empirical formula.

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Example Problem A compound is found to contain 68.5% carbon, 8.63% hydrogen, and 22.8% oxygen. The molecular weight of this compound is known to be approximately g/mol. Find the empirical and molecular formulas.

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**Hydrates Ionic compounds Water is bonded to the crystal structure**

Ex: CuSO3 • 7H2O The percentage of water in a hydrate can easily be calculated using the formula: % Water = Mass of water x 100 Mass of hydrate

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**Example Problem What is the percentage of water in CuSO3 • 7H2O?**

A 3.5 g sample of a hydrate is heated and only 1.7 g of the anhydrous salt remain. What is the percentage of water?

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**Law of Definite Proportions**

Formulas give the numbers of atoms or moles of each element Always a whole number ratio 1 molecule NO2 : 2 atoms of O for every 1 atom of N 1 mole of NO2 : 2 moles of O atoms to every 1 mole of N atoms

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**Law of Multiple Proportions**

When any two elements, A and B, combine to form more than one compound, the different masses of B that unite with a fixed mass of A have a small whole-number ratio Example: In H2O, the proportion of H:O = 2:16 or 1:8 In H2O2, H:O is 2:32 or 1:16

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**How Do We Determine Concentration?**

Molarity Molality

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**How do we make solutions?**

M1 = m1/V rearrange to M1V1 = m1 M2 = m2/V rearrange to M2V2 = m2 If m1=m2 then, M1V1 = M2V2

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M1V1 = M2V2 M1 = concentration of the first solution V1 = volume of the first solution M2 = concentration of the second solution V2 = volume of the second solution Let's consider a sample problem: You have 1 L of a M aqueous solution of table sugar. You want to dilute the solution to M. What do you do?

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Dilution To solve the problem, you simply plug in the three numbers you know: (0.125 M) (1 L) = (0.05 M) V2 2.5 L = V2 Using the equation, you determine that the volume of the diluted solution should be 2.5 L. So we simply add enough water to the first solution so that the solution's volume becomes 2.5 L.

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What is Saturation? A solution is saturated if it contains as much solute as can possibly be dissolved under the existing conditions of temperature and pressure Unsaturated: Has less than maximum amount of solute that can be dissolved Supersaturated: Contains more than maximum (How can this happen?)

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Chapter 7 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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