The Mole The Ten Commolements Thou shalt not use 6.02 x 1023 in vain. Thou shalt not use the term mole if thou has no true knowledge of the term mole.

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Presentation transcript:

The Mole

The Ten Commolements Thou shalt not use 6.02 x 1023 in vain. Thou shalt not use the term mole if thou has no true knowledge of the term mole. Thou shalt not kill a mole. Thou shalt not covet your neighbor's mole. Thou shall always remember to celebrate Mole Day. Thou shalt not disparage Mole Day. Thou shalt not use a mole out of season. Thou shalt always honor the one who introduced thou to Mole Day. You shalt always keep sacred 10/23. Thou shalt always remember these commolments or thou will never properly celebrate Mole Day.

1 mole of hockey pucks would equal the mass of the moon! 1 mole of pennies would cover the Earth 1/4 mile deep! 1 mole of basketballs would fill a bag the size of the earth!

The Mole 1 dozen = 1 gross = 1 ream = 1 mole = x There are exactly 12 grams of carbon-12 in one mole of carbon-12.

Avogadro’s Number 6.02 x is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ). Amadeo Avogadro I didn’t discover it. Its just named after me!

Calculating Formula Mass Calculate the formula mass of carbon dioxide, CO g + 2(16.00 g) = g  One mole of CO 2 (6.02 x molecules) has a mass of grams

Mole Relationships Mole Atoms or molecules Atoms or molecules Liters Grams 6.02 x AtomicMassAtomicMass L 22.4 L

Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li = g Li 1 mol Li 6.94 g Li 45.1

Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li = mol Li 6.94 g Li 1 mol Li 2.62

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol = atoms 1 mol 6.02 x atoms 2.07 x 10 24

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li = atoms Li 1 mol Li x atoms Li 1.58 x g Li1 mol Li (18.2)(6.022 x )/6.94

Standard Molar Volume Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro At STP (Standard Temperature and Pressure): 1 mole of a gas occupies 22.4 liters of volume At STP (Standard Temperature and Pressure): 1 mole of a gas occupies 22.4 liters of volume

Calculations with Moles: Using Standard Molar Volume How many moles of hydrogen are in 100 L of hydrogen at STP? 100 L = mol 22.4 L 1 mol 4.64

Calculations with Moles: Using Standard Molar Volume How many liters are occupied by 3 moles of oxygen gas at STP? 3 mol = L 1 mol 22.4 L 67.2

Some concepts to consider: mole is x items. 2 1 atom has a mass equal to its atomic mass in a.m.u.The atomic mass of Li (mass of 1 atom) is a.m.u. 3 1 mole of atoms has a mass equal to its atomic mass in grams. This is known as the molar mass. 1 mole of Li atoms have a mass of g. Molar mass = grams/mole 4 The formula mass of a molecule (or formula unit) equals the sum of the masses of its atoms in a.m.u. The formula mass of CO 2 (mass of 1 molecule) is = a.m.u. The formula mass of Na 2 O (mass of 1 formula unit) is = a.m.u. 5 1 mole of molecules (or formula units) has a mass equal to its formula mass in grams. This is known as the molar mass. 1 mole of molecules of CO 2 have a mass of g. Molar mass = grams/mole 1 mole of formula units of Na 2 O have a mass of g. Molar mass = grams/mole

In-depth Mole Calculations How many atoms of H are in 2.4 moles of water? 2.4 mol 1 mol 6.02 x mlcl x mlcl H 2 O 1.44 x mlcl H 2 O x 2 = 2.89 x 10 24

Given g of H 2 O : a. How many moles of H 2 O is that? b. How many grams of H are in the sample? c. How many grams of O are in the sample? d. How many molecules of H 2 O is that? e. How many moles of H are in the sample? f. How many moles of O are in the sample? g. What is the mass of two molecule of H 2 O? In-depth Mole Calculations 2 moles 4.04 g H 32.0 g O X mlcl 4 moles 2 moles a.m.u.

Percentage Composition the percentage by mass of each element in a compound

 100 = Percentage Composition %Cu = g Cu g Cu 2 S  100 = %S = g S g Cu 2 S % Cu 20.15% S Find the % composition of Cu 2 S.

%Fe = 28 g 36 g  100 =78% Fe %O = 8.0 g 36 g  100 =22% O Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. Percentage Composition

How many grams of copper are in a 38.0-gram sample of Cu 2 S? (38.0 g Cu 2 S)( ) = 30.3 g Cu Cu 2 S is % Cu Percentage Composition

 100 = %H 2 O = g g 24.51% H 2 O Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O? Percentage Composition

The Mole Empirical/Molecular Formulas

Empirical Formula C2H6C2H6 CH 3 Molecular formula Smallest whole number ratio of atoms in a compound Empirical formula

Empirical Formula 1.Find mass (or %) of each element. (may be given) 2. Find moles of each element. (g to moles) 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O g 1 mol g = 1.85 mol N 74.1 g 1 mol g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

Examples An organic compound is found to contain 92.25% carbon and 7.75% hydrogen. Determine the empirical formula. –Empirical Formula: CH What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? –Empirical Formula: Fe 2 S 3

Molecular Formula “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol g/mol = 2.00 empirical mass = g/mol (CH 2 ) 2  C 2 H 4