Magnet Chemistry Phillips 2016 The Mole Magnet Chemistry Phillips 2016

Can you imagine a mole of donuts? Measurements Dozen = 12 Baker’s dozen = 13 Gross= 144 Mole (mol) = 6.02x1023 A mole of water molecules has a volume of only 18 mL! Molecules are unbelievably small, so a lot of them doesn’t take up that much space. The mole is useful in chemistry because it links the microscopic world of atoms, molecules, and ions, to the macroscopic world Can you imagine a mole of donuts?

Avogadro’s Number 6.02x1023 is so important in chemistry it’s given its own name…Avogadro’s Number In numerical form it looks like this: 602,000,000,000,000,000,000,000 A mole of paper would reach past Pluto A mole of basketballs is the size of Earth A mole of rice would cover the land masses of Earth to a depth of 75 meters!

Scientific Notation Practice Using the mole requires that we are comfortable using scientific notation Write the following in scientific notation Write the following in standard notation 6000 0.0067 78,000,000 698,700 0.000009 0.0090076 6.00x103 6.70x10-3 7.80x107 6.99x105 9.00x10-6 9.01x10-3 6.02x104 9.03x109 7.77x10-2 9.21x10-7 60,200 9,030,000,000 0.0777 0.000000921

The mole The mole establishes a relationship between the atomic mass and the gram. Individual atoms and molecules are too small to be directly measured, so we refer to them using the unit amu. Using moles allows direct measurement. The mass in grams of 1 mole of a substance is equal to its atomic mass. 6.02x1023 atoms Cu = 63.5 g 6.02x1023 atoms H = 1.001 g 6.02x1023 atoms Fe = 55.8 g Mole was based on Carbon. How many g of C are in 1 mol?

What’s in a Mole? A mole of particles in an element is usually talking about atoms. The number of molecules in a mole of any molecular compound is 6.02x1023 How many atoms are in a molecule of ammonia (NH3)? 4 atoms 1 N atom and 3 H atoms Each molecule has 4 atoms So 1 mole of ammonia gas contains 1 mole of NH3 molecules, but four times as many atoms, or 4 moles of atoms (1 mol N atoms and 3 mol H atoms)

Mole Conversions Chemists measure amounts of substances by volume or mass. Because the mole measures both a mass and a number of particles (and volume, but that will be covered later), it is the central unit in converting the amount of substance from one type of measurement to another.

Moles to Molecules Molecules OR atoms The number of particles in 1 mole of any substance is always the same- Avogadro’s number. or How many atoms are in 3 mol of elemental Ne? 6.02x1023 particles 1 mol 1 mol 6.02x1023 particles 3 mol Ne 1 6.02x1023 atoms 1 mol 1.81x1024 atoms Ne

Unit Cancellation/ Dimensional Analysis Method (sometimes called ‘unit analysis’) Write down units asked for in answer to the right Write down the given value over 1 on the left Apply one or more unit factors to cancel units It’s as easy as 1-2-3!

Moles to Molecules (cont.) PARTICLES How many atoms of oxygen are in 6 mol of O2 molecules? How many moles of I2are in 8.02x1020 molecules of I2? Atoms Molecules? Use the subscript 6.02x1023 particles 1 mol Avogadro’s Number 1 mol 6.02x1023 particles MOLES 6 mol O2 1 6.02x1023 molecules O2 1 mol O2 2 atoms O 1 molecule O2 7.22x1024 atoms O 1 mol I2 6.02x1023 molecules I2 8.02x1020 molec. I2 1 1.33x10-3 mol I2

Molar Mass The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance. The atomic mass of iron is 55.85 amu. Therefore, the molar mass of iron is 55.85 g/mol. Since oxygen occurs naturally as a diatomic, O2, the molar mass of oxygen gas is 2 times 16.00 g or 32.00 g/mol.

Calculating Molar Mass The molar mass of a substance is the sum of the molar masses of each element and expressed as g/mol. What is the molar mass of magnesium nitrate, Mg(NO3)2? The sum of the atomic masses is: 24.31 + 2(14.01 + 16.00 + 16.00 + 16.00) = 24.31 + 2(62.01) = 148.33 amu The molar mass for Mg(NO3)2 is 148.33 g/mol.

Molar Mass Practice Ne = 20.18 g/mol O2 = 16.00*2 = 32.00 g/mol U = 238.03 g/mol NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol CO2 = 12.01 + 16.00*2 = 44.01 g/mol Al2(CO3)3 = 26.98*2 + (12.01*3) + (16.00*9) = 233.99 g/mol

Moles to Grams Conversions If you know the mass of a substance, you can calculate the number of moles. You have 11.2 g of NaCl. How many moles is that? 1. determine the molar mass of NaCl (using the periodic table) Na= 23.0 g/mol Cl= 35.5 g/mol 23.0 g/mol + 35.5 g/mol = 58.5 g/mol (get these #’s from the periodic table)

Moles to Grams Conversions Molar mass of NaCl = 58.5 g/mol Set up a conversion factor (a fraction whose value is equal to 1) The units we need should be in the numerator and the units you already know in the denominator.

Moles to Grams Conversions You can also determine the mass of a sample if you know the number of moles 2.50 mol of NaCl is how many grams? Set up a conversion factor 

Moles to Grams Conversions MASS How many moles are in 14 g LiOH? How many moles are in 15 g N2? moles x g = g mol Molar Mass g x mol = moles g MOLES 14 g LiOH 1 1 mol LiOH 23.95 g LiOH 0.58 mol LiOH 15 g N2 1 1 mol N2 28.02 g N2 0.54 mol N2

Moles to Grams Conversions How many grams are in 4 mol H2O2? How many grams are in 56 mol CaCO3? 4 mol H2O2 1 34.016 g H2O2 1 mol H2O2 136.06 g H2O2 56 mol CaCO3 1 100.09 g CaCO3 1 mol CaCO3 5605.04 g CaCO3

Mole Calculations II = 87.8 g Pb What is the mass of 2.55 × 1023 atoms of lead? We want grams, we have atoms of lead. Use Avogadro’s number and the molar mass of Pb 2.55 × 1023 atoms Pb × 1 mol Pb 6.02×1023 atoms Pb 207.2 g Pb 1 mole Pb × = 87.8 g Pb

Mole Calculations II 8.84 × 1021 molecules O2 How many O2 molecules are present in 0.470 g of oxygen gas? We want molecules O2, we have grams O2. Use Avogadro’s number and the molar mass of O2 0.470 g O2 × 1 mol O2 32.00 g O2 6.02×1023 molecules O2 1 mole O2 × 8.84 × 1021 molecules O2

Multistep Conversions You want to impress your date by boasting you know how many molecules of table sugar are in the cake you just made. You need 250 g of sugar (C12H22O11). How many sucrose molecules will be in the cake? Plan: Convert the mass to moles using the molar mass and then convert to moles using Avogadro’s number. MASS PARTICLES Avogadro’s Number Use molar mass MOLES 250 g C12H22O11 1 1 mol C12H22O11 342.3 g C12H22O11 6.02x1023 molecules 1 mol C12H22O11 4.4x1023 molec. C12H22O11

Multistep Conversions If you burned 4.0x1024 molecules of methane (CH4) during a laboratory experiment, what mass of methane did you use? Plan: Convert your given # of molecules to moles using Avogadro’s number, then convert the moles to grams using the molar mass of methane. (12.00 + 1.004*4) 4.0 x 1024 molec. CH4 1 1 mol CH4 6.02x1023 molec. CH4 16.016 g CH4 1 mol CH4 106.42 g CH4

Moles and Gases At the same temperature and pressure, equal volumes of gases contain the same number of gas particles. 1 mole of any gas at 0oC and 1 atm (Standard Temperature and Pressure; STP) has a volume of 22.4 L. This volume, 22.4 L/mol, is called molar volume.

Gas Density- not on this test The density of gases is much less than that of liquids. We can calculate the density of any gas at STP easily. The formula for gas density at STP is: = density, g/L molar mass in grams molar volume in liters

Calculating Gas Density not on this test What is the density of ammonia gas, NH3, at STP? First we need the molar mass for ammonia; 14.01 + 3(1.01) = 17.04 g/mol The molar volume NH3 at STP is 22.4 L/mol. Density is mass/volume: = 0.761 g/L 17.04 g/mol 22.4 L/mol

Molar Mass of a Gas 1.96 g 1.00 L 22.4 L 1 mole × = 43.9 g/mol We can also use molar volume to calculate the molar mass of an unknown gas. 1.96 g of an unknown gas occupies 1.00L at STP. What is the molar mass? We want g/mol, we have g/L. 1.96 g 1.00 L 22.4 L 1 mole × = 43.9 g/mol

Mole Unit Factors 1 mol = 6.02 × 1023 particles 1 mol = molar mass We now have three interpretations for the mole: 1 mol = 6.02 × 1023 particles 1 mol = molar mass 1 mol = 22.4 L at STP for a gas This gives us 3 unit factors to use to convert between moles, particles, mass, and volume.

Mole-Volume Calculation A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present? We want moles, we have volume. Use molar volume of a gas: 1 mol = 22.4 L 4.50 L CH4 × = 0.201 mol CH4 1 mol CH4 22.4 L CH4

Mass-Volume Calculation What is the mass of 3.36 L of ozone gas, O3, at STP? We want mass O3, we have 3.36 L O3. Convert volume to moles then moles to mass: 3.36 L O3 × × 22.4 L O3 1 mol O3 48.00 g O3 = 7.20 g O3

Molecule-Volume Calculation How many molecules of hydrogen gas, H2, occupy 0.500 L at STP? We want molecules H2, we have 0.500 L H2. Convert volume to moles and then moles to molecules: 0.500 L H2 × 1 mol H2 22.4 L H2 6.02×1023 molecules H2 1 mole H2 × = 1.34 × 1022 molecules H2

Moles and Gases: A Summary MASS PARTICLES You try it: A student fills a 1.0-L flask with CO2 at STP. How many molecules of gas are in the flask? Plan: 1) Convert from volume to moles using molar volume. 2) Then convert from moles to molecules using Avogadro’s number. Molar Mass Avogadro’s Number MOLES Molar Volume (22.4 L/mol) VOLUME of gases at STP

Moles and Gases Let’s see how you did… A chemical reaction produces 0.82 mole of oxygen gas. What volume will that gas occupy at STP? 1.0-L CO2 1 1 mol CO2 22.4 L CO2 6.02 x 1023 molec CO2 1 mol CO2 2.7 x 1022 molec. CO2 0.82 mol O2 1 22.4 L O2 1 mol O2 18 L O2

Law of Definite Composition The law of definite composition states that “Compounds always contain the same elements in a constant proportion by mass”. Sodium chloride is always 39.3% sodium and 60.7% chlorine by mass, no matter what its source. Water is always 11.2% hydrogen and 88.8% oxygen by mass.

Law of Definite Composition Figure: 04-08 Title: Law of Definite Composition Caption: A drop of water, a glass of water, and a lake of water all contain hydrogen and oxygen in the same percent by mass, that is, 11.2% hydrogen and 88.8% oxygen. Notes: The Law of Definite Composition states that no matter its source, a compound contains the same elements in the same percents by mass. A drop of water, a glass of water, and a lake of water all contain hydrogen and oxygen in the same percent by mass.

Percent Composition The percent composition of a compound lists the mass percent of each element. For example, the percent composition of water, H2O is: 11% hydrogen and 89% oxygen All water contains 11% hydrogen and 89% oxygen by mass.

Calculating Percent Composition There are a few steps to calculating the percent composition of a compound. Lets practice using H2O. Assume you have 1 mole of the compound. One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen. 2(1.01 g H) + 1(16.00 g O) = molar mass H2O 2.02 g H + 16.00 g O = 18.02 g H2O

Calculating Percent Composition Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water: 2.02 g H 18.02 g H2O × 100% = 11.2% H 16.00 g O 18.02 g H2O × 100% = 88.79% O

Percent Composition Problem TNT (trinitrotoluene) is a white crystalline substance that explodes at 240°C. Calculate the percent composition of TNT, C7H5(NO2)3.

Solution Percent Composition Problem TNT (trinitrotoluene) is a white crystalline substance that explodes at 240°C. Calculate the percent composition of TNT, C7H5(NO2)3. First, find molar mass of TNT: 7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N + 32.00 g O) = g C7H5(NO2)3 84.07 g C + 5.05 g H + 42.03 g N + 96.00 g O = 227.15 g C7H5(NO2)3.

Percent Composition of TNT 84.07 g C 227.15 g TNT × 100% = 37.01% C 1.01 g H 227.15 g TNT × 100% = 2.22% H 42.03 g N 227.15 g TNT × 100% = 18.50% N 96.00 g O 227.15 g TNT × 100% = 42.26% O

Try one more: Percent Composition What is the percent carbon in C5H8NO4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 %C b) 24.3 %C c) 41.1 %C

Chemical Formulas of Compounds Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). 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 If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

Types of Formulas Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. *Ionic formula is always empirical formula Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

To obtain an Empirical Formula 1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of moles of each element. 3. 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 through by the smallest number that will give all whole numbers * Be careful! Do not round off numbers prematurely

Sol’n: require mole ratios, so convert grams to moles A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. Sol’n: require mole ratios, so convert grams to moles moles of N = 2.34g of N = 0.167 moles of N 14.01 g/mole moles of O = 5.34 g = 0.334 moles of O 16.00 g/mole Formula: (HONORS only)

Calculation of the Molecular Formula A compound has an empirical formula of NO2. The colorless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?

Empirical Formula from % Composition A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams This will contain 28.60 grams of B and 10.60 grams H Determine the number of moles of each Determine the simplest whole number ratio

More Empirical & Molecular Formulas The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule. The molecular formula of benzene is C6H6 The empirical formula of benzene is CH. The molecular formula of octane is C8H18 The empirical formula of octane is C4H9.

Another Exp: Empirical Formulas We can calculate the empirical formula of a compound from its composition data. We can determine the mole ratio of each element from the mass to determine the formula. Ex: radium oxide, Ra?O?. A 1.640 g sample of radium metal was heated to produce 1.755 g of radium oxide. What is the empirical formula? We have 1.640 g Ra and 1.755-1.640 = 0.115 g O.

Calculating Empirical Formulas The molar mass of radium is 226.03 g/mol and the molar mass of oxygen is 16.00 g/mol. 1 mol Ra 226.03 g Ra 1.640 g Ra × = 0.00726 mol Ra 1 mol O 16.00 g O 0.115 g O × = 0.00719 mol O We get Ra0.00726O0.00719. Simplify the mole ratio by dividing by the smallest number. We get Ra1.01O1.00 = RaO is the empirical formula.

Another Exp: Empirical Formulas from Percent Composition We can also use percent composition data to calculate empirical formulas. Assume that you have 100 grams of sample. Benzene is 92.2% carbon and 7.83% hydrogen, what is the empirical formula. If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen.

Empirical Formulas from Percent Composition Calculate the moles of each element: 1 mol C 12.01 g C 92.2 g C × = 7.68 mol C 1 mol H 1.01 g H 7.83 g H × = 7.75 mol H The ratio of elements in benzene is C7.68H7.75. Divide by the smallest number to get the formula. 7.68 C = C1.00H1.01 = CH 7.75 H

Another Exp: Molecular Formulas The empirical formula for benzene is CH. This represents the ratio of C to H atoms of benzene. The actual molecular formula is some multiple of the empirical formula, (CH)n. Benzene has a molar mass of 78 g/mol. Find n to find the molecular formula. = CH (CH)n 78 g/mol 13 g/mol n = 6 and the molecular formula is C6H6.

Conclusions Avogadro’s number is 6.02 × 1023 and is one mole of any substance. The molar mass of a substance is the sum of the atomic masses of each element in the formula. At STP, 1 mole of any gas occupies 22.4 L.

Conclusions Continued We can use the following flow chart for mole calculations:

Conclusions Continued The percent composition of a substance is the mass percent of each element in that substance. The empirical formula of a substance is the simplest whole number ratio of the elements in the formula. The molecular formula is a multiple of the empirical formula.