Unit 6 Chapter 12 Chemical Quantities or "Our Friend the Mole"

Stoichiometry stoichiometry—using balanced chemical equations to obtain info

Particles of matter are too small and numerous to count 12.1 Counting Particles of Matter Particles of matter are too small and numerous to count SI unit of chemical quantity = the mole (abbreviated mol) The mole is a counting unit

How do you measure how much? You can measure mass = grams or volume = liters or you can count pieces = MOLES

Moles 1 mole = 6.02 x 1023 representative particles 602 000 000 000 000 000 000 000 Treat it like a very large dozen 6.02 x 1023 is called Avogadro's number

Representative particles = The smallest pieces of a substance representative particles = ATOMS IONS MOLECULES FORMULA UNITS

Representative particles For a molecular (covalent) compounds it is a molecule compound with all nonmetals CO, BF3 , Cl2 1 mole = 6.02 x 1023 molecules

Representative particles For an element it is an atom Unless it is diatomic one symbol, no charge: Br,Cs 1 mole = 6.02 x 1023 atoms

Representative particles For an ionic compound it is a formula unit compound with metal and nonmetal - KI, Na2SO4 1 mole = 6.02 x 1023 formula units

Representative particles For ions it is 1 mol ions one symbol with charge (monatomic) or more than one symbol with charge (polyatomic: Na+ , N3- , (C2H3O2) – 1 mole ions = 6.02 x 1023 ions

Molar Mass = The mass of 1 mole of an element in grams. We can make conversion factors from these. Example - We can write this as 12.01 g C = 1 mol To change grams of a compound to moles of a compound. Or moles to grams

Molar Masses the atomic masses on the periodic table have a unit of amu (atomic mass unit) GAM = gram atomic mass = the atomic mass (listed on the periodic table) written in grams 1 atom Xe = 131.30 u GAM of Xe = 131.301 g

molar mass—the mass, in g, of 1 mole of a substance Add up the gram atomic mass of all elements in compound to calculate molar mass of a substance

Find the molar mass of methane, CH4. CH4 = 1(12.0) + 4(1.0) = 12.0 + 4.0 = 16.0 g Find the molar mass of calcium hydroxide, Ca(OH)2. Ca(OH)2 = 1(40.1) + 2(16.0) + 2(1.0) = 74.1 g

2. What is the molar mass of a molecule of H2? 1. What is the molar mass of 1 mole of phosphorus? 2. What is the molar mass of a molecule of H2? 3. What is the molar mass of a formula unit of MgCl2? 30.97 g 2.016 g 94 g

MOLAR VOLUME of any gas at STP: 22.4 L = 1 mol Molar Volume: volume-to-mole and mole-to-volume conversions At STP, all gases occupy the same amount of space: MOLAR VOLUME of any gas at STP: 22.4 L = 1 mol

dimensional analysis steps for success: = using the units (dimensions) to solve problems steps for success: 1) identify unknown (read carefully) 2) identify known (read carefully) “Play checkers” with the units, moving them diagonally, canceling when appropriate. All units should cancel except those of the desired answer. 3) plan solution 4) calculate 5) check (sig.figs., units, and math)

CONVERSION FACTOR SUMMARY: 6.02 x 10^23 representative particles 1 MOLE &

CONVERSION FACTOR SUMMARY: MOLAR MASS (g) & 1 MOLE 1 MOLE MOLAR MASS (g) for a gas at STP: 22.4 L & 1 MOLE 1 MOLE 22.4 L

Calculation question How many molecules of CO2 are the in 4.56 moles of CO2 ? 4.56 moles CO2 X 6.02 x 1023 molecules CO2 1 1 mole CO2 = 27.5 x 1023 molecules CO2

For example How many moles is 5.69 g of NaOH? 5.69 g NaOH x 1 mole= 0.142 mol NaOH 1 40.0 gNaOH need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

Examples = 28.10 g C How much would 2.34 moles of carbon weigh? 2.34 mol C X 12.01 g C 1 1mol C = 28.10 g C

Calculation question How many moles of salt is 5.87 x 1022 formula units? 5.87 X 1022 formula units NaCl X 1 mole NaCl_ 1 6.02 x 1023 f.u. NaCl = 0.0975 moles NaCl

Examples 4.61 g Mg X _1 mol Mg 1 24.3 g Mg = 0.1897 mol Mg How many moles of magnesium in 4.61 g of Mg? 4.61 g Mg X _1 mol Mg 1 24.3 g Mg = 0.1897 mol Mg

example What is the volume, in L, of 0.495 mol of NO2 gas at STP? 0.495 mol NO2 x 22.4 L NO2 = 11.1 L NO2 1 1 mol NO2

How many moles are found in 84 L of neon gas at STP? 84 L Ne x 1 mol Ne = 3.8 mol Ne 1 22.4 L Ne

There are many types of mole problems: 1 step: r.p. mol & mol r.p. mass mol & mol mass 2 step: mass r.p. & r.p. mass mass volume & volume mass r.p. volume & volume r.p.

Moles of Compounds 1 mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula for the compound. Example: 1 mole of ammonia (NH3) has 1 mole of nitrogen atoms and 3 moles of hydrogen atoms.

Percent composition composition of a compound is the percent by mass of each element in the compound. Find the mass of each element divide by the total mass of compound.

Determine the percent composition of calcium chloride (CaCl2). Step 1: determine the molar masses of each element in compound and the compound Ca = 40.078 g Cl2 = 2 x 35.453 g = + 70.906 g CaCl2 = 110.984 g

****make sure they add up to 100 Step 2: divide the molar mass of each element by the molar mass of the compound and times by 100 ****make sure they add up to 100

Empirical Formula The lowest whole number ratio of elements in a compound. CH2 H2O

You can determine the empirical formula from percent composition and mole ratios *** percent means “parts per hundred” assume you have 100 g of the compound Step 1: change the percent sign to g (grams) Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N

Step 2:calculate the number of moles for each element by converting grams to moles using 1 mole = molar mass 38.67 g C x 1mol C = 3.220 mole C 1 12.01 g C 16.22 g H x 1mol H = 16.1 mole H 1 1.01 g H 45.11 g N x 1mol N = 3.220 mole N 1 14.01 g N

So empirical formula is CH5N Step 3: divide each number of moles by the smallest number of moles 3.220 mole C = 1 mol C 3.220 16.1 mole H = 5 mol H 3.220 mole N = 1 mol N So empirical formula is CH5N

Determine the empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O 38.4 g Mn X 1mol Mn = 0.699 mole Mn 1 55 g Mn 16.8 g C X 1mol C = 1.339 mole C 1 12 g C 45.7 g O X 1mol O = 2.798 mole O 1 16 g O

So formula is MnC2O4 0.699 mole Mn = 1 mol Mn 0.699 1.339 mole C = 2 mol C 2.798 mole O = 4 mol O So formula is MnC2O4

For many compounds, the empirical formula is not the true formula A molecular formula tells the exact number of atoms of each element in a molecule

Example: acetic acid molecular formula = C2H4O2 empirical formula = CH2O The molecular formula for a compound is always a whole-number multiple of the empirical formula.

To determine molecular formula you need to know the molar mass of the compound Divide the actual molar mass by the molar mass of the empirical formula. Multiply the empirical formula by this number. molar mass of compound molar mass of empirical formula

Example: A compound has an empirical formula of ClCH2 and a molar mass of 98.96 g/mol. What is its molecular formula? molar mass of ClCH2 = 49.0 g/mol 98.96 = 2.01 49.0 2 X (ClCH2) Empirical formula = ClCH2 Molecular formula = Cl2C2H4

Example A compound has an empirical formula of CH2O and a molar mass of 180.0 g/mol. What is its molecular formula?

Example Ibuprofen is 75.69 % C, 8.80 % H, 15.51 % O, and has a molar mass of about 207 g/mol. What is its molecular formula?

Unit 6 Chemical Quantities Continued Chapter 12.2 “Using the Mole"

remember our friend the mole? Recall that the mole should be treated like a dozen 1 dozen = 12 pieces & 1 mole = 602 000 000 000 000 000 000 000 pieces (or 6.02 X 1023 r.p)

remember our friend the mole? You still need to know how and when to use:  Avogadro’s number (6.02 x 1023), representative particles  Molar mass, grams in one mole,  Molar volume of a gas at STP (22.4 L)

Conversion factor review 1 mole = 6.02 x 1023 r.p 1 mole = molar mass (g) 1 mole = 22.4 L (gas @STP)

So what about this silly mole? Our useful friend the mole allows us to do calculations called stoichiometry—using balanced chemical equations to obtain info ex: 2C + O2 2CO

Mole - Mole (MOL – MOL) Conversions new conversion factor – # of mol A = # of mol B # = coefficients in balanced equation

I.Mole - Mole (MOL – MOL) Conversions the most important, most basic stoich calculation B. uses the coefficients of a balanced equation to compare the amounts of reactants and products C. coefficients are mole ratios

D. the way to go from substance A to substance B E. mol – mol is the only time the mole number in the conversion is not automatically 1. (Avogadro’s #, molar mass, and 22.4 L are all = to 1 mol) MOL – MOL : # mol A # mol B # mol B # mol A # = coefficients

***always start with a balanced chemical equation MOL – MOL Conversions # mol A # mol B # mol B # mol A ex: How many moles of carbon monoxide are produced when 0.750 mol of oxygen reacts with carbon? 2C + O2 2CO ***always start with a balanced chemical equation

How many moles of carbon monoxide are produced when 0 How many moles of carbon monoxide are produced when 0.750 mol of oxygen reacts with carbon? 2C + O2 2CO Step 1; determine given and unknown given = 0.750 mol O2 unknown = ? mol CO

How many moles of carbon monoxide are produced when 0 How many moles of carbon monoxide are produced when 0.750 mol of oxygen reacts with carbon? 2C + O2 2CO Step 2; setup conversion using mole ratios (coefficients) in balanced equation 0.750 mol O2 X 2 mol CO = 1.5 mol CO 1 1 mol O2

Step 1; determine given and unknown given = 0.661 mol Al2O3 Find the number of moles of the reactants, given 0.661 mol of product is formed. 4Al + 3O2 2Al2O3 Step 1; determine given and unknown given = 0.661 mol Al2O3 unknown = ? mol Al = ? mol O2

4Al + 3O2 2Al2O3 Step 2; setup conversion using mole ratios (coefficients) in balanced equation 0.661 mol Al2O3 X 4 mol Al = 1.32 mol Al 1 2 mol Al2O3 0.661 mol Al2O3 X 3 mol O2 = 0.992 mol O2 1 2 mol Al2O3

II. MASS – MASS Conversions – Using molar mass in stoich problems to predict masses of reactants and/or products a balanced chemical equation can be used to compare masses of reactants and products B. mass – mass cannot change which substance you are dealing with; only mol – mol can do that

PT = periodic table, molar mass # = coefficients, chemical equation MASS – MASS Conversions GIVEN g A X 1 mol A X # mol B X PT g B 1 PT g A # mol A 1 mol B PT = periodic table, molar mass # = coefficients, chemical equation

How many grams of hydrochloric acid are made from the reaction of 0 How many grams of hydrochloric acid are made from the reaction of 0.500 g of hydrogen gas with excess chlorine gas? H2 + Cl2 2HCl Known = 0.500g H2 Unknown = ? g HCl

H2 + Cl2 2HCl GIVEN g A X 1 mol A X # mol B X PT g B 0.500 g H2 X 1 mol H2 X 2 mol HCl X 36.5 gHCl 1 2.0 g H2 1 mol H2 1 mol HCl = 18 g HCl H2 + Cl2 2HCl GIVEN g A X 1 mol A X # mol B X PT g B 1 PT g A # mol A 1 mol B

Calculate the numbers of grams of oxygen formed when 25 Calculate the numbers of grams of oxygen formed when 25.0 g of sodium nitrate decomposes into sodium nitrite and oxygen. 2NaNO3 2NaNO2 + O2 25.0 g NaNO3 X 1 mol NaNO3 X 1 mol O2 X 32.0 g O2 1 85.0g NaNO3 2 mol NaNO3 1 mol O2 = 20.3 g O2

“mole – mass” (mass – mole) calculations “MASS – MOLE”: GIVEN g A X 1 mol A X # mol B PT g A # mol A “MOLE – MASS”: GIVEN mol A X # mol B X PT g B # mol A 1 mol B PT = periodic table, molar mass # = coefficients

How many g of water are produced from the complete combustion of 0 How many g of water are produced from the complete combustion of 0.6829 mol of C2H2? 2C2H2 + 5O2 4CO2 + 2H2O 0.6829 mol C2H2 X 2 mol H2O X 18.0 g H2O 1 2 mol C2H2 1 mol H2O = 12.3 g H2O

Using the equation 2C2H2 + 5O2 4CO2 + 2H2O how many moles of O2 would be needed to produce 56.09 g of CO2? 56.09 g CO2 X 1 mol CO2 X 5 mol O2 1 44.0 g CO2 4 mol CO2 = 1.59 mol O2

“mass – volume” (volume – mass) calculations – “MASS – VOLUME”: (gases @ STP) GIVEN g A X 1 mol A X # mol B X 22.4 L B 1 PT g A # mol A 1 mol B “VOLUME – MASS”: (gases @ STP) GIVEN L A X 1 mol A X # mol B X PT g B 1 22.4 L A # mol A 1 mol B PT = periodic table, molar mass # = coefficients

How many L of hydrogen are produced from the decomposition of 3 How many L of hydrogen are produced from the decomposition of 3.50 g of water at STP? 2H2O 2H2 + O2 3.50 g H2O X 1 mol H2O X 2 mol H2 X 22.4 L H2 1 18.0 g H2O 2 mol H2O 1 mol H2 = 4.36 L H2

Using the equation 2C2H2 + 5O2 4CO2 + 2H2O how many liters of water vapor are produced when 5.02 g of C2H2 undergoes complete combustion? 5.02 g C2H2 x 1 mol C2H2 x 2 mol H2O x 22.4 L H2O 1 26.0 g C2H2 2 mol C2H2 1 mol H2O = 4.32 L H2O

“volume – volume” calculations GIVEN L A x 1 mol A x # mol B x 22.4 L B 1 22.4 L A # mol A 1 mol B # = coefficients (SHORT CUT: compare coefficients!)

How many L of sulfur trioxide are produced from the reaction of 36 How many L of sulfur trioxide are produced from the reaction of 36.1 L of oxygen with sulfur dioxide at STP? 2SO2 + O2 2SO3 36.1 L O2 x 1 mol O2 x 2 mol SO3 x 22.4 L SO3 1 22.4 L O2 1 mol O2 1 mol SO3 = 72.2 L SO3 SHORTCUT: coefficient of O2 = 1 coefficient of SO3 = 2 so 36.1 L x 2 = 72.2 L SO3

SHORTCUT: coefficients = 1 mol HCl to 1 mol CO2. How many liters of CO2 are produced from 0.252 L of HCl reacting with NaHCO3? NaHCO3 + HCl NaCl + CO2 + H2O 0.252 L HCl x 1 mol HCl x 1 mol CO2 x 22.4 L CO2 1 22.4 L HCl 1 mol HCl 1 mol CO2 = 0.252 L CO2 SHORTCUT: coefficients = 1 mol HCl to 1 mol CO2. so 0.252 L HCl = 0.252 L CO2

mass –particle (particle – mass) calculations “MASS – PARTICLE”: GIVEN g A x 1 mol A x # mol B x 6.02 x 1023 r.p. B 1 PT g A # mol A 1 mol B “PARTICLE – MASS”: GIVEN r.p. A x 1 mol A x # mol B x PT g B 1 6.02 x 1023 r.p. A # mol A 1 mol B PT = periodic table, molar mass # = coefficients

How many molecules of NH3 are produced from reacting 2 How many molecules of NH3 are produced from reacting 2.07 g of H2 with excess N2? N2 + 3H2 2NH3 2.07 g H2 x 1 mol H2 x 2 mol NH3 x 6.02 x 1023 NH3 1 2.0 g H2 3 mol H2 1 mol NH3 = 4.2 x 1023 molecules NH3

Limiting Reactants Rarely are the reactants in a chemical reaction present in the exact mole ratios specified in the balanced equation. Usually, one or more of the reactants are present in excess, and the reaction proceeds until all of one reactant is used up.

The reactant that is used up is called the limiting reactant. The left-over reactants are called excess reactants.

Limiting Reactant To figure out the limiting reactant: You must do the stoichiometric calculation with all reactants. The one that produces the least amount of product is the limiting reactant.

Ex: 4Al + 3O2  2Al2O3 What is the limiting reactant when 35 g of aluminum reacts with 35 g of oxygen? How much aluminum oxide is formed in this reaction? 35 g Al X 1 mol Al X 2 mol Al2O3 = 1 27 g Al 4 mol Al 35 g O2 X 1 mol O2 X 2 mol Al2O3 = 1.1 mol Al2O3 1 16 g O2 4 mol O2 0.65 mol Al2O3 Aluminum is limiting reactant 0.65 mol Al2O3 X 162 g Al2O3 = 105.3 g Al2O3 1 1 mol Al2O3

Percent Yield theoretical yield —amount of product predicted by the math (theory) actual yield —amount of product obtained in lab

percent yield —percentage of product recovered; comparison of actual and theoretical yields % YIELD = ACTUAL YIELD X 100 THEORETICAL YIELD

35. 0 g of product should be recovered from an experiment 35.0 g of product should be recovered from an experiment. A student collects 22.9 g at the end of the lab. What is the percent yield? 22.9 g X 100 = 65.4% 35.0 g

= THEORETICAL YIELD = 3.19 g NaCl % YIELD = 2.89 g NaCl X 100 = 90.6% What is the percent yield if 2.89 g of NaCl is produced when 1.99 g of HCl reacts with excess NaOH? Water is the other product. HCl + NaOH NaCl + H2O Actual yield = 2.89 g NaCl Theoretical yield = ? 1.99 g HCl x 1 mol HCl x 1 mol NaCl x 58.5 g NaCl 1 36.5 g HCl 1 mol HCl 1 mol NaCl = THEORETICAL YIELD = 3.19 g NaCl % YIELD = 2.89 g NaCl X 100 = 90.6% 3.19 g NaCl

The End