Q: Why do chemists call helium, curium and barium the medical elements

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Q: Why do chemists call helium, curium and barium the medical elements Q: Why do chemists call helium, curium and barium the medical elements? A: Because if you can't helium or curium, you barium! Stoichiometry Ch. 12 The optimist sees the glass half full. The pessimist sees the glass half empty. The chemist see the glass completely full, half in the liquid state and half in the vapor state.

Stoichiometry Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction Stoichiometry = the calculation of quantities in chemical reactions It allows chemists to tally the amts. of reactants and products using ratios of moles or representative particles Why are chemists great for solving problems? They have all the solutions.

Interpreting Chemical Equations N2(g) + 3H2(g) → 2NH3(g) Number of Atoms 2 atoms of nitrogen and 6 of H on each side Number of Molecules 1 molecules nitrogen reacts w/3 molecules of hydrogen to make 2 molecules of ammonia Moles Coefficients are moles 1 mole nitrogen, 3 moles hydrogen, 2 moles ammonia MOST IMPORTANT Mass Mass of 1 mol N2 (28 g) + mass of 3 mol H2 (6 g) equals the mass of 2 mol NH3 (34 g) Volume 22.4 L N2 reacts w/ 67.2L (22.4 x 3) H2 to form 44.8 L (22.4 x 2) NH3

C.P. 12.1 pg. 358 P.P. 3-4 pg. 358 Q: Why was the mole of oxygen molecules excited when he walked out of the singles bar? A: He got Avogadro's number!

Molar Ratios Molar ratio = a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles Q: What did one ion say to the other? A: I've got my ion you. Q: What did the bartender say when oxygen, hydrogen, sulfur, sodium, and phosphorous walked into his bar? A: OH SNaP!

Mole-Mole Problems Mole-Mole problems = convert from moles of one substance to moles of another 1. Write out and balance chem. equation 2. Determine molar ratios indicated by equation 3. Set up a dimensional analysis problem to convert the # of moles from one substance to moles of another Ex: How many moles of N2O & H2O are produced from 2.25 moles NH4NO3? * C.P. 12.2, pp 11-12 pg. 360

Solving Stoichiometry Problems HOW TO SOLVE Quantity of given convert to moles moles of given Find molar ratio moles of unknown convert to desired units quantity of unknown 3 Types 1. Mass-Mass 2. Mass-Volume 3. Volume-Volume A small piece of sodium that lived in a test tube fell in love with a Bunsen burner. "Oh Bunsen, my flame," the sodium pined. "I melt whenever I see you," The Bunsen burner replied, "It's just a phase you're going through."

Q: What do you call a clown who's in jail? A: A silicon.

Stoichiometry Island Diagram Known Unknown Substance A Substance B Mass Mass 1 mole = molar mass (g) 1 mole = molar mass (g) Use coefficients from balanced chemical equation Volume Mole Mole Volume 1 mole = 22.4 L @ STP 1 mole = 22.4 L @ STP (gases) (gases) 1 mole = 6.022 x 1023 particles (atoms or molecules) 1 mole = 6.022 x 1023 particles (atoms or molecules) Particles Particles Stoichiometry Island Diagram

Mass-Mass Problems Mass-Mass problem = you are given the mass of one substance and asked to find the mass of another substance involved in the same reaction Ex: Glucose is burned in oxygen to produce carbon dioxide and water. What mass of water is produced from 1.5 grams of glucose? * SP 12.3, PP 13-14 pg. 361 SP 12.4, PP 15-16 pg. 364

Mass-Volume Problems Mass-Volume Problem = you are given the mass of one substance and asked to find the volume of another substance Ex: An air bag contains 125 g of sodium azide. What volume of nitrogen gas is produced at STP? 2NaN3 → 2Na + 3N2 * Old chemists never die, they just stop reacting.

Volume-Volume Problems Volume-Volume problems = you are given a volume and asked to find a volume ***Coefficients represent the ratio of moles and volume of gases both Ex: Ammonia synthesis N2 + 3H2 → 2NH3 Calculate volume of hydrogen that reacts w/ 15.5 L of nitrogen * SP 12.5, PP 17-18 pg. 365 SP 12.6, PP 19-20 pg. 366

Limiting Reactants(Reagents) Background: To mix pink paint you need a 3:1 ratio of white to red. You have 6 L of white & 1.5 L of red. How much pink paint can you make? 1.5 red + 4.5 white = 6.0 pink w/ 1.5 L white left You are limited by the amt. of red paint you have Most often reactants are available in nonstoichiometric proportions Limiting Reactant(reagent) = reactant that determines (limits) the amt. of product that can be formed by a chem. Reaction Will be completely used up Others will have some unchanged or leftover (excess)

Limiting Reactants cont… The quantities of products formed in a reaction are always determined by the quantity of the limiting reactant Whether or not the reaction is limited depends upon the molar ratios Identifying Limiting Reactants 1) write balanced chem. equation 2) calculate amt. of product formed by each amt. of reactant 3) smaller answer indicates limiting reactant & amt. of product

Method #2 Begin by writing a correctly balanced chemical equation Write down all quantitative values under equation (include units) Convert ALL reactants to units of moles Divide by the coefficient in front of each reactant The smallest value is the limiting reactant

Limiting Reactant Problems Ex: Determine limiting reactant in which 3.5 g copper added to a solution containing 6.0 g silver nitrate Cu +2AgNO3 → Cu(NO3)2 + 2Ag 3.5 g Cu(1 mol Cu/63.5 g Cu)(2 mol Ag/1 mol Cu)(108 g Ag/1 mol Ag) = 12 g Ag ***We can choose either product but Ag is easier molar mass*** 6.0 g AgNO3(1 mol AgNO3/169.9 g AgNO3)(2 mol Ag/2 mol AgNO3)(108 g Ag/1 mol Ag) = 3.8 g Ag AgNO3 leads to the smallest amt. of Ag so AgNO3 is our limiting reactant

Finding the Excess Reagent A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? First, we need to create a balanced equation for the reaction: 4 NH3(g) + 5 O2(g)4 NO(g) + 6 H2O(g) Next we can use stoichiometry to calculate how much product is produced by each reactant.  NOTE:  It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared.

Cont. The reactant that produces the lesser amount of product in this case is the oxygen, which is thus the "limiting reactant." Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen).

Cont. We're not finished yet though.  1.70 g is the amount of ammonia that reacted, not what is left over.  To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample.

More Problems Ex: The fizz produced when some antacid tablets are dropped into H2O is created by the production of CO2 during the reaction b/w sodium bicarbonate (NaHCO3) & citric acid (H3C6H5O7). Suppose 2.0 g of NaHCO3 & 0.5 g of citric acid are present. What is the limiting reactant and what volume of CO2 will be produced? 3NaHCO3 + H3C6H5O7 → 3CO2 + 3H2O + Na3C6H5O7 * Some NaHCO3 is left over. The amt. is determined by calculating the amt. of NaHCO3 would react w/ 0.5 g H3C6H5O7 & subtracting the amt. from the starting amt. SP 12.7, PP 25-26 pg. 370, SP 12.8, PP 27-28 pg. 371??

Percent Yield We learned how much product SHOULD be produced. This is not always what we get. Reasons: 1) some may not react 2) some may be lost in transferring 3) some may react w/ a side reaction Percent yield = the ratio of the actual yield to the theoretical yield expressed as a percent The percent yield is a measure of the efficiency of a reaction carried out in the laboratory Theoretical (expected) yield = max. amt. of product that should be produced from given amts. of reactants based on calculations Actual yield = amt. of product that is really obtained in the lab from a chem. reaction

Percent Yield Problems Theoretical yield is determined by doing a limiting reactant problem Theoretical yield probs: SP 12.9, PP 29-30 pg. 374 Percent yield probs: SP 12.10, PP 31-32 pg. 375

Stoichiometry Practice How many grams of KClO3 are required to produce 9.00 L of O2 at STP? (Decomp. KClO3) How many grams of silver will be formed from 12.0 g copper? Cu + 2 AgNO3  2 Ag + Cu(NO3)2 The compound diborane (B2H6) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10 kg of diborane completely? (The products are B2O3 and H2O). Solid aluminum react with chlorine gas to yield solid aluminum chloride. If 125 g of both chlorine and aluminum react, what is the limiting reactant and how much aluminum chloride is produced.

2 TiO2 + 4 Cl2 + 3 C CO2 + 2 CO + 2 TiCl4 How many moles of chlorine will react with 4.55 moles of carbon? 4 mol Cl2 x mol Cl2 = 4.55 mol C = 6.07 mol C 3 mol C How many grams of titanium (IV) oxide will react with 4.55 moles of carbon? 2 mol TiO2 80 g TiO2 x g TiO2 = 4.55 mol C = 243 g TiO2 3 mol C 1 mol TiO2 How many molecules of TiCl4 will react with 115 g TiO2? 1 mol TiO2 2 mol TiCl4 6.02x1023 molecules TiCl4 x molecules TiCl4 = 115 g TiO2 80 g TiO2 2 mol TiO2 1 mol TiCl4 = 8.66x1023 molecules TiCl4

Limiting Reactant Problems According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate? At STP, what volume of “laughing gas” (dinitrogen monoxide) will be produced from 50 g of nitrogen gas and 75 g of oxygen gas? Carbon monoxide can be combined with hydrogen to produce methanol, CH3OH. Methanol is used as an industrial solvent, as a reactant in some synthesis reactions, and as a clean-burning fuel for some racing cars. If you had 152 kg of carbon monoxide and 24.5 kg of hydrogen gas, how many kilograms of methanol could be produced? 4. How many grams of water will be produced from 50 g of hydrogen and 100 g of oxygen? Answers: 1. 7.1 x 1023 atoms Ag 2. 40 dm3 N2O 3. 174.3 kg CH3OH 4. 112.5 g H2O

% Yield Practice When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.