Stoichiometry © 2009, Prentice-Hall, Inc. Warmup – Mass Percent Find the percent by mass: 1.What is the mass % of silver in silver nitrate? (63.5%) 2. What is the mass % of N in ammonium carbonate? (29.2%)

Stoichiometry © 2009, Prentice-Hall, Inc. Finding Empirical Formulas

Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

Stoichiometry © 2009, Prentice-Hall, Inc. Rhyme Rhyme to remember order of steps to convert % composition  empirical formula: Percent to mass Mass to mole Divide by small Times till whole

Stoichiometry © 2009, Prentice-Hall, Inc. Sample Exercise 3.13 (p. 99) Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of vitamin C? (C 3 H 4 O 3 )

Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas Assuming g of ascorbic acid, C:40.92 g x = mol C H: 4.58 g x= 4.53 mol H O:54.50 g x = mol O 1 mol g 1 mol g 1 mol 1.01 g

Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= H:= 1.33 O:= Multiply all by 3: 3 C:4 H:3 O mol mol 4.53 mol mol

Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 3 H 4 O 3

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 1 (3.13) A g sample of phosgene, a compound used as a chemical warfare agent during World War I, contains g of carbon, g of oxgen, and g of chlorine. What is the empirical formula of this substance? (a)CO 2 Cl 6, (b)COCl 2, (c)C O Cl 0.044, (d)C 2 OCl 2

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 2 (3.13) A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance? (C 4 H 4 O)

Stoichiometry © 2009, Prentice-Hall, Inc. Molecular Formula from Empirical Formula The empirical formula (relative ratio of elements in the molecule) may not be the molecular formula (actual ratio of elements in the molecule). e.g. ascorbic acid (vitamin C) has empirical formula C 3 H 4 O 3. The molecular formula is C 6 H 8 O 6.

Stoichiometry © 2009, Prentice-Hall, Inc. Molecular Formula from Empirical Formula To get the molecular formula from the empirical formula, we need to know the molecular weight, MW. (“molecular molar mass”) The ratio of molecular weight (MW) to formula weight (FW) of the empirical formula must be a whole number. (“formula molar mass”)

Stoichiometry © 2009, Prentice-Hall, Inc. Sample Exercise 3.14 (p. 100) Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C 3 H 4. The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene? (Hint: you do not need to figure out the empirical formula first) (C 9 H 12 )

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 1 (3.14) Cyclohexane, a commonly used organic solvent, is 85.6% C and 14.4% H by mass with a molar mass of 84.2 g/mol. What is its molecular formula? (a)C 6 H, (b)CH 2, (c)C 5 H 24, (d)C 6 H 12, (e)C 4 H 8

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 2 (3.14) Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its (molecular) molar mass is 62.1 g/mol. a) What is the empirical formula of ethylene glycol? (CH 3 O) b) What is its molecular formula? (C 2 H 6 O 2 )

Stoichiometry © 2009, Prentice-Hall, Inc. Hydrates Anhydrous = dry, does not contain water Hydrate = a compound formed by the combination of water with some other substance in which the water retains its molecular state as H 2 O

Stoichiometry © 2009, Prentice-Hall, Inc. Determining Hydrate Formulas 1.Find the formula of a hydrate from experimental mass data: a) Calculate the number of moles of the anhydrous salt (mass  mole conversion) b) Calculate the mass of H 2 O by subtracting mass of anhydrous salt from mass of hydrated salt c) Calculate the moles of H 2 O (mass  mole conversion) d) Calculate ratio: # mol H 2 O:# mol anhydrous salt e)  correct formula = salt · x H 2 O (x = mole ratio)

Stoichiometry © 2009, Prentice-Hall, Inc. Determining Hydrate Formulas 2.Find the formula from mass % data: a) Use the method of using 100 g (from 100 %) to give mass in grams of the salt and water Use mass  mole conversion to calculate the # moles of the anhydrous salt and the # moles of water. b)Calculate the mole ratio of H 2 O:anhydrous salt

Stoichiometry © 2009, Prentice-Hall, Inc. Determining Hydrate Formulas 3.Find the % H 2 O in a hydrate from its formula: Use mass percent.

Stoichiometry © 2009, Prentice-Hall, Inc. Hydrate practice problems 1.A mass of 2.50 g of blue hydrated copper sulfate (CuSO 4 x H 2 O) is placed in a crucible and heated. After heating, 1.59 g white anhydrous copper sulfate (CuSO 4 ) remains. What is the formula for the hydrate? Name the hydrate. (CuSO 4 5 H 2 O)

Stoichiometry © 2009, Prentice-Hall, Inc. Hydrate practice problems 2.A hydrate is found to have the following percent composition: 48.8% MgSO 4 and 51.2% H 2 O. What is the formula and the name for this hydrate? (MgSO 4 7 H 2 O)

Stoichiometry © 2009, Prentice-Hall, Inc. Hydrate practice problems 3.The formula for a hydrate is SnCl 2 2 H 2 O. What is the mass percent of water in this hydrate? (15.97% H 2 O)

Stoichiometry © 2009, Prentice-Hall, Inc. Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this. –C is determined from the mass of CO 2 produced. –H is determined from the mass of H 2 O produced. –O is determined by difference after the C and H have been determined.

Stoichiometry © 2009, Prentice-Hall, Inc. Sample Exercise 3.15 (p. 101) Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g CO 2 and g H 2 O. Determine the empirical formula of isopropyl alcohol. (C 3 H 8 O)

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 1 (3.15) The compound dioxane, which is used as a solvent in various industrial processes, is composed of C, H, and O atoms. Combustion of a g sample of this compound produces g CO 2 and g H 2 O. A separate experiment shows that it has a molar mass of 88.1 g g/mol. Which of the following is the correct molecular formula for dioxane? (a)C 2 H 4 O, (b)C 4 H 2 O 2, (c)CH 2, (d)C 4 H 8 O 2

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 2 (3.15) a)Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a g sample of this compound produces g CO 2 and g H 2 O. What is the empirical formula of caproic acid? (C 3 H 6 O) b)Caproic acid has a molar mass of 116 g/mol. What is its molecular formula? (C 6 H 12 O 2 )

Stoichiometry © 2009, Prentice-Hall, Inc. Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O.

Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products.

Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculations Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams. C 6 H 12 O O 2  6 CO H 2 O

Stoichiometry © 2009, Prentice-Hall, Inc. Limiting Reactants

Stoichiometry © 2009, Prentice-Hall, Inc. How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients. Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

Stoichiometry © 2009, Prentice-Hall, Inc. How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.

Stoichiometry Stoichiometry Balloon Races © 2009, Prentice-Hall, Inc.

Stoichiometry © 2009, Prentice-Hall, Inc. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount. –In other words, it’s the reactant you’ll run out of first (in this case, the H 2 ).

Stoichiometry © 2009, Prentice-Hall, Inc. Limiting Reactants In the example below, the O 2 would be the excess reagent.

Stoichiometry © 2009, Prentice-Hall, Inc. Solving Limiting Reactant Problems Given: 8.51 g H 2 and 9.25 g O 2 2 H 2 + O 2  2 H 2 O Which reactant is limiting? How much excess is left over?

Stoichiometry © 2009, Prentice-Hall, Inc. One Method Given: 8.51 g H 2 and 9.25 g O 2 2 H 2 + O 2  2 H 2 O g A have  g B need  Is this more or less than what you have? 8.51 g H 2 (1 mol H 2 )(1 mol O 2 )(32.00 g O 2 ) = 67.4 g O 2 HAVE 2.02 g H 2 2 mol H 2 1 mol O 2 NEED 67.4 g O 2 > 9.25 g O 2 NEED > HAVE  O 2 is limiting

Stoichiometry © 2009, Prentice-Hall, Inc. Given: 8.51 g H 2 and 9.25 g O 2 2 H 2 + O 2  2 H 2 O Because O 2 is limiting, Step g O 2 (1 mol O 2 )(2 mol H 2 )(2.02 g H 2 ) = 1.17 g H 2 used g O 2 1 mol O 2 1 mol H 2 Step g H 2 HAVE g H 2 USED 7.34 g H 2 left over = excess reactant

Stoichiometry © 2009, Prentice-Hall, Inc. We can also use a tool called a RICE table R = reaction I = initial quantities C = change E = end quantities (later in the year = equilibrium

Stoichiometry © 2009, Prentice-Hall, Inc. R 2 H 2 + O 2  2 H 2 O I 4.21 mol mol 0 C mol mol mol E 3.63 mol 0 mol mol limiting theoretical reactant yield Take coefficients into account in this step Supporting calculations: 8.51 g H 2 (1 mol H 2 ) = 4.21 mol H g H g O 2 (1 mol O 2 ) = mol O g O 2 Resulting calculations: 3.63 mol H 2 (2.02 g H 2 ) = 7.34 g H 2 1 mol H 2 in excess mol H 2 O(18.02 g H 2 O) = 10.4 g H 2 O product

Stoichiometry © 2009, Prentice-Hall, Inc. Sample Exercise 3.18 (p. 108) The most important commercial process for converting N 2 from the air into nitrogen- containing compounds is based on the reaction of N 2 and H 2 to form ammonia (NH 3 ): N 2(g) + 3 H 2(g)  2 NH 3(g) How many moles of NH 3 can be formed from 3.0 mol of N 2, and 6.0 mol of H 2 ? (4.0 mol NH 3 )

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 1 (3.18) When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction 2 CH 3 OH (l) + 3 O 2(g)  2 CO 2(g) + 4 H 2 O (g), what is the excess reactant and how many moles of it remains at the end of the reaction? (a)9 mol CH 3 OH (l), (b)10 mol CO 2(g), (c)10 mol CH 3 OH (l), (d)14 mol CH 3 OH (l), (e)1 mol O 2(g)

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 2 (3.18) Consider the reaction 2 Al (s) + 3 Cl 2(g)  2 AlCl 3(s). A mixture of 1.50 mol of Al and 3.00 mol of Cl 2 are allowed to react. a) What is the limiting reactant? (Al) b) How many moles of AlCl 3 are formed? (1.50 mol) c) How many moles of the excess reactant remain at the end of the reaction? (0.75 mol Cl 2 )

Stoichiometry © 2009, Prentice-Hall, Inc. Sample Exercise 3.19 (p. 108) Consider the following reaction that occurs in a fuel cell: 2 H 2(g) + O 2(g)  2 H 2 O (l) This reaction, properly done, produces energy in the form of electricity and water. Suppose a fuel cell is set up with 150 g of hydrogen gas and 1500 grams of oxygen gas (each measurement is given with two significant figures). How many grams of water can be formed? (1300 g H 2 O)

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 1 (3.19) Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light- emitting diodes and solar cells:Ga (l) + As (s)  GaAs (s) If 4.00 g of gallium is reacted with 5.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction? (a)4.94 g As, (b)0.56 g As, (c)8.94 g Ga, (d)1.20 g As.

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 2 (3.19) A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur: Zn (s) + 2 AgNO 3(aq)  2 Ag (s) + Zn(NO 3 ) 2(aq) a)Which reactant is limiting? (AgNO 3 ) b)How many grams of Ag will form? (1.59 g) c)How many grams of Zn(NO 3 ) 2 will form? (1.39 g) d)How many grams of the excess reactant will be left at the end of the reaction? (1.52 g Zn)

Stoichiometry © 2009, Prentice-Hall, Inc. Theoretical Yield The theoretical yield is the maximum amount of product that can be made. –In other words it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures.

Stoichiometry © 2009, Prentice-Hall, Inc. Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield). Actual Yield Theoretical Yield Percent Yield =x 100

Stoichiometry © 2009, Prentice-Hall, Inc. Sample Exercise 3.20 (p. 106) Adipic acid, H 2 C 6 H 8 O 4, is used to produce nylon. It is made commercially by a controlled reaction between cyclohexane (C 6 H 12 ) and O 2 : 2 C 6 H 12(l) + 5 O 2(g)  2 H 2 C 6 H 8 O 4(l) + 2 H 2 O (g) a)Assume that you carry out this reaction starting with 25.0 g of cyclohexane, and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (43.5 g H 2 C 6 H 8 O 4 ) b)If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid? (77.0%)

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 1 (3.20) If 3.00 g of titanium metal is reacted with 6.00 g of chlorine gas, to form 7.7 g of titanium(IV) chloride in a combination reaction, what is the percent yield of the product? (a)65%, (b)96%, (c)48%, (d)86%

Stoichiometry © 2009, Prentice-Hall, Inc. Practice Exercise 2 (3.20) Imagine that you are working on ways to improve the process by which iron ore containing Fe 2 O 3 is converted into iron. In your tests you carry out the following reaction on a small scale: Fe 2 O 3(s) + 3 CO (g)  2 Fe (s) + 3 CO 2(g) a)If you start with 150 g of Fe 2 O 3 as the limiting reagent (reactant), what is the theoretical yield of Fe? (105 g Fe) b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield? (83.7%)