Chapter 4: Aqueous Reactions and Solution Stoichiometry

CHAPTER 3 STOICHIOMETRY: CHEMICAL EQUATIONS Stoichiometry: The area of study that examines the quantities of substances consumed and produced in chemical reactions. Remember: Atoms are neither created nor destroyed during any chemical reaction or physical process. CHEMICAL EQUATIONS

Chemical equation represents the chemical reaction . The arrow is called the yield sign. 2 2 BALANCING EQUATIONS A chemical equation must have an equal number of atoms of each element on each side of the arrow.

RULES/HINTS FOR BALANCING EQUATIONS 1) Use correct formulas. 2) The coefficients must be the lowest wholenumbers possible. 3) Do not change or add subscripts (number of atoms of each element in a molecule). 4) Balance H and O last. 5) If a polyatomic ion stays together on both sides of the arrow, then keep it together when balancing. 2HCl + Na2CO3 2NaCl + H2O + CO2 C2H5OH + 3O2 2CO2 + 3H2O

SAMPLE EXERCISE 3.1 The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms. a) Write the chemical formulas for the reactants and products. O2+ NO NO2(unbalanced) (b) Write a balanced equation for the reaction. O2+ 2NO 2NO2 (balanced) (c) Is the diagram consistent with the law of conservation of mass? Yes; because there are equal numbers of both N and O atoms in the two boxes

PRACTICE EXERCISE In the following diagram, the white spheres represent hydrogen atoms, and the blue spheres represent nitrogen atoms. To be consistent with the law of conservation of mass, how many NH3 molecules should be shown in the right box? Answer: Six NH3 molecules. STATES OF MATTER We use the symbols (g) for gas, (l) for liquid, (s) for

solid, and (aq) for aqueous to identify the states of matter of the reactants and products. Catalysts, photons (h), or heat () may stand above the reaction arrow (). CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) SAMPLE EXERCISE 3.2 Balance this equation: Na(s) + H2O(l) NaOH(aq) + H2(g) Answer: 2Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g) ____________________________________________________ PRACTICE EXERCISE:

Balance the following equations: a. Fe(s) + O2(g) Fe2O3(s) b. C2H4(g) + O2(g) CO2(g) + H2O(g) c. Al + HCl AlCl +H SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITY 1) Combination Reactions:

In synthesis/combination reactions two or more substances react to form one product. Generic Reaction: A + B AB Real Reaction: 2Mg + O2 2MgO Examples of Combination Reactions: N2 (g) + 3 H2 (g) 2 NH3 (g) C3H6 (g) + Br2 (l) C3H6Br2 (l) Examples of Combination Reactions:

Decomposition Reactions: In a decomposition reaction one substance undergoes a reaction to produce two or more products. Generic Reaction: AB A + B Real Reaction: CaCO3 CaO + CO2 Examples: CaCO3 (s) CaO (s) + CO2 (g) 2 KClO3 (s) 2 KCl (s) + O2 (g) 2 NaN3 (s) 2 Na (s) + 3 N2 (g) SAMPLE EXERCISE 3.3 Write the balanced equations for the following

reactions: a. The synthesis reaction that occurs when lithium metal and fluorine gas react. 2 Li(s) + F2(g) 2 LiF(s) b. The decomposition reaction that occurs when solid barium carbonate is heated. BaCO3(s) BaO(s) + CO2(g) PRACTICE EXERCISE

Write the balance equation for the following reactions: a. Solid mercury (II) sulfide decomposes into its elements when heated. b. The surface of aluminum metal undergoes combination reaction with oxygen in the air. Answer: (a) HgS(s) Hg(l) + S(s) (b)4 Al(s) + 3 O2(g) 2 Al2O3(s) a COMBUSTION REACTIONS Combustion reactions are rapid

reactions that produce a flame. They require air (O2) as a reactant. When hydrocarbons are combusted completely, the products are CO2 and H2O. Generic Reaction: Real Reaction: Examples: CH4 (g) + 2 O2 (g) CxHy + O2 CO2 + H2O C3H8 + 5O2 3CO2 + 4H2O

CO2 (g) + 2 H2O (g) SAMPLE EXERCISE 3.4 Write the balanced equation for the reaction that occurs when methanol, CH3OH(l), is burned in air. CH3OH(l) + O2(g) CO2(g) + H2O(g) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) The multiply by 2 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g) ______________________________________________________ Practice Exercise:

Write the balance equation for the reaction that occurs when ethanol, C2H5OH(l), is burned in air. FORMULA WEIGHTS The formula weight of a substance is the sum of the atomic weights of each atom in its chemical formula. Example: H2SO4 = 2(1 amu) + 32 amu + 4(16 amu) = 98 amu For molecular compounds: The formula weight is also called the molecular weight (MW). For ionic compounds: No separated molecules so it called specifically the formula weight (FW).

For Elements: In case of isolated atoms is called atomic weight (AW). FW of NaCl = 23 amu + 35.5 amu = 58.5 amu SAMPLE EXERCISE 3.5 Calculate the formula weight of a. sucrose (C12H22O11) MW or (FW) of sucrose (C12H22O11) = 12(12 amu) + 22(1 amu) + 11(16 amu) = 342 amu b. Calcium nitrate . FW Ca(NO3)2 = 40.1 amu) + 2[14 amu+3(16 amu)] =

164.1 amu __________________________________________________________ Practice Exercise: Calculate the formula weight of (a) Al(OH)3 and (b) CH3OH. Answer: (a) 78.0 amu, (b) 32.0 amu PERCENT COMPOSITION Percent Composition is the percentage by mass contributed by each element in a substance. Generally: (number of atoms)(atomic weight) % element =

x 100 (FW of the compound) Example: For C2H6 (MW = 30 amu) (2)(12.0 amu) %C = (30.0 amu) (6)(1.01 amu) %H =

(30.0 amu) x 100 = x 100 = 24.0 amu x 100 = 80.0% x 100

= 20.0% 30.0 amu 6.06 amu 30.0 amu SAMPLE EXERCISE 3.6 Calculate the percent of carbon, hydrogen, and oxygen (by mass) in C12H22O11. Solution: ___________________________________________________ Practice Exercise:

Calculate the percentage of nitrogen, by mass, in calcium nitrate. Answer: 17.1% AVOGADROS NUMBER AND THE MOLE The mole is just a number of things. In daily life: 1 dozen = 12 things 1 pair = 2 things 1 Cross = 144 things In Chemistry the unit for dealing with the number of atoms, ions, or molecules, any particles the mole, abbreviated mol.

1 mol = 6.02 x 1023 representative particles SAMPLE EXERCISE 3.7 Arrange the following samples in order of increasing numbers of carbon atoms: 12 g 12C, 1 mol C2H2, 9 x 1023 molecules of CO2. Solution: 12 g of 12C atoms.

= 1 mol of C atoms = 6.02 1023 C 1 mol of C2H2 = 6.02 1023 C2H2molecules = 2 6.02 1023 C atoms. 9 x 1023 molecules of CO2 = 9 x 1023 C atoms. The order: 12 g 12C (1 mol C) < 9 1023CO2 molecules (1.5 mol C) PRACTICE EXERCISE Arrange the following samples in order of increasing number of O atoms: 1 mol H2O, 1 mol CO2, 3 x 1023

molecules O3. Answer: 1 mol H2O (6 1023O atoms) < 3 1023molecules O3(9 1023O atoms) < 1 mol CO2(12 1023O atoms) Important Notes: 1 molecule of H2SO4 2 atoms of H 1 molecule of H2SO4 1 atom of S 1 molecule of H2SO4 4 atoms of O 1 mol of H2SO4 (6.022x 1023) molecules of H2SO4 1 mol of H2SO4 2 moles of H 2(6.022x 1023) atoms of H 1 mol of H2SO4 1 moles of S (6.022x 1023) atoms of S 1 mol of H2SO4 4 moles of O 4(6.022x 1023) atoms

of O SAMPLE EXERCISE 3.8 Calculate the number of H atoms in 0.350 mol of C6H12O6. Solution: ??? H atoms 0.350 mol C6H12O6. moles C6H12O6 molecules C6H12O6atoms H Another solution: PRACTICE EXERCISE

How many oxygen atoms are in: a. 0.25 mol of calcium nitrate? Answer: 9.0 1023. b. 1.50 mol of sodium carbonate? Answer: 2.71 1024 MOLAR MASS A mole is always the same number (6.02 x 1023).

But 1 mole sample of different substances will have different masses. A mol of an element = AW (amu) in grams of that element. The mole in grams of substance called the molar mass of the substance (g/mol). Each is 1 mole of different elements have different masses but all

contain same number of atoms = 6.02 x 1023 SAMPLE EXERCISE 3.9 What is the mass in grams of 1.000 mol of glucose, C6H12O6? Solution: MW glucose C6H12O6 = 6(12 amu) + 12(1 amu) + 6(16 amu) = 180 amu So 1 mole of glucose = 180 g And the molar mass = 180g/mol

_____________________________________________________ Practice Exercise: Calculate the molar mass of calcium nitrate. Answer: 164.1 g/mol SAMPLE EXERCISE 3.10 Calculate the number of moles of glucose in 5.380g of glucose. ?? mol Glucose 5.380 g Glucose Conversion factor: 1mol Gl 180 g Gl _______________________________________________________ Practice Exercise: How many moles of sodium bicarbonate are in 508g of sodium

bicarbonate (also known as sodium hydrogen carbonate)? Answer: 6.05 mol NaHCO3 SAMPLE EXERCISE 3.11 Calculate the mass, in grams. Of 0.433 mol of calcium nitrate. ?? g of Ca(NO3)2 0.433 mol Ca(NO3)2 mol Ca(NO3)2 164.1 g Ca(NO3)2 PRACTICE EXERCISE What is the mass, in grams, of a. 6.33 mol sodium bicarbonate?

b. 3.0 x 10-5 mol of sulfuric acid? 3g Answer: 532 g Answer: 2.9 10 SAMPLE EXERCISE 3.12 a) How many molecules are in 5.23g of glucose? ?? molecules 5.23 g glucose b. How many oxygen atoms are in this sample? ?? O atoms 5.23 g glucose

PRACTICE EXERCISE a) How many molecules are in 4.20g of nitric acid? Answer: 4.01 1022 molecules HNO3 b) How many O atoms are in this sample? Answer: 1.20 1023atoms O EMPIRICAL FORMULAS (EF) FROM ANALYSES An empirical formula is the lowest whole number ratio of elements in chemical formula (written as subscripts). 1. When given percentages, assume 100g so that

the percentages can be grams. 2. Convert grams into moles. 3. Divide all answers by the lowest number to get subscripts. 4. If you get a 0.5, or 1.5 then multiply all numbers by 2. If you get a 0.33 or 0.66, then multiply all numbers by 3 (in order to obtain smallest whole numbers). SAMPLE EXERCISE 3.13 Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.5% O by mass. What is the empirical formula of ascorbic acid?

Solution: Assume 100 g sample for simplicity: Divide on the smallest value: (3.406) Multiply by 3: The EF: PRACTICE EXERCISE A 5.325g sample of methyl benzoate, a compound used

in perfumes, contains 3.758g C, 0.316g H, and 1.251g of O. What is the empirical formula of this substance? Answer: C4H4O MOLECULAR FORMULA (MF) The molecular formula is the actual formula of a molecule. Sometimes it can be the same as the empirical formula. MF can be determined from the EF if we have the actual molecular weight. MF = n(EF) where n is whole number

Example: C2H6 = 2(CH3) The whole number obtained by: MW = 30 for C2H6 EW = 15 for CH3 So = 30/15 = 2 _________________________________________________________ In the previous example 3.13: If the actual

MW of ascorbic acid = 176 what is the MF of the acid? EF was : C3H4O3 EW [3(12) +4(1)+3(16) ]= 88 Actual MW 176 n = 176/88 = 2 MF = 2 (C3H4 O3) = C6H8O6 __________________________________________________ SAMPLE EXERCISE 3.14 Mesitylene, a hydrocarbon in crude oil, has an empirical formula of C3H4. The molecular weight of the substance is 121 amu. What is the molecular

formula of mesitylene? EF : C3H4 EW [3(12) +4(1)]=40 Actual MW 121 n = 121/40 = 3.02 MF = 3 (C3H4) = C9H12 PRACTICE EXERCISE Ethylene glycol, the substance used in antifreeze, is composed of 38.7%C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. a) What is the empirical formula of ethylene glycol? Answers: CH3O

b. What is the molecular formula for the previous question? Answers: C2H6O2 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 CO2 produced

H is determined from the mass of H2O produced O is determined by difference after the C and H have been determined SAMPLE EXERCISE 3.15 Isopropyl alcohol, a component of rubbing alcohol, is composed of C, H, and O. Combustion of 0.255g of

isopropyl alcohol produces 0.561g of CO2 and 0.306g H2O. Determine the empirical formula. Solution: We need to determine: 1) g C 2) g H 3) g O = mass of sample (g C + g H) 4) Change all grams to moles PRACTICE EXERCISE (b) Caproic acid, the smell in dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225g sample of this compound

produces 0.512g CO2, 0.209g H2O. What is the empirical formula of caproic acid? Answers: C3H6O (b) Caproic acid has a molar mass of 116 g/mol. What is the molecular formula? Answer: C6H12O2 QUANTITATIVE INFORMATION FROM BALANCED EQUATIONS

The coefficients in a balanced equation indicate both the relative numbers of moles in the reaction. A mole ratio is the ratio of the coefficients of two substances in a chemical equation. You must use the mole ratio to convert from one substance to another. C2H5OH + 3O2 2CO2 + 3H2O Conversion factors: 1 mol C2H5OH

3 mol O2 1 mol C2H5OH 2 mol CO2 1 mol C2H5OH 3 H 2O

3 mol O2 2 mol CO2 3 mol O2 3 mol H2O SAMPLE EXERCISE 3.16 How many grams of water are produced in the

oxidation of 1.00g of glucose (MW=180)? C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) 1.00 g ?? g 1mol C6H12O6(s) 6 mol H2O(g) PRACTICE EXERCISE How many grams of O2 can be prepared from 4.50g KClO3?

2KClO3(s) 2KCl(s) + 3O2(g) Answer: 1.77 g SAMPLE EXERCISE 3.17 Lithium hydroxide reacts with carbon dioxide to form lithium carbonate and water. How many grams of carbon dioxide (MW = 44) can be absorbed by 1.00g of lithium hydroxide (FW = 23.95)? Solution: 2 LiOH(s) 1.00g

+ CO2(g) Li2CO3(s) + H2O(l) ??g 2 mol LiOH(s) 1 mol CO2(g) Grams LiOH moles LiOH moles CO2 grams CO2 PRACTICE EXERCISE What mass of O2 is consumed in the combustion of 1.00g of propane (C3H8)? Answer: 3.64 g

LIMITING REAGENTS (LR) Limiting reactant or (limiting reagent): The reactant that is completely consumed in a reaction. The other reactants are called the excess reactants or reagents. The amount of product is always based on the limiting reagent. DETERMINE THE (LR) For the balanced equation:

2H2 Stiochimetry: Excess reactant: + O2 2 mol 5 mol 2H2O 1mol 2 mol 1mol 2

mol The reaction stops when O2 consumed; so O2 is called LR Excess of Reactant : 5 mol H2 2 mol H2 = 3 mol H2 Guide for determination of the (LR): 1. 2. 3. Reactions contain more than given amounts of reactants are LR reactions. Convert grams of reactants to moles.

Divide the results of each mole on its coefficient SAMPLE EXERCISE 3.18 How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2? N2(g) Given: + 3H2(g) 3 mol

6 mol 3.0/1 = 3 6/3 = 2 LR 2NH3(g) PRACTICE EXERCISE A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. 2Al(s) + 3Cl2(g) 2AlCl3(s)

a) Which is the limiting reagent? Answer: Al b) How many moles of AlCl3 are formed? Answer:1.50 mol How many moles of the excess reactant remain at the end of the reaction? Answer: 0.75 mol Cl2 SAMPLE EXERCISE 3.19 Suppose a fuel cell is set up with 150g of hydrogen gas and 1500g of oxygen gas. A)How many grams of water can be formed?

2H2(g) + O2(g) 2H2O(g) 150 g For H2 : 75/2 = 37.5 H2 is LR 1500 g ??g For O2: 47/1 =47 B) How many grams of reactant present in excess

remain after the reaction completed? 2H2(g) + O2(g) 150 g 2H2O(g) ?? g reacted Excess O2 = origin reacted = 1500-1200 = 300 g PRACTICE EXERCISE A strip of zinc metal with a mass of 2.00g is placed in an aqueous solution containing 2.50g of silver nitrate, causing the following reaction to occur:

Zn(s) + 2AgNO3(aq) 2Ag(s) + Zn(NO3)2(aq) a. Which reactant is limiting? b. How many grams of Ag will form? c. How many grams of Zn(NO3)2 will form? d. How many grams of the excess reagent will be left? PERCENT YIELD An important indicator of the efficiency of a

particular laboratory or industrial reaction. The percent yield of a reaction relates the actual yield or experimental yield (done in the lab) to the theoretical yield (calculated using stoichiometry). Actual yield 100% percent yield Theoretical yield Actual yield < Theoretical yield , because: 1- Incomplete reaction. 2- Side reactions.

3- Some product lost (gas or deposited solids). SAMPLE EXERCISE 3.20 Adipic acid, H2C6H8O4, is used to produce nylon. Adipic acid is made by the following reaction: 2C6H12(l) + 5O2(g) 2H2C6H8O4(l) + 2H2O(g) a. Assume that you carry out this reaction starting with 25.0g of cyclohexane and 100.0 g Oxygen. What is the theoretical yield of the adipic acid? 2C6H12(l) + 5O2(g) 2H2C6H8O4(l)

+ 2H2O(g) 25.0g 100.0 g (g ???) Th. Yield For C6H12 : 0.3/2 = 0.15 C6H12 is LR For O2: 3.12/5 = 0.63 b. If you obtain 33.5g of adipic acid from your reaction, what is the percent yield of adipic acid?

PRACTICE EXERCISE a. If you start with 150g of Fe2O3 as the limiting reagent, what is the theoretical yield of Fe? Fe2O3(s) + 3CO(g) 2Fe(s)

+ 3CO2(g) Answer: 105 g Fe b. If the actual yield of Fe in your test was 87.9g, what is the percent yield? Answer: 83.7% HW PROBLEMS 1, 4, 6, 8, 11, 15, 21, 25, 27, 32, 35, 38, 41, 47, 49, 52, 53, 55, 59, 63, 65, 70, 80,

81, 83, 85. . : . . : : :

: : : :

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