Stoichiometry (Chapter 3) Antoine Lavoisier eperimental observations lead to the development of the Law of Conservation of Mass (a.k.a. Law of Conservation of Matter). The updated version of the law of conservation of mass states: During an ordinary (non nuclear) chemical reaction, atoms are neither created nor destroyed. Stoichiometry is the study of the quantitative nature of chemical reactions. Chemical equations represent the qualitative and quantitative nature of chemical reactions. The equation: 2 H 2 (g) + O 2(g) 2 H 2 O (l) represents a chemical reaction in which two diatomic hydrogen (gas) molecules react with one diatomic oygen (gas) molecule react to yield two water (liquid) molecules. Hydrogen and oygen are the reactants (a.k.a. reagents) and water is the product. The symbol separates the reactant(s) from the product(s) and serves the same purpose as the = sign does in algebra. The total number of atoms of each element (and the sum of their respective masses) on the reactant side of the must be equal to the total number of atoms of each element (and the sum of their respective masses) on the product side. Chemical equations frequently contain additional symbols to represent the physical state (phase) of the substance(s). These symbols include: (S) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous (the substance is dissolved in water). In the equation: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) The small 4 (known as the subscript) behind the H in CH 4 (g) indicates that there are four atoms of hydrogen in each methane molecule. The large 2 (known as the coefficient) in front of the O 2 (g) indicates that there are two oygen molecules. If no coefficient is shown, it is understood to be one. Thus we have one molecule of methane (CH 4 (g) ) reacting with two oygen molecules to produce (yield) one carbon dioide and two water molecules. The coefficient is multiplied throughout the entire molecule whereas the subscript only applies to the element (or polyatomic ion in a parenthesis) it immediately follows. So in the equation: Na 2 SO 4 (aq) + BaCl 2 (aq) BaSO 4 (s) + 2 NaCl (aq) The Na 2 SO 4 (aq) represents one sodium sulfate molecule that contains two sodium atoms, one sulfur atoms, and four oygen atoms. The 2 NaCl (aq) represents two sodium chloride molecules each contains one sodium atom and one chlorine atom. Page 1 of 11
Checking the equation, we find that we have two sodium atoms, one sulfur atom, four oygen atoms, one barium atom and two chlorine atoms on the reactant side of the equation and one barium atom, one sulfur atom, four oygen atoms, two sodium atoms, and two chlorine atoms on the product side. Thus the total number of atoms of each element is unchanged they are just combined differently to form different substances. Since atoms were neither created or nor destroyed, the above balanced equations satisfy the law of conservation of mass. Since an atom s valence electrons are involved in the formation of chemical bonds, we can make some general predictions regarding an element s chemical properties based on its position on the Periodic Table. Elements in the same family (column) typically react in a similar manner. For eample: the Group 1 metal, sodium (Na) reacts vigorously with water to produce sodium hydroide and hydrogen: 2 Na (s) + H 2 O (l) 2 NaOH (aq) + H 2 (g) We would epect the other Group 1 (alkali metals) to undergo a similar reaction with water. Reaction Types A combustion reaction involves a substance reacting rapidly in the presence of oygen. For eamples, hydrogen burns to form water: 2 H 2 (g) + O 2(g) 2 H 2 O (l) Hydrocarbons (compounds containing only carbon and hydrogen) burn in the presence of sufficient oygen to form carbon dioide and water. Eamples: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Combination (a.k.a. composition or synthesis) reactions occur when two (or more) elements combine to form a single product: 2 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Decomposition (a.k.a. analysis) reactions occur when a single substance breaks down into two (or more) simpler substances: 2 KClO 3 (s) 2 KCl (s) + O 2 (g) See Sections 4.2-4.4 for a discussion of other types of chemical reactions: Echange (Metathesis) Reactions. Acid-Base Neutralization Reactions Acid-Base Reactions with Gas Formation Oidation-Reduction Reactions In each type of reaction, we must balance the equation using only coefficients to obtain the same number of atoms of each elements on both the reactant and products sides of the raction. Page 2 of 11
Atomic and Molecular Masses Atom, molecules, and their subatomic particles (i.e., electron, proton, and neutron) are etremely small with negligible mass. Rather than epressing their mass in terms of grams or kilograms, a special, more convenient, atomic mass unit (u) has been defined. The atomic mass unit is defined as 1 / 12 th the mass of a single atom of the most abundant isotope of carbon, 12 C. You do not need to memorize these numbers, but 1 u = 1.6605402 10-24 g and 1 g = 6.0221421 10 23 u. The atomic mass (a.k.a. atomic weight) for an element is epressed in terms of atomic mass units. Since most elements have multiple naturally occurring isotopes, their atomic masses are not whole numbers. An element s atomic mass is an average mass based on the mass and relative natural abundance of its various naturally occurring isotopes. The formula mass for a formula unit (ionic substances) or molecule (molecular substances) equals the sum of the atomic masses of the atoms in the formula. For eample, the formula mass for potassium chlorate, KClO 3 would be determined as follows: (Please refer to your Periodic Table for the atomic masses.) KClO 3 has one potassium atom (AM = 39.0983 u), one chlorine atom (AM = 35.4527 u), and three oygen atoms (AM = 15.9994 u), so the formula mass would be find by the equation 39.0983 u + 35.4527 u + 3 (15.9994 u) = 122.5492 u Al 2 (SO 4 ) 3 has two aluminum atoms (AM = 26.981538 u), three sulfur atoms (AM = 32.066 u), and 12 oygen atoms (AM = 15.9994 u), the formula mass would be found by: 2 (26.981538 u) + 3 (32.066 u) + 12 ( 15.9994 u) = 342.154 u Molecular mass refers to the formula mass of a single molecule of a molecular substance. Please note that the term molecular mass (molecular compounds) is often (incorrectly) used to refer to the formula mass of an ionic compound. It s a hard habit to break, so I ll probably be guilty of this several times throughout the semester. Percentage Composition The elemental percentage composition for compounds typically is determined by eperimental analysis. However, we can obtain some information from the compound s chemical formula. Please refer to the following steps: Page 3 of 11
To determine the percent composition of the elements in a compound, follow these steps: 1. Determine the formula mass for the substance. 2. Determine the total atomic mass for each element in the substance. 3. Divide the substance s formula mass into each of the respective total atomic masses and multiple each of the result by 100. Eample: Find the percentage composition for aluminum sulfate 1. The formula mass is 342.154 u (see above). 2. The total atomic masses of aluminum atoms: 2 (26.981538 u) = 53.963076 u Al sulfur atoms: 3 (32.066 u) = 96.198 u S oygen atoms: 12 ( 15.9994 u) = 191.993 u O 3. % Al = 53.963076 u Al 100 = 15.7716% Al 342.154 u Al2(SO4)3 % S = %O = 96.198 u S 342.154 u Al2(SO4)3 191.993 u O 342.154 u Al2(SO4)3 100 = 28.115% S 100 = 56.1130% O To check our work, we add the respective percentages: 15.7716% + 28.115% + 56.1130% = 99.9996% which when rounded to the required three decimal places gives us 100.000%. NOTE: Due to rounding errors, the total percent may not eactly equal 100%. Page 4 of 11
Masses in the laboratory Since atoms and molecules are so incredibly small, it is impossible to count them out individual. Instead we take advantage of the relationship that eists between the number of particles and their formula mass. By definition, the mass of a single atom of 12 C is 12 u. Since 1 g = 6.0221421 10 23 u, 6.0221421 10 23 12 C atoms have a mass of 12 g. This relationship between the number of particles and mass is defined has follows: The number of 12 C atom in eactly 12.0000 g of 12 C is the mole. The number of particles in a mole, 6.0221421 10 23, is known as Avogadro s number. The mole (abbreviated as mol) is the bridge between the number of particles and their mass in grams. Eamples: Eample 1: Convert 25.00 g H 2 O to mol 25.00 g H2O 18.01528 g H 2O = 1.388 mol H 2O Eample 2: Convert 1.388 mol H 2 O to molecules 1.388 mol H 2O 6.0221421 10 23 molecules H 2O = 8.358 10 23 molecules H 2O Eample 3: Convert 25.00 g H 2 O to molecules 25.00 g H2O 18.01528 g H 2O 23 6.0221421 10 molecules H 2O = 8.359 10 23 molecules H 2O Note the slightly different answers for eamples two and three. The round-off error occurred when the rounded answer from eample one were used in eample two. Eample three combined the steps into one calculation without the intermediate rounding. Empirical and Molecular Formulas The molecular formula represents the actual number of the atoms of the various elements in a molecule of a molecular compound. The empirical formula represents the simplest ratio of the atoms of the various elements in the compound. Page 5 of 11
Octane, the primary ingredient in gasoline, has a molecular formula of C 8 H 18. Since both 8 and 18 are divisible by 2, octane s empirical formula would be C 4 H 9. Ionic compounds do not consist of molecules, so their empirical formulas are used to represent them. A compound s empirical and/or molecular formula(s) can be determined from eperimentally determined percentage composition data. DETERMINING A COMPOUND S EMPIRICAL FORMULA: 1. Assume a 100.00 g sample so that the mass percents may be converted to grams. 63.15% C, 5.30% H, and the rest is O 63.15 g C. 5.30 g H, and 31.55 g O 2. For each element, convert its mass to moles. 63.15 g C 1 mol C/12.011 g C = 5.258 mol C 5.30 g H 1 mol H/1.00794 g H = 5.26 mol H 31.55 g O 1 mol O/15.9994 g O = 1.972 mol O 3. Divide the number of moles of each element by the smallest number of moles from above to get the mole ratio. 5.258 mol C/1.972 mol O = 2.666 = 8/3 5.26 mol H/1.972 mol O = 2.67 = 8/3 1.972 mol O/1.972 mol O = 1 = 3/3 3a. If the above ratios contain fractions, multiply each term by some whole-number to eliminate that fraction. Multiple by 3, the common denominator, to get: C 8 H 8 O 3 If molecular mass data is also available, one may determine a compound s molecular formula. Page 6 of 11
DETERMINING A COMPOUND S MOLECULAR FORMULA: Method 1 1. Determine the compound s empirical formula. Eample: Given: empirical formula C 4 H 9, MM = 114.2309 2. Determine the empirical formula s formula mass 4 * 12.011 + 9 * 1.00794 = 57.11546 3. Divide the empirical formula s formula mass by the compound s molar mass. 114.2309/57.11546 = 2 4. Multiply the empirical formula s subscripts by the number obtained in Step 3. (C 4 H 9 ) 2 --> C 8 H 18 Method 2 1. For each element, multiple the mass percent (as a decimal) for that element by the compound s molar mass and divide by the element s atomic mass..84117 C * 114.2309 / 12.011 = 8.158827 H * 114.2309 / 1.00794 = 18 thus, C 8 H 18 NOTE: We can determine the empirical formula for the molecular since both subscripts are divisible by a common factor (2) to yield C 4 H 9 Page 7 of 11
Combustion Analysis: Determing Empirical and/or Molecular Formulas The combustion analysis of a 5.982 g sample of a compound in the vitamin D family (containing C, H, and O) yielded 18.490 g of CO 2 and 6.217 g of H 2 O. If the compound s molar mass is 398.6726, what are its empirical and molecular formulas? The general reaction is: C H y O z(s) + O 2(g) --> CO 2(g) + H 2 O (g) 18.490 g CO 2 12.011 g C/44.0098 g CO 2 = 5.0462 g C 6.217 g H 2 O 2(1.00794 g H)/18.01528 g H 2 O = 0.69569 g H Why must we solve for the mass of oygen indirectly? Since some of the oygen is from the atmosphere, we must determine the amount of oygen in the original compound by subtracting the mass(es) of the other element(s) from the initial mass of the compound. g O = g sample - (g C + g H) = 5.982 g - (5.0462 + 0.69569) g = 0.2401 g O 5.0462 g C 1 mol C/12.011 g C = 0.4201 mol C 0.69569 g H 1 mol H/1.00794 g H = 0.6902 mol H 0.2401 g O 1 mol O/15.9994 g O = 0.015007 mol O 0.4201 mol C/0.015007 mol O = 28 0.6902 mol H/0.015007 mol O = 46 empirical formula = C 28 H 46 O formula mass = 398.6728 molar mass/formula mass = 398.6728/398.6728 = 1 molecular formula = C 28 H 46 O Empirical and Molecular Formula Practice Problems 1. Serotonin (MM = 176) is a compound that conducts nerve impulses in the brain and muscles. It is 68.2% C, 6.86% H, 15.9% N and 9.08% O. What is serotonin s molecular formula? 2. A sample of a certain inorganic insecticide contains 5.969 g Pb, 0.029 g H, 2.158 g As, and 1.844 g O. What is the empirical formula for this insecticide? Page 8 of 11
3. A compound contains only carbon, hydrogen, and oygen. Combustion of 10.68 mg of the compound yields 16.01 mg CO 2 and 4.37 mg H 2 O. The compound s molar mass is 176.1 g/mol. What are the compound s empirical and molecular formulas? 4. Coniine, a toic substance isolated from poison hemlock, contains only carbon, hydrogen, and nitrogen. Combustion analysis of a 5.024 mg sample yields 13.90 mg of CO 2 and 6.048 mg of H 2 O. What is coniine s empirical formula? 5. Ibuprofen, a headache remedy, is 75.69% C, 15.51% O, and 8.80% H. What is ibuprofen s empirical formula? 6. Ethylene glycol (MM = 62.0), commonly used as automotive antifreeze, contains only carbon, hydrogen, and oygen. Combustion analysis of a 23.46 mg sample yields 20.42 mg of H 2 O and 33.27 mg of CO 2. What are ethylene glycol s empirical and molecular formulas? 7. Menthol (MM = 156.3), a strong-smelling substance used in cough drops, is a compound of carbon, hydrogen, and oygen. When 0.1595 g menthol was subjected to combustion analysis, it produced 0.449 g CO 2 and 0.184 g H 2 O. What is menthol s empirical formula?. 8. An analysis of nicotine, a poisonous compound found in tobacco leaves, shows that it is 74.0348% C, 8.6980% H, and 17.2672 % N. Its molar mass is 162.2346 g/mol. What are the empirical and molecular formulas for nicotine? 9. Heachlorophene, a compound made up of atoms of carbon, hydrogen, chlorine, and oygen, is an ingredient in germicidal soaps. Combustion of a 1.000 g sample yields 1.407 g of carbon dioide, 0.134 g of water, and 0.5228 g of chlorine gas. What is heachlorophene s empirical formula? Balanced Equations and Masses of Reactants and Products Once again, according to the Law of Conservation of Mass, atoms can neither be created nor destroyed during a chemical reaction. We use balanced chemical equations to represent the quantitative relationships that eist between the reactants and products. Remember that the mole is the bridge that links the reactants to products. The general format is: Mass reactant Moles reactant Moles product Mass reactant Page 9 of 11
A chemical reaction can continue to produce products only as long as some of each reactant is available. Once one of the reactants is completely consumed, the reaction stops. The reactant that is consumed first the limiting reactant determines the maimum amount of product(s) the theoretical yield that can be produced under ideal conditions. Eample: Step 1: What mass of water can be produced when 10.00 g of hydrogen is reacted with 50.00 g of oygen? Write the balanced chemical equation for the reaction. 2 H 2 (g) + O 2 (g) 2 H 2 O (g) Step 2: Convert the given mass(es) of the reactant(s) to moles 10.00 g H2 1 mol H 2 2.01588 g H 2 = 4.961 mol H 2 50.00 g O2 1 mol O 2 31.9988 g O 2 = 1.563 mol O 2 Step 3: Identify the limiting reactant. Separately, for each reactant, we can convert mass of the reactant to moles, use the coefficient ratio to get moles of product and then convert to grams of product. The smallest calculated yield is the theoretical yield and the starting reactant is limiting. 10.00 g H2 1 mol H 2 2.01588 g H 2 2 mol H 2O 2 mol H 2 18.01528 g H 2O = 89.37 g H2O 50.00 g O 1 mol O 31.9988 g O 2 mol H 2O 1 mol O 2 18.01528 g H2O = 56.30 g H2O Since 56.30 g < 89.37 g, oygen was the limiting reactant and 56.30 g O 2 is the theoretical yield. Percent Yield = actual yield theoretical yield 100 = What would be the percent yield if the above reaction only produced an actual yield of 37.95 g H 2 O? % yield = 37.95 g 56.30 g 100 = 67.41% Page 10 of 11
An alternate approach From the balanced equation, we see that two moles of hydrogen are needed for each mole of oygen. To determine which reactant is consumed first, we can do the following compare the number of moles of each reactant divided by their respective coefficient from the balanced equation 4.961 mol H 2 mol H > 1.563 mol O 1 mol O 2.481 > 1.563 so oygen would be consumed first When we compare the two fractions, the > or < symbol always points to the limiting reactant in this case the oygen. Use the coefficient ratios from the balanced equation and formula masses to complete the calculations: 50.00 g O 1 mol O 31.9988 g O 2 mol H 2O 1 mol O 2 18.01528 g H2O = 56.30 g H2O What would be the percent yield if the above reaction only produced an actual yield of 37.95 g H 2 O? % yield = 37.95 g 56.30 g 100 = 67.41% PLEASE NOTE: Chemical reactions rarely (if ever) occur under ideal conditions, so the actual yield normally is always less than 100%. Many reactions may have very small percent yields. If your calculated percent yield eceeds 100%, then some combination of eperimental and/or calculation error(s) has occurred. Answers for the Empirical and Molecular Formula Practice Problems on Pages 8-9: 1. C 10 H 12 N 2 O 2. PbHAsO 4 3. C 3 H 4 O 3 C 6 H 8 O 6 4. C 8 H 17 N 5. C 13 H 18 O 2 6. CH 3 O C 2 H 6 O 2 7. C 10 H 20 O 8. C 5 H 7 N C 10 H 14 N 2 9. C 13 H 6 Cl 6 O 2 Page 11 of 11