Chapter 3: Stoichiometry Goal is to understand and become proficient at working with: 1. Avogadro's Number, molar mass and converting between mass and moles (REVIEW). 2. empirical formulas from analysis. (Combustion analysis is challenging for some. You must work to understand and practice until you are proficient!) 3. chemical equations (Balancing REVIEW). 4. quantitative information from balanced equations. 5. limiting reactants and percent yield. You must work to understand and practice until you are proficient! You should also understand molecular, empirical and structural formulas. Larson-Foothill College Stoichiometry 1 Formula Mass, Molecular Mass and Molar Mass Formula mass: Sum of atomic masses in amu for ionic compounds and elements. (1 amu = 1.6605 x 10-24 g) This is the mass of a single formula unit. Example: determine the formula mass of aluminum oxide. Molecular mass: Sum of atomic masses in amu for molecular compounds. This is the mass of a single molecule. Example: determine the molecular mass of methanol. Convert this mass to grams. Note: Formula weight or molecular weight are often used in place of mass. In fact, mass is more correct! The molar mass (M) of a substance is the mass per mole of its entities (atoms, molecules, or formula units). Molar mass (M): Sum of atomic masses in g units for anything! This is the mass of one mole of a any compound/element. Example: What are the molars masses of aluminum oxide and methanol? Stoichiometry Larson-Foothill College 2
Conversions Between Mass, Moles, Formula Units and Atoms (Ions) The molar mass is the bridge between mass (laboratory measurement) and amount in moles. We have no instruments in the lab that count moles. X = compound Mass of X Moles of # of X is the whole Mass % of an element Mass of an element Moles of an element # of atoms These are the parts of X 3 Determine: Molar Mass, formula units and atoms 1. The molar mass of zinc nitrate. 2. The number of grams in 2.75 mole of zinc nitrate. 3. The number of formula units in 72 mg of zinc nitrate. 4. A sample of zinc nitrate contains 3.6x10 20 oxygen atoms. What mass of zinc nitrate is present in mg? 4
Percent Composition by Mass The percent composition by mass of a compound is constant. The percent composition by mass can be calculated for each element in a compound using the compound formula and the atomic masses of each element. The sum of the percent composition of all elements in a compound must equal what? Examples Problems: 1. TNT (trinitrotoluene) is used as an explosive. Calculate the mass percent of carbon in TNT. Model of TNT 2. Calculate the mass in grams of carbon in 1.00 lb of TNT. 5 Types of Chemical Formulas Molecular formulas give the actual number of atoms of each element in one molecule of a compound. (Used only for covalent compounds.) Examples: CH3COOH Glucose* Empirical formulas give the lowest whole-number ratio of atoms of each element in a compound. Examples: CH3COOH Glucose* *Glucose is made during photosynthesis from water and carbon dioxide, using energy from sunlight. The reverse of the photosynthesis reaction, which releases this energy, is a very important source of power for cellular respiration. Note: For organic compounds, the standard order of the elemental symbols given in the molecular formula is C, followed by H, then the remaining symbols in alphabetical order. For organic acids, the acidic hydrogens are indicated by listing them first in the molecular formula. Compounds with different molecular formulas can have the same empirical formula. Empirical formulas are always used for ionic compounds. Do you know why? Larson-Foothill College 6
Molecular Structure and Isomers A molecular formula gives more information about a compound than its empirical formula. However, a molecular formula does not contain structural information. Let s compare ethanol with dimethyl ether, both with molecular formula C2H6O. Ethanol: (b.p. = 78.5 C, d= 0.789 g/ml at 20 C) Dimethyl Ether: (b.p. = 25 C, d = 0.00195 g/ml at 20 C) Isomers: Structural Isomers: Larson-Foothill College 7 Empirical Formulas from Mass Data If mass % or mass data is known - the empirical formula can be found for a compound. 1. Convert % composition to mass in g % A > g A > x mol A % B > g B > y mol B 2. Convert mass to moles 3. Find mole ratio(s) x mol A y mol B 4. Ratio(s) give empirical formula A x B y In the reaction Cr + O 2 > Cr x O y, 0.345 g of Cr produces 0.504 g of Cr x O y. What is the empirical formula of Cr x O y? 8
Molecular Formula From Empirical Formula 1. For molecular substances, the empirical formula may not be the molecular formula. However, they are related: (Molecular formula) = n(empirical formula). Where n is a whole number. 2. To determine the molecular formula from the empirical formula, we need the molecular mass (or molar mass): molar mass (g/mol) n = empirical formula mass (g/mol) Adipic acid is used to make nylon. It contains 49.31% carbon, 6.90% hydrogen and the remainder oxygen by mass. The molar mass is 146.1g. What is the molecular formula of adipic acid? 9 Combustion Analysis to find Empirical Formula The empirical formula of simple organic compounds can be determined by complete combustion and mass analysis of the products. C x H y O z + O 2 > CO 2 + H 2 O Use original sample mass to find g O g CO 2 g C x H y O Burn in O z 2 mol CO 2 mol C g C g O mol O g H 2 O mol H 2 O mol H g H empirical formula of C x H y O z 10
Practice Combustion Analysis Problem Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005-g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the empirical formula for menthol? If the compound has a molar mass of 156 g/mol, what is its molecular formula? Write the balanced combustion reaction. 11 Chemical Equations A chemical equation uses formulas to express the identities and quantities of substances involved in a physical or chemical change. The formation of HF gas on the macroscopic and molecular levels. Subscripts indicate the number of atoms of each element in a molecule or formula unit. Coefficients indicate the number of molecules (or formula units) or the number of moles of molecules (or moles formula units). 12
Balancing Chemical Equations - Steps 1. Write correct formulas for reactants and products. 2. Balance atoms using coefficients; formulas cannot be changed: Balance groups of atoms that are common to both reactants and products first, e.g. NO 3, SO 4 2, etc. Balance metals next. Balance nonmetals last. Check your answer. 3. Specify states of matter: (s), (l), (g) or (aq) Write the chemical formulas for reactants and products and then balance the following reactions, including phase symbols: Hydrogen Gas Reacts with oxygen gas to produce liquid water. Aqueous barium chloride reacts with aqueous sodium phosphate to produce solid barium phosphate and a solution containing sodium chloride. 13 Balancing Simple Organic Combustion Reactions Organic combustion reactions involve an organic compound and oxygen gas as reactants. Simple organic compounds contain only C, H and O atoms. The products of a simple organic combustion are only CO 2 and H 2 O. These reactions are always exothermic. You may have learned the term exothermic in a previous introductory course. If so, do you remember what it means? To balance these reactions: 1. Write correct formulas for reactants and products. 2. Balance the C atoms with CO 2. (All the C is turned into CO 2 ) 3. Balance the H atoms with H 2 O. (All the H is turned into H 2 O) 4. Balance the O atoms with O 2. 5. Check your answer. Balance the following combustion reaction: C 3 H 8 (g)+ O 2 (g) >? 14
Mass Relationships in Stoichiometry The coefficients in a chemical equation can be interpreted in two ways: 1. As individual formula units 2. As moles of formula units We will usually use the later - moles! The coefficients allow us to convert from moles of one substance into moles of another using mole ratios (stoichiometric factors). For the reaction 3A > 2B the following mole ratios are indicated: 3 moles reacted A 2 moles B produced or 2 moles B produced 3 moles A reacted Larson-Foothill College Stoichiometry 15 Examples of Stoichiometry 1. Automotive air bags inflate when sodium azide, NaN 3, rapidly decomposes to its component elements: Molar Masses: Na 22.99 g/mol N 2 28.02 g/mol NaN 3 65.02 g/mol grams NaN 3? grams Na, grams N 2 Molar Mass moles NaN 3 Balanced Equation Molar Mass moles Na, moles N 2 a) If 10.0 g of NaN 3 reacts, how many moles of Na form? What mass of sodium oxide can be obtained from this amount of Na? b) How many grams of NaN 3 are required to produce 0.283 m 3 of nitrogen gas if the N2 gas has a density of 1.25 g/l? Larson-Foothill College Stoichiometry 16
Limiting Reactant Calculations In this type of calculation two (or more) reactant amounts are given. One of the reactants is completely used up, leaving an excess of the other reactant(s). We call the used up reactant the limiting reactant. The limiting reactant determines the maximum amount of product(s) that can be formed. The trick is to determine which reactant is limiting! For this we use stoichiometry. Zn excess HCl limits Correct stoichiometric ratio Zn limits HCl excess Larson-Foothill College Zn(s) + 2 HCl(aq) > ZnCl2(aq) + H2(g) Stoichiometry 17 Practice Limiting Reactant Problem Silicon, the second most abundant element in the Earth s crust after oxygen, rarely occurs as the pure free element in nature. Pure silicon, required for computer ships and solar cells, can be produced by the following reaction: SiCl 4 (l) + 2Mg(s) > Si(s) + 2MgCl 2 (s) Starting with 225.0 g of silicon tetrachloride and 75.0 g of magnesium, how many grams of silicon can be produced? How many grams of each reactant remain? Silicon is a solid at room temperature, with relatively high melting and boiling points of 1414 and 3265 C, respectively. Like water, it has a greater density in a liquid state than in a solid state, and so, like water but unlike most substances, it does not contract when it freezes, but expands. Silicon conducts heat well and is a semiconductor. In its crystalline form, pure silicon has a gray color and a metallic luster. Like germanium, silicon is rather strong, very brittle, and prone to chipping. 18
Yields in Chemistry The theoretical yield is the maximum amount of product that can be produced. This is determined by a limiting reactant calculation. The actual yield is the amount of product actually produced in an experiment. This is determined experimentally and should be less then the theoretical yield. It should never be more than the theoretical yield! percent yield = actual yield theoretical yield x 100% Think about it. Why are percent yields often less that 100%? What would you think if you obtained a percent yield greater than 100%? 19 Yields in Chemistry From our practice problem of limiting reactants, if 33.6 g of Si is actually produced in an experiment, what is the percent yield? If this yield is typical, what mass of Mg is needed to produce 50.0 g of Si? Silicon is metalloid, readily either donating or sharing its four outer electrons. It typically forms four bonds. Like carbon, silicon s four bonding electrons give it opportunities to combine with many other elements or compounds to form a wide range of compounds. Unlike carbon, it can accept additional electrons and form five or six bonds Spectral Lines of Silicon 20
More Practice Problems 1. Without doing any detailed calculations, rank the following samples in order of increasing total number of atoms: 0.50 mol H 2 O 23 g Na 6.0 10 23 N 2 molecules. 2. One molecule of the antibiotic known as penicillin G has a mass of 5.342 10 21 g. What is the molar mass of penicillin G? 3. Hemoglobin, the oxygen-carrying protein in red blood cells, has four iron atoms per molecule and contains 0.340% iron by mass. Calculate the molar mass of hemoglobin. 21 4. An organic compound was found to contain only C, H, and Cl. When a 1.50-g sample of the compound was completely combusted in air, 3.52 g of CO2 was formed. In a separate experiment the chlorine in a 1.00-g sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound. 22
5. A 0.652 g sample of a pure strontium halide reacts with excess sulfuric acid. The solid strontium sulfate formed is separated, dried, and found to weigh 0.755 g. Assuming100% yield for the strontium sulfate, what is the identity of the halide? 23