CHEM 141 Lecture 5 PDF

CHEM 141 M FALL 2024 – LECTURE 5 SEPTEMBER 3, 2024 – GREYTAK Reading: ‒ Today: Brown Chpt. 3.2-3.5 (mass fractions; moles) ‒ Next time: Brown Chpt. 3.6-3.7 (reaction yield and limiting reagents) Homework: ‒ HW 2 due tonight!...

CHEM 141 M FALL 2024 – LECTURE 5 SEPTEMBER 3, 2024 – GREYTAK Reading: ‒ Today: Brown Chpt. 3.2-3.5 (mass fractions; moles) ‒ Next time: Brown Chpt. 3.6-3.7 (reaction yield and limiting reagents) Homework: ‒ HW 2 due tonight! A precisely machined sphere of 28Si used to define Avogadro’s number as 6.02214076×1023. (credit: National Institutes of Standards & Technology) CHEM 141 NAMING INORGANIC COMPOUNDS (2.8): COMMON CATIONS AND ANIONS 2 CHEM 141 TEST YOUR SKILL: BALANCED EQUATIONS (3.1) Balance the equation C4H9OH(l) + O2(g) → CO2(g) + H2O(g) 3 CHEM 141 TEST YOUR SKILL: BALANCED EQUATIONS Solid calcium phosphate and an aqueous sulfuric acid solution react to give calcium sulfate, which comes out of the solution as a solid. The other product is phosphoric acid, which remains in solution. Write a balanced equation for the reaction using complete formulas for the compounds with phase labels. Ca3(PO4)2(s) + 3H2SO4(aq) → 3CaSO4(s) + 2H3PO4(aq) 4 CHEM 141 THREE BASIC TYPES OF REACTIONS (3.2) Many types of chemical reactions: several types come up often! Combination reactions (*including formation reactions) Decomposition reactions Combustion reactions 5 CHEM 141  COMBINATION REACTIONS In combination reactions two or more substances react to form one product. A formation reaction is a (possibly hypothetical) reaction that forms a compound directly from the elements Some combination reactions: 2 Mg(s) + O2(g) 2 MgO(s)* N2(g) + 3 H2(g) 2 NH3(g) C3H6(g) + Br2(l) C3H6Br2(l) 6 CHEM 141  DECOMPOSITION REACTIONS In a decomposition reaction one substance breaks down into two or more substances. Examples: CaCO3(s) CaO(s) + CO2(g) 2 KClO3(s) 2 KCl(s) + O2(g) 2 NaN3(s) 2 Na(s) + 3 N2(g) Decomposition of sodium azide has been used to inflate airbags in cars. 7 CHEM 141 COMBUSTION REACTIONS Combustion reactions are generally rapid reactions that produce a flame. Combustion reactions most often involve oxygen in the air as a reactant. Examples: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) 8 CHEM 141 MOLECULAR WEIGHT (MW), FORMULA WEIGHT (FW), AND % COMPOSITION (3.3) A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C2H6, the molecular weight would be C: 2(12.011 amu) + H: 6(1.00794 amu) 30.070 amu 9 CHEM 141 PERCENT COMPOSITION One can find the mass percentage (% of the mass of a compound that comes from any of the elements) from the compound’s molecular or empirical formula as follows: (number of atoms)(atomic weight) % Element = × 100% (FW of the compound) 10 CHEM 141 EXAMPLE: PERCENTAGE CALCULATION What is the mass percentage composition of H, C, and O in oxalic acid (H2C2O4)? Answer: H, C, O = 2.24, 26.68, 71.08 % 11 CHEM 141 IONIC COMPOUNDS AND FORMULAS Remember, ionic compounds exist with a three-dimensional order of ions. There is no simple group of atoms to call a molecule. As such, ionic compounds use empirical formulas and formula weights (not molecular weights). 12 CHEM 141 FORMULA WEIGHT (FW) A formula weight is the sum of the atomic weights for the atoms in an ionic chemical formula. This is the quantitative significance of a formula. The formula weight of calcium chloride, CaCl2, would be Ca: 1(40.08 amu) + Cl: 2(35.453 amu) 110.99 amu 13 CHEM 141 THE MOLE (3.4) ‒ Dihydrogen oxide (water, H2O, molecular weight 18.0 amu) contains 16 grams of O for every 2.0 grams of H ‒ 18.0 g H2O has the same number of molecules as 2.0 grams of H2 ‒ Sodium hydride (NaH) contains 23.0 grams of Na for every 1.0 gram of H ‒ So 24 grams of NaH contains the same amount of H as 1.0 gram of H2. Wouldn’t it be nice if there was a convenient way to specify the number of molecules or number of atoms that’s about the same as there are H atoms in 1 gram of H? There is, and it’s called the “mole”: 14 CHEM 141 AVOGADRO’S NUMBER (NA) Chemical processes happen between atoms and molecules. But in a lab, if we want to know the number of atoms in anything of ordinary size, we usually count them by weighing. 6.02214 × 1023 atoms or molecules is an amount that brings us to lab size. It is ONE MOLE. 15 CHEM 141  THE MOLE: A CHEMIST’S “DOZEN” When we count large numbers of objects, we often use units such as ‒ 1 dozen objects = 12 objects. ‒ 1 gross objects = 144 objects (a dozen dozen). The chemist’s “dozen” is the mole (abbreviated mol). A mole is the measure of material containing 6.02214 × 1023 particles. 1 mole = 6.02214  10 23 particles This number is Avogadro’s number. 16 CHEM 141 THE MOLE First thing to understand about the mole is that it can specify Avogadro’s number of anything. For example, 1 mole of marbles corresponds to 6.02214 × 1023 marbles. 1 mol of sand grains corresponds to 6.02214 × 1023 sand grains. One mole of anything is 6.02214 × 1023 units of that thing. 17 CHEM 141 THE MOLE The second, and more fundamental, thing to understand about the mole is how it gets its specific value. The value of the mole is equal to the number of atoms in exactly 12 grams of pure 12C 𝟏𝟐 𝟏𝟐 𝒈 𝑪 = 𝟏 𝒎𝒐𝒍 𝑪 𝒂𝒕𝒐𝒎𝒔 = 𝟔. 𝟎𝟐𝟐𝟏𝟒 × 𝟏𝟎𝟐𝟑 𝑪 𝒂𝒕𝒐𝒎𝒔 18 CHEM 141 MOLAR MASS The molar mass is the mass of 1 mol of a substance (i.e., g/mol). The molar mass of any element is the atomic weight for that element taken from the periodic table. ‒ Watch out: the molar mass of a diatomic gas is 2x the molar mass of the element! The formula weight (in amu’s) will be the same number as the molar mass (in g/mol). 19 ‒ Remember, 1 amu was the 1/12 CHEM 141 the mass of one atom of 12C. MOLE RELATIONSHIPS One mole of atoms, ions, or molecules contains Avogadro’s number of those particles. One mole of molecules or formula units contains (Avogadro’s number × the number of atoms or ions in the formula) of each element. For example: 20 CHEM 141 STOICHIOMETRIC CALCULATIONS AND THE MOLE BRIDGE Mole Ratio Moles Moles Molar Molar Avogadro’s Avogadro’s Mass Mass Number (N) Number (N) Molarity Molarity (mol/L) (mol/L) Grams Grams # molecules or atoms # molecules or atoms Liters Liters 22 CHEM 141 DETERMINING A CHEMICAL FORMULA FROM EXPERIMENTAL DATA (3.5) Remember, the empirical formula is the simplest, whole- number ratio of the atoms or moles of elements in a compound, not a ratio of masses Can be determined from elemental analysis ‒ Percent composition ‒ Masses of elements formed when a compound is decomposed, or that react together to form a compound Concept of moles simplifies this process 23 CHEM 141 EXAMPLE: DETERMINING EMPIRICAL FORMULAS The compound para-aminobenzoic acid (PABA, a largely discontinued component of sunscreens) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. Tip: When given percentage data, assume the reaction was carried out with a 100-g sample. Answer: C7H7NO2 24 CHEM 141 DETERMINING EMPIRICAL FORMULAS One can determine the empirical formula from the percent composition by following these three steps. 25 CHEM 141 FINDING AN EMPIRICAL FORMULA 1. Convert the percentages to grams. a. Assume you start with 100 g of the compound. b. Skip if it is already in grams. 2. Convert grams to moles using the molar mass of each element. 3. Write a pseudoformula using moles as subscripts. 4. Divide all by the smallest number of moles. a. If the result is within 0.1 of a whole number, round to the whole number. 5. Multiply all mole ratios by a number to make all whole numbers. a. If ratio is.5 (about ½), multiply all by 2. b. If ratio is.33 or.67 (about 1/3 or 2/3), multiply all by 3. c. If ratio is 0.25 or 0.75 (about ¼ or ¾), multiply all by 4, etc. d. Skip if ratios are already whole numbers. 26 CHEM 141 TRY ON YOUR OWN: DETERMINING EMPIRICAL FORMULAS What is the empirical formula of a chromium oxide that is 68.4% Cr and 31.6% O? How about a magnesium oxide that appears to be 60.3% Mg? Answers: Cr2O3, MgO 27 CHEM 141 DETERMINING A MOLECULAR FORMULA Remember, the number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula. If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula. 28 CHEM 141 EXAMPLE: MOLECULAR FORMULA An empirical formula calculation based on a combustion analysis experiment shows a new compound has the empirical formula C4H8O. A mass spectrometry experiment determines that the molar mass of the compound is 216 g/mol. What is the molecular formula? Answer: C12H24O3 29 CHEM 141 COMBUSTION ANALYSIS A common technique for analyzing compounds is to burn a known mass of a compound and weigh the products. ‒ This is generally used for organic compounds containing C, H, and O. By knowing the mass of the products and composition of the constituent element in the product, the original amount of the constituent element can be determined. ‒ All the original C forms CO2, the original H forms H2O, and the original mass of O is found by subtraction. Once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found. 30 CHEM 141 COMBUSTION ANALYSIS 31 CHEM 141 EXAMPLE: COMBUSTION ANALYSIS A compound contains only C, H, and O. A 0.1000 g-sample burns completely in oxygen to form 0.0930 g of water (H2O) and 0.227 g of CO2. Calculate the mass of each element in this sample, then determine the empirical formula. Answer: C3H6O 32 CHEM 141 RE-DEFINING THE KILOGRAM? Avogadro’s # has been defined as a ratio of 12 grams (exactly 0.012 kg) to 12 amu New measurements make it possible to estimate the number of atoms of silicon in a polished, spherical single crystal to extremely high precision This could enable the kg to be defined based on the mass of a particular number of 28Si atoms A precisely machined sphere of 28Si How many moles of 28Si is 1 kg if (credit: National Institutes of Standards & Technology) NA=6.02214076×1023? 33 CHEM 141 THANKS!