CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1) 1. Just as we can talk about mass of one atom of an element as its Atomic Mass (or Atomic Weight), we can also extend that to mass of one molecule of a substance, the Molecular Mass (or Molecular Weight). Molecular Mass = sum of the Atomic Masses in a molecule Example: Molecular Mass of water is 18.0 amu 2. If the substance is ionic, we talk about its Formula Mass (or Formula Weight). Formula Mass = sum of the Atomic Masses in its formula Example: Formula Mass of CaF2 is 78.1 amu B. The Mole Concept. (Sections 3.2) 1. Atoms combine together most fundamentally by number. (Their fixed mass-combining ratios are simply a byproduct of this) When atoms interact, they COUNT each other, they don’t WEIGH each other. 2. Problem: a laboratory sized sample of matter contains an enormously large number of atoms or molecules. 1 Liter of air contains ~ 3 x 1022 molecules 3. It is inconvenient to handle these large numbers in our chemical calculations, and yet we must continually keep track of numbers of molecules. 4. Solution: we have adopted the mole concept. a. Just like a baker uses the dozen to describe 12 of something, or a hardware store uses a gross to refer to 144 of something. Chapter 3 Page 1 b. Better yet the hardware store weighs out nails rather than counting them. Chemists do the same thing, weigh samples but know how to determine the number of atoms or molecules in that mass. c. Chemist uses the mole to refer to number of atoms in one gram of hydrogen. (original definition) Now many atoms is that? Turns out to be: 6.02 x 1023 of them. Since Carbon atoms are 12 times the mass of Hydrogen atoms, 12 grams of carbon would be one mole of Carbon atoms. 16 grams of Oxygen would be one mole of Oxygen atoms, and so forth. d. Modern definition of the mole is based on Carbon and not Hydrogen, because Carbon is easier to handle. 1 mole = 6.02 x 1023 particles of anything countable. D. The Mole. (Section 3.2) 1. One mole = amount of any substance that contains as many particles of that substance as there are atoms of C in 12 g pure Carbon-12. 1 mole = 6.0221367 x 1023 particles = Avogadro’s number One mole of Fe is 6.022 x 1023 atoms of Fe 23 One mole of H2O is 6.022 x 10 molecules of H2O One mole of NaCl is 6.022 x 1023 formula units of NaCl 2. Abbreviation for mole is mol. 3. Why is mole defined this way? Because now: 1 mole of element X = Atomic Mass of X in grams Examples: 1 mole of C = 12.011 grams C 1 mole of Na = 22.99 grams Na Chapter 3 Page 2 4. Atomic Mass (Atomic Weight) of an element = number of grams in 1 mole of that element (55.85 = number of grams in mol of Fe): 26 Fe 55.85 Abbreviate Atomic Mass as AM 5. So now: 6.022 x 1023 atoms of Fe = 1 mol Fe = 55.85g Fe ÷ 6.022 x 1023 x AM of X atoms of X moles of X grams of X x 6.022 x 1023 ÷ AM of X 6. Problems: a. How many Na atoms in 10.00g Na? 1mol Na 6.022 ×1023atoms 10.00gNa × × 22.99g Na 1mol = 2.62 ×1023 Na atoms b. A reaction used up 5.00g Ni. How many moles Ni were used up? € 1 mol Ni −2 5.00gNi × = 8.52 ×10 mol Ni 58.69g Ni = 0.0852mol Ni c. Which weighs more, a mole of C or a mole of H? By what factor? € mol of C weighs ≈ 12 times a mol of H Chapter 3 Page 3 d. Calculate the mass of a Hydrogen atom in grams. 1 mol H 1.008g H 1atom H × 23 × 6.022 ×10 atoms H mol H =1.674 ×10−24 grams B. Molar Mass. (Section 3.2) €1. Molar Mass of a substance = mass of one mole of particles of that substance. 2. Molar Mass of a substance = the number of grams of that substance equal to the sum of Atomic Masses of elements in the formula. 3. Example: Molar Mass of CaF2. = 40.078 + 2 x 19.00 = 78.08 grams/mol 4. Example: Molar Mass of NaOH. = 23.00 + 16.00 + 1.01 = 40.01 grams/mol Therefore: 40.01g NaOH = 1 mol NaOH 23 = 6.022 x 10 NaOH units 23 + 23 - = 6.022 x 10 Na ion plus 6.022 x 10 OH ions 5. Example: Molar Mass of H2O. = 2 x 1.008 + 16.00 = 18.02 grams/mol 23 Thus, 18.02g H2O = 1 mole H2O = 6.022 x 10 H2O molecules. 23 And contains 6.022 x 10 oxygen atoms 23 12.04 x 10 hydrogen atoms 6. Example: Molar Mass of H2 = 2 x 1.008 = 2.016 grams/mol 23 Thus, 2.016 g H2 = 1 mole H2 = 6.022 x 10 H2 molecules in which there are 12.04 x 1023 H atoms present. Chapter 3 Page 4 7. For a molecular substance S (e.g. S = H2, H2O, CH4, etc...) ÷ 6.022 x 1023 x MolarMass molecules of S moles of S grams of S x 6.022 x 1023 ÷ MolarMass 8. How many molecules of ozone in 24.0g ozone? 23 1 mol O3 6.022 ×10 molecules O3 24.0g O3 × × 48.00g O 1 mole O 3 3 23 = 3.01×10 molecules O3 9. How many H atoms are present in 48.0g methane (CH4)? € 23 1 mol CH4 6.022 ×10 molecules CH4 48.0g CH4 × × 16.04g CH4 1 mole CH4 24 =1.80 ×10 molecules CH4 24 4 H for every molecule CH4, so (times 4) = 7.21 x 10 H atoms. € 10. How many moles of NaF are in 147 grams of it? 11. I want to measure out 2.5 moles of NiSO4. How many grams do I measure out? Chapter 3 Page 5 Part Two: Determining Chemical Formulas A. Percent Composition from Compound Formulas. (Section 3.3) 1. If formula of compound is known, its percent composition by mass of each element can be found, and vice versa. ÷ Atomic Masses (AM) compound percent composition formula in terms of elements x Atomic Masses (AM) 2. First we’ll go from formula to percent composition. Example: Calculate the percent composition of the elements in ammonium nitrate: given formula = NH4NO3 Formula Mass (FM) = 80.0 g/mol 3 elements present; what is % N, % H, % O? mass of N % N = ×100% mass of compound 2 × AM of N = ×100% FM of compound 2 ×14.00 g/mol = ×100% = 35.0% N 80.0 g/mol 4 × AM of H % H = ×100% FM of compound € 4 ×1.008 g/mol = ×100% = 5.04% H 80.0 g/mol 3 × AM of O % O = ×100% FM of compound € 3 ×16.00 g/mol = ×100% = 60.0% O 80.0 g/mol Chapter 3 € Page 6 3. Note that the sum of percentages equals 100%. 35.0 % N + 5.04 % H + 60.0 % O = 100% This makes a good check. B. Chemical Formula from Percent Composition. 1. Knowing the Percent Composition from experiments, we can determine the Chemical Formula. 2. Example: A certain acidic substance has the following composition by mass: 2.45 % H, 39.06 % S, and 58.48 % O. Determine the compound formula. a. We want the mole ratio of the elements present, not their masses. b. Convert masses to moles by dividing by atomic mass (AM). c. For convenience, work with 100 g sample of compound and find moles of each element: 1 mol H 2.45g H × = 2.43 mol H 1.008g H 1 mol S 39.0g S × =1.218 mol S 32.06g S 1 mol O 58.48g O × = 3.655 mol O 16.00g O d. Now take this mole ratio and convert it to simplest whole-number ratio; easiest to first try dividing all 3 numbers by the smallest: € 2.43 mol H =1.99 mol H ≈ 2H 1.218 1.218 mol S =1 mol S ≈1S 1.218 3.655 mol O = 3.001 mol O ≈ 3O 1.218 formula = H2SO3 (sulfurous acid) € Chapter 3 Page 7 3. This method produces only the simplest or empirical formula of a compound. Illustration: 4. Example: Analysis shows that a compound of Hydrogen and Oxygen is 5.93 % H and 94.07 % O. Calculate it’s formula: 1 mol H 5.93g H × = 5.883 mol H 1.008g H 1 mol O 94.07g O × = 5.879 mol O 16.00g O Now divide through by smallest number: € 5.883 mol H =1.001 mol H 5.879 5.879 mol O =1 mol O 5.879 formula = HO € See, this predicts simplest or empirical formula. Actual chemical formula is H2O2, hydrogen peroxide. 5. Percent compositions alone give only empirical formula for a compound. 6. This is generally all that is needed for ionic compounds. 7. For molecular substances we also need to know the Molar Mass. 8. Example: In the preceding problem, if we were also given that the Molar Mass of the compound is 34.0 amu, then we could find the true molecular formula: empirical formula is HO (Formula Mass = 17.0 g/mol) convert to formula H2O2 with Molar mass = 34.0 g/mol Chapter 3 Page 8 9.