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Stoichiometry: Calculations with Chemical Formulas and Equations Chapter 3 Chapter 3 Chemical Equations Balancing, Types of Equations Compound Composition Atom mass; Atomic and formula weight % Composition Empirical Formulas (we talked about this already) Mole Amounts; Molar Mass Mole Calculations Amounts in chemical reactions Chemical Equations are used to describe chemical reactions (mass balance) BEFORE AFTER Consider the combustion reaction shown below where one methane molecule is burned by 2 oxygen molecules to produce 1 carbon monoxide and 2 water molecules Note that this is a balanced equation; the number of atoms in the reactant side equals the number of atoms in the product side Combustion Reactions in air (burning of a hydrocarbon with O2) Hydrocarbons react with oxygen to give carbon dioxide and water: CH3CH3(g) + O2(g) CO2(g) + H2O(g) ??? (For balancing equations the text outlines a simple procedure that you may find useful) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) Combination Reactions (combining of reagents A and B to produce a compound C) Combination Reactions: A + B C 2Mg(s) + O2(g) 2MgO 2Na (s) + Cl2 (g) 2 NaCl (s) Decomposition Reactions: A compound decomposes or disproportionates into its constituents atoms or other stable entities Decomposition Reactions: C A + B Sodium azide decomposes to its constituent elements: NaN3(s) 3/2N2 (g) + Na(s) 2NaN3(s) 3N2 (g) + 2Na(s) (N3- is a polyatomic ion) CaCO3(s) CaO (s) + CO2(g) Decomposition Reactions Some compounds decompose to more stable compounds: 2H2O2(l) 2H2O(l) + O2(g) 2HgO(s) 2Hg(l) + O2(g) Chemical Composition The quantitative significance of formulas Atomic Weight Molecular Weight / Formula Weight Percent Composition Empirical Formulas What is the mass of an atom? We use the Atomic Mass Unit Scale to describe masses of atoms. C-12 atom 1.99848 x 10-23 g 12.0000 amu H-1 atom 1.6735 x 10-24 g 1.0078 amu 1 atomic mass unit (amu) = 1.66054 x 10-24 g What is the Mass of an Atom (atomic weight is based on the natural occurrence of isotopes ? When we describe the mass of atoms of an element, we must consider all of the naturally occurring isotopes. Carbon is composed of 98.8925 % C-12 (12.00000 amu) 1.1080 % C-13 (13.00335 amu) Its weighted average is 12.011 amu. Its atomic weight is 12.011 amu. What is the Atomic Weight of Copper? Atomic weight is different than mass number. (62.94 amu)(0.6917) = 43.5356 (64.93 amu)(0.3083) = 20.0179 63.55 amu Atomic Weights 29 Cu 63.55 Atomic Weight Atomic Weight Boron is comprised of 10B and 11B. Which isotope is more abundant? Mole Scale We usually measure substances in terms of gram (kilogram) scales which are billions upon billions of times larger than the amu scale. 1 amu = 1.66054 x 10-24 g 1 gram = 6.02214 x 1023 amu The term mole is used to correlate a huge number of particles (atoms, molecules...ions ) to a manageable amount (weight) of a sample 1 mole has 6.02214 x 1023 members (atoms molecules ions..) 1 dozen has 12 members Mole Scale The mass of one C-12 atom is exactly 12 amu. The mass of 6.02214x1023 C-12 atoms is 12.0000 grams. The mass of exactly 1 mole of C-12 atoms is exactly 12 grams. 6.02214 x 1023 a carbon sample weighs 12.01 grams Avogadro’s Number 6.02214199 x 1023 Avogadro’s number: 6.02214x1023 particles per mole (of anything) Avogadro’s Number 6.02214199 x 1023 1 mole = 6.022 x 1023 C-12 atoms = 6.022 x 1023 C atoms = 6.022 x 1023 H2O molecule = 6.022 x 1023 NaCl formula units Avogadro’s Number Use Avogadro’s number as a conversion factor to convert between moles and number molecules or formula units Also number of molecules or formula units and mass 1 mole = 6.022x1023 molecules or formula units Moles 1 mole of H2O Contains 6.022x1023 H2O molecules (18 grams) Contains 2x(6.022x1023) H atoms (2 grams) Contains 1x(6.022x1023) O atoms (16 grams) Molar Mass 1 mole of C-12 has a mass of 12.00 g of C has a mass of 12.01 g of H has a mass of 1.008 g of H2O has a mass of 18.02 g of Ca(NO3)2 has a mass of 164.098 g Molar mass = mass of 1 mole of that substance Molar Mass If you write the units on molar mass as X g/mol, then molar mass can be used as a conversion factor for converting between moles of a substance and grams of a substance. What is the mass (in grams) of 0.500 mole of oxygen gas? Convert 5.7 g CCl4 to grams to mole of CCl4 (MW CCl4 = 153.81 g/mol) (5.7 grams of CCl4) x (1 mole CCl4/153.81) = 0.037 mol CCl4 Convert 10.50 grams of N2 gas to number of N2 molecules. 0.3747 mol N2 2.257x1023 N2 molecules Use the concept of the mole to go from mass to mummer of particles Percent Composition (#Atoms in formula)(MM of atom) x 100 MM of Compound What is the % by mass composition of Cl or Mg in MgCl2? 25.53% Mg Weight of Mg in MgCl2/total weight Percent Composition What is the mass %Mg in MgCl2? Group Work How would you go about determining the mass of carbon in 5.0 g CCl4? Group Work How would you go about determining the % of carbon in 5.0 g CCl4? (12.011 grams of C in 153.811 g of CCl4) How many in 5 grams of CCl4 (5 g) (12.011/ 153.811) = (0.078089)(5.0g) = 0.39 g % Composition What is the percent carbon in C2H4? (24/28) ??? Why % C = 85.63% What is the percent carbon in C3H6? % C = 85.63% Empirical Formula Different compounds with the same percent composition have the same empirical formula. The empirical formula shows the ratio of atoms of each element as the smallest whole-number ratio. What is the empirical formula for Benzene (C6H6)? Determining the Empirical Formula Can we determine the formula for a compound from the % composition? Empirical formula % Composition (mole ratio) (mass ratio) Given that the % composition of a compound is 30.45% N and 69.56% O, what is the formula for the compound? Assume 100 g of compound: 30.45 g N and 69.56 g O Convert g to moles: 30.45 g N /14 = 2.173 mol N (69.56 g O)/(16.00g O) = 4.348 mol O Divide each by the smallest number to get whole-number subscripts: (2.173 mol N)/(2.173) = 1 mol N (4.348 mol O )/(2.173) = 2 mol O NO2 Determining the Molecular Formula for a Compound Now that we know that the empirical formula is NO2, can we determine the molecular formula? Only if we know the molecular weight. Determining the Molecular Formula for a Compound If the molecular weight of the compound is given to us as 46.01 amu, then the molecular formula is NO2. If the molecular weight of the compound is given to us as 92.02 amu, then the molecular formula is N2O4. Models of NO2 and N2O4 Same % Composition Same Empirical Formula Different molecular weights

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