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Chapter 7 Mass Relationships in Chemical Reactions Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Introduction In the margin of your notes record two things a chemical formula can tell us A chemical formula indicates: 1. elements present in a compound 2. relative number of atoms/ions of each element present in a compound Introduction Chemical formulas also allow chemists to calculate a number of other characteristic values for a compound: formula mass molar mass percentage composition Formula mass is the sum of the atomic masses of all the atoms in a compound, in atomic mass units (amu) How do we find the atomic mass of an element? • recall masses on P.T. are based on 12C Formula mass for SO2 1S SO2 2O SO2 32.07 amu + 2 x 16.00 amu 64.07 amu 1 molecule SO2 = 64.07 amu NO!! Need to be able to relate amu to a macroscale (grams) Discovery of the MOLE Amedeo Avogadro Italian Scientist The mole: the amount of a substance that contains as many particles as there are atoms in exactly 12.00 grams of 12C 1 mol = 6.02 x 1023 atoms, formula units, molecules How do we measure an amount? 1 mol = 6.02 x 1023 particles 1 mole C atoms = 12.00 g C 1 mole C atoms = 6.022 x 1023 atoms 1 C atom = 12.00 amu Relating Mass to Numbers of Particles Molar Mass the mass in grams of one mole (or approx. 6.02 × 1023 particles) of a substance • calculated by adding the masses of the elements present in a mole (same as recorded values on P.T. compound’s molar mass is numerically equal to its formula mass For any element atomic mass (amu) = molar mass (grams/mol) 1S SO2 2O SO2 32.07 grams/mol + 2 x 16.00 grams/mol 64.07 grams/mol 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 Chemical Formulas Express Composition It is often useful to know the percentage by mass of a particular element in a chemical compound. Mining for Diamonds! Mining for Bauxite: Al(OH)3 Percentage Composition of Iron Oxides The mass percentage of an element in a compound is the same regardless of the sample’s size. There are 2 ways to determine percent composition of a compound: • From a chemical formula • From experimental data Percent Composition: The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100 percent. Percent composition of an element in a compound = n x molar mass of element x 100% molar mass of compound n = the number of moles of the element in 1 mole of the compound Find the percent composition of the elements in water (H2O). H: 2 x 1.01 = 2.02 O: 1 x 16.00 = 16.00 18.02 g/mol H: 2 x1.01 g/mol H 18.02 g/mol H2O X 100 = 11.21 % H O: 16.00 g/mol O 18.02 g/mol H2O X 100 = 88.79 % O Find the percent composition of the elements in Al2(SO4)3. Calculate Total Mass: Al: 2 x 26.98 = 53.96 S: 3 x 32.07 = 96.21 O: 12 x 16.00 = 192.00 342.17 g/mol Find the percent composition of the elements in Al2(SO4)3. - continued. Al : 53.96 g/mol Al X 100 = 15.77 % Al 342.17 g/mol Al2(SO4)3 S: 96.21 g/mol S X 100 = 28.12 % S 342.17 g/mol Al2(SO4)3 O: 192.00 g/mol O 342.17 g/mol Al2(SO4)3 X 100 = 56.11 % O Find the percent composition of a compound that contains 0.9480 g of C, 0.1264 g of O, and 0.0158 g H. C = 0.9480 g O = 0.1264 g H = 0.0158 g 1.0902 g C: 0.9480 g 1.0902 g X 100 = 86.96 % O: 0.1264 g 1.0902 g X 100 = 11.59 % H: 0.0158 g 1.0902 g X 100 = 1.45 % Mole Conversions Mass to Mole Conversions MOLAR MASS 11.2 g NaCl x 1 mol NaCl 58.44 g NaCl Na: 1 X 22.99 = 22.99 Cl: 1 X 35.45 = 34.45 58.44 g/mol 0.192 mol NaCl Mole to Mass Conversions MOLAR MASS 3.2 mol Zn(NO3)2 Zn: 1 X 65.39 = 65.39 N: 2 X 14.07 = 28.14 O: 6 X 16.00 = 96.00 189.53g/mol x 189.53 g Zn(NO3)2 1 mol Zn(NO3)2 606 g Zn(NO3)2 Particle to Mole Conversions AVOGADRO’S NUMBER 8.74 x 1023 atoms CaCO3 x 1 mol CaCO3 6.02 x 1023 atoms CaCO3 1.45 mol CaCO3 Mole to Particle Conversions AVOGADRO’S NUMBER 0.36 mol Al x 6.02 x 1023 atoms Al 1 mol Al 2.2 x 1023 atoms Al MASS Mole Map MOLE 6.02 x 1023 PARTICLES What conversion factor do I use??? Multistep Conversions 250 g C12H22O11 X 1 mol C12H22O11 342.34 g C12H22O11 C: 12 X 12.01 = 144.12 H: 22 X 1.01 = 22.22 O: 11 X 16.00 = 176.00 342.34 g/mol 1 mol C12H22O11 X 6.02 x 1023 molec. C12H22O11 342.34 g C12H22O11 1 mol C12H22O11 4.4 x 1023 molecules C12 H22 O11 Remember Empirical Formula: A chemical formula that gives the simplest whole-number ratio of the elements in the formula. - Subscripts are used for these ratios. Example Problem Determine the empirical formula of a compound found to have 13.5 g of Ca, 10.8 g of O, and 0.675 g of H. 1. 2. 3. 4. Assume 100 g Convert to mole Divide by the smallest Multiply ‘til whole Ca : 13.5 g Ca x 10.8 g O x 1 mol 16.00 g = 0.675 mol O 0.675 g H x 1 mol 1.01 g = 0.668 mol H O: H: 1 mol = 0.337 mol Ca 40.08 g Writing the complete formula: a) Round to whole numbers b) Put parentheses around polyatomic ions c) Re-write the final formula. Ca(OH)2 Remember Molecular Formula: A chemical formula that gives the actual number of the elements in the molecular compound. Example: C2H4 C6H12O6 NOT CH2 NOT CH2O Comparing Empirical and Molecular Formulas • There is a direct relationship between empirical and molecular formulas. • There is a direct relationship between the empirical formula mass and the molecular formula mass. FIND THE COMMON MULTIPLE! The correct ratio can be found by: dividing the experimental formula mass by the empirical formula mass Example Problem The empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this actual compound is 283.89 g/mol. What is the compound’s molecular formula? 1. Find Molar Mass of empirical formula. P: 2 x 30.97 = 61.94 O: 5 x 16.00 = 80.00 141.94 g/mol 2. Divide the experimental formula mass by the empirical formula mass. 283.89 g/mol 141.94 g/mol = 2.00 3. Multiply the subscripts by the common multiple. 2 × (P2O5) = P4O10