Lecture 7. Professor Hicks Inorganic Chemistry (CHE151) Empirical Formula

Transcription

1 Lecture 7 Professor icks Inorganic hemistry (E151) Empirical Formula Experimentally determined chemical formula as smallest whole number ratio of atoms - For most ionic compounds it is the formula unit - For molecular compounds this is generally not the chemical formula NO 2 nitrogen dioxide Examples: N 2 O 4 dinitrogen tetroxide 2 2 acetylene NO 2 and N 2 O 4 have different chemical formula but same empirical formula NO (acetylene )and 6 6 (benzene) different chemical formulas same empirical formula 6 6 benzene 1

2 Empirical formula empirical formula determined experimentally by decomposing a compound chemical formula ethane = 2 6 empirical formula is the chemical formula with the smallest whole number ratio of the elements as they are in the compound 50.0 g ethane 40.0 g 10.0 g convert to moles or 3 mole ratio of elements in ethane divide all subscripts by the smallest one empirical formula of ethane converted to smallest whole number ratio 3.33 moles 10.0 moles there is no way to determine how many of each atom were in the molecule from decomposing and measuring masses all we can do is measure the ratio of atoms Empirical formula chemical formula expressed with smallest whole number ratio of atoms mass compound O calculated divide by atomic mass 1) chemical formula with ratios of moles mass mass O mass moles moles O moles mass element divide by atomic mass moles of element 2) convert smallest whole number ratios 3) round to integers O O O divide all by smallest 2

3 mass compound O O 2 mass mass O mass alculated divide by molar mass moles moles O moles O

4 Percent omposition (aka percent by mass) Ratio of part of a thing to the whole to which it belongs whole The whole could be mixture - Parts are elements or compounds The whole could be a compound - Parts are elements part 1 part 2 part 3 Example: A 10% solution of salt in water has 10 g salt for every 100 grams of the solution 4

5 hemical formulas conversion factors imagine decomposing 1 mole O 2 mass description 1 mole O 2 = 44.0 g O 2 it would form 1 mole = 12.0 g 2 mole O = 2 16 = 32.0 g O 44.0 grams O 2 (mass 1 mole O 2 ) O g g O g O g O 2 visualize a decomposition of compound into elements O O % mass in O 2 % mass O in O g atoms grams O atoms (mass 1 mole ) (mass 2 mole O) chemical formulas conversion factors 44.0 grams O 2 O 2 % mass in O 2 if you know a mass of O 2 this factor will tell you how much carbon is in it g atoms O O grams O atoms mass O 2 g x g O g = mass g % mass in O 2 converts mass of O 2 into mass of 5

6 ombustion reactions compound ( O) + O 2 (g) O 2 (g) + 2 O (l) Example: The combustion reaction of ethanol, 2 6 O. 2 6 O (l) + 3 O 2 (g) 2 O 2 (g) O (l) notice all the is in O 2 notice all the is in 2 O A combustion reaction begins the process of decomposing a substance into its elements by separating and into O 2 and 2 O ombustion Analysis mass O = mass compound mass mass mass compound O use % in O 2 as conversion factor O 2 combustion separates by subtraction done by experiment 2 O use % in 2 O as conversion factor done by calculation mass mass O mass 1) convert to moles 2) empirical formula 6

7 3.108 When g of Mg is heated strongly in a nitrogen (N 2 ) atmosphere, a chemical reaction occurs. The product of the reaction weighs g. alculate the empirical formula of the compound containing Mg and N. 7

8 Molar Mass from experimental data Volatile liquids can be evaporated to a gas Properties of the gas P, V, T measured Number of moles (n) calculated from n = PV/RT Gas is condensed and its mass measured Molar mass calculated as Molar mass = Mass sample Number of moles 8

9 Mass spectrometer Molar Mass + Empirical formula Molecular Formula The molar mass is a small whole number multiple of the mass of the empirical formula If the EF and MM are determined how many empirical formula units in the molecular formula can be calculated Molecular formula calculated Example: A compound is found to have the EF 2. In a separate experiment its molar mass is found to be 56 g/mol. What is its molecular formula? EF weighs 14 g/mol 56/14 =

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11 If the molar mass of butryic acid is determined to be approximately 88 g/mol, what is its molecular formula? Notice this problem is not from textbook 11

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