Empirical & Molecular Formulas Chapter 8 Empirical & Molecular Formulas

Types of Formulas The molecular formula of a compound indicates the actual number of each atom present in a compound. An empirical formula of a compound indicates the smallest whole number ratio of the elements of each atom present in a compound.

Empirical and Molecular Formulas Substance Molecular Formula Empirical Formula Benzene C6H6 CH Acetylene C2H2 Glucose C6H12O6 CH2O Water H2O

Chemical Formulas of Compounds Chemical formulas give the relative numbers of atoms or moles of each element in a compound - always a whole number ratio. Example CO2 2 atoms of O for every 1 atom of C or 2 moles of O for every 1 mole of C If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

To Calculate an Empirical Formula 1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of moles of each element. 3. Divide each answer from step 2 by the smallest number of moles to obtain the simplest whole number ratio. If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers * Be careful! Do not round off numbers prematurely

A sample of a brown gas, a major air pollutant, is found to contain 2 A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance.

A sample of a brown gas, a major air pollutant, is found to contain 2 A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N 5.34g O

A sample of a brown gas, a major air pollutant, is found to contain 2 A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N mol = 0.167 mol N 14.0g 5.34g O mol = 0.334 mol O 16.0g

A sample of a brown gas, a major air pollutant, is found to contain 2 A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N mol = 0.167 mol N 14.0g N0.167O0.334 5.34g O mol = 0.334 mol O 16.0g

A sample of a brown gas, a major air pollutant, is found to contain 2 A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N mol = 0.167 mol N 14.0g N0.167O0.334 5.34g O mol = 0.334 mol O 16.0g 0.167 mol N 0.334 mol O = 1 mol N = 2 mol O 0.167 0.167

A sample of a brown gas, a major air pollutant, is found to contain 2 A sample of a brown gas, a major air pollutant, is found to contain 2.34g N and 5.34g O. Determine a formula for this substance. 2.34g N mol = 0.167 mol N 14.0g N0.167O0.334 5.34g O mol = 0.334 mol O 16.0g 0.167 mol N 0.334 mol O = 1 mol N = 2 mol O 0.167 0.167 NO2

To Calculate a Molecular Formula 1. Determine the empirical formula and calculate its mass. 2. Divide the molecular mass by the mass of the empirical formula you determined in step 1. Your answer should be rounded to the nearest whole number. 3. Multiply the empirical formula by the whole number answer found in step 2.

A colorless liquid, used in rocket engines has a molar mass of 92 A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molecular formula of the compound if the empirical formula of NO2?

A colorless liquid, used in rocket engines has a molar mass of 92 A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molecular formula of the compound if the empirical formula of NO2? NO2 = 14.0 + 2(16.0) = 46.0 g/mol

NO2 = 14.0 + 2(16.0) = 46.0 g/mol 92.0 g/mol = 2 46.0 g/mol A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molecular formula of the compound if the empirical formula of NO2? NO2 = 14.0 + 2(16.0) = 46.0 g/mol 92.0 g/mol = 2 46.0 g/mol

NO2 = 14.0 + 2(16.0) = 46.0 g/mol 92.0 g/mol = 2 46.0 g/mol A colorless liquid, used in rocket engines has a molar mass of 92.0 g/mol. What is the molecular formula of the compound if the empirical formula of NO2? NO2 = 14.0 + 2(16.0) = 46.0 g/mol 92.0 g/mol = 2 46.0 g/mol 2 x NO2 = N2O4

What is the empirical formula of a compound that contains 53 What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?

What is the empirical formula of a compound that contains 53 What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? 53.73g Fe 46.27g S

What is the empirical formula of a compound that contains 53 What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? 53.73g Fe mol 46.27g S mol = 0.963 mol Fe = 1.44 mol S 55.8g 32.1g

What is the empirical formula of a compound that contains 53 What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? 53.73g Fe mol 46.27g S mol = 0.963 mol Fe = 1.44 mol S 55.8g 32.1g 0.963 mol Fe 1.44 mol S = 1 Fe = 1.5 S 0.963 mol 0.963 mol

What is the empirical formula of a compound that contains 53 What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S? 53.73g Fe mol 46.27g S mol = 0.963 mol Fe = 1.44 mol S 55.8g 32.1g 0.963 mol Fe 1.44 mol S = 1 Fe = 1.5 S 0.963 mol 0.963 mol FeS1.5 x 2 = Fe2S3

What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram formula mass of 624 g/mol?

What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram formula mass of 624 g/mol? Fe6S9

3 x Fe2S3 = Fe6S9 2Fe + 3S = 2(55.8) + 3(32.1) = 207.9 g/mol 624 g/mol What is the molecular formula of a substance with the empirical formula Fe2S3 and a gram formula molecular mass of 624 g/mol? 2Fe + 3S = 2(55.8) + 3(32.1) = 207.9 g/mol 624 g/mol 207.9 g/mol 3 x Fe2S3 = Fe6S9 ≈ 3

What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the percent composition of 75.7% C, 8.80% H, and 15.5% O?

What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the percent composition of 75.7% C, 8.80% H, and 15.5% O? 75.7g C 8.80g H 15.5g O

What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the percent composition of 75.7% C, 8.80% H, and 15.5% O? 75.7g C mol C = 6.31 mol C 12.0g C 8.80g H mol H = 8.80 mol H 1.0 g H 15.5g O mol O = 0.969 mol O 16.0g O

What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the percent composition of 75.7% C, 8.80% H, and 15.5% O? 75.7g C mol C 6.31 mol C = 6.31 mol C = 6.5 C 12.0g C 0.969 mol 8.80g H mol H 8.80 mol H = 9 H = 8.80 mol H 1.0 g H 0.969 mol 15.5g O mol O 0.969 mol H = 1 O = 0.969 mol O 16.0g O 0.969 mol

What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the percent composition of 75.7% C, 8.80% H, and 15.5% O? 75.7g C mol C 6.31 mol C = 6.31 mol C = 6.5 C 12.0g C 0.969 mol 8.80g H mol H 8.80 mol H = 9 H = 8.80 mol H 1.0 g H 0.969 mol 15.5g O mol O 0.969 mol H = 1 O = 0.969 mol O 16.0g O 0.969 mol C6.5H9O x 2 = C13H18O2

Therefore the molecular formula is also C13H18O2 What are the empirical and molecular formulas for Ibuprofen if the molar mass is 206 g/mole and the percent composition of 75.7% C, 8.80% H, and 15.5% O? C13H18O2 13(12.0) + 18(1.0) + 2(16.0) = 206g/mol Therefore the molecular formula is also C13H18O2

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