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CHAPTER 3-C Percent Composition, Molecular Formulae, Combustion Analysis

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## CHAPTER 3-C Percent Composition, Molecular Formulae, Combustion Analysis

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**CHAPTER 3-C**Percent Composition, Molecular Formulae, Combustion Analysis © 2012 by W. W. Norton & Company**Percent Composition: Identifies the elements present in a**compound as a mass percent of the total compound mass. The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage. Percentage Composition**Empirical Formula**• Theempirical formula gives the ratio of the number of atoms of each element in a compound. CompoundFormulaEmpirical Formula Hydrogen peroxide H2O2 OH Benzene C6H6 CH Ethylene C2H4 CH2 Propane C3H8 C3H8**Percentage Composition**• Glucose has the molecular formula C6H12O6. What is its empirical formula, and what is the percentage composition of glucose? Empirical Formula = smallest whole number ratio CH2O**Percentage Composition**CH2O Total mass = 12.01 + 2.02 + 16.00 = 30.03 %C = 12.01/30.03 x 100% = 39.99% %H = 2.02/30.03 x 100% = 6.73% %O = 16.00/30.03 x 100% = 53.28%**Percentage Composition**Saccharin has the molecular formula C7H5NO3S. What is its empirical formula, and what is the percentage composition of saccharin? Empirical Formula is same as molecular formula MW = 183.19 g/mole %C = (7 x 12.011)/183.19 x 100% = 45.89% etc.**A compound’s empirical formula can be determined from its**percent composition. A compound’s molecular formula is determined from the molar mass and empirical formula. Empirical Formula**A compound was analyzed to be 82.67% carbon and 17.33%**hydrogen by mass. What is the empirical formula for the compound? Assume 100 g of sample, then 82.67 g are C and 17.33 g are H Empirical Formula**Convert masses to moles:**82.67 g C x mole/12.011 g = 6.88 moles C 17.33 g H x mole/1.008 g = 17.19 mole H Find relative # of moles (divide by smallest number) Empirical Formula**Convert moles to ratios:**6.88/6.88 = 1 C 17.19/6.88 = 2.50 H Or 2 carbons for every 5 hydrogens C2H5 Empirical Formula**Empirical Formula is: C2H5**Formula weight is: 29.06 g/mole If the molecular weight is known to be 58.12 g/mole Then the molecular formula must be: C4H10 Empirical Formula**Empirical Formula**• Combustion analysis is one of the most common methods for determining empirical formulas. • A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram. • It is particularly useful for analysis of hydrocarbons.**Combustion Analysis**Combustion Analysis: the technique of finding the mass composition of an unknown sample (X) by examining the products of its combustion. X + O2 → CO2 + H2O 0.250 g of compound X produces: 0.686 g CO2 and 0.562 g H2O**Combustion Analysis**X + O2 → CO2 + H2O 1. Find the mass of C & H that must have been present in X (multiply masses of products by percent composition of the products). C: 0.686 g x 12.01 g/44.01 g = 0.187 g C H: 0.562 g x (2 x 1.008 g)/18.02 g = 0.063 g H**Combustion Analysis**X + O2 → CO2 + H2O 0.187 g C + 0.063 g H = 0.250 g total So compound X must contain only C and H!! 2. Find the number of moles of C and H C: 0.187 g x mole/12.01 g = 0.0156 moles C H: 0.063 g x mole/1.008 g = 0.063 moles H**Combustion Analysis**X + O2 → CO2 + H2O 3. Find the RELATIVE number of moles of C and H in whole number units (divide by smallest number of moles) C: 0.0156/0.0156 = 1 H: 0.063/0.0156 = 4 If these numbers are fractions, multiply each by the same whole number.**Combustion Analysis**X + O2 → CO2 + H2O 3. Write the Empirical Formula (use the relative numbers as subscripts) CH4**Combustion Analysis**Summary • Find the mass of C and H in the sample. • Find the actual number of moles of C and H in the sample. • Find the relative number of moles of C and H in whole numbers. • Write the empirical formula for the unknown compound.**Combustion Analysis**NOTE! In step # 1 always check to see if the total mass of C and H adds up to the total mass of X combusted. If the combined mass of C and H is less than the mass of X, then the remainder is an unknown element (unless instructed otherwise). If a third element is known, calculate the mass of that element by subtraction (at the end of step 1), and include the element in the remaining steps.**Combustion Analysis**Combustion Analysis provides the Empirical Formula. If a second technique provides the molecular weight, then the molecular formula may be deduced. • Calculate the empirical formula weight. • Find the number of “formula units” by dividing the known molecular weight by the formula weight. • Multiply the number of atoms in the empirical formula by the number of formula units.**Combustion Analysis**The molecular weight of glucose is 180 g/mole and its empirical formula is CH2O. Deduce the molecular formula. • Formula weight for CH2O is 30.03 g/mole • # of formula units = 180/30.03 = 6 • Molecular formula = C6H12O6