# PERCENT COMPOSITION EMPIRICAL MOLECULAR FORMULAS 1 Percent Composition

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PERCENT COMPOSITION, EMPIRICAL & MOLECULAR FORMULAS 1

Percent Composition o. New food labels are required to describe the ingredients using percents of the daily reccommended allowance • These numbers tell what part of the total # calories can be obfrom a product • AKA percent composition of tained 2

Percent Composition o To get the information found on food labels the chemists had to know what fraction of the whole was each component • Component/total and then multiply by 100 • There a couple of procedures used to calculate percent compositions 3

Calculating PC given formula What percentage of Hydrogen and Oxygen is in Water (H 2 O)? Assume you have 1 mole of water, and calculate its molar mass (2 • 1. 008 g) + (1 • 15. 994 g) = 18. 01 g 4

Calculating PC given formula o. There are 2 mols of H atoms for every 1 mol of Water molecules o. How much do 2 mols of H atoms weigh? H: (2 • 1. 008 g)= 2. 016 g H o Percent of H in Water? 2. 016 g H X 100%= 11. 2% 18. 01 g H 2 O 5

Calculating PC given formula o. There is 1 mol of O atoms for every 1 mol of Water molecules o. How much does 1 mol of O atoms weigh? O: (1 • 15. 994 g)= 15. 994 g O o Percent of O in Water? 15. 994 O X 100%= 88. 8% 18. 01 g H 2 O 6

Percent Composition o Another method of calculating the percent composition is by experimental analysis. • the overall mass of the sample is measured. • then the sample is decomp-osed or separated into its component elements 7

Percent Composition o The masses of the component elements are then determined and the percent composition is calculated as before • by dividing the mass of each element by the total mass of the sample • then multiplying by 100 8

Calculating PC given sample Find the percent composition of a compnd that contains 1. 94 g of carbon, 0. 48 g of Hydrogen, and 2. 58 g of Sulfur in a 5. 0 g sample of the compnd. 9

Calculating PC given sample o Calculate the percents for each element much like you would calculate the percents for anything. C: 1. 94 g/5. 0 g X 100% = 38. 8% H: 0. 48 g/5. 0 g X 100% = 9. 6% S: 2. 58 g/5. 0 g X 100% = 51. 6% 10

Empirical Formulas o Once the percent compositions are determined then they can be used to calculate a simple chem formula for the compnd • key is to convert the percents by mass into amounts in moles • Then, compare the moles using ratios to determine coefficients 11

Calculating Empirical Formulas What is the empirical formula of a compound that is 80%C and 20%H by mass o. Since we have been given percents rather than masses we need to make an assumption. • Let’s suppose we have a total sample that weighs 100 g. 12

Calculating Empirical Formulas o This allows us to say that if we had a 100 grams of sample, • 80 g is Carbon • 20 g is Hydrogen o Now that we have a set of masses we need to convert them to moles • Divide by the molar masses from the Periodic Table 13

Calculating Empirical Formulas 80 g C 20 g H 1 mole C 12 g C 1 mole H 1 g. H = 6. 7 mol C = 20 mol H • Now calculate the simplest ratio of each by dividing both values by the smallest value 14

Calculating Empirical Formulas Divide each mole value by the smaller of the two values: C: 6. 7/6. 7=1 H: 20/6. 7 = 2. 98 3 Ratio is 1 C’s for every 3 H’s; so the formula is = CH 3 15

Calculating Empirical Formulas Determine the empirical formula of a compound containing 25. 9 g of N and 74. 1 g of O. Notice we have masses this time not percents, we can convert masses directly to moles 16

Calculating Empirical Formulas 25. 9 g N 74. 1 g O 1 mol N 14 g N = 1. 85 mol N 1. 85 mol 1 mol O = 4. 63 mol O 16 g O 1. 85 mol 17

Calculating Empirical Formulas Is the final answer N 1 O 2. 5? Of course not! We need a whole number ratio… Each part of the ratio is multiplied by a number that converts the fraction to a whole number N 2(1)O 2(2. 5)= N 2 O 5 18

Molecular Formulas o The empirical formula indicates the simplest ratio of the atoms in the compnd • However, it does not tell you the actual numbers of atoms in each molecule of the compnd • For instance, glucose has the molecular formula of C 6 H 12 O 6 • Empirical form would be CH 2 O 19

Molecular Formulas o The empirical formula of CH 2 O, could be several compnds. • C 2 H 4 O 2 or C 3 H 6 O 3 or C 100 H 200 O 100 o It’s more important to know the exact numbers of atoms involved §The numbers of atoms define the properties of the compnd 20

Molecular Formulas o The molecular formula is always a whole-number multiple of the emp. formula o In order to calculate the molecular formula you must have 2 pieces of information • Empirical formula • Molar mass of the unknown compound (must be given) 21

Calculating Molecular Formulas Find the molecular formula of a compound that contains 56. 36 g of O and 54. 6 g of P. If the molar mass of the compound is 189. 5 g/mol. 1) Find the Empirical Formula 2) Find the MM of the Emp. Form. 3) Find the ratio of the 2 molar masses (Mol MM/Emp MM) 22

1)Find the Empirical Formula 1 mol O 56. 36 g O 16 g O 54. 6 g P 1 mol P 31 g P = 3. 5 mol O 1. 8 mol = 1. 8 mol P 1. 8 mol Empirical formula: P 1 O 2 23

2) Find the MM of the Emp Form. MM of PO 2: (1 • 31 g P) + (2 • 16 g O) = 63 g/mol 3)Find the ratio of the 2 molar masses (mol MM/emp MM) GIVEN 189. 5 g/mol = 3. 00 CALCULATED 63 g/mol 24

Calculating Molecular Formulas o So the Molecular formula is 3 times heavier than the Empirical formula • Therefore, the molecular formula has 3 times more atoms than the emp. formula P 3(1)O 2(3)= P 3 O 6 25

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