PERCENT COMPOSITION, EMPIRICAL & MOLECULAR FORMULAS

1. PERCENT COMPOSITION, EMPIRICAL & MOLECULAR FORMULAS

2. Percent Composition • New food labels are required to describe the ingredients using percents of the daily reccom- mended allowance • These numbers tell what part of the total # of calories can be obtained • from a product • AKA percent composition

3. Percent Composition • 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 are a couple of procedures used to calculate percent compositions

4. Calculating PC given formula What percentage of Hydrogen and Oxygen is in Water (H2O)? Assume you have 1 mole of water, and calculate its molar mass (2•1.008g) + (1•15.994g) = 18.01g

5. 2.016g H Calculating PC given formula • There are 2 mols of H atoms for every 1 mol of Water molecules H: (2•1.008g)= 2.016g H • Percent of H in Water? 11.2% X 100%= 18.01 g H2O

6. 15.994 O Calculating PC given formula • There is 1 mol of O atoms for every 1 mol of Water molecules O: (1•15.994g)= 15.994g O • Percent of O in Water? 88.8% X 100%= 18.01 g H2O

7. Percent Composition • Another method of calculating the percent composition is by experimental analysis. • the overall mass of the sample is measured. Then the sample is separated into its component elements • The equation is the SAME as before!

8. Percent Composition • 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

9. Calculating PC given sample Find the percent composition of a compound that contains 1.94g of carbon, 0.48g of Hydrogen, and 2.58g of Sulfur in a 5.0g sample of the compound.

10. Calculating PC given sample • Calculate the percentage for each element much like you would calculate the percentage for anything. C: 1.94g/5.0g X 100% = 38.8% H: 0.48g/5.0g X 100% = 9.6% S: 2.58g/5.0g X 100% = 51.6%

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

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

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

14. = = 6.7mol C 20 mol H Calculating Empirical Formulas • Now calculate the simplest ratio of each by dividing both values by the smallest value 1 mole C 80g C 12 g C 1 mole H 20g H 1 g H

15. 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 = CH3

16. Calculating Empirical Formulas Determine the empirical formula of a compound containing 25.9g of N and 74.1g of O. Notice we have masses this time not percents, we can convert masses directly to moles

17. = = 1.85 mol N 4.63 mol O 1 N = 2.5 O's = Calculating Empirical Formulas 1 mol N 25.9g N 14 g N 1.85 mol 1 mol O 74.1g O 16 g O 1.85 mol

18. Calculating Empirical Formulas Is the final answer N1O2.5? Of course not! We need a wholenumberratio… Each part of the ratio is multiplied by a number that converts the fraction to a whole number N2(1)O2(2.5)= N2O5

19. Molecular Formulas • 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 C6H12O6 • Empirical form would be CH2O

20. Molecular Formulas • The empirical formula of CH2O, could be several compnds. • C2H4O2 or C3H6O3 or C100H200O100 • It’s more important to know the exact numbers of atoms involved • The numbers of atoms define the properties of the compnd

21. Molecular Formulas • The molecular formula is always a whole-number multiple of the emp. formula • 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)

22. 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. • Find the Empirical Formula • Find the MM of the Emp. Form. • Find the ratio of the 2 molar masses (Mol MM/Emp MM)

23. 2 O's = 1 P = • Find the Empirical Formula 1 mol O 56.36g O = 3.5 mol O 16 g O 1.8 mol 1 mol P = 1.8 mol P 54.6g P 31g P 1.8 mol Empirical formula: P1O2

24. 63 g/mol • Find the MM of the Emp Form. MM of PO2: (1•31g P) + (2•16g O) = 63g/mol • Find the ratio of the 2 molar masses (mol MM/emp MM) GIVEN 189.5 g/mol = 3.00 CALCULATED

25. Calculating Molecular Formulas • 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 P3(1)O2(3)= P3O6