1. T HE M OLE N OTES PART 2 Percent Comp Empirical Formulas Molecular Formulas

2. P ERCENT C OMP : The percent by mass of each element in a compound Percent – a part over the whole x 100 Part x 100 Whole

3. P ERCENT C OMP : The mass of each element in a compound divided by the entire mass of the compound then multiplied by 100.. Part = the mass of the element you are wanting the % mass for. Whole = the molar mass of the entire compound

4. P ERCENT C OMP : If we have Mg 3 (PO 4 ) 2 and we want to know the mass % of Magnesium: How many magnesium atoms are there in 1 mol? _____ What is the molar mass of just Mg? 3 x 24g = 72g of Mg in Mg 3 (PO 4 ) 2 Part: 72g Mg Whole: what is the molar mass of the whole compound? 72 + (31 x 2) + (16 x 8) = 263g/mol Part/whole = 72/263 x 100 = 27%

5. P ERCENT C OMP : Now lets calculate the Mass percent of Phosphorous and Oxygen in Mg 3 (PO 4 ) 2 : Phosphorous: 2 phosphorous = 2 x 31 = 62 62g/263g x 100 = 23.9% = 24% Oxygen: 8 oxygens= 8 x 16 = 128 128g/263g x 100 = 48.7% = 49%

6. HINT: H OW TO CHECK YOUR WORK To check your work add up all the Percent Masses of each element: Mg – 27% P – 24% 100 O – 49% It should be right around 100…within a number or 2 (this is because we are rounding to the nearest whole number)

7. P RACTICE : Now complete the following problems on your Notes handout: 1. Lithium Sulfate 2. Iron (III) Oxide

8. E MPIRICAL AND M OLECULAR F ORMULAS : Empirical formula: the formula for a compound with the smallest whole number ratio of elements Molecular formula: The formula that specifies the actual number of atoms of each element in a substance (may or may not be the same as the empirical) CH 4 is the empirical formula for: CH 4, C 2 H 8 C 3 H 12 (these are molecular formulas)

9. E MPIRICAL F ORMULAS : 1. Determine the number of moles of each substance present in the compound. (Use percent comp) *This will typically be given to you. 2. Determine the grams *Assume that there is a hundred grams of compound, therefore, the percentage is the same as the number of grams. 1. Convert the grams to moles 2. Determine the ratio 3. Write the Empirical Formula

10. E MPIRICAL F ORMULAS : S TEPS 1 AND 2 1. A compound has 40% sulfur and 60% oxygen. 2. How many grams of each? Sulfur = 40% = 40g Oxygen = 60% = 60g

11. E MPIRICAL F ORMULAS : S TEP 3 Convert grams to moles: (S=40g and O=60g) 40 g S 1 mol S = 1.25 moles S 32 g S 60 g O 1 mol O = 3.75 moles O 16 g O

12. E MPIRICAL F ORMULAS : S TEP 4 Determine the ratio 1.25 moles S and 3.75 moles O 1. Choose the smallest number (1.25 moles S) 2. Divide all by the smallest number 1.25 moles S = 1 mol S 3.75 moles O = 3 mol O 1.25 1.25

13. E MPIRICAL F ROMULA : S TEP 5 You came up: 1 mol S & 3 mols O You use these numbers to write the empirical formula: SO 3

14. P RACTICE : Write the empirical formulas for the following problem. Complete this on your notes handout. Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 49% carbon, 8% hydrogen, and 43% oxygen.

15. M OLECULAR F ORMULA : 1. Find the empirical formula 2. Find the molar mass of the empirical formula 3. Divide the given molar mass by the EF’s molar mass 4. Multiply each subscript of the EF by the number you go in Step 3 5. Write the formula

16. M OLECULAR F ORMULA : S TEP 1 Find the empirical formula: A compound has the following composition: 77% carbon, 12% hydrogen, and 11% oxygen. Its molar mass is 282 g/mol, what is its molecular formula. Assume there are 100g. 77g C 1 mol C = 6.42 mols C 12g C 12 g H 1 mol H = 12 mols H 1 g H 11 g O 1 mol O =.688 mols O 16 g O

17. M OLECULAR F ORMULA : S TEP 1 C ONT. Find the empirical formula 6.42 mols C= 912 mols H = 17.688 mols O = 1.688 mols O.688 mols O.688 mols O EF: C 9 H 17 O

18. M OLECULAR F ORMULA : S TEP 2 Find the molar mass of the EF: EF: C 9 H 17 O Carbon: (9 x 12) = 108 Hydrogen: (17 x 1) = 17 Oxygen: (1 x 16) = 16 108 + 17 + 16 = 142g/mol

19. M OLECULAR F ORMULA : S TEP 3 Divide the molar mass of the given by the molar mass of EF. Given Molar Mass: 282 g/mol EF Molar Mass: 142g/mol 282 ÷ 142 = 1.98 (rounds to 2)

20. M OLECULAR F ORMULA : S TEP 4 Multiply the subscripts of the EF by the number you came up with in step 3. EF: C 9 H 17 O Answer from step 3: 2 C 9 : 9 x 2 = 18 H 17 : 17 x 2 = 34 O: 1 x 2 = 2 *these now become the subscripts for the molecular formula. Molecular Formula: C 18 H 34 O 2

21. P RACTICE : M OLECULAR F ORMULA Complete the following problem on your notes problem page. A colorless liquid composed of 47% nitrogen and 53% oxygen, has a molar mass of 60 g/mol. What is the empirical and molecular formula?