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The Mole Notes part 2

The Mole Notes part 2. Percent Comp Empirical Formulas Molecular Formulas. Percent Comp:. The percent by mass of each element in a compound Percent – a part over the whole x 100 Part x 100 Whole. Percent Comp:.

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The Mole Notes part 2

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  1. The Mole Notes part 2 Percent Comp Empirical Formulas Molecular Formulas

  2. Percent Comp: • The percent by mass of each element in a compound • Percent – a part over the whole x 100 Part x 100 Whole

  3. Percent Comp: • 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. Percent Comp: • If we have Mg3(PO4)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 Mg3(PO4)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. Percent Comp: • Now lets calculate the Mass percent of Phosphorous and Oxygen in Mg3(PO4)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: How 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. Practice: • Now complete the following problems on your Notes handout: • Lithium Sulfate • Iron (III) Oxide

  8. Empirical and Molecular Formulas: • 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) • CH4 is the empirical formula for: • CH4, C2H8 C3H12 (these are molecular formulas)

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

  10. Empirical Formulas: Steps 1 and 2 • A compound has 40% sulfur and 60% oxygen. • How many grams of each? Sulfur = 40% = 40g Oxygen = 60% = 60g

  11. Empirical Formulas: Step 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. Empirical Formulas: Step 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. Empirical Fromula: Step 5 • You came up: 1 mol S & 3 mols O You use these numbers to write the empirical formula: SO3

  14. Practice: • 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. Molecular Formula: • Find the empirical formula • Find the molar mass of the empirical formula • Divide the given molar mass by the EF’s molar mass • Multiply each subscript of the EF by the number you go in Step 3 • Write the formula

  16. Molecular Formula: Step 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. Molecular Formula: Step 1 Cont. • Find the empirical formula 6.42 molsC = 9 12 molsH = 17 .688 molsO = 1 .688 molsO .688 molsO .688 mols O EF: C9H17O

  18. Molecular Formula: Step 2 • Find the molar mass of the EF: EF: C9H17O Carbon: (9 x 12) = 108 Hydrogen: (17 x 1) = 17 Oxygen: (1 x 16) = 16 108 + 17 + 16 = 142g/mol

  19. Molecular Formula: Step 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. Molecular Formula: Step 4 • Multiply the subscripts of the EF by the number you came up with in step 3. EF: C9H17O Answer from step 3: 2 C9: 9 x 2 = 18 H17: 17 x 2 = 34 O: 1 x 2 = 2 *these now become the subscripts for the molecular formula. Molecular Formula: C18H34O2

  21. Practice: Molecular Formula • 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?

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