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Warm Up:

Warm Up:. Molecular Formula. the formula of a compound in which the subscripts give the actual number of each element in the formula C 2 H 4 O 2. Empirical Formula. the formula of a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula

thomasvscott
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Warm Up:

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  1. Warm Up:

  2. Molecular Formula • the formula of a compound in which the subscripts give the actual number of each element in the formula • C2H4O2

  3. Empirical Formula • the formula of a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula • C2H4O2 • CH2O = empirical

  4. Notice two things: 1. The molecular formula and the empirical formula can be identical.2. You scale up from the empirical formula to the molecular formula by a whole number factor.

  5. Convert percentages to empirical formula Percent to massMass to moleDivide by smallMultiply 'til whole

  6. Sample Problem • A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Step 1:Change percents to grams 72.2 g Mg 27.8 g N

  7. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Step 2: Convert to moles Mg: 72.2 g x 1 mol = 2.97 mol 24.3 g N: 27.8 g x 1 mol = 1.98 mol 14 g

  8. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Step 3: Divide each by smallest amount of moles Mg: 2.97 1.98 N: 1.98 1.98 = 1.5 = 1

  9. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Step 4:Multiply if not all whole numbers Mg: 1.5 x 2=3 N: 1 x 2 = 2 Mg3N2

  10. A compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the empirical formula?

  11. 1. 69.58% Ba = 69.58 g Ba 6.090% C = 6.090 G C 24.32% O = 24.32 G O 2. 69.58g Ba x 1 mol = 0.507 mol 137.3 g 6.090g C x 1 mol = 0.501 mol 12 g 24.32g O x 1 mol = 1.5 mol 16 g

  12. 3. Ba= 0.507 mol / 0.501 mol = 1.01 C = 0.501 mol / 0.501 mol = 1 O = 1.5 mol / 0.501 mol = 2.99 4. Already whole number BaCO3

  13. Molecular Formula The simplest formula for vitamin C is C3H4O3. Experimental data indicates that the molecular mass of vitamin C is about 180. What is the molecular formula of vitamin C? Step 1:Find molar mass (aka GFM) of empirical formula Step 2: Divide number given by molar mass of empirical Step 3: Multiply each subscript by this whole number factor

  14. C3H4O3 180/88= 2.05 C6H8O6

  15. A substance is decomposed and found to consist of 53.2% C, 11.2% H, and 35.6% O by mass. Calculate the molecular formula of the unknown if its molar mass is 90 g/mol

  16. Question 10 1: Assume 100 g total. Thus: 53.2 g C, 11.2 g H, and 35.6 g O 2: C: 53.2 g ¸ 12.01 g/mol = 4.430 mol C H: 11.2 g ¸ 1.01 g/mol = 11.09 mol H O: 35.6 g ¸ 16.00 g/mol = 2.225 mol O 3: 4.430 11.09 2.225 4.43/2.225 = 1.99 11.09/2.225 = 4.98 2.225/2.225 = 1 4: the empirical formula is C2H5O

  17. C2H5O 5. Molar mass: 45 6. 90/45=2 7: Molecular formula: C4H10O2

  18. Find the empirical and molecular formulas for this compound 51.28 percent C 9.40 percent H 27.35 percent O 11.97 percent N MW=234

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