 # 1 Empirical Formulas Honors Chemistry. 2 Formulas The empirical formula for C 3 H 15 N 3 is CH 5 N. The empirical formula for C 3 H 15 N 3 is CH 5 N.

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1 Empirical Formulas Honors Chemistry

2 Formulas The empirical formula for C 3 H 15 N 3 is CH 5 N. The empirical formula for C 3 H 15 N 3 is CH 5 N. you are basically dividing by the greatest common factor, which in this case is 3, for each element subscript. you are basically dividing by the greatest common factor, which in this case is 3, for each element subscript. Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

3 Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

4 Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11 (Correct formula) (Lowest whole number ratio)

5 Calculating Empirical Formula   We can get a ratio from the percent composition. 1) 1)Assume you have a 100 g sample - the percentage become grams (75.1% = 75.1 grams) 2) 2)Convert grams to moles. 3) 3)Find lowest whole number ratio by dividing each number of moles by the smallest value.

6 Example   Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.   Step 1 : Assume 100gram sample   Step 2 : Convert grams to moles for each element   38.67 g C 1mol C = 3.22 mole C 12.01 g C   16.22 g H 1mol H = 16.06 mole H 1.01 g H   45.11 g N 1mol N = 3.22 mole N 14.01 g N Now divide each value by the smallest value

7 Example   The ratio is 3.22 mol C = 1 mol C= 1 3.22 mol N 1 mol N   The ratio is 16.06 mol H = 4.988 mol H = 5 3.22 mol N 1 mol N Empirical Formula is = C 1 H 5 N 1 which is = CH 5 N

8 Empirical to molecular   Since the empirical formula is the lowest ratio, the actual molecule would weigh more.   By a whole number multiple.   Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula

9 Empirical to molecular True Formula Mass = Whole # Empirical Formula Mass ratio The whole number ratio is what you multiply each subscript in the empirical formula by to get the true molecular formula.

10 and its empirical formula is C 4 H 5 N 2 O, Caffeine has a molar mass of 194 g. and its empirical formula is C 4 H 5 N 2 O, what is its molecular formula? n Find the empirical formula molar mass. n 97 g/mole n Set up ratio 194g = 2 n 97g n Multiply each subscript in the empirical formula by 2. n C 8 H 10 N 4 O 2 is the molecular formula.

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