Empirical Formula

Finding Chemical Formulas The percent composition of a compound can be used to find the simplest chemical formula of the compound. This simplest formula is called the empirical formula.  Empirical Formula – the lowest whole number ratio of elements in a compound. The empirical formula may be the same as the molecular formula or the molecular formula may be a multiple of the empirical. 

 Ex1

CO2 empirical and molecular formula the same  Ex2 N2H4 has empirical formula NH2

Steps For Determining Empirical Formula 1.) Start with the # of grams of each element, given in the problem. If percentages are given, assume the total mass is 100 grams so that the mass of each element = the percent given. 2.) Convert the mass of each element to moles using the molar mass from the periodic table.

Steps For Determining Empirical Formula 3.) Divide each mole by the smallest number of moles calculated. 4.) Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula.

Example Problem #1 A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound? 13.5 g Ca 1 10.8 g O

1 mole 40.08 g Ca 1 mole

1 0.675 g H 1

= 0.337 mol Ca

= 0.675 mol O

16.00 g O 1 mole 1.01 g H

= 0.668 mol H

Example Problem #1 Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number. 0.337 mol Ca

=

1 Ca

=

2.00 = 2 O

=

1.98 = 2 H

0.337 mol 0.675 mol O 0.337 mol 0.668 mol H 0.337 mol

Example Problem #1 This is the mole ratio of elements and is represented by subscripts in the empirical formula. 1 – Ca 2–O 2–H CaO2H2 = Ca(OH)2

Example Problem #2 NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of NutraSweet. 57.14 g C 1

1 mole

= 4.76 mol C

12.01 g C

6.16 g H

1 mole

= 6.09 mol H

1 9.52 g N

1.01 g H 1 mole

= 0.680 mol N

1 27.18 g O

14.01 g N 1 mole

= 1.70 mol O

1

16.00 g O

Example Problem #2 Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number. 0.680 mol N

=

1N

0.680 mol 6.09 mol H

=

8.95 = 9 H

0.680 mol 1.70 mol O

=

2.5 O

0.680 mol 4.76 mol C

=

7C

0.680 mol

Example Problem #2 If the number is too far to round, then multiply each solution by the same factor to get the lowest whole number multiple. This is the mole ratio of elements and is represented by subscripts in the empirical formula. 2.5 x 2 = 5 – O 7 x 2 = 14 – C 9 x 2 = 18 – H 1x2=2–N C14H18N2O5

Now, we can find the molecular formula by finding the mass of the empirical formula and setting up a ratio: C14H18N2O5 = (14 x 12.01) + (18 x 1.01) + (2 x 14.01) + (5 x 16.00) = 294.00 g/mol Molecular Formula = molar mass empirical formula mass

= 294.30 = 1 294.00

So the empirical formula is the molecular formula

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