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 Empirical and Molecular Formula Calculations

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Empirical formula is the smallest whole number ratio of moles of each element in a compound.

CaCl2 --> there is 1 mole of calcium for every 2 moles of chlorine

Level 1 Simple Empirical formula questions

What is the empirical formula of the following compounds? (so reduce the formula if you can)

 molecular formula empirical formula C2H4 CH2 C11H22O11 CH2O H2O H2O C25H50 CH2

Level 2  Empirical Formula Calculation Steps

Step 1 If you have masses go onto step 2.

If you have %.  Assume the mass to be 100g, so the % becomes grams.

Step 2 Determine the moles of each element.

Step 3 Determine the mole ratio by dividing each elements number of moles by the smallest value from step 2.

Step 4 Double, triple … to get an integer if they are not all whole numbers

The question should have included a molecular mass.

Step 5 Determine the mass of your empirical formula

Step 6 Divide the given molecular mass by your E.F. mass in step 5

Step 7 Multiply the atoms in the empirical formula by this number

Examples-Caffeine has an elemental analysis of 49.48% carbon, 5.190% hydrogen, 16.47% oxygen, and 28.85% nitrogen. It has a molar mass of 194.19 g/mol. What is the molecular formula of caffeine?

(Hint-Save the molar mass 194.19g/mol until the end)

49.48% C, 5.190%H,  16.47% O and 28.85% N

Step 1 Assume a mass of 100g so % becomes grams

49.48g C, 5.190gH,  16.47g O and 28.85g N

Step 2 determine the moles of each element

49.48g  C x ( 1 mole/12.0 g C) = 4.123moles C
5.190g H x (1 mole / 1.0 g H) = 5.190 moles H
16.47g  O x (1 mole /16.0 g O ) =1.029moles O
28.85g N x ( 1 mole/ 14.0 g N) = 2.061 moles N

Step 3 determine the mole ratio by dividing each elements number of moles by the smallest

Dividing by the smallest (1.029) we get

C: 4.123 / 1.029 = 4.007
H: 5.190 / 1.029 = 5.044
O: 1.029 / 1.029 =1.000
N: 2.061 / 1.029 = 2.002

Step 4 Double, triple .. to get an integer is they are not all whole numbers

The values are all really close to whole numbers.

Empirical Formula= C4H5ON2

Example- Molecular Formulas (Steps 5-7)

It has a molar mass of 194.19 g/mol.

Step 5 After you determine the empirical formula, determine its mass.

Empirical Formula= C4H5ON2

(4 carbon x 12.0) + (5 hydrogen x1.0) + (1 oxygen x 16.0) + (2 nitrogen x 14.0) =97.0g/mol

Step 6 Determine how many times greater the molecular mass is compared to the mass of the empirical formula.

molecular mass/ empirical formulas mass

194.19g/mol  / 97.0g/mol = 2

Step 7 Multiply the empirical formula by this number

2x C4H5ON2 =C8H10O2N4

better==> C8H10N4O2

***note

if step 6 does not work out to be a whole number your empirical formula is wrong or your teacher screwed up

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