Custom Search

Back to Percent Composition by Mass

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)

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

Molecular Formula (additional steps)

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}

Back to Percent Composition by Mass

Chemical Demonstration Videos