Empirical and Molecular Formula Calculation
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 formulaempirical formulaC2H4CH2C11H22O11CH2OH2OH2OC25H50CH2
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.
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