Emperical & Molecular Formula
Table of Contents
1. Emperical
1.1. Definition
Reduced or simplified of a molecular formula, or a simplier ration of elements.
1.2. Example
- Hydrogen Peroxide is &H2O2&, it's emperical form is \(HO\). (A 2 was factored out.)
2. Molecular
2.1. Definition
True chemical formula for a compound. It's property is that it can be further simplified to an emperical formula. But if it does not, then it does not have an emperical formula.
2.2. Example
- Water is \(H_{2}O\), it's molecular formula cannot be simplified, hence it has no emperical formula
- \(H_{8}O_{4}\) and \(H_{12}O_{6}\) both have similar emperical forms, but has a differenct molecular
3. Conversions
3.1. Ratio to Emperical
- Assume the compound has a mass of 100g.
- Convert the percentage-ed values into grams. (Just change the percentage sign to grams, since the assumed mass is 100g)
- Divide the values by their respective molar mass.
- Divide both the resulting values by the smallest value.
- Round.
3.1.1. Example
- Compound contains 35.98% Aluminium and 64.02% sulfur
- Convert
- \(Al=35.98g\)
- \(S=64.02\)
- Divide
- \(Al=\frac{35.98}{26.98}=1.334\)
- \(S=\frac{64.02}{32.07}=1.996\)
- Divide both by smallest number
- \(Al=\frac{1.334}{1.334}=1\)
- \(S=\frac{1.996}{1.334}=1.496\)
- Round
- \(Al=1=1\)
- \(S=1.496=1.5\)
- Chemeistry does not allow non-integers!!
- \(Al=1*2=1\)
- \(S=1.5*2=3\)
3.2. Emperical to Molecular
Assuming you know the molecular mass' molar mass.
- Find the molar mass of the emperical formula.
- Divide the molar mass by the emperical formula's molar mass
- Multiply all the emperical form's subscripts by the resulting product.
3.3. Molecular to Emperical
- Find the mass of the compound.
- Divide the molar mass by the given mass
- Round the divided mass to the nearest integer.
- Multiply all the subscripts by said HCF (post-rounding).
3.3.1. Example
- \(C_{3}H_{6}N_{2}\) is the EF of a compound, find it's MF (assuming mm of 214).
- It's mass is \(70.11\).
- \(\frac{214}{70.11}=3.05\)
- Round down \(3.05\) to \(3\)
- Multiply the subscripts of each element by \(3\)
- Answer is \(C_{9}H_{18}N_{6}\)