Empirical Formula and Molecular Formula

Empirical Formula Empirical formula is the simplest whole

number ratio of atoms in a formula

Empirical information is always based on lab data

Molecular Formula Molecular formula is the actual ratio of

the atoms that form a molecule of a substance.

Sometimes the empirical and molecular are the same but they do not have to be!

More molecular…. Example H2O for water empirical and molecular

formula are the same.

H 2 : O 1 is the simplest whole number ratio of atoms in the formula and it is also how the molecule actually forms.

Example glucose C6H12O6 for glucose the molecule is 6:12:6 This is not the simplest whole number ratio of atoms in the formula.

Molecular and Empirical Formulas If glucose were analyzed in the

laboratory its empirical ratio would be 1:2:1. Additional test would have to be run to determine the mass of the substance

CalculationsStep one= List individual elements in the

formula. Change the given information from grams or percent into moles

Step two= Decide which mole value is the smallest on the list.

Step three= Divide all mole values by that smallest value

more Calculations…… Step four= If a decimal of .5 or .3

appear in the ratio you obtained do not round up or off!

Ratios with .5 must be multiplied by two.Ratios with .3 must be multiplied by three.

Formula of Hydrates The three basic steps are the same for

hydrates. But with hydrates…… The two parts being compared are:

Moles of anhydrous in the sample vs. Moles of water in the sample

Elements are not listed separately!

Example Empirical /Molecular An organic compound was found to

contain 92.25 % carbon and 7.75 % hydrogen. If the molecular mass is 78.0 grams, please calculate both the empirical and molecular formulas for this substance.

Step 1

C 92.25g/12.01 g = 7.681 moles H 7.75g/1.01 g= 7.67 moles Divide through by the smaller value! 1 : 1 ratio C1H1 is empirical formula Now we must check for molecular

formula!

Step 2 C1H1 add up the empirical mass of this

substance. 12.01+ 1.01g = 13.02grams Divide the given molecular mass by the

empirical mass. 78.0/ 13.02 = 6 Molecular formula is C6H6

Final Answer Empirical formula C1H1

Molecular formula C6H6

Modern Chemistry example p.246 Quantitative analysis shows that a

compound contains 32.38% sodium, 22.65%sulfur, and 44.99% oxygen. Please find the empirical formula for this compound.

Hydrate Example Strontium chloride hexahydrate 158.52g plus 108.12g = 266.64g 108.12/266.64 x 100 = 40.5 % H2O

Hydrate Calculation CuSO4* ? H2O Crucible&cover 111.55g Crucible& cov& hydrate131.53g Crucible/cov& anhydrous 124.32g

Hydrates Subtract C&C from anhydrous (after

heating) this will equal grams of CuSO4.

Change grams to moles.

Hydrates If you subtract hydrate from anhydrous

you will find the mass of the water removed from the sample.

Change grams water to moles.

Hydrates Look at mole values. Decide which mole value is smaller Divide both values by the smaller to get

a one to something ratio

Taa! Daa! you are done.

The End!

Go forth and calculate!