December 9, 2010 - 4C

Lab to determine :
1. the percentage of water in an unknown hydrate
2. the moles of water present in each mole of this unknown hydrate, when given the molar mass of the anhydrous salt
3.write the empirical formula of the hydrate

We found the mass of hydrate, water given off, and mass of anhydrous salt.

from this we could find the empirical formula of the hydrate

December 1 2010 - Empirical and molecular formula

Empirical formulas give the lowest term ratio of atoms (or moles) in the formula.
*All ionic compounds are empirical formula.
Ex. -C4H10 (Butane) is a molecular formula.
-C2H5 is the empirical formula

To find empirical formula:
1.Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages).
2.Convert the mass of each element to moles of each element using the atomic masses.
3.Find the ratio or the moles of each element by dividing the number of moles of each by the smallest number of moles.
4.Use the mole ratio to write the empirical formula.

Example
1: A 14.0 g sample of a compound that contains only carbon and hydrogen is completely burned. The resulting products are 20.49 g of CO2 and 12.59 g of H2O. If the molar mass of the compound is known to be 30 g/mol, then what is the empirical and molecular formula of the compound?

Step 1: Calculate the number of moles of CO2 that are formed. This is the same as the number of moles of C in the original compound.

CO2 molar mass = C + 2O = (12.01) + 2(16.00) = 44.01 g/mol

(20.49 g CO2)(1mol/44.01 g) = 0.4656 mol CO2 therefore 0.4656 mol C in the original compound.

Step 2: Calculate the number of moles of H2O that is formed. Since there are 2 H atoms per H2O molecule, this is 1/2 the amount of H in the original compound.

H2O molar mass = 2H + O = 2(1.01) + 16.00 = 18.02 g/mol

(12.59 g H2O)(1 mol/18.02 g) = 0.6987 mol H2O therefore twice as much H (1.397 mol) in the original compound.

Step 3: Substitute the moles into a trial formula C0.4656H1.397

Step 4: Divide by the smallest trial subscript, so that you can simplfy to a whole number ratio

Therefore, CH3 is the empirical formula.

Step 5: Calculate a trial molar mass for the empirical formula

CH3 molar mass = 1C + 3H = (12.01) + 3(1.01) = 15.04 g/mol.

Step 6: Divide the actual molar mass of the compound by the trial molar mass, to find how many formula units there are in the real compound

(30 g/mol)/(15.04 g/mol) = 1.995 which is almost 2

So there are 2(CH3) units in the actual molecular formula, C2H6


To find molecular formula:
1. Determine the molecular mass in grams.
2. Divide the molecular mass of the compound by the molecular mass of the empirical formula.
3. Round the quotient to the closest integer.
4. Multiply the rounded number by all the subscripts, using the product as the new subscripts.


Example:

A compound has an empirical formula of CH2 and a molecular mass of 42g. Determine its molecular mass.

Step 1: Carbon = 12 + Hydrogen = 2(1.01)
Molecular Mass (g) = 14.02
Step 2: 42 ÷ 14.02 = 2.999
Step 3: 3
Step 4: CH2 • 3 = C3H6