## Tuesday, December 14, 2010

### Formula of a Hydrate Lab (by Bev)

What is a hydrate?

A hydrate is a molecule that has a definite number of water molecules in its crystal structure.  The crystals seem like a dry solid, but they can be decomposed into an anhydrous salt (without water) and water when heated.  A colour change often happens when the hydrate is decomposed

ex. calcium chloride hexahydrate: CaCl2*6H2O
6mol H2O are chemically combined to 1mol CaCl2

What are hydrates used for?

Grains of anhydrous CaCl2 can be used to in damp rooms to absorb the moisture out of he air, thereby forming a hydrate.

Lab 4C: Formula of a hydrate

The mass of a dry crucible was found on a centigram balance & the hydrate used in this experiment, MgSO4*7H2O, was heated in a dry crucible over a Bunsen burner and weighed as well.  Using this information, and mass of the anhydrous salt & water was calculated.

%H2O in hydrate: mass of H2O/mass of hydrate = 2.23g/5.02g x 100% = 44.4%

# of mol of MgSO4 in hydrate:
mass after heating-mass of empty crucible = mass of anhydrous salt
28.90g-26.11g = 2.79g

1mol anhydrous salt = mass anhydrous salt x mol/120.4g = 2.79g x mol/120.4g = 0.02mol

mass of hydrate = mass of crucible & hydrate - mass of crucible = 31.13g-26.11g = 5.02g

mass of H2O = mass hydrate - mass of salt = 5.02g - 2.79g = 2.23g

1mol H2O =  mass of H2O x mol/18g = 2.23g x mol/18g = 0.1239mol

mass of salt/1mol salt = 2.79g/120.4g/mol = 0.023mol

mol H2O/mol salt = ratio of H2O to anhydrous salt = 0.1239mol/0.23mol = .539

## Thursday, December 2, 2010

### Empirical and Molecular Formula - Isabelle Cheng

Isabelle Cheng
Block 2-2 Chemistry 11
Ms. Chen
Empirical and Molecular Formula
Empirical - smallest whole number ratio of atoms which represent the molecular composition of a species. In other words “species” means elements. It is where a number the numbers can be divided by a certain number. For example, C8H16 it would be reduced down to C2H4. That is the lowest terms in this case.
Here is one example of doing an Empirical Formula question.
Say that you have 54.09% of Ca, 43.1% of O and 2.73% H. What is the empirical formula?
The first step:
Separate them into elements.
Ca
O
H
The second step:
Now you want to convert grams into moles. This means you have to write the mass on the side now.
Ca =  54.09
O = 43.1
H = 2.73
The third step:
Now you have to multiply it by one mole and divide it by the mass of the element (species).
Ca = 54.09/40.1 = 1.349
O = 43.16/ 16= 2.699
H = 2.73/1 = 2.73

The fourth step:
Now divide both by the smallest molar amount.
Ca = 1.349/1.349 = 1
O = 2.699/1.352 = 2
H = 2.73/1.352 = 2
The fifth step:
Now put them together:
Ca(OH)2 which is Calcium Hydroxide!
Now moving onto molecular formula!
Molecular Formula - it is a multiple of many empirical formula and has the real number of atoms that combine to form a molecule!
The formula for this is:
Let N = the WHOLE NUMBER MULTIPLE OF THE EMPIRICAL MASS:
multiple N =       Molar mass
Empirical mass
One example of the Molecular Formula is this:
A hydrocarbon is 84.25% carbon and 15.75% hydrogen and has a molecular weight of 114. What is the molecular formula?
C = 84.25/12 = 7.021
H = 15.75/1 = 15.75
Then you go:
7.021/7.021 = 1
15.78/7.021 = 2.25
Next step:

C = 1 X 4 = 4
H = 2.25 X 4 = 9
It is.....:
4 C = 48
9 H = 9

## Wednesday, December 1, 2010

### Percent Composition (by Bev)

Percent composition is the percentage by mass of a species in a chemical formula

ex. 1: percent composition of AgOH (assume there is 1mol)

total MM (molar mass): 107.9 + 19.0 + 1.0 = 127.9
MM of Ag: 107.9 (same numerical value as the atomic mass)
MM of O: 16.0
MM of H: 1.0

% Ag: 107.9g/mol / 127.9g/mol x 100% = 84%
% O: 19.0g/mol / 127.9g/mol x 100% = 15%
% H: 1.0g/mol / 127.9g/mol x 100% = 1%

Check: 84% + 15% + 1% = 100%*
*the percentages should add up to or very close to 100%

ex. 2: percent composition of CO2 (assume there is 1mol)

total MM: 12.0 + 16.0(2) = 44.0g/mol
MM of C: 12.0g/mol
MM of O: 32.0g/mol

% C: 12.0g/mol / 44.0g/mol x 100% = 27.3%
% O: 32.0g/mol / 44.0g/mol x 100% = 72.2%

Check: 27.3% + 72.2% = 100%

ex. 5: percent composition of  O in Cu2CO3

total MM: 123.5g/mol
MM of O: 48.0g/mol

% O: 48.0g/mol / 123.5g/mol x 100% = 39%

ex. 4: a compound contains 5.1 g of Cl, 22.0g of c, an unknown mass of O, & the total mass is 44.1g.  Calculate % composition.

Mass of O: 44.1g-5.1g-22.0g = 17.0g

% Cl: 5.1g / 44.1 g x 100% = 11.6%
% O: 17.0g / 44.1g x 100% = 38.6%
% C: 22.0g / 44.1g x 100% = 49.9%

Check: 11.6% + 38.6 % + 49.9% = 100.1 %