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

November 28 2010 - Percent Composition

The percent composition of a compound is a relative measure of the mass of each element present in the compound.  


To calculate the percent composition (percentage composition) of a compound
1.Calculate the molecular mass (molecular weight, formula mass, formula weight), MM, of the compound
2.Calculate the total mass of each element present in the formula of the compound
3.Calculate the percent compositon (percentage composition): % by weight (mass) of element
    = (total mass of element present ÷ molecular mass) x 100


Eg.
Calculate the molecular mass (MM):MM = (2 x 22.99) + 32.06 + (4 x 16.00) = 142.04


Calculate the total mass of Na present:2 Na are present in the formula, mass = 2 x 22.99 = 45.98


Calculate the percent by weight of Na in Na₂SO₄:%Na = (mass Na ÷ MM) x 100 = (45.98 ÷ 142.04) 
x 100 = 32.37%
Calculate the total mass of S present in Na₂SO₄:1 S is present in the formula, mass = 32.06


Calculate the percent by weight of S present:%S = (mass S ÷ MM) x 100 = (32.06 ÷ 142.04) x 100 
= 22.57%


Calculate the total mass of O present in Na₂SO₄:4 O are present in the formula, mass = 4 x 16.00 = 64.00


Calculate the percent by weight of O in Na₂SO₄:%O = (mass O ÷ MM) x 100 = (64.00 ÷ 142.04) x 100 = 45.06%


The answers above are probably correct if %Na + %S + %O = 100, that is,     32.37 + 22.57 + 45.06 = 100

November 19, 2010

Mole Conversions


1. 6.022 * 10^23 particles = 1 mole
a) particles to moles
Particles / (6.022*10^23) = moles
Ex. (3.01*10^24 particles) * (1mole / 6.022*10^23 particles) = 5.00 moles with significant figures
b) moles to particles
moles * 6.022*10^23 = particles
Ex. 0.75 moles * (6.022*10^23 molecules) / 1mole = 4.5 * 10^23 molecules with significant figures
c) grams to moles
mass of something * (1mole / atomic mass)
Ex. 3.45g Carbon to moles
3.45g * (1mole / 12.0g [Carbon's atomic mass]) = 0.288 moles Carbon
Ex. 6.2g of MgCl2 to moles
6.2g * (1mole / 95.3g [MgCl2 atomic mass])  = 0.0650 moles MgCl2

November 1 2010 - Density

A materials density is defined as its mass per unit volume.  It is, essentially, a measurement of how tightly matter is compacted together.  

To calculate the density of an object, you take the mass and divide it by the volume.  The equation will look like this : P=M/V  where P is the Density, M is the mass and V is the volume. 

The unit for solids are g/Cm³ and the unit for liquids are g/mL

1 Cm³ of water = 1mL

The density of water is 1.0g/mL or 1000g/L

If the density of an object is greater than the density of a liquid, then it will sink

If the density of an object is less than the density of a liquid, then it will float.

=) DENSITY VIDEO!!!!!!!!!

Oct 26, Significant Digits

After the test that I have failed, today we have learned significant digits.

Significant Digits(Figures)
-The last digit in a measurement is uncertain.
-Trailing zeros after the decimal point are counted
-Leading zeros aren't counted
-Trailing zeros without a decimal point aren't counted

Example : 2.567 = 4 significant digits
               0.0097 = 2 significant digits
               10000 = 1 significant digits
               10000. = 5 significant digits
Exact Numbers
-Same quantities are defined as exactly a certain amount and no rounding is required and have infinite significant digits
 Ex. Number of people, animals, items etc
Rounding Rules
-if the number you are rounding from is greater than 5 you round up
-if the number you are rounding from is smaller than 5 you keep the number in front
-when the number you are rounding to is 5 then:
                 -if the number you are rounding from is 5, round to make the last digit even
                 -if the number you are rounding from isn't 5 then you round up
Example:
     Rounding to the tenth
1.01 = 1.0
1.06 = 1.1
1.00 = 1.0
4.95493 = 5.0
4.85493 = 4.8
1.05 = 1.1