(November 13, 2009) Today, we learned that empirical formulas are the simplest formulas. It is like the lowest common denominator of a molecular formula. Therefore, the empirical formula is the reduced of a molecular formula. For example, the molecular formula of tricarbon hexaoxide is C3O6.This can be reduced to CO2 (reduced by 3), this is the empirical formula.
Empirical formulas are the simplest whole number ratios in a compound. Molecular Formulas show the actual atoms and bonds.
- Example - : A sample of an unknown compound is analyzed and found to contain 8.4 g of "C", 21.0 g of "H", and 5.1 g of "O". Find the empirical formula.
To do this, you must fill in a chart:
You must find the amount of moles of each element by dividing its molar mass by the given mass.
(November 10, 2009) Today we learned how to calculate the percentage of mass of elements in compounds. Recall that molar mass is grams over moles: g/mol and molar volume is liters over moles: l/mol.
To find the percentage mass, first find the total molar mass of the compound. Then, for each element in the compound, find its individual molar mass. Divide the element mass with the compound mass:
% of Element = (Element Mass / Compound Mass) x 100
- Example - Compound: Fe3N2
In iron (II) nitride, it's molar mass is 195.4 g. *Note that iron has a mass of 55.8g and nitrogen has a mass of 14.0g: 3(55.8) + 2(14.0) = 195.4 g
Next, the individual mass of iron is 167.4 g (3 x 55.8 g), nitride is 28.0 g (2 x 14.0 g)
Now find the percentage mass by taking element's mass and divide it by the compound's mass.
% of Fe = (167.4 g / 195.4 g) x 100 = 85.7%
% of N = (28.0 g / 195.4 g) x 100 = 14.3 %
To ensure the percentages are accurate, add them together. If their sum adds up to 100, it is accurate. If their sum adds up to a couple of decimals off, that is because of the rounding. Try to adjust the decimals so that your sum will equal 100.
To find the mass of an element in a given sample, first find the total molar mass of the compound. Then, for each element in the compound, find its individual molar mass. Divide the element mass with the compound mass. Lastly, multiply the quotient with the mass of the given sample:
- Example - Compound: 62.0 g sample of ZnO
In zinc oxide, the molar mass is 81.4 g. *Note that zinc has a mass of 65.4 g and oxygen has a mass of 16.0 g: 1(65.4 g) + 1(16.0 g) = 81.4 g
Next, the individual mass of zinc is 653.4 g (1 x 65.4 g) and oxygen is 16.0 g (1 x 16.0)
Now divide the individual masses with the compound mass
Zn = (65.4 g / 81.4 g) = .8034
O = (16.0 g / 81.4 g) = .1966
Then multiply the quotients with the mass of the given sample (62.0 g).
Zn = (.8034)(62.0 g) = 49.8 g
O = (.1966)(62.0 g) = 12.2 g
Therefore, in a 62.0 g sample of ZnO, there are 49.8 g of zinc and 12.2 g of oxygen.
To ensure masses are correct, add them together so that they equal the mass of the given sample. If they do not equal the mass of the sample, that may because of the rounding. Adjust the decimals so that the sum will equal the given mass of the sample.
Today's class (November 4), We learned how to calculate the density of gases at STP (Standard Temperature and Presssure)
-density equals the mass per unit volume -an easy way to rememeber how to calculate Mass, Density and Volume is to picture it as a triangle with Mass on top and Density and Volume on the bottom -to calculate the density of gases at STP we follow this equation
DensitySTP=Mass of 1 mole\Volume of 1 mole=Molar Mass\22.4 L=g per mol\L per mol
Examples:
1. Calculate the density of O2 @ STP
Dstp=32 g per mol\22.4 L per mol=1.43 g per L
2. The density of a gaseous mixture of C8H18 @STP=?
D=M\V D=114\22.4 D=5.1 g per L
3. Calculate the density of Dihydrogen Phosphate
D=M\V D=97\22.4 D=4.3 g per L
D&V-mass=d x v-MASS-molar mass-MOLE-6.02x10 to the 23-MOLECULES-subscripts-ATOMS
During class on October 27, 2009, the whole class participated in a lab. Through this experiment, it's main objectives were to:
Determine the number of moles of copper produced in the reaction of iron and copper(II) chloride
Determine the number of moles of iron used up in the reaction of iron and copper(II) chloride
Determine the ratio of moles of iron to moles of copper
Determine the number of atoms and formula units involved in the reaction
The procedure of this particular lab consisted of:
Finding the mass of a clean, empty, dry 250 mL beaker. Record the mass to the nearest 0.01g.
Add aprrox. 8 grams of copper (II) cloride crystals to the beaker. Find the mass and record it in your notebook.
Add 50mL of distilled water to the beaker. Swirl th ebeaker around to dissolve all the copper(ii) chloride crystals.
Obtain two clean, dry nails. Find the mass of the nails and record it in your notebook.
Place the nails into the copper(II) chloride solution. Leave them undisturbed for approx. 20 minutes.
Use the tongs to carefully pick up the nails, one at a time. Use distilled water in a wash bottle to rinse off any remaining copper from the nails before removing them completely from the beaker.
After the nails are completely dry, find the mass of the nails and record it in your notebook.
Decan the liquid from the solid.
Rinse the solid again with about 25 mL of distilled water. Decant again. Repeat this step three or four times.
Wash the solid with about 25 mL of hydrochloric acid. Decant again; then, once more, clean the solid with 25 mL of distilled water.
After final washing, place the copper in a drying oven to dry.
Find the mass of the beaker plus the copper and record it in your notebook.
