November 16th Class

On November 16th we learned about empirical formula's and how to find them.

Empirical Formula's
- give the whole number of ratio's of elements in a compound while molecular formula's give the actual number.

After learning what empirical formula's are we learned how to find them.

Finding Empirical Formula's

Example:
A sample of an unknown compound is analyzed and found to contain 8.4 g of carbon 2.1 g of hydrogen and 5.6 g of oxygen.

* The easiest way to answer this question is to organize the information that is given into a chart format.


C2H6O

If the ratio ends in a decimal of... 0.5 then multiply by 2.
If the ratio ends in a decimal of... 0.33 or 0.66 then multiply by 3.
If the ratio ends in a deciaml of... 0.25 or 0.75 then multiply by 4.
If the ratio ends in a decimal of... 0.2, 0.4 or 0.6 then multiply by 5.

Nov.12 Class

In the beginning of class we reviewed 10-2 questions and everything up to the mole conversion is on the midterm.

molar mass -> g/mol

molar volume -> L/mol -> mL/mol

Example:
A sample of an unknown gas contains 0.0554mol and occupies a volume of 602.0mL. Determine the molar volume.

Percent Mass of Elements in Compunds

Find the % carbon by mass in Glucose. (C6H12O6)
C = 6(12.o) + H = 12(1.0) + O = 6(16.0) = 180 g/mol
% of Carbon = 72g/180g = 0.4 -> 40%
% of Hydrogen = 12g/180g = 0.0666... -> 7%
% of Oxygen = 96g/180g = 0.5333... -> 53%

Percent Composition
- means the % mass of each element in a compound.

Find the composition of NO3
N = 1(14.0) + O = 3(16.0) = 62.0g/mol
% of Nitrogen = 14g/62g = 0.228 -> 23%
% of Oxygen = 48g/62g = 0.77 -> 77%

Finding the mass of an element in a given mass of a compound.

Find the mass of Carbon contained in a 30.0g sample of CO3.
(First find the percent. Then take the percentage as a decimal and multiply it by the given amount.)
C = 1(12.0) + O = 3(16.0) = 60g/mol
% C = 12g/60g = (0.200) (30.0) = 6g
% O = 48g/60g = (0.800)(30.0) = 24g

Find the mass of K,C, and O contained in K2CO3 if the sample is 500.0g
K = 2(39.1) = 78.2 + C = 12 + O = 16(3) = 48 = 138.2g/mol
% of K = 78.2/138.2 = 56.6% -> 0.566(500g) = 283.0g
% of C = 12/ 138.2 = 8.7% -> 0.087(500g) = 43.5g
% of O = 48/138.2 = 34.7% -> 0.347(500g) = 173.5g
Total = 500.0g

Density & Moles - Continued: Nov9 Class

Today in class, we drew out the flow chart for all the calculations we could do with this unit of moles. We did 3 examples and the last one is one made up.

Example: 1.25L of an unknown gas has a mass of 3.47g. What is the molar mass if it is at STP?
Example 2: A 5.00g sample of solid lead contains 0.247 mol of Pb. Calculate the density of Lead.

Example 3: 250mL of a gas which is known to contain one sulfur atom and an unknown number of fluorides has a mass of 1.63 g at STP
SFx
Step 1: Find the molar mass (g/mol)

Step 2: Find the chemical forumla


Example 4: A 8.00mL sample of (solid) Tin contains 0.632 mol of Sn. Calculate the density of Tin.


We spent the last 20-25 minutes of class, finishing questions #11-30 on the worksheet.

Density & Moles- Nov 5

On Thursday class we learned about density of gases at STP and how to convert into mole from different units.

Density ---> mass per unit volume


Density of gases at STP
-1 mole of gas

Calculate the density of O2 at STP


Example:

A mystery gas has a density of 1.696g/L at STP. It is a diatomic element. Identify the gas.
MM = (1.696 g/L)(22.4 L/mol) = 38 g/mol--->19 g/mol---> Flourine, F2


The density of Boron (solid) is 2.34 g/mL. How many molecules are in a 60.0 mL piece?



Liquid Mercury has a density of 13.55 g/mol. Find the volume occupied by 1.806 x 10^27 atoms of G

Atoms & Molecules- Oct. 30

On Friday we learned about Atoms & Molecules:



For mono atomic elements
a molecule = an element
Eg. Ne=Ne

Diatomic elements
Molecule, Element
Cl2, Cl

Molecules of compounds
H-O-H


Molecule
2 'H' atoms in 1 molecule
1 'O' atoms in 1 molecule


Example:
Write Ammonium Carbonate ---> (NH4)2 CO3
N: 2
H: 8
O: 3
C: 1

Moles <-----> Molecules
6.02 x 10^23 molec/ 1 mol


Example:
How many molecules are in 0.25 mol of CO2?
0.25 mol x 6.02 x 10^23/ 1 mol = 1.505 x 10^23 molec


Example:
5.1772 x 10^24 molecules of H2O = ? moles
5.1772 x 10^24 molecules 1 mol/6.02 x 10^23 =


Example:
Find the number of 'H' atoms in 4.0 mol of ammonia(NH3)?
moles--->molecules--->H atoms
4.0 mol x 6.02 x 10^23/ 1 mol = 2.41 x 10^24 molecules x 3 = 7.22 x 10^24 'H' atoms

Mole Ratio Lab - October 28 Class

Today in class, we completed the Mole Ratio Lab. Our objective of the lab was to find out the ratio of moles of iron, to moles of copper.

The Pre-Lab consisted of:
1) 2Fe + 3CuCl2 -> 3Cu 2FeCl3.
2) The ratio of moles of copper produced to moles of iron consumed is 3:2

The materials we used were:
Apparatus:
Beakers (250 mL)
Wash bottle
Stirring Rod
Crucible tongs
Centigram balance
drying oven
safety glasses
lab apron
plastic gloves
sand paper or emery cloth
face shield
Reagents:
copper (II) chloride
2 iron nails (approx. 5 cm)
1 M hydrochloric acid
distilled water

Our procedure:
1. Find the mass of 250mL beaker. Record the mass to the nearest hundredth of a decimal (g)
2. Add 8g of copper (II) chloride crystals to the beaker. Find and record the mass.
3. Add 50mL of distilled water to the beaker. To dissolve the crystals, swirl the beaker around.
4. Clean and dry 2 nails. Use sand paper if needed. Then, find and record the mass of the nails.
5. Put the nails into the solution and leave them there for 20 min. Observe the formation of copper and some of the iron that will be used up in the beaker.
6. Pick up the nails one by one using the tongs. Before removing the nails from the beaker, use distilled water to rinse off any remaining copper. Use a stirring rod to scrape any excess copper if needed. Let the nails dry on a paper towel.
7. Find and record the mass of the nails after they are completely dry.
8. (Decant- pour off only the liquid from a container that is holding both liquid and solid.)
Decant the liquid from the solid by putting the liquid into another beaker.
9. After decanting, rinse the solid with 25mL of distilled water. Decant again. Repeat this step 4 more times.
10. Wash the solid with 25mL of 1M hydrochloric acid. Decant again (twice). Then clean the solid with 25mL of distilled water.
11. Place the oven in a drying oven to dry.
12. Let the copper dry, then find and record the mass of the beaker + copper.
13. Make sure to clean up your lab and wash your hands properly.

Our observations:
Mass of empty dry beaker = 159.34g
Mass of beaker + copper (II) chloride = 167.34g
Mass of 2 iron nails = 5.45g
Mass of 2 iron nails (after) = 4.95g / 5.0g
Mass of beaker with copper (after) = 160.72g

We discovered that 0.5g of iron was used in the reaction and 4.95 g of copper was produced.
Our final ratio was 1.5:1 and we had a percent error of 56%. This is because we could have spilled some of the liquids and there was too much liquid in the copper and it didn't dry out fully.

Gases & Moles - Oct. 26 Class

Gases and Moles

The volume of a balloon occupied by a certain gas depends on the temperature and pressure.

Standard Pressure & Temperature (STP)

* 0°C & 101.3 kPa

(273 K)

Standard Ambient Temperature & Pressure (SATP) (24.8L/mol)
* 25°C & 100kPa

(298 K)

The volume of 1.0 mole of any gas at STP is 22.4

The molar volume at STP is 22.4L

Example

1) Find the volume occupied by 0.060 mol of CO2 gas at STP

2) Find the number of moles in a 264.0 mL sample of NO2 at 0° and 101.3 kPa(STP)

3) Find the volume occupied by 22.0g of CO2 (g) at STP