Molar Volume of A Gas at STP

GOOD EVENING READERS-WHO-ARE-EXTREMELY-TIRED-AND-WANT-SLEEP i curtsy in your general direction!

mother nature is an ironic vicious woman. it's a shame, she was such a nice lady before we came along

Before we begin our adventure into the land of  calculating molar volume, we must first question the blogger:

What is STP?

• standard temperature (0°C) or (273.15 k) and pressure (1 atmosphere)
• the standard condition to compare volume of gases
• where one mole of gas occupies 22.4L 

What is Molar Volume?

• the volume occupied by one mole of gas 
• 22.4L (=22.4dm³) 

IN A NUTSHELL; THIS POST IS ON CONVERSION

-note: you can only convert from mole to molar volume and vice versa-

From Moles to Molar Volume

the volume occupied by 15.85 moles of N2 is 335.0L.

From Molar Volume to Moles

There are 0.379 moles contained in the sample of CO2 gas.

TA DA!

simple post wasn't it?

OH NO!

I MUST LEAVE YOU LOVELY READERS!

i must save the aussies!

*bonus: mole/molarity review game

JY

Diluting Solutions to Prepare Workable Solutions

To ship chemicals around the world they are concentrated to their lowest form. This means they contain hardly any water. It would be pointless if they contained water because you would just be spending money on shipping water about. Because of this we need to be able to prepare solutions by adding the correct amount of water to receive the desired molarity.

Eg, I have 1L of 18.0M HCl           I need 0.600L of 3.00M HCl

**The moles you start with are the moles you end with***

Formula M1L1=M2L2
 this is because the moles stay the same and each side is  formula for calculating # Moles.

Therefore. 1x 18= 18 moles               0.600x 3= 1.8 moles

1.8/18 = 0.1

That means we need 0.1L of the original solution plus water to to make the total volume equal 0.800L

Here is a process of what you do once you've calculated how much water you need!

http://www.youtube.com/watch?v=bnQJ2q36d2E&feature=related

And if that went to fast for you here is another one

http://www.youtube.com/watch?v=A2YyIo8vSCA&feature=related

Actually the send one is far more detailed than the first one, but i like the music in the first one.

Here is a picture depicting this process

Molar Concentration of Molarity of Solutions

FIRST OFF, WELCOMEE BACK! THIS IS THE VERY FIRST BLOG OF 2011! I HOPE ALL OF YOU HAVE AN AWESOME BREAK! I KNOW YOU GUYS ALL MISSED THE ATOMIC THREE! WELL TIME TO GET TO WORK!


So last class we discussed material about molar concentration or known as molarity of solutions. Here are some notes that are a summary of last days class.


Molar concentration is concentration measured by the number of moles of solute per liter of solution.


Formula:
Molar concentration (M) =  Moles of solute (n)   /   Volume of solution (V in liter)


This formula can be flipped to find the volume and or the number of moles. 


moles of solute= molarity X volume of solution 
Volume of solution= moles of solute/molarity

Examples:

#1 Calculate the molarity when NaOH has 0.510 moles and the volume is 1.400. Be sure to use the right number of significant figures. 

M= 0.510moles of NaOH/ 1.400L

=0.364 moles/ L

#2
Q: Find out the molar concentration of acetic acid (CH₃COOH) solution when  1.24 grams of acetic acid (CH₃COOH) dissolved in 155.0 ml solution?
Solution:  Given,
mass of acetic acid (w)  =  1.24 g
volume of acetic acid (V) = 155.0 ml
                                      = (155.0 / 1000) L         {Because  1 Liter  = 1000 ml}
                                     = 0.155 L
Molar mass of acetic acid (W)  = atomic mass of C atom  + atomic mass of 2 O atoms  + atomic mass of 4 H atoms
                                                        = (12.0  + 2 * 16.00  + 4 * 1.0) g / mol                                                  
                                                        = 48.0 g / mol
And now by the help of molar concentration (molarity) formula we can determine molar concentration of  acetic acid,


Molar concentration of acetic acid (M)  = mass of acetic acid (w)   /   molar mass of acetic acid (W) *  volume of acetic acid (V)
                                                                      =  1.24 g   /  (48.0 g / mol  *  0.155 L)
                                                                      = 0.167 M
So when  1.24 grams of acetic acid (CH₃COOH) dissolved in 155.0 ml solution, molar concentration of this solution is 0.167 M.

 See you guys next class :)!

Formula of a Hydrate

Today we did a laboratory experiment to determine the empirical formula of a hydrate.
The lab can be found on page 45 of the Essential Experiments for Chemistry lab book.

Glossary:
Hydrate: A compound containing water (H20) in it's crystal structure. The general formula is AB xH2O
Anhydrous: The form of a compound without water.
Carbohydrate: Organic compound with the general formula Cx(H2O)y

The sutructure of a hydrate

Objectives:
To determine the percentage of water in a hydrate, the moles of water present in each mole of the hydrate and to write the empirical formula of the hydrate.

Procedure:
Summarized for full procedure refer to lab book

• Put on safety equipment and set up bunsen burner.
• Heat a crucible for 3 min, then let it cool
• Determine the mass of the crucible
• Place the hydrate in the crucible, determine the mass, and then heat for 5 min once the bottom of the crucible turns red.
• Let it cool, then determine it's mass
• Repeat the previous two steps to verify your results.
• Add a few drops of water and record the changes that appear

Results:

1. Determine the mass of the water by subtracting the after heating mass from the before heating.

Then divide the mass of the water by the mass of the hydrate. Multiply your answer by 100 to get a percent.

2. Calculate the mass of the salt by subtracting the mass of the water from the hydrate. Then divide the mass of the salt by the molar mass of the hydrate which is 120.4g. The will give you the number of moles of the salt.

3. Calculate the # of moles of water present in the hydrate by dividing the mass of the water by the molar mass of water, which it 18.0g. 

4. Divide both molar amounts by the smaller molar amount. This should equal one for the salt and seven for water

5. Substitute these numbers into the general empirical formula AB xH2O.

Answer:

AB 7H2O

Here is a video summarizing the above lab
http://www.youtube.com/watch?v=wNqzkNW72rM&feature=related

Empirical Formula of Organic Compounds

salutations drowsy readers!

BE AWAKE AS ARSENIC BASED LIFE FORMS EXIST! wait.

Before we begin what the title implies, what must brush up on our limited knowledge of what we call organic compounds!

(because embedding is disabled, i must give a link)

if the point was vague,organic compounds are basically anything that contains carbon

AND TO BEGIN NOW OUR LESSON!

We can find the empirical formula of an organic compound by BURNING it. When we BURN the organic compound (reacting with oxygen), this leaves us with the BURNT(okay i'll stop) product. From the mass of the products, the moles of each element in the original UNBURNT(last one, i swear!) reactants can be calculated. 

we can do this because of the wonderful law of conservation of mass!
-the law states that the mass of the product is the same as the mass of the reactants-
it is also assumed that all carbon and all oxygen are used up during the combustion.

Sounds confusing? PERHAPS! Let us try an example!

A 10g sample is burned, producing 20g of CO2 and 8.0g of H2O. What is the empirical formula?

• convert the grams into moles
• Mol CO2 = 20g x 1mol CO2 / 44.0g CO2 = 0.455 mol CO2
• Mol H2O = 8g x 1mol H2O / 18.0g H2O = 0.444 mol H2O

• find how many moles of C and H, these are the elements that make up the organic compound
• Mol C = (0.455 mol CO2 x 1 mol C) / 1 mol CO2 = 0.455 mol C
• Mol H = (0.444 mol H2O x 2 mol H) / 1 mol H2O = 0.888 mol H

• find the empirical mass
• divide both moles by the smallest molar amount
• C = 0.455 / 0.455= 1
• H = 0.888 / 0.455 = 2

• SUM IT ALL UP!
• the empirical formula is CH2

• don't forget to check your answer!
• convert from moles to grams, it should add up to 10g
• IF your answer does not add up, remember that oxygen could be a component of the compound
• mass of O = mass of compound - mass of c + mass of H

and as this concludes the BURNT lesson, worry not readers with a BURNING passion to learn, as I leave you a feeling of toasty(wait, what?) security!

-JY

Empirical and Molecular Formula

The Empirical Formula gives the lowest term ration of atoms/ moles in the formula
**all ionic compounds are given in the empirical formula


Ex. C8H12-------> C2H3
     Molecular to empirical
General case of converting grams of substances to empirical formula:
As we have already learned to do convert from grams to moles for each measurement, by divide it by it's atomic mass.
Then dived each molar amount by the smallest molar amount to receive a ration. Round these ratios to whole numbers and the make them your subscripts of your equation.
An example is provided in the video below.

General Case of converting percentage ratios into empirical formula:
Assume you have 100g of the substance. Convert each percentage into grams. Divide the amounts by their atomic mass. It will give you the ratio's which you'll need to round off and then use as your subscripts.



Molecular Formula: Is a multiple of the empirical formula which shows the actual number of atoms that form  each specific compound

Molar Mass / Molar Mass of empirical formula = Molecular Formula

Examples are in the video below!

This video does a far better job of explaining than I ever could so please actually watch it! It's worth it trust me!

http://www.youtube.com/watch?v=gfBcM3uvWfs

Percent Composition

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

These are the rules on how to calculate the percent composition;

Calculate the molecular mass of the compound

Calculate the total mass of each element present in the formula of the compound

Calculate the percent compositon (percentage composition): % by weight (mass) of element
= (total mass of element present ÷ molecular mass) x 100

Examples.
1.Calculate the percent by weight of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl)
Calculate the molecular mass (MM):
MM = 22.99 + 35.45 = 58.44


Calculate the total mass of Na present:
1 Na is present in the formula, mass = 22.99


Calculate the percent by weight of Na in NaCl:
%Na = (mass Na ÷ MM) x 100 = (22.99 ÷ 58.44) x 100 = 39.34%


Calculate the total mass of Cl present:
1 Cl is present in the formula, mass = 35.45


Calculate the percent by weight of Cl in NaCl:
%Cl = (mass Cl ÷ MM) x 100 = (35.45 ÷ 58.44) x 100 = 60.66%


Note: If the answers add up to 100 they are probably correct.

2.A compund has a total molar mass of 18g/mol and a % composition of oxygen and 11.1% hydrogen. How many oxygen atoms are there in the compound?
11.1/100 x 18g/mol = 2.0g/mol = 2 hydrogen

100-11.1=88.9/100 x 18g/mol = 16 = 1 oxygen

There is one oxygen atom in the compound

3.A school boy buys a bag of candy. Each piece of candy contains 5g of glucose (C6 H12 O6). A) find the % composition of glucose B) if the bag contains 150g of candy, does the percent composition of the glucose change? Explain.

a)MMg=180g/mol

C6 = 72/180 x 100 =40%

H12 = 12/180 x 100 = 7%

O6 =96/180 x 100 = 53%

b)No, because the % composition is based on each individual compound and will not affect the candy as a whole. The % composition will not change because if each piece of candy is the same than the composition will stay the same because the numbers will be larger but will have the same ratio.

4.A molecule has a mass of 142g. It contains 5 atoms of one element and 2 atoms of another. What is the other element if one of them is oxygen and oxgyen has five atoms? What is the percentage composition of both elements?

Oxygen: 16 X 5 = 80g

142g-80g=62g

62g/2=31g

**Check for a component with the atomic mass of 31g. 

Phosphorus=31g. 

Check by firguring out the percent composition. 

62/142=43.7%            43.7+56.3=100g.

80/142=56.3%

Answer= P2O5

5.Calculate the molecular mass (MM) of (NH4)3PO4:


MM = 3x[14.01 + (4 x 1.008)] + 30.97 + (4 x 16.00) = 3 x [14.01 + 4.032] + 30.97 + 64.00 = (3 x 18.042) + 30.97 + 64.00 = 54.126 + 30.97 + 64.00 = 149.096


Calculate total mass of N and then the percent:
3 N are present, mass = 3 x 14.01 = 42.03
%N = (mass N ÷ MM) x 100 = (42.03 ÷ 149.096) x 100 = 28.19%


Calculate total mass of H, and then the percent:
12 H are present in the formula, mass = 12 x 1.008 = 12.096
%H = (mass H ÷ MM) x 100 = (12.096 ÷ 149.096) x 100 = 8.11%


Calculate the total mass of P and the percent:
1 P is present in the formula, mass = 30.97
%P = (mass P ÷ MM) x 100 = (30.97 ÷ 149.096) x 100 = 20.77%


Calculate the total mass of O and the percent:
4 O are present in the formula, mass = 4 x 16.00 = 64.00
%O = (mass O ÷ MM) x 100 = (64.00 ÷ 149.096) x 100 = 42.93%


The answers above are probably correct if %N + %H + %P + %O =100, that is,
28.19 + 8.11 + 20.77 + 42.93 = 100

Check out this video!
http://www.youtube.com/watch?v=xbEeyT8nK84
Search up some problems to help you through this!