TITRATIONS

WHAT IS TITRATION?


A Titration is a common laboratory experimental technique used to determine the concentration of an unknown solution! Because volume measurements play a key role in titration, it is also known as volumetric analysis.


So what is needed to conduct a titration experiment?


1. Buret

2. Stopcock

3. Pipet (Glass tube) & Erlenmeyer Flask

4. Indicators- Used to identify the end point of filtration

5. Stock solution - Known Solution

All together your lab should look like ours, ready to go!!! (Although we're conducting titration experiments next class, this is how i envision our lab to look like lol)

Okay so today in class, we learned how to use the information given to us to determine the concentration of such and such substances:

Dora the ultimate explorer completed a titration of 0.330 M NaOH with 15.00 mL samples of HI of unknown concentration. The data he gathered is below. Determine the concentration of HI.

Since we aren't given the volume used, we subtract the final reading from the initial reading to get it. This has already been done for you.


Next we add up 11.2 mL + 10.9 mL + 11.4 mL = divide by 3 = 11.16666667 mL
*Note we don't have to use 8.9 because it's not a close number to the others


Next we change 11.16 mL into L

11.16 x 1L
            1000 mL


= .0111L


Next we take our concentration and change that into Moles of HI

0.330 M x .011 L = .00363 x 1 = 0.00363 mol
    L                                         1


Finally, we take the moles we got (0.00363)
and divide it by the litres we initially received (15.00 mL)
*remember we must change mL into units of L!


15.00 mL x 1 L
                  1000 mL
= .015 L

Now we have moles and litres, and Concentration is Moles/Litres so all we have to do is divide the 2!

0.00363 mol

.015 L

= .246 M

Fairly straight forward right?

If you want to know more about what Titration is, here is youtube video to watch!

Post by Renren

Solution Stoichiometry (Feb. 24th)

YES, MORE STOICHIOMETRY! GOTTA LOVE IT <3

heres a lil re-cap from last class (feb. 22nd):


water is the universal solvent.
-solutions are the homogenous mixtures composed of a solute and a solvent.
-solute is the chemical present in lesser amount (whatever mis dissolved)
-solvent is the chemical present in the greater amount (whatever does the dissolving)

-------------------------------------------------------------------------------------------------------------

Solution Stoichiometry (feb. 24th):

example: 100mL of 0.250 M iron (II) chloride reacts with excess coper. how many grams of iron are produced?

remember :

1.) WRITE A BALANCE EQUATION: 

2.) CHANGE INTO MOLES:

3.) NEED / HAVE:

4.) CONVERT MOLES INTO MASS:

- how many moles of copper (II) chloride are produced?

- determine [CuCl2]

post by:ivyg

SOLUTION STOICHIOMETRY. Feb 22

Today we learned about Solution Stoichiometry!

H2O = A universal solvent

• 
• Solutions are homogeneous(composed of similar or identical parts or elements) mixtures
• Homogeneous mixtures are composed of a SOLUTE and a SOLVENT
• A Solute is the chemical present in lesser amount (dissolved)
• A Solvent is the chemical present in the greater amount (whatever does the dissolving)
• CHEMICALS DISSOLVED IN WATER ARE AQUEOUS! ie. H2SO4(aq) ---> dissolved in H2o. How much? This is the CONCENTRATION

Basically, molarity is MOLES/ VOLUME

MOL/L = M (MOLARITY)

Getting it? Good.

Examples: 

DETERMINE THE CONCENTRATIONS OF THE SOLUTIONS BELOW:

a) 0.118 mol of water in 2.50 L

0.118 mol

2.50 L 

= 4.7 x 10^-2 M or mol/L

b) 12.5 g of PbCl2 in 30 mL of water

*First change grams into moles, than change mL into L, than take your moles and divide by your amount of liters!

   12.5 g x 1 mol = 0.0448 mol

278.2 g  

                             30 mL x 1 L = .03 L

                                    1000mL

SOLUTION: 0.0448 mol

          .03 L

= 1.49 M

= 1.5 M

Now lets crank up the chemistry and do so more difficult questions!

How many litres of solution are required to make the following solutions?

a) 0.250 mol dissolved to create a 1.75 M solution of Na2S

0.25 mol x 1 L = 0.143 L

  1.75 mol

*for these types of equations, you can flip mol/L to become L/mol since you're trying to find L!

b) 35.0 g of I2 used to create a 0.150 M solution

35.0 g x 1mol = .138 mol x   1L    = .919 L

253.8g            0.150 M

Pretty easy stuff no?? HERE IS A VIDEO TO HELP EXPLAIN TO YOU THE CONCEPT OF MOLARITY IF YOU'RE HAVING SOME TROUBLE! Go around the 4 minute mark because the guy sneezes haha...LOL!

Post by Ren Flores

PERCENT YEEEEEYEE YIELD

The theoretical yield of a reaction is the amount of products that SHOULD be formed. The actual amount depends on the experiment. The percent yield is like a measure of success.
                                                             -how close is the actual amount to the predicted amount?


actual / theoretical = percent % yield           theoretical-actual / theoretical = percent % error








                                      actual amount of product
percentage yield = ------------------------------------------- x 100
                                      expected amount of product

Example:  A student makes a single displacement reaction that produces 2.755 grams of copper. He determines that 3.150 grams of copper should have been produced. Find the student's percentage yield.

                     actual amount of product
percentage yield = ------------------------------------------- x 100
                           expected amount of product

  

                    2.755g

percentage yield = --------------- x 100
                           3.150g  

percentage yield = 87.4603174 %

percentage yield = 87.46 %

*remember!! 1. Recall definition of percent yield.

2. Substitute the actual and predicted yields.

3. Answer: what is the percent yield

post by ivyg

Other Conversions (Volume and Heat)

So today we learned about Conversions using Volume and Heat! We already learned all those mass to moles to mass to mass shenanigans works, so we shall start fresh! D'accord?


A few points before we begin:

• Heat can be included as a separate term in a chemical reactions (Enthalpy)
• Reactions that release heat are exothermic A + B ---> C + energy
• Reactions that absorb heat are endothermic A + B + energy ---> C
• BOTH can be used in stoichiometry

LETS COMMENCE SHALL WE.......

Examples:

How many liters of oxygen are necessary for the combustion of 425 g of sulfur, assuming that the reaction occurs at STP? The balanced equation is S + O2 ---> SO2

425 g of S x 1 mol = 13.2 mol x 1 O2 x 22.4 L = 297 L

                         32.1 g                           1 S       1 mol

easy enough right? try the one on the bottom and see if you get it. The answer is underneath but you must highlight it with your mouse

How many liters of hydrogen are produced if 225 g of iron reacts with hydrochloric acid, assuming that the reaction occurs at STP? The balanced equation is:

Fe + 2HCL ---> FeCl2 + H2

Answer:  90.2 L

noice work, okay now for the iffy part... HEAT

KJ= Heat (energy)

Find the amount of heat released when 5.0 mol of H2 are consumed according to the reaction

N2 + 3 H2 ---> 2NH3 + 1( 46.2 kJ)

• first you can imagine that 46.2 kJ has a coefficient of 1 infront of it

5.0 mol x 1 (kJ) = 1.66 x 46. 2 kJ = 77 kJ

                    3                              

one more:

Find the amount of heat released when 250 g of Ammonia form according to the reaction:

N2 + 3H2 ---> 2NH3 + 1( 46.2 kJ)

Watch and learn:

250 g x 1 mol = 14.7 x 1 = 7.35 mol x 46.2 kJ 

                  17 g                       2 

= 340 kJ

THAT IS ALL! 

Post by Ren again

Today

 (Biology)

• Worked on our Stoichiometry Review Worksheet (preparing for next weeks test)
• G block Chemistry cut short because of Science olympics
• Senior Chemistry won first place!!!!!!! Congrats to them. And Bio won third (no big deal) 

LIMITING REACTIONSSSS

Today we learned about Limiting Reactions.

What is limiting reactions you may ask?

In Chemical reactions, usually one chemical gets used up first before the other, right? The Chemical used up first in a chemical reaction is called the limiting reactant (reagent).

Once it is used up, the reaction comes to a hold! Limiting reactants determines the quantity of products formed.

QUESTION: So how do we find the limiting reactants in a chemical equation? 

ANSWER: To find the Limiting Reactant, assume one reactant is used up. Determine how much of this reaction is required!

kk... you ready?

Question: Nitrogen gas reacts with Hydrogen gas to produce ammonia. How many moles of Ammonia are produced when 0.65 moles of nitrogen gas react with 0.45 moles of hydrogen gas?

Balanced Chemical equation: N2+3H2 ---> 2NH3

N2= 28.0 g/mol

H2= 2.0 g/mol

NH3= 17.0 g/mol

REMEMBER= WHAT YOU NEED

                     WHAT YOU HAVE

Gets it????????????

Now lets try some theoretical yield questions.

Theoretical yield is ACTUAL divided by THEORETICAL times by 100.

This will leave you with a percent, and the theoretical yield of a reaction is the amount of products that SHOULD be formed.

Question: Determine the percent yield for the reaction between 3.73 g of Na and excess O2 if 5.34 g of Na2O2 is recovered.

2Na+ O2---> Na2O2

It's actually very simple :) Here's a video if you need a little extra help! Watch it!

Post by Ren Flores

Mass to Mass Conversions

Today we learned about Mass to Mass problems and converting between them. It was actually really fun!

Example: Questions:

How many grams of O2 are produced from the decomposition of 3.0 g of Potassium Chlorate?


This question gives us grams, but we must change it to grams of another element. First we must write a balanced Chemical equation:


2KClO3 ----> 2K +  Cl2 + 3O2


Than we take what we are given, which is 3.0 grams. Change 3.0 grams into moles, than find what you need over what you have, than with the moles you are given, find grams!


3.0 g x 1 mol  = 0.024 mol x 3 O2 (what you need)
             122.6 g                        2KClO3 (what you have)


= .036 mol x 32.0 g
                      1 mol
= 1.2 g


There you have it!

Let's try one more because this stuff is easy eh?


Question: Determine the mass of lithium hydroxide produced when 0.38 g of lithium nitride reacts with water according to the following equation:

Li3N +3H2O ---> NH3+ 3LiOH

0.38 g x 1 mol = 0.011 mol x 3(LiOH) = 0.033 g x 23.9 g 

                                                         34.7 g                         1mol                            1 mol

=0.78mol

Nice! Okay carry on with your lives

Post by ren ren again

MOLE TO MASS AND MASS TO MOLE

Don't let the title fool you! Today's class was really not that hard!

• Some Questions will give you an amount of moles and ask you to determine the mass. Converting moles to mass only requires ONE ADDITIONAL STEP!

Example: When 6.5 mol of Potassium chlorate decomposes, Potassium Chloride and oxygen gas are produced. What mass of potassium chloride is produced?

... What do we do from here? If you're like me, chances are your face looks like the girl above! But do not worry! Let's take it step by step!

a) First write a balanced Chemical equation. You're given potassium chlorate, and its a decomposition reaction to produce potassium Chloride and oxygen gas.

2KClO3 ---> 2KCl + 3O2

b) Next, find the mass of Potassium Chloride produced.

Watch the steps:

Given 6.5 mol of Potassium Chlorate. Trying to find mass of Potassium chloride.

6.5 mol x 2KCL (what you need) = 6.5 mol x 74.6 g(Potassium Chloride)

                  2KClO3(what you have)                    1mol

= 484.9 g

=4.8 x 10^2 g

Getting it? That was an equation from Moles to mass. Let's try one more before we move on

Example: How many grams of silver nitrate are needed to produce 1.02 mol of Silver Chloride?

a) Write a balanced Equation

2 AgNO3 + BaCl2 ---> Ba(NO3)2 + 2AgCl

b) Next find how many grams of silver nitrate are needed to produce 1.02 mol of silver chloride:

1.02 x 2mol AgNO3 (what you need)= 1.02 x 169.9 g

            2mol AgCl                                                 1mol

= 173.30 g

=173 g

Are you starting to get it? GREAT LETS TRY 1 MORE AND SEE HOW YOU DO!

Since we've learned how to convert from moles to mass, lets try from MASS TO MOLES;)

Question: In the reaction between Nitrogen Dioxide and water, Nitric acid and nitrogen monoxide are formed. How many moles of Nitrogen dioxide react if 120.6 g of Nitric Acid are produced?

a) first make a balanced chemical equation:

3 NO2 + H20------> 2HNO3 + NO

b) Now use what you have and solve (given 120.6 g... change that into moles using the steps)

120.6 g x     1 mol     = 1.914 mol x 3NO2(What you need)      

              63 g (HNO3)                         2HNO3 (what you have)

=2.87 mol of NO2

GREAT JOB! 

POST BY REN REN REN REN

ENERGY AND PERCENT YIELD

My face in class today (due to learning something new!)

Today we learned about Energy and Percent Yield.

• Enthalpy is the energy stored in chemical bonds
• The Symbol of Enthalpy is H, Units of Joules (J)
• Change in Enthalpy is ∆H

It is important to remember what Exothermic and Endothermic reactions are. (wikipedia definitions)

EXOTHERMIC: 

The term exothermic ("outside heating") describes a process or reaction that releases energy from the system, usually in the form of heat, but also in the form of light (e.g. a spark, flame, or explosion), electricity (e.g. a battery), or sound (e.g. burning hydrogen).

 

ENDOTHERMIC:  The word endothermic ("within-heating") describes a process or reaction in which the system absorbs energy from the surroundings in the form of heat. Its etymology stems from the Greek prefix endo-, meaning “inside” and the Greek suffix –ther, meaning “heat”.

IN EXOTHERMIC REACTIONS, ENTHALPY DECREASES (HEAT IS BEING RELEASED)

IN ENDOTHERMIC REACTIONS, ENTHALPY INCREASES

Calorimetry: To experimentall determine the heat released we need to know 3 things:

1. Temperature Change (∆T)
2. Mass (m)
3. Specific heat capacity (C)

These are related by the equation: ∆H = mC∆T or ∆H= mC (Tf-Ti)

An example:

Calculate the heat required to warm a cup of 400 g of water (C= 41.8J/g°C) from 20.0°C to 50.0°C.

STEPS:

∆H = mC∆T

∆H =(400g)(4.18J/g°C)(50.0°C-20.0°C)

=50160 J 

Pretty simple stuff right? Too bad the homework is pretty difficult! Wish us luck!

Post by RenRen

MOLE TO MOLE CONVERSIONS

Some chemistry questions...

REN FLORES asks: Mr. Doktor? When are we ever going to need mole to mole conversions?

IVY GLORIA: Yeah Mr. Doktor! I dun geddit. Mole to Mole? How is that a conversion?

MR DOKTOR: Well, you see, let's say you know how many moles of a compound is reacting in an equation. Once you have the reactions and a balanced equation, you may need to know how many moles you end up with. It's useful to have when you need to know the end result of a chemical reactions. The process is easier to explain in examples than in words.

ADRIENNE ROSS *hand raised, skin glowing*: Oo! Oo! Me! Allow me to show them some brilliant examples! (and quickly so I can go back to watching and starring in pretty little liars!!!)

MOLE TO MOLE CONVERSIONS

Coefficients in balanced equations tell us the # of moles reacted or produced

They can also be used at conversion factors

THIS IS VERY VERY VERY IMPORTANT. ALWAYS REMEMBER:

*****WHAT YOU NEED OVER WHAT YOU HAVE******

IVY GLORIA: Goodness! Wha does dat mean?

ADRIENNE ROSS *huge white smile, skin glowing*: Well, here's an example. If you know you have 0.15 moles of methane consumed in a combustion reaction, and you want to know how many moles of CO2 are produced, start off by writing out the equation...

First of all, you need to know that methane is CH4...

Next, balance out the equation...

Now, here is where the WHAT YOU NEED OVER WHAT YOU HAVE part comes in *cute, but mature sounding, giggle*.

We NEED to know how many moles of CO2 is produced, so we take the coefficient of CO2 and put it over (or divide it by) what we HAVE. What we HAVE is the moles of methane, so take the coefficient of methane and divide the coefficient of CO2 by it.

So it's pretty easy from here on out. We multiply 0.15 moles of methane by 1/1 and we get moles of CO2!

REN FLORES: Oh I get it now!

ADRIENNE ROSS *huge smile, glowing skin*: Great! Here are 2 more examples for you to practice with, then get more practice by doing your homework worksheets!