Lab 4C- Formula of a Hydrate

OK, so last Tuesday, all we did was have our quiz on percent composition, empirical formula and molecular formula.
Then last Thursday, we did Lab 4C- Formula of a Hydrate. This lab was done to determine the percentage of water in an unknown hydrate and to determine the moles of water presnt in this unknown hydrate.

First of all, we heated the crucible with a bunsen burner for a couple of minutes to make sure it was dry.

Then, we let the crucible cool down and weighed the empty crucible. We then added the hyrdate to the crucible and weighed it again. After, we heated the crucible until it was dull red on the bottom and waited for it to cool down and weighed it again. We repeated this step again- the second reading of the mass hydrate should have been within 0.03 grams of the first reading.

Finally, we added a few drops of water into the crucible and observed the changes.

Calculating the empirical Formula of Organic Compounds!

Yummmm, organic beets!

The empirical formula of an organic compound can be found by COMBUSTING the compound (reacting it with oxygen). The mass of the products can then be measured to find how much of each reactant was present. Recall: The Law of Conservation of Mass: the mass of the reactants will always equal the mass of the products!

Let's try an easy example: 

A 7.30 gram sample of a hydrocarbon is burned to give 23.8 grams of CO₂ and 7.30 grams of H₂O. What is the empirical formula?  

_{Notice that all of the C's and all of the H's went into making the carbon dioxide and water!} 

Step 1: Convert the grams of carbon dioxide and water into moles

Mol CO2 = 23.8g CO2 x (1Mol CO2)/(44.0g CO2)* = 0.541 mol CO2

* This is the molar mass of CO2

Mol H2O =7.30g H2O x (1Mol H2O)/(18.0g H2O)* = 0.406 mol H2O

* This is the molar mass of H2O

After this step, you can conclude that 0.541 moles of CO2 and 0.406 moles of H2O were produced.

Step 2: We want to isolate the carbon in CO2 and the hydrogen in H2O, these are the elements that make up organic compounds.

Mol C = 0.541 mole CO2 x (1 mole C)*/(1 mole CO2) = 0.541 mol
*Number of moles of C in 1 mole of CO2

Mol H = 0.406 mole H2O x (2 mole H)*/(1 mole CO2) = 0.812 mol

*Number of moles of H in 1 mole of H2O

After this step, you can conclude 0.541 moles of C and 0.812 moles of O were in the original organic substance.

Step 3: Recall how to find the empirical mass.

Divide both number of moles by the smallest molar amount (in this case it's C > 0.541)

C> 0.541/0.541 = 1 (Now read this: We have to multiply this because we multiplied 1.5)

H> 0.812/0.541 = 1.5 (Read this first: We have to multiply by 2 to make this into a whole number)

So, the empirical formula would be C2H3

Step 4: Check your answers!

Convert the moles of C and the moles of H to grams - they should add up to 7.30g

0.541 moles C x (12.0g C)/(1 mole C) = 6.49g C

0.812 moles H x (1.0g H)/(1 mole H) = 0.812g H

6.49g C + 0.812g H = 7.30g

If after checking your answer, you find that your masses do not add up, realize there must be a component of oxygen present in the compound. 

The mass of O = Mass of the compound - Mass of C + Mass of H

The mass of oxygen can then be converted to moles. Reapply step 3.

Some easy review for the quiz - http://lhs2.lps.org/staff/sputnam/practice/UnitV_EmpForm.htm

Here is a step by step tutorial on how to find empirical formula and molecular formula or the "real formula."

 

Empirical & Molecular Formula

Empirical formula: is the lowest ratio of atoms/moles in a formula.

Note: All ionic compounds are in its empirical formula

An example would be...

H2F10 is a molecular formula which reduces into HF5 which is the empirical formula!

For example, we were given  58.5% of carbon, 7.3% of hydrogen, and 34.1% nitrogen. What is the empirical formula?

* Assume that there's 100.0g*

1) Convert grams --> moles

C: 58.5 g x 1 mol = 4.88 mol               
                     12.0 g
H: 7.3 g x 1 mol  = 7.3 mol
                   1.0 g
N: 34.1 g x 1 mol= 2.44 mol
                      14.0g
2) Divide all 3 by smallest molar amount.
C: 4.88/2.44= 2
H: 7.3/2.44= 2.99 --> 3 (Round it only if it's super close!)
N: 2.44/2.44= 1
Empirical Formula= C2H3N
3) Scale ratios to whole numbers if needed. (For this example, it isn't needed..but if for example you ended up with 1.5, multiple it by 2,3,4,5, etc until you get a whole number.)

Here's a nice little flow chart to show you the steps:

 

Molecular Formula (MF): multiples of an empirical formula & shows the actual number of atoms that combine to form a molecule.
To calculate the MF: 
                         n=      molar mass of compound       
                               molar mass of empirical formula 
 n= a whole number multiple of the empirical mass
So then... MF= N x empirical formula
Ex. A molecule has an empirical formula of HO and a molar mass 51.0 g. What's the molecular formula?
Mass of HO= 17.0 g
51.0 g= 3
17.0g                               
Using the above formula of MF=N x empirical formula --> 3x(HO)= H303
Ex. A compound contains 8.96 g of Nitrogen and 5.19 g of Oxygen. The molar mass of the compound is 88.0 g. What's the molecular mass?
N: 8.96g x 1 mol = 0.64 mol

                    14.0g

O: 5.19g x 1 mol  =0.32 mol
                    16.0g
Dividing by the smallest amount:
N: 0.64/0.32= 2                     Therefore the empirical formula is N2O.
O: 0.32/0.32= 1
Then using the molecular formula:
Molar mass of N2O= 44.0 g/mol
88.0 g/mol= 2                       MF= 2 x N20= N402
44.0 g/mol
Note: When writing down the molecular formula, if it doesn't form an ionic compound, write it alphabetically. 

Time for some random chemistry jokes? Kekekeeke.
Q: What is the chemical formula for the molecules in candy?
A: Carbon-Holmium-Cobalt-Lanthanum-Tellurium or CHoCoLaTe 
Ooooh..trickytricky. ;P
Q: What does a teary-eyed, joyful Santa say about chemistry?
A: HOH, HOH, HOH!
Q: What did the bartender say when oxygen, hydrogen, sulfur, sodium, and phosphorous walked into his bar?
A: OH SNaP!