Lab 4C "Formula of hydrate"

Hydrate:  A compound that contains water. Written in the formula: AB x C(H20)

Anhydrous:  Without water.

Here is the procedure for our lab:

We found the mass of the water by subtracting the mass of the anhydrous salt after heating from the original hydrate. Then, by dividing the mass of the water by the mass of the hydrate and multiplying by 100, we got a percent of the water in the hydrate. We then determined the number of moles in the salt by dividing the mass of the salt by the molar mass of the hydrate (120.4g). By dividing the mass of the water by the molar mass of the water, we determined the number of moles present in the hydrate.


By:  JZ

Calculating the Empirical Formula of a ORGANIC Compound

C_xH_2y =(burned in O₂)=> x CO₂ + y H₂O

• From the weight of CO₂ and H₂O produced by burning a definite amount of the substance, you can figure out the percent of C and H in the compound.
• Percentage of O is usually obtained by subtracting all percentages of C, H, and N, if the compound does not contain any other element.

Review Empirical Formula:

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

For example:

Calculate the empirical formula of a compound that is 45.99 % C, 5.38% H and the rest is O.

1. Find the moles of the compound

C:  45.99g x  1 mol / 12 g  =  3.8325 moles

H:  5.38g x 1 mol  /  1.0 g  =  5.38 moles

O:   

• 45.99 + 5.38 = 51.37
• 100 - 51.37 = 48.63  -->  O

48.63 x 1 mol  /  16 g  =  3.039375 moles

2. Divide each answer with the lowest number, which is the oxygen.

C:  3.8325 / 3.039375 = 1.260950031 -->  1.27 x 4 = 5
H:  5.38 / 3.039375 = 1.770100761-->  1.77 x 4 = 7
O:  3.03975 / 3.03975  =  1-->  1 x 4 = 4

ANS:  C₅H₇O₄

By: Hikari Hoshika

Empirical & Molecular Formulas!!!

Empirical Formula :  help state the lowest ratio of atoms(or moles) inside a formula

*ALL IONIC COMPOUNDS ARE EMPIRICAL FORMULAS*

For example: C₁₂H_{10 Acnaphthene's empirical formula would be C6H5}

To convert molecular formulas to empirical formulas, use this :

1. Convert grams to moles

2. Divide each mole by the lowest molar amount

3. Scale the ratios to whole numbers

An example would be like this:

Calculate the empirical formula of a compound that contains 65% nitrogen and 35% fluorine.

*assume that it adds up to 100 grams*

Step 1: N:  65g x 1 mol    = 4.643

                                     14 g

            _{F: 35g x 1 mol     = 1.842}

                           _19g

_{Step 2:              4.643   = 2.52}

                         1.842

                          1.842   = 1

                          1.842

Step 3: Since 2.52 isn't a whole number, we need to scale it. Multiply each number by a whole number until 2.52 is a whole number .

In this case, 2.52 x 2 = 5.04<----since it's so close you can round down to 5.

Don't forget to multiply 2 to the other number as well . 2 x 1 =2

N5F2 is the empirical formula.

Tip: if the numbers are already whole, just use them as the empirical formula.

Molecular Formula (MF): is the multiple of an empirical formula that shows the actual # of atoms combined to form a molecule.

Formula :   n=      molar mass of compound 

                         molar mass of empirical formula 

Which means that MF= empirical formula x n

Example: A compound contains a empirical formula of C2H3 with a molar mass of 54 g /mol. Find the molecular formular.

Molar mass of C2H3= 27 g/mol

n=   54g/mol  =2

       27g/mol

MF= 2(C2H3) = C4H6


By : JN