Molecular Formulas

• A molecular formula represents the actual numbers of atoms of different elements elements in one molecule of a compound.

Ex.

• An example is NO2. Experimental data indicates that the molecular mass of this compound is  about 92.02g/mol.  What is the molecular formula of this compound?
• First, calculate the sum of the atomic masses for NO2 . Look up the atomic masses for the elements from the Periodic table . The atomic masses are found here:
• Nitrogen is 14.0
• Oxygen  is 16.0

• Plugging in these numbers, the sum of the atomic masses for NO2 is:
•                     (14.0) + 2(16.0) = 46

• This means the formula mass of NO2 is 46.0. Compare the formula mass (46.0) to the approximate molecular mass (92.02). The molecular mass is twice the formula mass (92.02/46 = 2.0), so the simplest formula must be multiplied by 2 to get the molecular formula:
• The answer: 2 x NO2 = N204

- George Spencer

Empirical Formulas

• Empirical formulas are the simplest formula of a compound
• They only show the simplest , not actual atoms

Ex.

• The empirical formula for chlorine gas is Cl
• Dinitrogen tetraoxide ≠ N2O4

To determine the empirical formula we need to know the ratio of  each element.

To determine the ratio, fill in the table below for each problem.

Ex:

Atoms	Mass	Molar Mass	Moles	Moles/smallest mole	Ratio
C	8.4	12.0	0.7	2	2
H	2.1	1.0	2.1	6	6
O	5.6	16.0	0.35	1	1

C2H6O
Determine the empirical formula of a compound containing 24.74% potassium, 34.76% manganese, and 40.50% oxygen.

Atoms	Mass	Molar Mass	Moles	Moles/smallest mole	Ratio
K	24.74	39.1	0.63	1	1
Mn	34.76	54.9	0.63	1	1
O	40.50	16.0	2.53	4	4

KMnO4

• The simplest ratio may be decimals. For certain decimals you need to multiply everything by a common number.

Decimal	Multiplying coefficient
0.5	2
0.33, 0.66	3
0.25, 0.75	4
0.2, 0.4, 0.6, 0.8	5

Ex:
Determine the empirical formula of a compound that is 50.5% carbon, 5.26% hydrogen, and 44.2% nitrogen.

Atoms	Mass	Molar Mass	Moles	Moles/smallest mole	Ratio
C	50.5	12.0	4.21	1.33 X 3	4
H	5.26	1.0	5.26	1.66 X 3	5
O	44.2	16.0	3.16	1 X 3	3

C4H5O3

Here is a video that explains molecular and empirical formulas. 

-Benedict Suratos

Percent Composition

What is Percentage Composition? 

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

How to Solve for Percentage Composition? 

To calculate the percentage composition of a compound:
1) Calculate the molecular mass of the compound
2) Calculate the total mass of each element present in the formula of the compound
3) Calculate the percentage composition: 

    % by weight (mass) of element = (total mass of element present ÷ molecular mass) x 100 

Example One: 

Find the percentage composition of a compound that contains 17.6 g of iron and 10.3 g of sulphur. The total mass of the compound is 27.9 g 

         Iron: (17.6 g/ 27.9 g) = 63.1%

         Sulfur: (10.3 g/ 27.9 g) = 36.9% 

         Fe = 63.1 %

           S = 36.1 %

                100.0% 

Example Two: 

Find the percentage composition of a compound that contains 30.2 g of bromine and 4.9 g of magnesium.  

         32.0g 

      +   4.9g

         36.9g (The total mass of the compound) 

         Bromine: (32.0 g/ 36.9 g) = 86.7%

         Magnesium: (4.9 g/ 36.9 g) = 13.3% 

         Br = 86.7 %

        Mg =13.3 %

                100.0% 

For a more insightful look into the world of percentage composition, feel free to watch the video below.

 

-Simon Sierra

Density and Moles

As mentioned in an earlier post and hopefully common knowledge by this point, density is a measure of mass per volume. It is best defined by this formula:

As density is a quantitive measure of both mass or volume, the units used are g/L or g/mL. 

In molar conversions, density becomes an extension of mass that is solved either dividing or multiplying by it. Several examples are below. 

Example One: 

Water has a density of 1.0 g/mL. Determine the mass of 11.5 mL of water. 

        11.5 mL x (1g/1mL) = 11.5 grams  

How many moles are in 11.5 mL of water? 

        11.5g x (1 mol/18 g) = 0.639 mole  

Example Two:

If a gold ring has a volume of 7.50 mL and contains 0.736 mol of Gold determine the density of gold. 

        0.736 mol x (197.0 g/1 mol) = 145 g 

      

        D = (145 g/7.50 mL) 

            = 19 g/mL  

-Simon Sierra

Multistep Conversions

Multistep conversion take all the information mentioned in the previous molar-related posts and applies the processes to a question/exercise such as:

• 11.5g of H2 gas are placed in a balloon at STP. Determine the volume of the balloon? 

In such a scenario, one that you may encounter on any given day, you are being asked to convert between two units of factors with no direct relation (Mass cannot be converted in a single step into Volume; they represent separate measurements). Therefore, it is necessary to follow this convenient "road map": 

 

This is simplified by the following points: 

• Mass to Moles, and vice versa, uses Molar Mass - Atomic Mass of Element or Compound g/mol 
• Volume (at STP) to Moles, and vice versa, uses Molar Volume - 22.4 L/mol   
• Particles (Molecules/Atoms) to Moles, and vice versa, uses Avogadro's Number - 6.02 x 10^3   

Using such information, one can solve the aforementioned problem as follows: 

11.5 g (always begin with the given number) x 1 mole/ 2 grams = 5.75 moles 

5.75 moles x 22.4 Liters/ 1 mole = 129 Liters 

It is important that all the data and units are inserted and cancelled properly and for that, here's another "road map". 

  

And here is another example: 

• A sample of oxygen gas contains 3.5 x 10^21 molecules. How many grams of oxygen is this?

              3.5 x 10^21 molecules x 1 mole/ 6.02 x 10^23 molecules 

           =5.8 x 10^-3 moles 

             5.8 x 10^-3 moles x 2 (16.0) g */ 1 mole

           =0.19 grams of oxygen gas 

*Pay attention to isolating diatomic elements as their atomic mass becomes that of two molecules of that element*

The universality of the application of multi-step conversions, in that it is common in dimensional analysis and recurring in chemistry, is made even simpler by this video: 

-Simon Sierra

Iron & Copper Reaction Lab

This lab really tested not only our skills in chemistry, but also our skills in patience. It was time consuming but allowed us to see how quickly rust can form, although that was not the objective; the lab was designed so that we can use different conversion factors and data of varying data to analyze the chemical reaction of iron and copper. Iron Man approves! Better yet, Captain Kirk does!

He's eighty.  This is him at thirty-two:

Like a fine wine.

-George Spencer

Molar Volume Lab

This class was spent doing a lab to measure the molar volume of a gas. By filling a immersed cylinder with butane gas through the means of a lighter, we could interpret the volume of both cylinder filled and lighter emptied.

-Simon Sierra