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