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C

The term 'molecular formula' is closely related to the empirical formula; the latter represents the simplest ratio of elemental atoms of a compound in the form of positive integers. Both these expressions might be same in few cases; for example, water (HC

_{169723}H_{270464}N_{45688}O_{52243}S_{912}is the longest molecular formula discovered till now. It belongs to a protein called*Titan*, and consists of a mind-boggling 1,89,819 letters!_{2}O) has the same molecular as well as empirical atomic ratios. In all the sub-branches of chemistry, the molecular formula is of extreme importance, and hence, the method to find it should be known to every chemistry student.

Before calculating the atomic ratio of elements that comprises the molecular expression, it is necessary that one finds out the empirical formula of the studied compound. To do this, one needs to know the percentage of elemental composition of the particular compound, and the individual values should total up to 100%, for ease of calculation. These percentages are equated as weight in grams of the elements. Now, by dividing each weight by the respective atomic masses, molar values of every element are acquired. They are divided by the smallest molar value, including itself. The result is written in the form of a ratio (for example, X:Y:Z = 1:2:3), and the ratio numbers are either rounded off to the nearest digit or multiplied by a positive integer to get whole numbers. These are then written as the units of the empirical formula.

Steps for Calculation

Step 1

Multiply all the elemental atomic weights with their respective suffixes that are present in the empirical formula, and add all these values to get the empirical formula weight/empirical unit weight. Make sure that you include at least two decimal points of the atomic weights for accuracy. If you do not know the empirical formula, you need to find it out as described above.

Step 2

Note down the molecular weight of the compound, and divide it by the empirical unit weight (M

_{w}). This will give you the number of empirical units. If this value is in fraction, either multiply it by a positive integer or round it off to the nearest whole number. The molecular weight will be given in most cases.

Step 3

If the weight of the compound is only given in grams, then the number of moles need to be found out (for gaseous compounds). Use the formula given below for this step:

n = (P × V) ÷ (R × T)

Where,

n = no. of moles

P = atmospheric pressure

V = volume occupied

R = gas constant

T = temperature in Kelvins

By dividing weight in grams by value of 'n', we get the molecular weight (M

_{w}) of the compound.

Step 4

The number of empirical units correspond to the molecular units. Thus, if the value of the former is '2', then the molecular formula suffixes are twice of the ones present in the empirical formula. Write the molecular formula in a linear form - for example, X

_{2}Y

_{5}Z

_{11}.

Illustrative Example

Let's say a gaseous compound consists of 70.26% carbon, 9.53% oxygen, and 20.21% hydrogen. It has a volume of 1.5 liters at a temperature of 35.0 C (308 K) and 1 atmospheric pressure. The weight and volume of this compound is 16 grams and 1.5 liters, respectively. Calculate the molecular formula of this compound. The gas constant is given as 0.082 L x atm/mol K.

Firstly, the empirical formula is calculated. The atomic weights can be referred from a periodic table. By equating the percent values as weight in grams and dividing them by the atomic weights, we get the molar amounts:

C = 70.26 ÷ 12.01 = 5.85 mol

H = 20.21 ÷ 1.008 = 20.04 mol

O = 9.53 ÷ 16 = 0.59 mol

H = 20.21 ÷ 1.008 = 20.04 mol

O = 9.53 ÷ 16 = 0.59 mol

The smallest value is 0.59 that belongs to oxygen, which is used as a denominator for dividing other values, including itself. The results are then rounded off and written as suffixes to get the empirical formula. This is represented as:

C = 5.85 ÷ 0.59 = 9.91 = 10

H = 20.04 ÷ 0.59 = 33.96 = 34

O = 0.59 ÷ 0.59 = 1

H = 20.04 ÷ 0.59 = 33.96 = 34

O = 0.59 ÷ 0.59 = 1

Thus, the empirical formula is:

C

_{10}H_{34}O_{}The compound is a gas and its molecular weight is not known. To calculate this weight value, we need to find out the number of moles using the ideal gas law equation, which is given in step no. 3. Divide the weight in grams by the 'n' value to get the molecular weight. This is represented as:

n = (1 × 1.5) ÷ (0.082 × 308)

= 3 ÷ 25.25

= 0.11 mol (number of moles)

= 3 ÷ 25.25

= 0.11 mol (number of moles)

The molecular weight is given as:

M

= 16 ÷ 0.11

= 145.45 g/mol

_{w}= weight in grams ÷ no. of moles= 16 ÷ 0.11

= 145.45 g/mol

Now, add the number of atoms accordingly, by multiplying the atomic weights with the empirical unit suffixes. This is written as:

Empirical weight (E

= 120.1 + 34.272 + 16

= 170.37 gm

_{w}) = 10 (12.01) + 34 (1.008) + 1 (16)= 120.1 + 34.272 + 16

= 170.37 gm

Divide the molecular weight by the empirical weight to get the empirical units, with the following step:

M

= 0.85

_{w}÷ E_{w}= 145.45 ÷ 170= 0.85

As this value is near to a whole number, it can be rounded off to 1. Hence, by multiplying the subscripts of the empirical formula by 1, the molecular formula remains the same as that of the former, which is:

C

_{10}H_{34}O_{}Practice Problems

Calculate the molecular formula of a compound whose molecular weight is 300 g/mol, and has the following components: 60.80% carbon, 20.74% hydrogen, and 18.46% oxygen by weight. The atomic weights are: C (12.01), H (1.008), and O (16).

Calculate the molecular formula of a compound with molecular weight 107.47 g/mol. It consists of 60.42% phosphorus and 39.58% oxygen. The atomic weight of phosphorus is 30.97.

A compound consists of 54.20% C, 13.27% H, 5.47% N, and 27.06% O. Calculate the molecular formula if the molar mass of the compound is 300 g/mol.

Acetic acid consists of 40.50% C, 23.30% H, and 36.2% O. Determine the molecular formula of this acid solution. The molecular weight is 136 g/mol.

Whenever you calculate such formulas, it is necessary to refer to a periodic table, as it will provide you with the exact values of atomic weights, along with elemental symbols. Solve as many problems as you can to be an expert in finding out the answers to such formulas.