## Class 11 Determining Chemistry Empirical and Molecular Formula of Compounds

In this article, the topic “*Determining Empirical and Molecular Formula of Compounds”* discussed. This is the **11th Science** part of Unit 1 ** Chemistry of class 11**.

__Empirical formula__

The **empirical formula** of a compound is defined as the formula that shows the ratio of elements present in the compound, but not the actual numbers of atoms found in the molecule. The ratios are denoted by subscripts next to the element symbols.

The empirical formula is also known as the **simplest formula** because the subscripts are the smallest whole numbers that indicate the ratio of elements.

__Molecular formula__

It describes the **exact number and type of atoms** in a single molecule of a compound. The constituent elements are represented by their chemical symbols, and the number of atoms of each element present in each molecule is shown as a subscript following that element’s symbol.

__Steps for determining an Empirical Formula__

- Start with the number of grams of each element, given in the problem.

If percentages are given, assume that the total mass is 100 grams so that

the mass of each element = the percent given.

- Convert the mass of each element to moles using the molar mass from the periodic table.
- Divide each mole value by the smallest number of moles calculated.
- Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula.
- If the number is too far to be round of then multiply each solution by the same factor to get the lowest whole number multiple.

e.g. If one solution is 1.5, then multiply each solution in the problem by 2 to get 3.

e.g. If one solution is 1.25, then multiply each solution in the problem by 4 to get 5.

- Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the compound is known. Simply calculate the mass of the empirical formula and divide the
**molar mass of the compound**by the**mass of the empirical formula**to find the ratio between the molecular formula and the empirical formula. Multiply all the atoms (subscripts) by this ratio to find the molecular formula.

**Example 1: **A compound is analyzed and calculated to consist of 13.5 g Ca, 10.8 g O, and 0.675 g H. Find the empirical formula of the compound.

**Solution: **Start by converting the mass of each element into moles by looking up the atomic numbers from the periodic table. The atomic masses of the elements are 40.1 g/mol for Ca, 16.0 g/mol for O, and 1.01 g/mol for H.

13.5 g Ca x (1 molCa / 40.1 g Ca) = 0.337 molCa

10.8 g O x (1 mol O / 16.0 g O) = 0.675 mol O

0.675 g H x (1 mol H / 1.01 g H) = 0.668 mol H

Next, divide each mole amount by the smallest number or moles (which is 0.337 for calcium) and round to the nearest whole number:

0.337 mol Calcium / 0.337 = 1.00 mol Calcium

0.675 mol Oxygen / 0.337 = 2.00 mol Oxygen

0.668 mol Hydrogen / 0.337 = 1.98 mol Hydrogen which rounds up to 2.00

Now we have the subscripts for the atoms in the empirical formula: **CaO _{2}H_{2}**

Finally, apply the rules of writing formulas to present the formula correctly. The cation of the compound is written first, followed by the anion. The empirical formula is properly written as **Ca(OH) _{2}**

__Finding Empirical and Molecular Formula from Percent Composition__

**Percent composition**** = (element mass / compound mass) X 100**

If the % composition of a compound is given, then steps for finding the empirical formula are:

- Assume that there are 100 grams of sample. This makes the calculation simple because the percentages will be the same as the number of grams. For example, if 40% of the mass of a compound is oxygen then we can calculate oxygen to be 40grams.
- Convert grams to moles. The empirical formula is a comparison of the number of moles of a compound so the values should be in moles. Using the oxygen example again, there are 16.0 grams per mole of oxygen so 40 grams of oxygen would be 40/16 = 2.5 moles of oxygen.
- Compare the number of moles of each element to the smallest number of moles we got and divide by the smallest number.
- Round of ratio of moles to the nearest whole number as long as it is close to a whole number. In other words, we can round 1.992 up to 2, but we can’t round 1.33 to 1. We need to recognize common ratios, such as 1.333 being 4/3. For some compounds, the lowest number of atoms of an element might not be 1! If the lowest number of moles is four-thirds, then multiply all ratios by 3 to get rid of the fraction.
- Write the empirical formula of the compound. The ratio numbers are subscripts for the elements.
- To find the molecular formula, we need to know the molar mass of the compound. With the help of molar mass, we can find the ratio of the actual mass of the compound to the empirical mass.
**If the ratio is one (as with water, H**_{2}O), then the empirical formula and molecular formula are the same. - If the ratio is 2 (as with hydrogen peroxide, H
_{2}O_{2}), then multiply the subscripts of the empirical formula by 2 to get the correct molecular formula.

**Example 2: **During the excessive physical activity, lactic acid (Molar mass M=90.08 g/mol), is formed in the muscle tissue which causes muscle soreness. Elemental analysis shows that this compound contains 40.0 mass% of Carbon, 6.71 mass% of Hydrogen and 53.3 mass% of Oxygen. Find the Empirical and Molecular Formula of the compound.

**Solution: **

**1. Determination of Empirical Formula**

Mass percent means the mass of elements in 100grams of the lactic acid.

So, the mass of Carbon in 100grams of lactic acid= (40/100) x100= 40grams of carbon.

Similarly, mass of Hydrogen in 100 g of lactic acid= (6.71/100) x100=6.71 grams of Hydrogen

In the same way, mass of Oxygen = (53.3/100) x100= 53.3 g of Oxygen

Now we calculate number of moles of each element-

__For Carbon__ 40 x 1/12 =3.33 moles of carbon

__For Hydrogen__ 6.71 x 1/1.008 = 6.66 moles of Hydrogen

__For Oxygen__ 53.3 x 1/16 = 3.33 moles of Oxygen

Thus the formula comes out to be C_{3.33}H_{6.66}O_{3.33}

Converting the subscripts to integer value we divide the subscripts of each integer by 3.33

Thus we get the formula as C_{3.33/3.33}H_{6.66/3.33}O_{3.33/3.33}

C_{1}H_{2}O_{1}

**The Empirical Formula comes out to be CH _{2}O**

**2. Determination of Molecular Formula**

The Molecular Formula subscripts are whole number multiples of the Empirical Formula. To find this multiple, we divide the given molar mass of the compound (which is known 90.08g/mol) by the Empirical Formula mass (CH_{2}O = 30.03g/mol)

Whole number multiple= 90.08 g/mol / 30.03g/mol

It comes out to be 3.00

Now we multiply the subscripts of the Empirical formula by 3 we get the Molecular Formula

**3. Molecular Formula = C _{3}H_{6}O_{3}**

To verify, we find the molar mass from the Molecular Formula

**Molar mass of Lactic Acid** = (3 x Molar mass of Carbon) + (6 x Molar mass of Hydrogen) + (3 x Molar mass of Oxygen)

**Molar mass of Lactic Acid = **(3 x 12.01 g/mol) + (6 x 1.008g/mol) + (3 x 16.00)

**Molar mass of Lactic Acid = 90.08 g/mol, which was given in the question.**

## In **11 Science** you can visit another notes, NCERT Solutions for Chemistry Class 11

**Chemistry Class 11** : **Molar Mass**

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