Chemistry Empirical Formula, Stoichiometry and Limiting Reagent
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Topics Covered :

`●` Empirical and Molecular Formula
`●` Stoichiometry and Stoichiometric Calculations
`●` Limiting Reagent

Empirical Formula for Molecular Formula :

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`

`color{blue} ul(mathtt ("EMPIRICAL FORMULA"))` : It represents the simplest whole number ratio of various atoms presents in a compound.

`color{blue} ul(mathtt ("MOLECULAR FORMULA"))` Molecular Formula : It is the exact number of different types of atoms present in molecule of a compound.

Method for determining the empirical and molecular formula can be learnt from the following examples :

`=>` Difference Between molecular mass and formula mass

1. Formula mass of a molecule is the sum of the atomic weights of atoms in its empirical formula.

2. Molecular mass of a molecule is its average mass that is calculated by adding together the atomic weights of atoms in the molecular formulae.

3. So, the main difference between the above two depends on whether we use empirical formula or molecular formula.

`●` The formula mass & molecular mass of `H_2O` are one & the same, where the formula & molecular mass of Glucose are different from other.
Q 2663191045

A compound contains `4.07 %` hydrogen, `24.27 %` carbon and `71.65 %` chlorine. Its molar mass is `98.96` g. What are its empirical and molecular formulas?


Step 1. Conversion of mass per cent to grams.
Since we are having mass per cent, it is convenient to use `100 g` of the compound as the starting material. Thus, in the `100 g` sample of the above compound, `4.07g` hydrogen is present, `24.27g` carbon is present and `71.65 g` chlorine is present.

Step 2. Convert into number moles of each element

Divide the masses obtained above by respective atomic masses of various elements.

Moles of hydrogen ` = (4.07 g)/(1.008 g) = 4.04`

Moles of carbon ` = (24.27 g)/(12.01 g) = 2.021`

Moles of chlorine ` = (71.65 g)/(35.453 g) = 2.021`

Step 3. Divide the mole value obtained above by the smallest number

Since 2.021 is smallest value, division by it gives a ratio of `2:1:1` for `H:C:Cl` . In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient
Step 4. Write empirical formula by mentioning the numbers after writing the symbols of respective elements.

`CH_2Cl` is, thus, the empirical formula of the above compound.

Step 5. Writing molecular formula
(a) Determine empirical formula mass Add the atomic masses of various atoms present in the empirical formula.

For `CH_2Cl` empirical formula mass is `12.01 + 2 xx 1.008 + 35.453 = 49.48 g`

(b) Divide Molar mass by empirical formula mass `text(Molar Mass )/text(Empirical formula mass) = (98.96 g)/(49.48 g)`

` 2 = (n)`

(c) Multiply empirical formula by n obtained above to get the molecular formula

Empirical formula `= CH_2Cl, n = 2`. Hence
molecular formula is `C_2H_4Cl_2`.
Q 1933445342

Determine the empirical formula of an oxide of iron which has `69.9%` iron and `30.1%` dioxygen by
Class 11 Exercise 1 Q.No. 3

`:.` Empirical formula `= Fe_2O_ 3`.
Q 1903556448

Determine the molecular formula of an oxide of
iron in which the mass per cent of iron and
oxygen are `69.9` and `30.1` respectively.
Class 11 Exercise 1 Q.No. 8

Empirical formula mass of `Fe_2O_3`

`=2 xx 55.85+3 xx 16.00=159.7 g mol^(-1)`

`n=(text(Molar mass))/(text(Empirical formula mass))=159.8/159.7=1`

Hence, molecular formula is same as empirical
formula, viz., `Fe_2O_ 3`.

Stoichiometry and Stoichiometric Calculations :

The word Stoichiometry is derived from Greek word.

Steoicheion meaning `=>` element

Metron `=>` measure

So, it deals with the calculation of masses (sometimes volumes also) of the reactants and product involved in a chemical reaction.

e.g. Consider the combustion of methane.

The balanced equation for this is :

`CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(g)`

1 molecule 2 molecule 1 molecule 2 molecule

N molecule 2N molecule N molecule 2N molecule

1 mole 2 mole 1 mole 2 mole

16 gram 64 gram 44 gram 36 gram

22.4 L 44.8 L 22.4 L 44.8 L

Here, `CH_4` and `O_2` are reactants

and `CO_2` and `H_2O` are products

Since all the reactants and products are gas therefore it is indicated by letter (g).

Note: In case of solids and liquid, (s) and (l) are written.

The coefficients 2 for `O_2` and 1 for `H_2O` are called stoichiometric coefficients.

Note : These stoichiometric coefficients represent the number of molecules or moles participating in a reaction.

According to the above chemical reaction, we can say that

(i) 1 mole of `CH_4(g)` reacts with 2 moles of `O_2` and give 1 mole of `CO_2`(g) and 2 mole of `H_2O`(g).

We can replace 1 mole by 1 molecule or 1 mole by 22.7 L or 1 mole by 16 g for `CH_4`

Similarly, this can be done for `O_2`, `CO_2` and `H_2O` also.

So, from this we can say that mass, mole and no. of molecules can be interconverted.

Mass ⇋ moles ⇋ no. of molecules

`text(Mass)/text(volume)` = density
Q 2603191048

Calculate the amount of water (g) produced by the combustion of 16 g of methane.


The balanced equation for combustion of methane is

`CH_4 (g) +2O_2(g) → CO_2 (g) +2H_2O (g)`

(i)16 g of `CH_4` corresponds to one mole.
(ii) From the above equation, 1 mol of

`CH_4 (g)` gives 2 mol of `H_2O (g).`

`2` mol of water `(H_2O) = 2 xx (2+16)`

`= 2 xx 18 = 36 g`

1 mol `H_2O = 18 g H_2O => (18 g H_2 O)/(1 mol H_2 O)`

Hence `2` mol `H_2 O xx ( 18 g H_2 O)/(1 mol H_2O)`

` = 2xx18 g H_2O = 36 g H_2O`
Q 2663291145

How many moles of methane are required to produce `22 g CO_2 (g)` after combustion?


According to the chemical equation,

`CH_4 (g) +2O_2 (g) → CO_2 (g) +2H_2O(g)`

`44g CO_2(g)` is obtained from `16 g CH_4 (g)`

[ `because 1` mol `CO_2`(g) is obtained from 1 mol of `CH_4`(g) ]

mole of `CO_2 (g)` ` = 22 g CO_2 (g) xx (1 mol CO_2 (g))/(44 g CO_2 (g))`

`= 0.5 mol CO_2 (g)`

Hence, `0.5` mol `CO_2 (g)` would be obtained from `0.5` mol `CH_4 (g)` or `0.5` mol of `CH_4 (g)` would be required to produce `22`g `CO_2 (g).`
Q 1983356247

How much copper can be obtained from `100g ` of
copper sulphate `(CuSO_4)` ?
Class 11 Exercise 1 Q.No. 7

One mole of `CuSO_4` contains `1` mole (1 g atom) of
Cu Molar mass of `CuSO_4` molecules `= 63.5 + 32 + 4xx 16`
`=159.5 g mol^(-1)`
Thus, `Cu` that can be obtained from `159.5 g` of
`CuSO_4=63.5 g`
Cu that can be obtained from `100 g` of `CuSO_4`
`=(63.5)/(159.5) xx 100 g=39.81 g`
Q 1903645548

Calculate the amount of carbon dioxide that
could be produced when
(i) `1` mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in `16 g` of

(iii) `2` moles of carbon are burnt in `16 g` of
Class 11 Exercise 1 Q.No. 4

The balanced equation for the combustion of
carbon in dioxygen/air is

`C(s)+O_2(g)-> CO_2(g)`
1 mole 1 mole 1 mole
(12 g) (32 g) (44 g)

(i) In air, combustion is complete. Therefore,
`CO_2` produced from the combustion of `1`
mole of carbon `= 44g`.

(ii) As only `16 g` of dioxygen is available. it can
combit.e only with `0.5` mole of carbon, i.e.,
dioxygen is the limiting reactant. Hence. `CO_2` ·
produced `= 22 g`.

(iii) Here again, dioxygen is the limiting reactant.
`16 g` of :lioxygen can combine only with `0.5`
mole ofcarbon. `CO_2` produced again is equal
to `22 g`.

Limiting Reagents :

Sometimes reactants are not present in the reaction as required by a balanced chemical reaction. In that case, one reactant is in excess over the other. The reactant which is present in the lesser amount gets consumed and no reaction takes place whatever be the amount of the other reactant presents. Therefore the reactant which gets consumed limits the amount of product and is called the limiting reagents.
Q 2663391245

`50.0 kg` of `N_2 (g)` and `10.0 kg` of `H_2 (g)` are mixed to produce `NH_3 (g).` Calculate the `NH_3 (g)` formed. Identify the limiting reagent in the production of `NH_3` in this situation.


A balanced equation for the above reaction is written as follows : Calculation of moles

`N_2 (g)+3H_2 (g) ⇌ 2NH_3 (g)`

moles of `N_2`

`= 50 .0kg N_2 xx (1000 g N_2)/(1 kg N_2) xx (1 mol N_2)/(28.0 g N_2)`

` = 17.86xx10^2 mol`

moles of `H_2`

` = 10.00 kg H_2 xx (1000 g H_2)/(1 kg H_2) xx (1 mol H_2)/(2.016 g H_2)`

` = 4.96×10^3 mol`

According to the above equation, `1 mol N_2 (g)` requires `3 mol H_2 (g)`, for the reaction. Hence, for `17.86×10^2 mol` of `N_2`, the moles of `H_2 (g)` required would be

`17.86xx10^2 mol N_2 xx (3 mol H_2 (g))/(1 mol N_2 (g))`

` = 5.36xx10^3 mol H_2`

But we have only `4.96×10^3 mol H_2`. Hence, dihydrogen is the limiting reagent in this case. So `NH_3(g)` would be formed only from that amount of available dihydrogen i.e., `4.96 × 10^3 mol`

Since `3 mol H_2(g)` gives `2 mol NH_3(g)`

`4.96×10^3 mol H_2 (g) xx (2 mol NH_3 (g))/(3 mol H_2 (g))`

`= 3.30×10^3 mol NH_3 (g)`

`3.30×10^3 mol NH_3 (g)` is obtained.

If they are to be converted to grams, it is done as follows

`1 mol NH_3 (g) = 17.0 g NH_3 (g)`

`3.30×10^3 mol NH_3 (g) xx (17.0 g NH_3 (g))/(1 mol NH_3 (g))`

` = 3.30xx10^3 xx 17 g NH_3 (g)`

` = 56.1 Kg NH_3`
Q 1943291143

In a reaction

Identify the limiting reagent, if any, in the
following reaction mixtures.

(i) 300 atoms of A+ 200 molecules of B
(ii) 2 mol A+3 mol B
(iii) 100 atoms of A+ 100 molecules of B
(iv) 5 mol A+ 2.5 mol B
(v) 2.5 mol A+ 5 mol B
Class 11 Exercise 1 Q.No. 23

(i) According to the given reaction, 1 atom of
A reacts with 1 molecule of B. Therefore
200 molecules of B will react with 200 atoms
of A and 100 atoms of A will be left
unreacted. Hence, B is the limiting reagent
while A is the excess reagent.

(ii) According to the given reaction, 1 mol of A
reacts with 1 mol of B. Therefore 2 mol of A
·will react with 2 mol of B. Hence, A is the
limiting reactant

(iii) No limiting reagent

(iv) 2.5 mol of B will react with 2.5 mol of A.
Hence B is the limiting reagent.

(v) 2.5 mol of A will react with 2.5 mol of B .
Hence A is the limiting reagent.