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?

Solution:

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
mass.
Class 11 Exercise 1 Q.No. 3
Solution:

:. 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
Solution:

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.

Solution:

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?

Solution:

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 22g 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
Solution:

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
dioxygen.

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

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.

Solution:

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
A+B_2->AB_2

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
Solution:

(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.