sir i m tring very hard to understand mole concept.bt i m nt getting it at all..sir plz help me out of this prob.actually i have joined rao iit .i m finding it difficult to solve sums based on mole concept .plz help me out sir.

neha bhandare, 8 years ago

Grade:11

1 Answers

deapan

131 Points

8 years ago

study the notes given by me

MOLE CONCEPT

Formula Weight (FW) or Formula Mass

The formula weight of a substance is the sum of the atomic weights of all atoms in a formula unit of the compound, whether molecular or not. Sodium chloride, NaCl, has a formula weight of 58.44 amu (22.99 amu from Na plus 35.45 amu from Cl). NaCl is ionic, so strictly speaking the expression "molecular weight of NaCl" has no meaning. On the other hand, the molecular weight and the formula weight calculated from the molecular formula of a substance are identical.

Solved Problem

1.Calculate the formula weight of each of the following to three significant figures, using a table of atomic weight (AW): (a) chloroform CHCl3 (b) Iron sulfate Fe2 (SO4)3.

2. Calculate the molecular mass (formula mass) of C₆H₁₂O₆?

3.Calculate the molecular mass of C₁₂H₂₂O₁₁?

Solution

a.chloroform

1 x AW of C = 12.0 amu

1 x AW of H = 1.0 amu

3 x AW of Cl = 3 x 35.45 = 106.4 amu

Formula weight of CHCl3 = 119.4 amu

The answer rounded to three significant figures is 119 amu.

b. Iron(III)Sulfate

2 x Atomic weight of Fe = 2 x 55.8 = 111.6 amu

3 x Atomic weight of S = 3 x 32.1 = 96.3 amu

3 x 4 Atomic weight of O =12x16 = 192.0 amu

Formula weight of Fe2 (SO4)3 = 399.9 amu

The answer rounded to three significant figures is 4.00 x 10² amu.

6 × AW of C = 72

12 × AW of H = 12

6 × AW of O = 96

= 72 amu + 12amu + 96 amu = 180 amu = molecular mass of C₆H₁₂O₆

12 × AW of C = 144

22 × AW of H = 22

11 × AW of O = 176

= 144 amu + 22 amu + 176 amu = 342 amu = molecular mass of C₁₂H₂₂O₁₁

Problems for Practice

Calculate the formula masses of the following compounds

a. NO2 b. glucose (C6H12O6) c. NaOH d. Mg(OH)2

e. methanol (CH3 OH) f. PCl3 g. K2 CO3

Gram Atomic Weight

The amount of a substance whose mass in grams is numerically equal to its atomic mass is called Gram atomic mass of the substance. Ex : gram atomic mass of oxygen = 16 grams

Gram Molecular Weight

The amount of a substance whose mass in grams is numerically equal to its molecular mass (formula mass) is called gram molecular mass of that substance .

Molecular mass of O₂ is 32u(amu) Gram molecular mass of oxygen is 32 g

Problems

How many grams of ‘S’ are present in 49 g H₂SO₄?
How many grams of phosphorus are present in 490 grams of H₃PO₄?
Calculate the gram molecular weight of
Cl₂ b) KOH c) BeCl₂

d) FeCl₃ e) BF₃ f) CCl₂F₂

g) Mg(OH)₂ h) UF₆ i) SO₂

j) H₃PO₄ k) (NH₄)₂SO₄ l) CH₃COOH

Solution

98g of H₂SO₄has 32g of ‘S’

32 × 49

therefore 49g of H₂SO₄ has ----------- = 16 g of ‘S’

Avogadro's Number (N_A)

The number of atoms in a 12-g sample of carbon - 12 is called Avogadro's number (to which we give the symbol N_A). Recent measurements of this number give the value 6.0221367 x 10²³, which is 6.023 x 10²³.

A mole of a substance contains Avogadro's number of molecules. A dozen eggs equals 12 eggs, a gross of pencils equals 144 pencils and a mole of ethanol equals 6.023 x 10²³ ethanol molecules.

Significance

The molecular mass of SO₂ is 64 g mol-1. 64 g of SO₂contains 6.023 x 10²³ molecules of SO₂. 2.24 x 10^-2m³ of SO₂ at S.T.P. contains 6.023 x 10²³ molecules of SO2.

Similarly the molecular mass of CO₂is 44 g mol-1. 44g of CO₂contains 6.023 x 10²³ molecules of CO₂. 2.24 x 10^-2m³ of CO₂ at S.T.P contains 6.023 x 10²³ molecules of CO2.

Mole concept

While carrying out reaction we are often interested in knowing the number of atoms and molecules. Some times, we have to take the atoms or molecules of different reactants in a definite ratio.

Eg. Consider the following reaction

2 H₂ + O₂ = 2H₂O

In this reaction one molecule of oxygen reacts with two molecules of hydrogen. So it would be desirable to take the molecules of H₂ and oxygen in the ratio 2:1, so that the reactants are completely consumed during the reaction. But atoms and molecules are so small in size that is not possible to count them individually. In order to overcome these difficulties, the concept of mole was introduced. According to this concept number of particles of the substance is related to the mass of the substance.

Definition

The mole may be defined as the amount of the substance that contains as many specified elementary particles as the number of atoms in 12g of carbon - 12 isotope.

one mole of an atom consists of Avogadro number of particles.

One mole = 6.023 x 1023 particles

One mole of oxygen molecule = 6.023 x 1023 oxygen molecules

One mole of oxygen atom = 6.023 x 1023 oxygen atoms

One mole of ethanol = 6.023 x 1023 ethanol molecules

In using the term mole for ionic substances, we mean the number of formula units of the substance. For example, a mole of sodium carbonate, Na₂CO₃ is a quantity containing 6.023 x 10²³ Na₂CO₃units. But each formula unit of Na₂CO₃ contains 2 x 6.023 x 10²³ Na+ ions and one CO₃^2-ions.

When using the term mole, it is important to specify the formula of the unit to avoid any misunderstanding.

Eg. A mole of oxygen atom (with the formula O) contains 6.023 x 1023 Oxygen atoms. A mole of oxygen molecule (formula O2) contains 6.023 x 1023 O2 molecules (i.e) 2 x 6.023 x 1023 oxygen.

Molar mass

The molar mass of a substance is the mass of one mole of the substance. The mass and moles can be related by means of the formula.

Mass

-------------- = mole

Molar mass

Eg. Carbon has a molar mass of exactly 12g / mol.

Problems

Solved Problems

1. What is the mass in grams of a chlorine atom, Cl?

2. What is the mass in grams of a hydrogen chloride, HCl?

Solution

1) The atomic weight of Cl is 35.5 amu, so the molar mass of Cl is 35.5 g/mol. Dividing 35.5 g by 6.023 x 10²³ gives the mass of one atom.

35.5 g

Mass of a Cl atom = ------------- = 5.90 x 10^-23 g

6.023 x 10²³

= 5.90 x 10^-23 g

2) The molecular weight of HCl equal to the atomic weight of H, plus the atomic weight of Cl, (ie) (1.01 + 35.5) amu = 36.5 amu. Therefore 1 mol of HCl contains 36.5 g HCl

36.5 g

Mass of an HCl molecule = ------------------ = 6.06x10^-23g

6.02 x10²³

= 6.06x10^-23g

Problems For Practice

1. What is the molar mass in grams of a calcium atom, Ca?

2. What is molar mass in grams of an ethanol molecule, C₂H₅OH?

3. Calculate the molar mass (in grams) of each of the following species.

a. Na atom

b. S atom

c. CH₃Cl molecule

d. Na₂SO₃ formula unit

Calculate the molar mass of
Cl₂ b) KOH c) BeCl₂

d) FeCl₃ e) BF₃ f) CCl₂F₂

g) Mg(OH)₂ h) UF₆ i) SO₂

j) H₃PO₄ k) (NH₄)₂SO₄ l) CH₃COOH

m) MgO

Mole Calculations

To find the mass of one mole of substance, there are two important things to know.

i. How much does a given number of moles of a substance weigh?

ii. How many moles of a given formula unit does a given mass of substance contain.

Both of them can be known by using dimensional analysis. To illustrate, consider the conversion of grams of ethanol, C₂H₅OH, to moles of ethanol. The molar mass of ethanol is 46.1 g/mol, So, we write

1 mole C₂H₅OH = 46.1 g of C₂ H₅OH

Thus, the factor converting grams of ethanol to moles of ethanol is 1mol C₂ H₅OH /46.1g C₂ H₅OH. To covert moles of ethanol to grams of ethanol, we simply convert the conversion factor (46.1 g C₂ H₅OH /1 mol C₂ H₅OH).

Again, suppose you are going to prepare acetic acid from 10.0g of ethanol, C2H5OH. How many moles of C₂ H₅OH is this?

you convert 10.0g C₂ H₅OH to moles C₂ H₅OH by multiplying by the appropriate conversion factor.

1 mol C₂ H₅OH

10.0g C₂ H₅OH x ---------------------

46.1 g C₂ H₅OH

= 0.217 mol C₂ H₅OH

Converting Moles of Substances to Grams

Solved Problems

1. ZnI₂, can be prepared by the direct combination of elements. A chemist determines from the amounts of elements that 0.0654 mol ZnI₂ can be formed.

Solution

The molar mass of ZnI₂is 319 g/mol. (The formula weight is 319 amu, which is obtained by summing the atomic weight in the formula) Thus

319 g ZnI₂

0.0654 mol ZnI₂ x -------------------

1 mol ZnI₂

= 20.9 gm ZnI₂

Problems for Practice

H₂O₂ is a colourless liquid. A concentrated solution of it is used as a source of oxygen for Rocket propellant fuels. Dilute aqueous solutions are used as a bleach. Analysis of a solution shows that it contains 0.909 mol H₂O₂ in 1.00 L of solution. What is the mass of H₂O₂ in this volume of solution?
Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?.
CS₂ is a colourless, highly inflammable liquid used in the manufacture of rayon and cellophane. A sample contains 0.0205 mol CS₂. Calculate the mass of CS₂ in the sample.
What mass in grams is represented by

a) 0.40 mole of CO₂

b) 3.00 mole of NH₃

c) 5.14 mole of H₅IO₆

how many grams of Na₂CO₃ are to be weighted to get 0.1 mole Na₂CO₃?
What is the mass of 5 moles of sodium carbonate (Na₂CO₃)?
How much does 4.2 moles of Ca(NO₃)₂ weigh?

Converting Grams of Substances to Moles

In the preparation of lead(II)chromate PbCrO₄, 45.6 g of lead(II)chromate is obtained as a precipitate. How many moles of PbCrO₄ is this?

The molar mass of PbCrO₄ is 323 g/mol (i.e) 1 mol PbCrO₄ = 323 g PbCrO₄ Therefore,

45.6 g PbCrO₄ x 1 mol.PbCrO₄

----------------------------------------------- = 0.141 mol PbCrO₄

323 g PbCrO₄

= 0.141 mol PbCrO₄

Problems for Practice

1) Nitric acid, HNO₃ is a colourless, corrosive liquid used in the manufacture of Nitrogen fertilizers and explosives. In an experiment to develop new explosives for mining operations, a 28.5 g sample of HNO₃ was poured into a beaker. How many moles of HNO₃ are there in this sample of HNO₃?

How many number of moles are present in 540 g of glucose?
How many moles are present in34 grams of Cu(OH)₂
Obtain the moles of substances in the following.

a) 3.43 g C b) 7.05 g Br₂

c) 76 g C₄H₁₀ d) 35.4 g Li₂CO₃

e) 2.57 g As f) 7.83 g P₄

g) 41.4 g N₂H₄ h) 153 g Al₂(SO₄)₃

Calculation of the Number of Molecules in a Given Mass

Solved Problem

How many molecules are there in a 3.46 g sample of hydrogen chloride, HCl?

Note: The number of molecules in a sample is related to moles of compound (1 mol HCl = 6.023 x 10²³ HCl molecules). Therefore if you first convert grams HCl to moles, then you can convert moles to number of molecules.

Solution

1 mole HCl 6.023 × 10²³ HCl molecules

3.46g HCl × ----------------- × ------------------------------------

36.5g HCl I mole HCl

= 5.71 × 10²² HCl molecules

Problems for Practice

1. How many molecules are there in 56mg HCN?

2. Calculate the following

a. Number of molecules in 43g NH₃

b. Number of atoms in 32.0 g Br₂

c. Number of atoms in 7.46 g Li

3. Calculate the number of molecules present in 31.6g of KMnO₄

4. what is the total number of molecules present in 18 g of glucose (C₆H₁₂O₆)

5. Calculate the number of atoms of each element present in 0.5 moles of MgSO₄

6. how many gram atoms are present in 144 g of magnesium

7. Calculate the number of molecules of sulphur (S₈) in 16 g of solid sulphur

8. How many moles are present in 2.45 × 10²³ molecules of CH₄

9. how many grams are there in 3.4 × 10²⁴ molecules of NH₃

10) How many molecules are there in 24 grams of FeF₃

11) How many molecules are there in 450 grams of Na₂SO₄

12) How many grams are there in 2.3 × 10²⁴ atoms of silver

13) How many grams are there in 7.4 × 10²³ molecules of AgNO₃

14) How many grams are there in 7.5 × 10²³ molecules of H₂SO₄

15) How many grams are there in 9.4 × 10²⁵ molecules of H₂

16) How many molecules are there in 122 grams of Cu(NO₃)₂

17) How many molecules are there in 230 grams of CoCl₂

Calculation of Empirical Formula from Quantitative Analysis and Percentage composition

Empirical Formula

"An empirical formula (or) simplest formula for a compound is the formula of a substance written with the smallest integer subscripts". For most ionic substances, the empirical formula is the formula of the compound. This is often not the case for molecular substances. For example, the formula of sodium peroxide, an ionic compound of Na⁺ and O₂^2-, is Na₂O₂. Its empirical formula is NaO. Thus empirical formula tells you the ratio of numbers of atoms in the compound.

Steps for writing the Empirical formula

The percentage of the elements in the compound is determined by suitable methods and from the data collected, the empirical formula is determined by the following steps.

Divide the percentage of each element by its atomic mass. This will give the relative number of moles of various elements present in the compound.
Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements.
Multiply the figures, so obtained by a suitable integer of necessary in order to obtain whole number ratio.
Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.

Solved Problem

A compound has the following composition Mg = 9.76%, S = 13.01%, O = 26.01, H₂O = 51.22, what is its empirical formula?

[Mg = 24, S = 32, O = 16, H = 1]

Solution

Element	Percentage	Relative number of moles	Simple ration moles	Simplest whole number ratio
Magnesium	9.76	9.76 --------- = 0.406 24	0.406 ---------- = 1 0.406	1
Sulphur	13.01	13.01 ------------ = 0.406 32	0.406 ---------- = 1 0.406	1
Oxygen	26.01	26.01 ----------- = 1.625 16	1.625 ---------- = 4 0.406	4
Water	51.22	51.22 ---------- = 2.846 18	2.846 ---------- = 7 0.406	7

Hence the empirical formula is Mg SO₄.7H₂O.

Problems for Practice

Determine the empirical formula
C₆H₆ b) C₂H₆ c) C₃H₈

d) Fe₃(CO)₉ e) C₂H₄O₂ f) N₂H₄

g) CaBr₂ h) C₂H₂ i) Na₂SO₄

j) C₆H₅N k) P₄O₁₀ l) Re₂Cl₆

m) Se₃O₉ n) LiCl

A substance on analysis, gave the following percentage composition, Na = 43.4%, C = 11.3%, 0 = 43.3% calculate its empirical formula [Na = 23, C = 12, O = 16].

Ans:- Na₂CO₃

What is the simplest formula of the compound which has the following percentage composition: Carbon 80%, hydrogen 20%.

Ans:- CH₃

A compound on analysis gave the following percentage composition: C - 54.54%, H = 9.09%, O = 36.36%

Ans:- C₂H₄O

Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present in it. Carbon = 10.06 %, hydrogen = 0.84 %, chlorine = 89.10 %. Calculate the empirical formula of the compound

Ans :- CHCl₃

The percentage of nitrogen in ammonia is

14 × 100

Ans:- ------------

Determine the empirical formula from the present composition for each of the following
92.24% C 7.76% H

36.48% Na 25.44% S 38.08%O
49.99% C 5.61% H 44.40%O
38.76% Ca 19.97%P 41.27%O

Calculate the percentage composition for the following compounds

Cr₂O₃
Ca₃(PO₄)₂

Molecular Formula from Empirical Formula

The molecular formula of a compound is a multiple of its empirical formula.

Example

The molecular formula of acetylene, C₂H₂ is equivalent to (CH)₂, and the molecular formula of benzene, C₆H₆ is equivalent to (CH)₆. Therefore, the molecular weight is some multiple of the empirical formula weight, which is obtained by summing the atomic Weights from the empirical formula. For any molecular compound

Molecular Weight = n x empirical formula weight.

Where `n' is the whole number of empirical formula units in the molecule. The molecular formula can be obtained by multiplying the subscripts of the empirical formula by `n' which can be calculated by the following equation

Molecular Weight

n = -------------------------------------------

Empirical formula Weight

Steps for writing the molecular formula

Calculate the empirical formula
Find out the empirical formula mass by adding the atomic mass of all the atoms present in the empirical formula of the compound.
Divide the molecular mass (determined experimentally by some suitable method) by the empirical formula mass and find out the value of n which is a whole number.
Multiply the empirical formula of the compound with n, so as to find out the molecular formula of the compound.

Solved Problem

A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09%, O = 36.36 %. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.

Solution

Calculation of empirical formula

Element	Percentage	Relative number of moles	Simple ratio moles	Simplest whole number ratio
C	54.54	54.54 ------- - = 4.53 12	4.53 -------- = 2 2.27	2
H	9.09	9.09 -------- = 9.09 1	9.09 -------- = 4 2.27	4
O	36.36	36.36 ---------- = 2.27 16	2.27 --------- = 1 2.27	1

Calculation of Molecular formula

Empirical formula mass = 12 x 2 + 1 x 4 + 16 x 1 = 44

Molecular mass = 2 x Vapour density

= 2 x 44 = 88

Molecular mass 88

n = ------------------------------------ = --------- = 2

Empirical Formula mass 44

= C₂H₄O x 2

= C₄H₈O₂

2)A compound on analysis gave the following percentage composition: Na=14.31%, S = 9.97%, H = 6.22%, O = 69.5%, calculate the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16].

Solution :- Calculation of empirical formula

Element	Percentage	Relative number of moles	Simple ratio moles	Simplest whole number ratio
Na	14.31	14.31 ------- = 0.62 23	0.62 ------ = 2 0.31	2
S	9.97	19.97 -------- = 0.31 32	0.31 ------ = 1 0.31	1
H	6.22	6.22 ------ = 6.22 1	6.22 ----- = 20 0.31	20
O	69.5	69.5 -------- = 4.34 16	4.34 ------ = 14 0.31	14

The empirical formula is Na₂SH₂₀O₁₄

Calculation of Molecular formula

Empirical formula mass = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322

Molecular mass 322

n = ---------------------------------------- = ------------ = 1

Empirical formula mass 322

Hence molecular formula = Na₂SH₂₀O₁₄

Since all hydrogens are present as H₂O in the compound, it means 20 hydrogen atoms must have combined. It means 20 hydrogen atoms must have combined with 10 atoms of oxygen to form 10 molecules of water of crystallisation. The remaining (14 - 10 = 4) atoms of oxygen should be present with the rest of the compound.

Hence, molecular formula = Na₂SO₄.10H₂O.

Problems for Practice

An organic compound was found to have contained carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour - density was found to be 29.5. What is the molecular formula of the compound?

Ans:- C₂H₅NO

A compound contains 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75. Calculate the empirical and molecular formula.

Ans:- C₂H₃O₃, C₄H₆O₆

An acid of molecular mass 104 contains 34.6% carbon, 3.85% hydrogen and the rest is oxygen. Calculate the molecular formula of the acid.
What is the simplest formula of the compound which has the following percentage composition: carbon 80%, Hydrogen 20%, If the molecular mass is 30, calculate its molecular formula.
A compound analyzes as 79.08% C, 5.54% H and 15.38%N. what is the molecular formula if the molar mass is 273.36g/mol?
Determine the molecular formula
A compound has an empirical formula of NO₂ and a molar mass of 92g/mol
A compound has an empirical formula of C₂H₃O and a molar mass of 172g/mol
Ibuprofen, a common headache remedy, has an empirical formula of C₇H₉O and a molar mass of 215g/mol

COMPREHENSION

For questions 1 to 5

The term mole was introduced by Ostwald in 1896. A mole (mol) is defined as the number of atoms in 12.01g of carbon-12. The number of atoms in 12g of carbon-12 has been found experimentally to be 6.02 × 10²³. This number is also known as Avogadro’s number named in honour of Amedeo Avogadro. A mole of oxygen atoms contains 6.02 × 10²³oxygen atoms and a mole of oxygen molecules contains 6.02 × 10²³ oxygen molecules. Therefore, a mole of oxygen molecules is equal to two moles of oxygen atoms, i.e., 2 × 6.02 × 10²³ oxygen atoms. The mass of one mole atoms of any element is exactly equal to the atomic mass in grams (gram-atomic mass or gram atom) of that element. Similarly, the mass of 6.02 × 10²³ molecules (1 mole) of a substance is equal to its molecular mass in grams or gram-molecular mass or gram molecule. It has also been established by Avogadro’s hypothesis that one gram-molecular mass of any gaseous substance occupies a volume of 22.4 litres at NTP. Thus, one mole, i.e., 6.02  1023 molecules of any gaseous substance occupies 22.4 litres as volume at NTP.

Weight of a substance in grams number of molecules

Number of moles = --------------------------------------- = -------------------------------------

Gram molecular weight Avogadro number

The molecular formula of a substance may be determined from the empirical formula if the molecular mass of the substance is known. The molecular formula is always a simple multiple of empirical formula and the value of simple multiple is obtained by dividing molecular mass with empirical formula mass. Hence, empirical formula of a compound can be defined as the simplest whole number ratio formula of the compound.

Consider the following table for compound ‘X’ :

Element	Percentage	Atomic Mass	Relative No. of atoms	Simplest ratio
Carbon	66.67%	12	66.67 -------- = 5.55 12	5.55 ------ = 3 1.85
Hydrogen	7.4%	1	7.4 ---- = 7.4 1	7.4 ------ = 4 1.85
Nitrogen	25.9%	14	25.9 ----- = 1.85 14	1.85 -------- = 1 1.85

Now based on the data given above, answer the following questions.

1) The empirical formula of the compound ‘X’ (as per given data) is

(A) C₁₂H₁₆N₄ (B) C₆H₈N₂

2) If the molecular mass of the compound ‘X’ is found to be 108 g/mol, then its molecular formula is:

(A) C₁₂H₁₆N₄ (B) C₉H₁₂N₃

3) Calculate the total number of atoms present in 54 g of the compound ‘X’ (N_A=avogadro’s number)

(A) N_A (B) 16 N_A

4) Upon complete combustion of the 54 g of the compound ‘X’, the total amount of carbon dioxide released will be

(A) 264 g (B) 88 g

5) The released CO₂ (as Q. No. 18) is mixed with the same amount of laughing gas (N₂O) at NTP. Then, the total volume occupied by the mixture at NTP is

(A) 67.2 L (B) 44.8 L

Solution

C₃H₄N is the simplest whole number ratio formula of the compound.
Empirical formula mass = 12 × 3 + 1 × 4 + 14 = 54

n = 108

-------- = 2

Molecular formula = [C₃H₄N] 2 = C₆H₈N₂

1 mole = N_A molecules of C₆H₈N₂ = 16 N_A atoms

54g of the compound = 54

-------- = 0.5 mole

108

= (16 N_A) × 0.5 atoms = 8 NA

1 mole of C₆H₈N₂ upon complete combustion gives 6 moles of CO₂
moles of C₆H₈N₂upon combustion gives = 3 moles of CO₂

= 3 × 44 = 132 g

3 moles CO₂ + 3 moles of N₂O

= 6 × 22.4 = 134.4 L

FORMULAS FOR MOLE CONCEPT

Weight of a substance in grams number of molecules

Number of moles = --------------------------------------- = -------------------------------------

Gram molecular weight Avogadro number

Volume of a gas at STP

Number of moles of a gaseous substance = -----------------------------------

Gram molar volume

Gram molecular weight

Weight of a single molecule = ---------------------------------

Avogadro number

Molecular weight

Atomic weight of a gaseous element = ---------------------------------

Atomicity

Gram atomic weight

Absolute weight of an atom of any element is = --------------------------------------------

Avogadro number

Weight of the substance = Number of moles × gram molecular weight

Weight of the substance

Number of molecules = number of moles × Avogadro number = ------------------------------ × N_A

Gram molecular weight

Mass

-------------- = mole

Molar mass

Mass

-------------- = Molar mass

mole

Molecular formula = empirical formula × n (where n is a small whole number)

Molecular mass

n = -------------------------------------

empirical formula

Molecular mass = 2 x Vapour density

MOLE CONCEPT

Formula Weight (FW) or Formula Mass

Solved Problem

1.Calculate the formula weight of each of the following to three significant figures, using a table of atomic weight (AW): (a) chloroform CHCl3 (b) Iron sulfate Fe2 (SO4)3.

2. Calculate the molecular mass (formula mass) of C₆H₁₂O₆?

3.Calculate the molecular mass of C₁₂H₂₂O₁₁?

Solution

a.chloroform

1 x AW of C = 12.0 amu

1 x AW of H = 1.0 amu

3 x AW of Cl = 3 x 35.45 = 106.4 amu

Formula weight of CHCl3 = 119.4 amu

The answer rounded to three significant figures is 119 amu.

b. Iron(III)Sulfate

2 x Atomic weight of Fe = 2 x 55.8 = 111.6 amu

3 x Atomic weight of S = 3 x 32.1 = 96.3 amu

3 x 4 Atomic weight of O =12x16 = 192.0 amu

Formula weight of Fe2 (SO4)3 = 399.9 amu

The answer rounded to three significant figures is 4.00 x 10² amu.

6 × AW of C = 72

12 × AW of H = 12

6 × AW of O = 96

= 72 amu + 12amu + 96 amu = 180 amu = molecular mass of C₆H₁₂O₆

12 × AW of C = 144

22 × AW of H = 22

11 × AW of O = 176

= 144 amu + 22 amu + 176 amu = 342 amu = molecular mass of C₁₂H₂₂O₁₁

Problems for Practice

Calculate the formula masses of the following compounds

a. NO2 b. glucose (C6H12O6) c. NaOH d. Mg(OH)2

e. methanol (CH3 OH) f. PCl3 g. K2 CO3

Gram Atomic Weight

The amount of a substance whose mass in grams is numerically equal to its atomic mass is called Gram atomic mass of the substance. Ex : gram atomic mass of oxygen = 16 grams

Gram Molecular Weight

The amount of a substance whose mass in grams is numerically equal to its molecular mass (formula mass) is called gram molecular mass of that substance .

Molecular mass of O₂ is 32u(amu) Gram molecular mass of oxygen is 32 g

Problems

How many grams of ‘S’ are present in 49 g H₂SO₄?
How many grams of phosphorus are present in 490 grams of H₃PO₄?
Calculate the gram molecular weight of
Cl₂ b) KOH c) BeCl₂

d) FeCl₃ e) BF₃ f) CCl₂F₂

g) Mg(OH)₂ h) UF₆ i) SO₂

j) H₃PO₄ k) (NH₄)₂SO₄ l) CH₃COOH

Solution

98g of H₂SO₄has 32g of ‘S’

32 × 49

therefore 49g of H₂SO₄ has ----------- = 16 g of ‘S’

Avogadro's Number (N_A)

A mole of a substance contains Avogadro's number of molecules. A dozen eggs equals 12 eggs, a gross of pencils equals 144 pencils and a mole of ethanol equals 6.023 x 10²³ ethanol molecules.

Significance

The molecular mass of SO₂ is 64 g mol-1. 64 g of SO₂contains 6.023 x 10²³ molecules of SO₂. 2.24 x 10^-2m³ of SO₂ at S.T.P. contains 6.023 x 10²³ molecules of SO2.

Similarly the molecular mass of CO₂is 44 g mol-1. 44g of CO₂contains 6.023 x 10²³ molecules of CO₂. 2.24 x 10^-2m³ of CO₂ at S.T.P contains 6.023 x 10²³ molecules of CO2.

Mole concept

While carrying out reaction we are often interested in knowing the number of atoms and molecules. Some times, we have to take the atoms or molecules of different reactants in a definite ratio.

Eg. Consider the following reaction

2 H₂ + O₂ = 2H₂O

Definition

The mole may be defined as the amount of the substance that contains as many specified elementary particles as the number of atoms in 12g of carbon - 12 isotope.

one mole of an atom consists of Avogadro number of particles.

One mole = 6.023 x 1023 particles

One mole of oxygen molecule = 6.023 x 1023 oxygen molecules

One mole of oxygen atom = 6.023 x 1023 oxygen atoms

One mole of ethanol = 6.023 x 1023 ethanol molecules

When using the term mole, it is important to specify the formula of the unit to avoid any misunderstanding.

Eg. A mole of oxygen atom (with the formula O) contains 6.023 x 1023 Oxygen atoms. A mole of oxygen molecule (formula O2) contains 6.023 x 1023 O2 molecules (i.e) 2 x 6.023 x 1023 oxygen.

Molar mass

The molar mass of a substance is the mass of one mole of the substance. The mass and moles can be related by means of the formula.

Mass

-------------- = mole

Molar mass

Eg. Carbon has a molar mass of exactly 12g / mol.

Problems

Solved Problems

1. What is the mass in grams of a chlorine atom, Cl?

2. What is the mass in grams of a hydrogen chloride, HCl?

Solution

1) The atomic weight of Cl is 35.5 amu, so the molar mass of Cl is 35.5 g/mol. Dividing 35.5 g by 6.023 x 10²³ gives the mass of one atom.

35.5 g

Mass of a Cl atom = ------------- = 5.90 x 10^-23 g

6.023 x 10²³

= 5.90 x 10^-23 g

2) The molecular weight of HCl equal to the atomic weight of H, plus the atomic weight of Cl, (ie) (1.01 + 35.5) amu = 36.5 amu. Therefore 1 mol of HCl contains 36.5 g HCl

36.5 g

Mass of an HCl molecule = ------------------ = 6.06x10^-23g

6.02 x10²³

= 6.06x10^-23g

Problems For Practice

1. What is the molar mass in grams of a calcium atom, Ca?

2. What is molar mass in grams of an ethanol molecule, C₂H₅OH?

3. Calculate the molar mass (in grams) of each of the following species.

a. Na atom

b. S atom

c. CH₃Cl molecule

d. Na₂SO₃ formula unit

Calculate the molar mass of
Cl₂ b) KOH c) BeCl₂

d) FeCl₃ e) BF₃ f) CCl₂F₂

g) Mg(OH)₂ h) UF₆ i) SO₂

j) H₃PO₄ k) (NH₄)₂SO₄ l) CH₃COOH

m) MgO

Mole Calculations

To find the mass of one mole of substance, there are two important things to know.

i. How much does a given number of moles of a substance weigh?

ii. How many moles of a given formula unit does a given mass of substance contain.

1 mole C₂H₅OH = 46.1 g of C₂ H₅OH

Again, suppose you are going to prepare acetic acid from 10.0g of ethanol, C2H5OH. How many moles of C₂ H₅OH is this?

you convert 10.0g C₂ H₅OH to moles C₂ H₅OH by multiplying by the appropriate conversion factor.

1 mol C₂ H₅OH

10.0g C₂ H₅OH x ---------------------

46.1 g C₂ H₅OH

= 0.217 mol C₂ H₅OH

Converting Moles of Substances to Grams

Solved Problems

1. ZnI₂, can be prepared by the direct combination of elements. A chemist determines from the amounts of elements that 0.0654 mol ZnI₂ can be formed.

Solution

The molar mass of ZnI₂is 319 g/mol. (The formula weight is 319 amu, which is obtained by summing the atomic weight in the formula) Thus

319 g ZnI₂

0.0654 mol ZnI₂ x -------------------

1 mol ZnI₂

= 20.9 gm ZnI₂

Problems for Practice

H₂O₂ is a colourless liquid. A concentrated solution of it is used as a source of oxygen for Rocket propellant fuels. Dilute aqueous solutions are used as a bleach. Analysis of a solution shows that it contains 0.909 mol H₂O₂ in 1.00 L of solution. What is the mass of H₂O₂ in this volume of solution?
Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?.
CS₂ is a colourless, highly inflammable liquid used in the manufacture of rayon and cellophane. A sample contains 0.0205 mol CS₂. Calculate the mass of CS₂ in the sample.
What mass in grams is represented by

a) 0.40 mole of CO₂

b) 3.00 mole of NH₃

c) 5.14 mole of H₅IO₆

how many grams of Na₂CO₃ are to be weighted to get 0.1 mole Na₂CO₃?
What is the mass of 5 moles of sodium carbonate (Na₂CO₃)?
How much does 4.2 moles of Ca(NO₃)₂ weigh?

Converting Grams of Substances to Moles

In the preparation of lead(II)chromate PbCrO₄, 45.6 g of lead(II)chromate is obtained as a precipitate. How many moles of PbCrO₄ is this?

The molar mass of PbCrO₄ is 323 g/mol (i.e) 1 mol PbCrO₄ = 323 g PbCrO₄ Therefore,

45.6 g PbCrO₄ x 1 mol.PbCrO₄

----------------------------------------------- = 0.141 mol PbCrO₄

323 g PbCrO₄

= 0.141 mol PbCrO₄

Problems for Practice

How many number of moles are present in 540 g of glucose?
How many moles are present in34 grams of Cu(OH)₂
Obtain the moles of substances in the following.

a) 3.43 g C b) 7.05 g Br₂

c) 76 g C₄H₁₀ d) 35.4 g Li₂CO₃

e) 2.57 g As f) 7.83 g P₄

g) 41.4 g N₂H₄ h) 153 g Al₂(SO₄)₃

Calculation of the Number of Molecules in a Given Mass

Solved Problem

How many molecules are there in a 3.46 g sample of hydrogen chloride, HCl?

Solution

1 mole HCl 6.023 × 10²³ HCl molecules

3.46g HCl × ----------------- × ------------------------------------

36.5g HCl I mole HCl

= 5.71 × 10²² HCl molecules

Problems for Practice

1. How many molecules are there in 56mg HCN?

2. Calculate the following

a. Number of molecules in 43g NH₃

b. Number of atoms in 32.0 g Br₂

c. Number of atoms in 7.46 g Li

3. Calculate the number of molecules present in 31.6g of KMnO₄

4. what is the total number of molecules present in 18 g of glucose (C₆H₁₂O₆)

5. Calculate the number of atoms of each element present in 0.5 moles of MgSO₄

6. how many gram atoms are present in 144 g of magnesium

7. Calculate the number of molecules of sulphur (S₈) in 16 g of solid sulphur

8. How many moles are present in 2.45 × 10²³ molecules of CH₄

9. how many grams are there in 3.4 × 10²⁴ molecules of NH₃

10) How many molecules are there in 24 grams of FeF₃

11) How many molecules are there in 450 grams of Na₂SO₄

12) How many grams are there in 2.3 × 10²⁴ atoms of silver

13) How many grams are there in 7.4 × 10²³ molecules of AgNO₃

14) How many grams are there in 7.5 × 10²³ molecules of H₂SO₄

15) How many grams are there in 9.4 × 10²⁵ molecules of H₂

16) How many molecules are there in 122 grams of Cu(NO₃)₂

17) How many molecules are there in 230 grams of CoCl₂

Calculation of Empirical Formula from Quantitative Analysis and Percentage composition

Empirical Formula

Steps for writing the Empirical formula

The percentage of the elements in the compound is determined by suitable methods and from the data collected, the empirical formula is determined by the following steps.

Divide the percentage of each element by its atomic mass. This will give the relative number of moles of various elements present in the compound.
Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements.
Multiply the figures, so obtained by a suitable integer of necessary in order to obtain whole number ratio.
Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.

Solved Problem

A compound has the following composition Mg = 9.76%, S = 13.01%, O = 26.01, H₂O = 51.22, what is its empirical formula?

[Mg = 24, S = 32, O = 16, H = 1]

Solution

Element	Percentage	Relative number of moles	Simple ration moles	Simplest whole number ratio
Magnesium	9.76	9.76 --------- = 0.406 24	0.406 ---------- = 1 0.406	1
Sulphur	13.01	13.01 ------------ = 0.406 32	0.406 ---------- = 1 0.406	1
Oxygen	26.01	26.01 ----------- = 1.625 16	1.625 ---------- = 4 0.406	4
Water	51.22	51.22 ---------- = 2.846 18	2.846 ---------- = 7 0.406	7

Hence the empirical formula is Mg SO₄.7H₂O.

Problems for Practice

Determine the empirical formula
C₆H₆ b) C₂H₆ c) C₃H₈

d) Fe₃(CO)₉ e) C₂H₄O₂ f) N₂H₄

g) CaBr₂ h) C₂H₂ i) Na₂SO₄

j) C₆H₅N k) P₄O₁₀ l) Re₂Cl₆

m) Se₃O₉ n) LiCl

A substance on analysis, gave the following percentage composition, Na = 43.4%, C = 11.3%, 0 = 43.3% calculate its empirical formula [Na = 23, C = 12, O = 16].

Ans:- Na₂CO₃

What is the simplest formula of the compound which has the following percentage composition: Carbon 80%, hydrogen 20%.

Ans:- CH₃

A compound on analysis gave the following percentage composition: C - 54.54%, H = 9.09%, O = 36.36%

Ans:- C₂H₄O

Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present in it. Carbon = 10.06 %, hydrogen = 0.84 %, chlorine = 89.10 %. Calculate the empirical formula of the compound

Ans :- CHCl₃

The percentage of nitrogen in ammonia is

14 × 100

Ans:- ------------

Determine the empirical formula from the present composition for each of the following
92.24% C 7.76% H

36.48% Na 25.44% S 38.08%O
49.99% C 5.61% H 44.40%O
38.76% Ca 19.97%P 41.27%O

Calculate the percentage composition for the following compounds

Cr₂O₃
Ca₃(PO₄)₂

Molecular Formula from Empirical Formula

The molecular formula of a compound is a multiple of its empirical formula.

Example

Molecular Weight = n x empirical formula weight.

Molecular Weight

n = -------------------------------------------

Empirical formula Weight

Steps for writing the molecular formula

Calculate the empirical formula
Find out the empirical formula mass by adding the atomic mass of all the atoms present in the empirical formula of the compound.
Divide the molecular mass (determined experimentally by some suitable method) by the empirical formula mass and find out the value of n which is a whole number.
Multiply the empirical formula of the compound with n, so as to find out the molecular formula of the compound.

Solved Problem

A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09%, O = 36.36 %. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.

Solution

Calculation of empirical formula

Element	Percentage	Relative number of moles	Simple ratio moles	Simplest whole number ratio
C	54.54	54.54 ------- - = 4.53 12	4.53 -------- = 2 2.27	2
H	9.09	9.09 -------- = 9.09 1	9.09 -------- = 4 2.27	4
O	36.36	36.36 ---------- = 2.27 16	2.27 --------- = 1 2.27	1

Calculation of Molecular formula

Empirical formula mass = 12 x 2 + 1 x 4 + 16 x 1 = 44

Molecular mass = 2 x Vapour density

= 2 x 44 = 88

Molecular mass 88

n = ------------------------------------ = --------- = 2

Empirical Formula mass 44

= C₂H₄O x 2

= C₄H₈O₂

Solution :- Calculation of empirical formula

Element	Percentage	Relative number of moles	Simple ratio moles	Simplest whole number ratio
Na	14.31	14.31 ------- = 0.62 23	0.62 ------ = 2 0.31	2
S	9.97	19.97 -------- = 0.31 32	0.31 ------ = 1 0.31	1
H	6.22	6.22 ------ = 6.22 1	6.22 ----- = 20 0.31	20
O	69.5	69.5 -------- = 4.34 16	4.34 ------ = 14 0.31	14

The empirical formula is Na₂SH₂₀O₁₄

Calculation of Molecular formula

Empirical formula mass = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322

Molecular mass 322

n = ---------------------------------------- = ------------ = 1

Empirical formula mass 322

Hence molecular formula = Na₂SH₂₀O₁₄

Hence, molecular formula = Na₂SO₄.10H₂O.

Problems for Practice

An organic compound was found to have contained carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour - density was found to be 29.5. What is the molecular formula of the compound?

Ans:- C₂H₅NO

A compound contains 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75. Calculate the empirical and molecular formula.

Ans:- C₂H₃O₃, C₄H₆O₆

An acid of molecular mass 104 contains 34.6% carbon, 3.85% hydrogen and the rest is oxygen. Calculate the molecular formula of the acid.
What is the simplest formula of the compound which has the following percentage composition: carbon 80%, Hydrogen 20%, If the molecular mass is 30, calculate its molecular formula.
A compound analyzes as 79.08% C, 5.54% H and 15.38%N. what is the molecular formula if the molar mass is 273.36g/mol?
Determine the molecular formula
A compound has an empirical formula of NO₂ and a molar mass of 92g/mol
A compound has an empirical formula of C₂H₃O and a molar mass of 172g/mol
Ibuprofen, a common headache remedy, has an empirical formula of C₇H₉O and a molar mass of 215g/mol

COMPREHENSION

For questions 1 to 5

Weight of a substance in grams number of molecules

Number of moles = --------------------------------------- = -------------------------------------

Gram molecular weight Avogadro number

Consider the following table for compound ‘X’ :

Element	Percentage	Atomic Mass	Relative No. of atoms	Simplest ratio
Carbon	66.67%	12	66.67 -------- = 5.55 12	5.55 ------ = 3 1.85
Hydrogen	7.4%	1	7.4 ---- = 7.4 1	7.4 ------ = 4 1.85
Nitrogen	25.9%	14	25.9 ----- = 1.85 14	1.85 -------- = 1 1.85

Now based on the data given above, answer the following questions.

1) The empirical formula of the compound ‘X’ (as per given data) is

(A) C₁₂H₁₆N₄ (B) C₆H₈N₂

2) If the molecular mass of the compound ‘X’ is found to be 108 g/mol, then its molecular formula is:

(A) C₁₂H₁₆N₄ (B) C₉H₁₂N₃

3) Calculate the total number of atoms present in 54 g of the compound ‘X’ (N_A=avogadro’s number)

(A) N_A (B) 16 N_A

4) Upon complete combustion of the 54 g of the compound ‘X’, the total amount of carbon dioxide released will be

(A) 264 g (B) 88 g

5) The released CO₂ (as Q. No. 18) is mixed with the same amount of laughing gas (N₂O) at NTP. Then, the total volume occupied by the mixture at NTP is

(A) 67.2 L (B) 44.8 L

Solution

C₃H₄N is the simplest whole number ratio formula of the compound.
Empirical formula mass = 12 × 3 + 1 × 4 + 14 = 54

n = 108

-------- = 2

Molecular formula = [C₃H₄N] 2 = C₆H₈N₂

1 mole = N_A molecules of C₆H₈N₂ = 16 N_A atoms

54g of the compound = 54

-------- = 0.5 mole

108

= (16 N_A) × 0.5 atoms = 8 NA

1 mole of C₆H₈N₂ upon complete combustion gives 6 moles of CO₂
moles of C₆H₈N₂upon combustion gives = 3 moles of CO₂

= 3 × 44 = 132 g

3 moles CO₂ + 3 moles of N₂O

= 6 × 22.4 = 134.4 L

FORMULAS FOR MOLE CONCEPT

Weight of a substance in grams number of molecules

Number of moles = --------------------------------------- = -------------------------------------

Gram molecular weight Avogadro number

Volume of a gas at STP

Number of moles of a gaseous substance = -----------------------------------

Gram molar volume

Gram molecular weight

Weight of a single molecule = ---------------------------------

Avogadro number

Molecular weight

Atomic weight of a gaseous element = ---------------------------------

Atomicity

Gram atomic weight

Absolute weight of an atom of any element is = --------------------------------------------

Avogadro number

Weight of the substance = Number of moles × gram molecular weight

Weight of the substance

Number of molecules = number of moles × Avogadro number = ------------------------------ × N_A

Gram molecular weight

Mass

-------------- = mole

Molar mass

Mass

-------------- = Molar mass

mole

Molecular formula = empirical formula × n (where n is a small whole number)

Molecular mass

n = -------------------------------------

empirical formula

Molecular mass = 2 x Vapour density

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sir i m tring very hard to understand mole concept.bt i m nt getting it at all..sir plz help me out of this prob.actually i have joined rao iit .i m finding it difficult to solve sums based on mole concept .plz help me out sir.

sir i m tring very hard to understand mole concept.bt i m nt getting it at all..sir plz help me out of this prob.actually i have joined rao iit .i m finding it difficult to solve sums based on mole concept .plz help me out sir.

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sir i m tring very hard to understand mole concept.bt i m nt getting it at all..sir plz help me out of this prob.actually i have joined rao iit .i m finding it difficult to solve sums based on mole concept .plz help me out sir.

sir i m tring very hard to understand mole concept.bt i m nt getting it at all..sir plz help me out of this prob.actually i have joined rao iit .i m finding it difficult to solve sums based on mole concept .plz help me out sir.

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