A sample of ozone gas is found to be 40% dissociated into oxygen. The average molecular mass of sample should be :

Gram Molecular Mass:

The molecular mass of a substance (Molecule) express in gram is called gram molecular mass of the substance.

OR

Mass of 6.022×10^23 molecules of a substance is define as gram molecular mass or GMM.

OR

Mass of one mole  of a molecule is called GMM.

For example for O2 molecule:

Molecular mass of O2 molecule = mass of one  OF molecule = 2 × mass of one "O" atom = 2× 16 amu = 32 amu

GMM= mass of 6.022×10^23 "O2" molecules =

32 amu × 6.022×10^23 "O2"  molecules.

GMM= 32  × 6.022×10^23 × 1.67 ×10 ^-24 gram

GMM = 32 amu

Note : (6.022×10^23 × 1.67 ×10 ^-24  =1)

Gram Atomic Mass (GAM):

The atomic mass of an element express in gram is called gram atomic mass of the element.

OR

Mass of 6.022×10^23 atoms of an element is define as gram atomic mass or GAM.

OR

Mass of one mole atoms of an element is called GAM.

For example for Sodium atom:

Atomic mass of Sodium (Na) atom = mass of one 'Sodium"atom = 23 amu

GAM= mass of 6.022×10^23 "Na" atoms 

23 amu × 6.022×10^23 "Na" atoms 

GAM= 23  × 6.022×10^23 × 1.67 ×10 ^-24 gram

GAM = 23 amu

Note : (6.022×10^23 × 1.67 ×10 ^-24  =1)

Average atomic mass of magnesium is 24.31 amu. This magnesium is composed of 79 mole % of 24 Mg and remaining 21 mole % of 25Mg and 26 Mg. Calculate mole % of 26Mg.

If an element Z exist in two isotopic form Z^50 and Z^52. The average atomic mass of Z is 51.7. Calculate the abundance of each isotopic forms

Boron has two stable isotopes, (19%) and (81%). The atomic mass that should appear for boron in the periodic table is

The mass composition of universe may be 90% , and 10% . The average molecular of universe should be

Atomic mass:

(1) Atomic mass:

Atomic mass represent the mass of single isotopes or single atoms.

"Atomic mass and atomic weight do not mean the same:

Atomic mass is equal to the mass of total number of protons , neutrons  and electrons ( mass of electrons are negligible hence ignore)

Atomic mass = total mass of nucleons (total number of nucleons × mass of ons nucleons.

Atomic mass is independent of the atomic masses of isotopes.

Calculated  value of the atomic mass is whole number

Unit of atomic mass: the unit of atomic mass is atomic mass unit (amu) or unified atomic mass (u) and amu also known as Dalton(Da)

It is representation of unit of atomic weight is a.m.u. and  it is the smallest unit of mass and is define as;

We know that one mole (6.022x10^23) atoms C^12 atoms has  12 gm mass.

Hence mass of one C^12 atom is = 12/N_A=  12/6.022x 10^23 gm =1.99 x 10^-23 gm.

 The real mass of one atom of C^12 has been determined as 1.9924 × 10^-23 gram.

Conclusively we can summarise as

1 amu is = mass of one nucleons (protons or  nutrons)

1 amu is = 1.67×10^27 kg 

1 amu is = mass of one proton or nutron ( 1.67×10^27 kg )

(2) Atomic weight or relative average Atomic mass:

(3) Gram-atomic weight: When atomic weights of elements are expressed in grams, they are called gram atomic weights. 

Thus, the weight of an element in grams which is equal to its atomic weight is known as the gram atomic weight (GAW) of that element. 

  For example, atomic weight of sodium is 23 and its gram atomic weight is 23 grams. Similarly, the gram atomic weights of hydrogen, sulphur, nitrogen are 1.008 gram, 32 gram and 14 gram respectively.

Thus

1 GAW P = 31.0 gram phosphorus,

1 GAW K = 39.0 gram potassium,

1 GAW Ca = 40.0 gram calcium.

 (4) Gram atom or mole atom: The atomic weight of an element taken in grams is one gram-atom or mole atom of that element. 

For instance:

one gram atom of oxygen represents 16 gram oxygen.

One gram atom of sulphur represents 32 gram sulphur. 

Number of gram atoms of an element

  "One gram atomic weight of every element contains 6.023 × 10^23 atoms of the element.”

Other Questions:

(1) Calculate the average atomic weight of an element (X) containing three isotopes X56, X57 and X59 and natural occurrence of these isotopes are 90%, 8% and 2% respectively.

(2) For the reaction Initially 2.5 moles of Fe(NO3)3 and 3.6 moles of Na2CO3 are taken. If 6.3 moles of Na2CO3 is obtained the % yield of given reaction is ?

(3) 3.9 gm Al(OH)3 is reacted with excess of HCl , 6.5 gm AlCl3 is formed determined % yield of reaction ?

(4) 200 gm of CaCO3 on heating produces 11.2 litre of CO2 (g) at STP calculate % yield of reaction?

(5) How many grams of Fe2O3 formed by heating of 18 gm FeO with O2.

(6) How many moles O2 are required to needed produced 5 moles of Fe2O3.

(7) 367.5 gm KClO3 (122.5) when heated then calculate. (1) How many gram of O2 is produced? (2) How Many litre of O2 is produced at STP?

(8) Air contain 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³(ml) of acetylene?

(9) The hydrated salt, Na2SO4.nH2O undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be?

(10) Haemoglobin contains 0.33% of Iron by weight. The molecular weight of it is approx. 67200. The numbers of iron atoms (Atomic wt of Fe=56 u) present in one molecule of Haemoglobin are?

Atomic weight or relative average Atomic mass:

Atom: The smallest particle of an element which cannot exist in free form is called an atom.

Atoms have a tendency to exist as clusters (group of atoms or molecules) of same or different atoms. They take part in a chemical reaction without undergoing any decomposition. 

An atom is very small (atomic size of the order 10^–6 cm) and its real mass is very low.

The smallest and lightest atom is of hydrogen. The mass of the hydrogen atom is 1.008 a.m.u.

Weights of atoms of other elements are in the range of 2 to 260 a.m.u.

Atomic weight: The atomic weight of an element is a number which indicates how many times heavier is an average atom of that element as compared with 1/12  of the mass of an atom of carbon (12) (C^12).

* Atomic weights of elements are a absolute quantity and absolute quantity having an unit average relative weights.

* Both atomic weight and molecular weight are unit less quantities. These quantities are also regarded as relative atomic mass and relative molecular mass.

Average (relative) atomic mass: 

Most often the atomic weight of elements are fractional. It is due to the existence of isotopes of the elements which are atoms with different masses. 

Since an element is a mixture of all of its isotopes, the atomic weight of an element is the average of weights of all its isotopes. So, atomic weight have non –integral (fractional) values.

"Atomic weight (Average atomic weight or relative atomic mass) can be define as average weight of an elements with respect to all its isotopes and their relative abundance.

Calculation of average (relative) atomic mass: 

(A) If an elements exists in two isotopes having atomic masses ‘M1’ and ‘M2’ in the ratio  X: Y, Then   

                         or

(B) If an elements exists in two isotopes having atomic masses ‘M1’ and ‘M2’  and their abundant  percentage (%)  are  X % and Y % then 

(C) the calculated value may be or may not be a whole number.

Exercise: (1) Find the average atomic weight of mixture of containing 25% by mol Cl^35 and 75% by mol Cl^37.

Ans key: Cl^35 = 75%, Cl^37 = 35%

Exercise: (2) Boron have two isotopes B^11 and B^A and their relative abundant %  are 80% and 20% respectively and average atomic weight of Boron is 10.8 find the value of  “A”

Ans key: A = 10

Exercise: (3) Silver contains two isotopes Ag^107 and Ag^109 and average atomic weight of silver is 108.5 find percentage abundant of  isotopes of silver.

Ans key: Ag^107 = 25%, Ag^109 = 75%

Exercise: (4) Calculate the % abundance of Ag^109  it is know that silver contains two isotopes Ag^107 and Ag^109 and average atomic weight of silver is 108.5.

 Ans key: Ag^107 = 25%, Ag^109 = 75%

Exercise: (5) An element (X) have three isotopes X^20, X^21 and X^22. The percentage abundance of X^20 is 90% and it's average atomic mass is of element is 20.18. the percentage of abundance of X^21 should be.

Ans key: X^21 = 2%, X^22 = 8%

Exercise: (5) Calculate the average atomic weight of an element (X) containing three isotopes X56, X57 and X59 and natural occurrence of these isotopes are 90%, 8% and 2% respectively.

Other topic:

(1) What is the % composition of each element in (Mg(OH)2) magnesium hydroxide?

(2) Haemoglobin contains 0.33% of Iron by weight. The molecular weight of it is approx. 67200. The numbers of iron atoms (Atomic wt of Fe=56 u) present in one molecule of Haemoglobin are?

(3) The hydrated salt, Na2SO4.nH2O undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be?

(4) Air contain 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³(ml) of acetylene?

Calculate the average atomic weight of an element (X) containing three isotopes X56, X57 and X59 and natural occurrence of these isotopes are 90%, 8% and 2% respectively.

Write the POAC equations for all the atoms in the following reaction.

Illustrative solution:

Applying POAC on (P) atoms:

                                                    P à H₃PO₄

Numbers of moles of (P) atoms in P₄ = Numbers of moles of (P) atoms in H₃PO₄

                                               4x nP₄= 1x nH₃PO₄

Applying POAC on (H) atoms:

                                       1x nHNO3 = 3 x nH₃PO₄ + 2 x nH₂O

Applying POAC on (N) atoms:

                                       1 x nHNO₃ = 1 x nNO₂

Applying POAC on (O) atoms:

                                       3 x nHNO₃ = 4 x nH₃PO₄ +2 x nNO₂ + 1nH₂O

Related Questions:

Principle of atom conservation (PAOC):

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative example (4) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

Illustrative example (5) For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

Illustrative example (6) In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of Ammonium and magnesium ions to precipitate magnesium Ammonium phosphate (Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Which is weighed. A solution of NaH2PO4 yield 3.33 gm of Mg2P2O7. What is weighed of NaH2PO4 was present originally.

In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of Ammonium and magnesium ions to precipitate magnesium Ammonium phosphate (Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Which is weighed. A solution of NaH2PO4 yield 3.33 gm of Mg2P2O7. What is weighed of NaH2PO4 was present originally.

Related Questions:

Principle of atom conservation (PAOC):

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative example (4) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

Illustrative example (5) For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

 

Related Questions:

Principle of atom conservation (PAOC):

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative example (4) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

 Illustrative solution:

Related Questions:

Principle of atom conservation (PAOC):

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative solution:

By applying POAC on (C) atoms:

Related Questions:

Principle of atom conservation (PAOC):

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

Illustrative example (4) For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

Illustrative example (5) In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of Ammonium and magnesium ions to precipitate magnesium Ammonium phosphate (Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Which is weighed. A solution of NaH2PO4 yield 3.33 gm of Mg2P2O7. What is weighed of NaH2PO4 was present originally.

27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

 Here all the carbon of K2CO3 converted into a complex hence POAC can apply

Related Questions:

Illustrative example (2)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative example (3) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

Illustrative example (4) For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

Illustrative example (5) In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of Ammonium and magnesium ions to precipitate magnesium Ammonium phosphate (Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Which is weighed. A solution of NaH2PO4 yield 3.33 gm of Mg2P2O7. What is weighed of NaH2PO4 was present originally.

Mole concept and Stiochiometry:

 (1) SOME BASIC CONCEPT OF CHEMISTRY:
(1) BASIC TOPICS :
1.1.1 DALTON'S ATOMIC THEORY:
1.1.2 ATOMIC MASS AND MOLECULAR MASS:
1.1.3 PERCENTAGE COMPOSITION OF COMPOUNDS:
1.1.4 DETERMINATION OF EMPIRICAL AND MF:
(2) Mole concept and Stiochiometry
2.1.1 MOLE-MOLE ANALYSIS
2.1.2 CONCEPT OF LIMITING REAGENTS
2.1.3 PRINCIPLE OF ATOM CONSERVATION (PAOC)
2.1.4 GRAVIMETRY-WEIGHT WEIGHT ANALYSIS
2.1.5 EUDIOMETRY-VOLUME VOLUME ANALYSIS OF GAS
2.1.6 DEGREE OF HARDNESS OF HARD WATER
2.1.7 OLEUM AND ITS PERCENTAGE(%) LABBELING
2.1.8 VOLUME STRENGTH OF H2O2(aq)
2.1.9 PERCENTAGE (%) AVAILABLE CHLORINE IN BLEACHING POWDER
2.1.10 PERCENTAGE YIELD OF CHEMICAL REACTION:
2.1.11 PERCENTAGE PURITY OF GIVEN SAMPLE:
(3) law of chemical combinations:
3.12 (1) Law of conservation of mass: 

3.12 (2) Law of constant composition 

3.12 (3) Law of multiple proportion:

3.12 (4) Law of reciprocal proportion:

 

3.12 (5) Gay Lussac’s law of Combining Volumes:

 Related Questions:

(1):Calculate the mole fraction of solute in the 12 molal aqueous solution of CaCO3.

(2):The mole fraction of glucose in aqueous solution is 0.2 , The molality of solution will be.

(3):If the mole fraction of a solute is changed from 1/4 to 1/2 in the 800 gm of solvent . then the ratio of molality will be .

(4):Mole fraction of I2 (iodine) in benzene (C6H6) is 0.2 . Calculate the molality of iodine (I2) in benzene.

(5):Calculate the amount of water that should be added to 16 gm methanol to make mole fraction of methanol as 0.25.?

(6) What is relation between mole fraction and molality of solute ?

 

For the reaction Initially 2.5 moles of Fe(NO3)3 and 3.6 moles of Na2CO3 are taken. If 6.3 moles of Na2CO3 is obtained the % yield of given reaction is ?

 

3.9 gm Al(OH)3 is reacted with excess of HCl , 6.5 gm AlCl3 is formed determined % yield of reaction ?

200 gm of CaCO3 on heating produces 11.2 litre of CO2 (g) at STP calculate % yield of reaction?