ATOMIC MASS, MOLECULAR MASS AND GRAM ATOMIC MASS

Atomic Mass: As atoms are very tiny particles, their absolute masses are difficult to measure. However it is possible to determine the relative masses of different atoms if small unit of mass is taken as standard (previously, this standard was mass of one atom of hydrogen and taken as unity. Later on it was $\text{1/1}{\text{6}}^{\text{th}}$part of oxygen atom and now it is $\text{1/1}{\text{2}}^{\text{th}}$part of C-12 atom).

The atomic mass of an element can be defined as the number which indicates how many times the mass of one atom of the element is heavier in comparison to the mass of one atom of hydrogen.

Atomic Mass Unit: The quantity $\frac{1}{12}$mass of an atom of carbon-12 is known as the atomic mass unit and is abbreviated as amu. The actual mass of one atom of carbon-12 is

$\text{1}\text{.9924}\times \text{1}{\text{0}}^{\text{-23}}\text{g}$ or $\text{19924}\times \text{1}{\text{0}}^{\text{-26}}\text{kg}$

Thus 1 amu$=\frac{1.9924\times {10}^{-23}}{12}=1.66\times {10}^{-24}g$

Atomic mass of an element $=\frac{\text{Mass of one atom of the element}}{\text{1 amu}}$

Actual mass of an element $\text{= Atomic mass (amu)}\times \text{1}\text{.66}\times \text{1}{\text{0}}^{\text{-24}}\text{g}$

Determination of atomic mass

(i) Applying Dulong and Petit's law.

(ii) Cannizzaro's methods

(iii) By mitscherlich's law of isomorphism.

(iv) By measurement of V.D. of volatile chloride or bromide.

(i) Dulong & Petits Law: The product of specific heat of pure element and atomic mass of the element is equal to 6.4.

i.e. Atomic mass ´ specific heat = 6.4 (approx)

But this law is not applicable to lighter element like boron, carbon, silicon. To obtain correct atomic mass of element first of all equivalent mass of the element is known by any other method and their atomic mass = eq. weight ´ valency

In which valency has whole number value which can be deduced by dividing approximate by equivalent mass.

Dulongs and Petit's Law: $\text{Atomic mass}\times \underset{\left(\text{in cal/gm}\right)}{\mathrm{Specific}}\text{ heat = 6}\text{.4}$

Illustration1. The specific heat of a metal of atomic mass 32 is likely to be:

(A) 0.25 (B) 0.24

(C) 0.20 (D) 0.15

Solution: Specific heat = $\frac{6.4}{atomicmass}=\frac{6.4}{32}=0.2$

Hence (C) is correct.

(ii) Cannizzaro's methods

If an element has several compound with other same or different elements of known atomic mass then the compound that has minimum presence of former element indicate the atomic mass of former element.

Procedure

(i) First of all the molecular mass of all compounds known by applying

V.D × 2 = mol. weight

(ii) By analysis the presence of the desired element in each compound is known.

(iii) The mass that is lowest among all the compound indicate the atomic mass of that element.

Illustration 2. Estimate the atomic mass of nitrogen given that vapour density of NH₃ = 8.5, Nitrous oxide = 22, Nitric oxide = 15, Nitrogen peroxide = 23, Nitrogen

trioxide = 38. .

(iii) Law of Isomorphism

When two or more compound forms similar type of crystals or able to form mixed crystals, they are known as isomorphs. For examples: MgSO₄.7H₂O, ZnSO₄.7H₂O and FeSO₄.7H₂O are isomorphs of each other as their crystals posses same shape.

According to Mitscherlich [year 1819].

The valency of elements that are similarly placed to that of other elements in their isomorphs are always same.

In the above example Fe, Zn and Mg have same valency [2] and equal ratio of water molecule in each isomorphs.

If equivalent mass of one element is known then atomic mass can be calculated by knowing the valency of other isomorphs key element.

Illustration 3. Which pair of the following substances is said to be isomorphous?

(A) White vitriol and blue vitriol (B) Epsom salt and Glauber's salt

(C) Blue vitriol and Glauber's salt (D) White vitriol and Epsom salt

Solution: Epsom salt (MgSO₄.7H₂O) and White vitriol (ZnSO₄.7H₂O) contains divalent cation Mg²⁺ and Zn²⁺ and same number of water molecules as water of crystallization which hold criteria for isomorphism.

Hence (D) is correct.

(iv) Atomic mass from vapour density of a chloride:

The following steps are involved in this method

n Vapour density of chloride of the element is determined

n Equivalent mass of the element is determined

Let the valency of the element be x. The formula of its chloride will be$\text{MC}{\text{l}}_{\text{x}}$

Molecular weight of chloride = Atomic mass of M + 35.5 x

$\text{Atomic mass = x}\times \text{E}$

So ${\text{M}}_{\text{w}}\text{ = E}\times \text{x + 35}\text{.5x = 2 V}\text{.D}\text{.}$

$x=\frac{2VD}{E+35.5}$ Þ $\text{Atomic weight = x }\times \text{E}$

Illustration 4. One gram of chloride was found to contain 0.835g of chlorine. Its vapour density is 85. Calculate its molecular formula.

Solution: Mass of metal chloride = 1g

Mass of chlorine = 0.835

Mass of metal = 1 – 0.835 = 0.165g

E mass of metal$=\frac{0.165}{0.835}\times 35.5=7.01g$

Valency of metal$=\frac{2V.D.}{E+35.5}=\frac{2\times 85}{7.01+35.5}=4$

Formula of chloride$\text{= MC}{\text{l}}_{\text{4}}$

AVERAGE ATOMIC MASS

Elements are found in different isotopic forms (atoms of same elements having different atomic mass), so the atomic mass of any element is the average of all the isotopic mass within a given sample.

Average atomic mass $=\frac{\sum \%abundance\times \text{atomic mass}}{100}$

Illustration 5. Use the date given in the following table to calculate the molar mass of naturally occurring argon.

Solution: Molar mass of Ar

= 35.96755 × 0.071 + 37.96272 × 0.163 + 39.96924 × 0.766

= 39.352 g mol^-1

GRAM ATOMIC MASS OR GRAM ATOM

Atomic weight of an element in grams is called as Gram atomic mass of an element. Gram atomic mass is the weight of 1 mole atom $\text{(6}\text{.023}\times \text{1}{\text{0}}^{\text{23}}\text{ atoms)}$of the element. It is also called as 1 gram atom. e.g. AW of C = 12, GAW of $\text{C = 12g = 6}\text{.023}\times \text{1}{\text{0}}^{\text{23}}$atom of carbon = 1 mole atom of

C = 1 gram atom of carbon.

Illustration 6. Calculate the weight of following samples

(a) 100 atoms of carbon (b) 0.5 mole atoms of sulphur

(c) 8 gram atom of oxygen (d) 25 amu of chlorine

(e) $1.5\times 10^{20}$atoms of N

Solution: (a) 1 mole atom of C = 12 g

1 atom of C$=\frac{12}{6.023\times {10}^{23}}g$

100 atom of C$=\frac{12\times 100}{6.023\times {10}^{23}}g=1.99\times {10}^{-21}g$

(b) AW of S = 32

GAW of S = 32g = 1 mole atom

0.5 mole atom$\text{= 0}\text{.5}\times \text{32g = 16gms}$

(c) GAW = 1 gm atom

GAW of oxygen = 16 g = 1 gm atom of oxygen

8 gm atom of oxygen $\text{= 8}\times \text{16 = 128g}$

(d) 25 amu of chlorine

$\text{1 amu = 1}\text{.66}\times \text{1}{\text{0}}^{\text{-24}}\text{g}$

$\text{25 amu = 1}\text{.66}\times \text{1}{\text{0}}^{\text{-24}}\times \text{25g = 4}\text{.15}\times \text{1}{\text{0}}^{\text{-23}}\text{g}$

(e) $\text{1}\text{.5}\times \text{1}{\text{0}}^{\text{20}}$atom of N

1 atom of N = 14 amu

$1.5\times {10}^{20}atom=14\times 1.5\times {10}^{20}amu=14\times 1.5\times \frac{{10}^{20}}{6.023\times {10}^{23}}g$ $=\frac{14}{4}\times {10}^{-3}gm=3.5\times {10}^{-3}g$

Molecular Mass

Number of times a molecule of a compound is heavier than one atom of hydrogen or $\frac{1}{12}th$part of C-12 is molecular mass of the compound. It is the sum of the atomic mass of atoms present in a molecule.

For example Molecular mass of $\text{C}{\text{O}}_{\text{2}}\text{ = 44}$

i.e 1 molecule of $\text{C}{\text{O}}_{\text{2}}$is 44 times heavier than one atom of hydrogen or $\text{1/1}{\text{2}}^{\text{th}}$ part of C-12.

Gram Molecular Mass

Molecular mass expressed in grams is called as gram molecular mass. It is the weight of one mole molecules of a compound. It is also called as one gram molecule.

For example GMW of $\text{C}{\text{O}}_{\text{2}}\text{ = 44g = 1 mole}$molecule of $\text{C}{\text{O}}_{\text{2}}\text{ = 6}\text{.023}\times \text{1}{\text{0}}^{\text{23}}\text{ molecule }$

= 1 gram molecule of $\text{C}{\text{O}}_{\text{2}}$

Illustration 7. Calculate the actual mass of one molecule of $C{O}_2$

Solution: Molecular mass of $\text{C}{\text{O}}_{\text{2}}\text{= 44}$

Weight of one molecule of $\text{C}{\text{O}}_{\text{2}}\text{= 44 amu}$

Actual mass of $\text{C}{\text{O}}_{\text{2}}$molecule $\text{= 44}\times \text{1}\text{.66}\times \text{1}{\text{0}}^{\text{-24}}\text{g = 7}\text{.304}\times \text{1}{\text{0}}^{\text{-23}}\text{g}$

EMPIRICAL FORMULA

It is the formula which expresses the smallest whole number ratio of the constituent atom within the molecule. Empirical formula of different compound may be same. So it may or may not represent the actual formula of the molecule. It can be deduced by knowing the weight % of all the constituent element with their atomic masses for the given compound.

For example: C₆H₁₂O₆, CH₃COOH, HCHO

All have same empirical formula CH₂O, but they are different.

The empirical formula of a compound can be determined by the following steps:

$\bullet$Write the name of detected elements in column-1 present in the compound.

$\bullet$Write the corresponding atomic mass in column-2.

$\bullet$Write the experimentally determined percentage composition by weight of each element present in the compound in column-3.

$\bullet$Divide the percentage of each element by its atomic weight to get the relative number of atoms of each element in column-4.

$\bullet$Divide each number obtained for the respective elements in step (3) by the smallest number among those numbers so as to get the simplest ratio in column-5.

$\bullet$If any number obtained in step (4) is not a whole number then multiply all the numbers by a suitable integer to get whole number ratio. This ratio will be the simplest ratio of the atoms of different elements present in the compound. Empirical formula of the compound can be written with the help of this ratio in column-6.

Illustration 8. A compound contains C = 71.23%, H = 12.95% and O = 15.81%. What is the empirical formula of the compound?

MOLECULAR FORMULA

The formula which represents the actual number of each individual atom in any molecule is known as molecular formula.

For certain compounds the molecular formula and the empirical formula may be same.

$\text{Molecular formula = (Empirical formula}{\text{)}}_{\text{n}}$

$\text{Molecular weight = Empirical formula weight }\times \text{ n}$

If the vapour density of the substance is known, its molecular weight can be calculated by using the equation.

$\text{2 }\times \text{ Vapour density = Molecular weight}$

Illustration 9.The empirical formula of a compound is$C{H}_{2}O$. Its molecular weight is 90. Calculate the molecular formula of the compound. (Atomic weights C = 12, H = 1, O = 16)

Solution: Empirical formula $\text{= C}{\text{H}}_{\text{2}}\text{O}$

Empirical formula weight = (12 + 2 + 16) = 30

$n=\frac{molecularweight}{empiricalformulaweight}$

$\therefore n=\frac{90}{30}=3$

The molecular formula ${\text{(C}{\text{H}}_{\text{2}}\text{O)}}_{\text{3}}\text{ = }{\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$