Atomic Mass, Molecular Mass And Gram Atomic Mass

Q: What is the difference between vapour density and molecular mass?

Vapour density is the ratio of mass of a gas to the mass of the same volume of hydrogen at the same T and P. Molecular mass equals 2 times vapour density.

Q: Can two compounds have the same molecular formula but different structures?

Yes. They are called isomers. Ethanol and dimethyl ether both have molecular formula C2H6O but different structures.

Q: When does molecular formula equal empirical formula?

When n equals 1, i.e. molecular mass equals empirical formula mass. Water, ammonia, methane, and carbon dioxide all have n equal to 1.

Q: Why is Dulong-Petit's law not applicable to non-metals like carbon and boron?

Light elements like C, B, Si, Be have very high vibration frequencies. Quantum effects reduce their specific heats well below classical predictions, so the product atomic mass times specific heat is much less than 6.4.

Q: How do you find the molecular formula if only percentage composition and vapour density are given?

Find the empirical formula from percentages, compute the empirical formula mass, calculate molecular mass as 2 times V.D., then n equals molecular mass over empirical formula mass.

Q: What is the difference between mole, gram atom, and gram molecule?

Mole means 6.022 times 10 to the 23 entities. Gram atom is 1 mole of atoms of an element. Gram molecule is 1 mole of molecules of a compound. Mole covers both.

ChemistrySome Basic Concepts of ChemistryFor NEET aspirants

Atomic mass tells you how heavy an atom is on a standard scale. The modern reference is the carbon-12 atom, which is assigned a mass of exactly $12$ atomic mass units. One atomic mass unit (amu, or dalton) equals $\frac{1}{12}$ the mass of a $^{12} C$ atom, or $1.66 \times 1 0^{- 24} g$ . This concept covers atomic mass, methods to determine it (Dulong-Petit, Cannizzaro, isomorphism, vapour density of chloride), average atomic mass from isotopic abundance, gram atomic mass, molecular mass, gram molecular mass, empirical formula, and molecular formula - the toolkit for every stoichiometry problem in JEE and NEET.

Key Formulas - Quick Reference

$1 amu = \frac{1}{12} \times mass of one^{12} C atom = 1.66 \times 1 0^{- 24} g$
Atomic mass $= \frac{mass of one atom of the element}{1 amu}$
Dulong-Petit: $atomic mass \times specific heat (cal g^{- 1} K^{- 1}) \approx 6.4$ (for solid metals)
Vapour density method: molecular mass $= 2 \times V.D.$
Chloride vapour density: valency $x = \frac{2 \times V.D.}{E + 35.5}$ , atomic mass $= x \times E$
Average atomic mass $= \frac{\sum ( % abundance \times isotopic mass )}{100}$
Gram atomic mass $=$ atomic mass in grams $= 6.022 \times 1 0^{23}$ atoms
Molecular formula $= (empirical formula)_{n}$ , where $n = \frac{molecular mass}{empirical formula mass}$

1. Atomic Mass and the Atomic Mass Unit

Atoms are far too small to weigh individually on any ordinary balance. So instead of absolute masses, chemists use relative masses: the mass of one atom of an element compared to a chosen standard.

Historical reference standards: the standard has changed twice.

Originally: mass of $1$ hydrogen atom $= 1$ unit.
Then: $\frac{1}{16}$ of the mass of $^{16} O$ atom $= 1$ unit.
Since 1961: $\frac{1}{12}$ of the mass of a $^{12} C$ atom $= 1 amu$ (dalton, Da).

^{12}

C C-12 = 12 amu actual mass = 1.9924 x 10-23 g divided by 12 1 atomic mass unit 1 amu = 1.66 x 10-24 g (1 Dalton, Da)

Figure 1: The atomic mass unit (amu) is defined as

\frac{1}{12}

of the mass of a

^{12} C

atom.

Atomic mass unit (amu): the actual mass of one

^{12} C

atom is

1.9924 \times 1 0^{- 23} g

, so

1 amu = \frac{1.9924 \times 1 0 ^{- 23}}{12} = 1.66 \times 1 0^{- 24} g

The atomic mass of any element is then:

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

And to get the actual mass of one atom in grams: actual mass $=$ atomic mass $\times 1.66 \times 1 0^{- 24} g$ .

2. Methods to Determine Atomic Mass

2.1 Dulong and Petit's Law

For most solid elements, the product of atomic mass and specific heat (in

cal g^{- 1} K^{- 1}

) is approximately

6.4

atomic mass \times specific heat \approx 6.4

This gives an approximate atomic mass. It fails for light elements like $B$ , $C$ , and $Si$ (they have unusually low specific heats). To pin down the exact atomic mass, combine Dulong-Petit's approximate value with the equivalent mass:

$valency = \frac{approximate atomic mass}{equivalent mass} (rounded to nearest integer)$

$exact atomic mass = equivalent mass \times valency$

Solved Example 1

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:

By Dulong and Petit's law:

$specific heat = \frac{6.4}{atomic mass} = \frac{6.4}{32} = 0.20 cal g^{- 1} K^{- 1}$

Answer: (C)

2.2 Cannizzaro's Method

If an element forms several volatile compounds with elements of known atomic mass, the smallest mass of that element found in one molecule of any of its compounds is taken as its atomic mass (or an integer multiple of it).

The procedure:

Determine the molecular mass of each compound from its vapour density: $M = 2 \times V.D.$
Find the mass of the target element present in one molecule of each compound (by percentage analysis).
The smallest such mass is the atomic mass (or a whole-number multiple of it).

Solved Example 2

Estimate the atomic mass of nitrogen given: V.D. of

NH_{3} = 8.5

, nitrous oxide

= 22

, nitric oxide

= 15

, nitrogen peroxide

= 23

, nitrogen trioxide

= 38

. Nitrogen content by mass:

NH_{3}

has

82.35%

N_{2} O

has

63.64%

NO

has

46.67%

NO_{2}

has

30.43%

N_{2} O_{3}

has

36.84%

Solution:

First find molecular mass $= 2 \times$ V.D.:

Compound	V.D.	Molar mass	% N	Mass of N in 1 molecule
$NH_{3}$	$8.5$	$17$	$82.35$	$14$
$N_{2} O$	$22$	$44$	$63.64$	$28$
$NO$	$15$	$30$	$46.67$	$14$
$NO_{2}$	$23$	$46$	$30.43$	$14$
$N_{2} O_{3}$	$38$	$76$	$36.84$	$28$

The smallest mass of nitrogen appearing in a molecule is $14$ , so the atomic mass of nitrogen is $14$ .

2.3 Mitscherlich's Law of Isomorphism (1819)

Two or more substances that form similarly shaped crystals (isomorphs) contain elements of the same valency in corresponding positions.

Classic examples of isomorphs are the heptahydrate sulphates of divalent metals:

Epsom salt: $MgSO_{4} \cdot 7 H_{2} O$
White vitriol: $ZnSO_{4} \cdot 7 H_{2} O$
Green vitriol: $FeSO_{4} \cdot 7 H_{2} O$

All contain $M^{2 +}$ ions and $7$ water molecules of crystallisation, so $Mg$ , $Zn$ , and $Fe$ all have valency $2$ in these salts. If the equivalent mass of one is known, its atomic mass can be found using the valency shared with a known isomorph.

Solved Example 3

Which pair of 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 $MgSO_{4} \cdot 7 H_{2} O$ and white vitriol $ZnSO_{4} \cdot 7 H_{2} O$ both have divalent cations and the same number of waters of crystallisation. They form identical crystal shapes.

Answer: (D)

2.4 Atomic Mass from Vapour Density of a Chloride

For a volatile metal chloride $MCl_{x}$ with vapour density V.D., valency $x$ , and equivalent mass $E$ :

Molecular mass $= 2 \times V.D. = E \times x + 35.5 x$

Solving for the valency:

$x = \frac{2 V.D.}{E + 35.5}$

Then atomic mass $= x \times E$ .

Solved Example 4

One gram of a metal chloride contains

0.835 g

of chlorine, and its vapour density is

85

. Find its molecular formula.

Solution:

Mass of chlorine $= 0.835 g$ , so mass of metal $= 1 - 0.835 = 0.165 g$ .

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

Valency $x = \frac{2 \times 85}{7.01 + 35.5} = \frac{170}{42.51} \approx 4$ .

Atomic mass of metal $= 4 \times 7.01 = 28.04$ (silicon, $Si$ ).

Formula: $MCl_{4}$ , i.e. $SiCl_{4}$ .

3. Average Atomic Mass

Most elements occur as a mixture of isotopes - atoms of the same element with different numbers of neutrons and hence different masses. The atomic mass listed on the periodic table is a weighted average over the natural abundances.

Average atomic mass = \frac{\sum ( % abundance \times isotopic mass )}{100}

Solved Example 5

Natural chlorine consists of

75.77%

^{35} Cl

(mass

34.969

) and

24.23%

^{37} Cl

(mass

36.966

). Calculate the average atomic mass of chlorine.

Solution:

$\overset{ˉ}{M} = \frac{75.77 \times 34.969 + 24.23 \times 36.966}{100}$

$= \frac{2649.6 + 895.7}{100} = 35.45 amu$

This matches the periodic-table value of $35.45$ - and explains why chlorine's atomic mass is not a whole number.

Solved Example 6

Natural argon has three isotopes with masses

35.96755

37.96272

39.96924

and abundances

0.337%

0.063%

99.600%

respectively. Calculate the molar mass of naturally occurring argon.

Solution:

$\overset{ˉ}{M} = \frac{35.96755 \times 0.337 + 37.96272 \times 0.063 + 39.96924 \times 99.600}{100}$

$= \frac{12.12 + 2.39 + 3980.94}{100} = 39.95 g mol^{- 1}$

4. Gram Atomic Mass (Gram Atom)

Gram atomic mass is the atomic mass of an element expressed in grams. It equals the mass of

1

mole (

6.022 \times 1 0^{23}

atoms) of the element. It is also called

1

gram atom.

Examples:

Atomic mass of C $= 12 amu$ ; gram atomic mass of C $= 12 g$ $= 6.022 \times 1 0^{23}$ atoms of C.
Atomic mass of Na $= 23 amu$ ; gram atomic mass of Na $= 23 g$ $= 6.022 \times 1 0^{23}$ atoms of Na.

Solved Example 7

Calculate the mass of each sample:
(a)

100

atoms of carbon
(b)

0.5

mole atoms of sulphur
(c)

8

gram atoms of oxygen
(d)

25

amu of chlorine
(e)

1.5 \times 1 0^{20}

atoms of nitrogen

Solution:

(a) $6.022 \times 1 0^{23}$ atoms of C weigh $12 g$ , so $100$ atoms weigh $\frac{12 \times 100}{6.022 \times 1 0 ^{23}} = 1.99 \times 1 0^{- 21} g$ .

(b) $1$ mole S $= 32 g$ , so $0.5$ mole $= 16 g$ .

(d) $1 amu = 1.66 \times 1 0^{- 24} g$ , so $25 amu = 25 \times 1.66 \times 1 0^{- 24} = 4.15 \times 1 0^{- 23} g$ .

(e) $14 g$ of N contains $6.022 \times 1 0^{23}$ atoms, so $1.5 \times 1 0^{20}$ atoms weigh $\frac{14 \times 1.5 \times 1 0 ^{20}}{6.022 \times 1 0 ^{23}} = 3.49 \times 1 0^{- 3} g$ .

5. Molecular Mass

Molecular mass is the number of times one molecule of a compound is heavier than

\frac{1}{12}

of the mass of a

^{12} C

atom. It equals the sum of the atomic masses of all atoms in the molecule.

Example: molecular mass of $CO_{2}$ $= 12 + 2 (16) = 44 amu$ . So one $CO_{2}$ molecule is $44$ times heavier than $1 amu$ .

Gram Molecular Mass

Gram molecular mass is the molecular mass expressed in grams. It equals the mass of

1

mole (

6.022 \times 1 0^{23}

molecules) of the compound.

For $CO_{2}$ : gram molecular mass $= 44 g$ $= 6.022 \times 1 0^{23}$ molecules $= 1$ gram molecule.

Solved Example 8

Calculate the actual mass of one molecule of

CO_{2}

Solution:

Molecular mass of $CO_{2} = 44 amu$ .

Actual mass $= 44 \times 1.66 \times 1 0^{- 24} g = 7.30 \times 1 0^{- 23} g$ .

Cross-check: $\frac{44 g mol ^{- 1}}{6.022 \times 1 0 ^{23} mol ^{- 1}} = 7.31 \times 1 0^{- 23} g$ . Same answer.

6. Empirical Formula

Empirical formula gives the simplest whole-number ratio of atoms of each element in a compound.

Different compounds can share the same empirical formula:

Formaldehyde $HCHO$ , acetic acid $CH_{3} COOH$ , and glucose $C_{6} H_{12} O_{6}$ all have empirical formula $CH_{2} O$ .
Benzene $C_{6} H_{6}$ and acetylene $C_{2} H_{2}$ both have empirical formula $CH$ .

Steps to Determine Empirical Formula

List the elements present in one column and their atomic masses in the next.
Write the experimentally-determined percentage of each element by mass.
Divide each percentage by the atomic mass to get the moles of each element (relative number of atoms).
Divide each result by the smallest of these numbers to get the simplest ratio.
If any ratio is a fraction close to a simple form (e.g. $1.5$ , $2.5$ ), multiply all ratios by the smallest integer that makes them whole (e.g. multiply by $2$ ).
Write the empirical formula using these whole numbers as subscripts.

Solved Example 9

A compound contains C

= 71.23%

, H

= 12.95%

, O

= 15.81%

. Find its empirical formula.

Solution:

Element	%	Atomic mass	Moles $= \frac{%}{atomic mass}$	Simplest ratio
C	$71.23$	$12$	$5.94$	$\frac{5.94}{0.988} = 6.01 \approx 6$
H	$12.95$	$1$	$12.95$	$\frac{12.95}{0.988} = 13.11 \approx 13$
O	$15.81$	$16$	$0.988$	$\frac{0.988}{0.988} = 1$

Empirical formula: $C_{6} H_{13} O$ .

7. Molecular Formula

Molecular formula shows the actual number of atoms of each element in one molecule of the compound.

The molecular formula is always a whole-number multiple of the empirical formula:

$molecular formula = (empirical formula)_{n} where n = \frac{molecular mass}{empirical formula mass}$

For volatile compounds the molecular mass can be found from the vapour density:

$molecular mass = 2 \times V.D.$

Solved Example 10

The empirical formula of a compound is

CH_{2} O

and its molecular mass is

90

. Find the molecular formula. (Atomic masses: C

= 12

, H

= 1

, O

= 16

Solution:

Empirical formula mass $= 12 + 2 (1) + 16 = 30$ .

$n = \frac{molecular mass}{empirical formula mass} = \frac{90}{30} = 3$

Molecular formula $= (CH_{2} O)_{3} = C_{3} H_{6} O_{3}$ (e.g. lactic acid).

Solved Example 11

A hydrocarbon has

92.3%

carbon and vapour density

39

. Determine its molecular formula.

Solution:

Percentage of H $= 100 - 92.3 = 7.7%$ .

Moles of C : moles of H $= \frac{92.3}{12} : \frac{7.7}{1} = 7.69 : 7.70 = 1 : 1$ .

Empirical formula: $CH$ ; empirical mass $= 13$ .

Molecular mass $= 2 \times V.D. = 78$ ; so $n = \frac{78}{13} = 6$ .

Molecular formula: $C_{6} H_{6}$ (benzene).

Common Mistakes to Avoid

Watch out

Forgetting to divide by $12$ when converting between amu and grams. $1 amu = 1.66 \times 1 0^{- 24} g$ , not $1 \times 1 0^{- 24} g$ .
Applying Dulong-Petit to light elements (B, C, Si, Be). Their specific heats deviate significantly at ordinary temperature; the law gives wrong atomic masses.
Confusing molecular mass with vapour density. Molecular mass $= 2 \times$ V.D., not equal to V.D.
Rounding the mole ratios too aggressively in empirical formula problems. $1.33$ is $\frac{4}{3}$ , so multiply everything by $3$ , not by $1$ .
Treating average atomic mass as an isotopic mass. The periodic-table value is a weighted average; no single atom of chlorine actually weighs $35.45 amu$ .
Using percentage abundance directly without dividing by $100$ in the average atomic mass formula.
Confusing empirical and molecular formula. Two very different compounds can share the same empirical formula; the molecular formula uniquely (well, usually) identifies the compound.

Frequently Asked Questions

Why is carbon-12 chosen as the reference for atomic mass?

Carbon-12 was chosen in 1961 because it is stable, abundant, easy to purify, and gives atomic masses of most other elements that are close to whole numbers. It also gave the amu (dalton) a modern, precisely reproducible definition. Since 2019, the mole is defined by fixing Avogadro's number, which keeps the amu and gram atomic mass consistent.

Why is the atomic mass of chlorine 35.45 and not a whole number?

Because natural chlorine is a mixture of two isotopes: about $75.77%$ $^{35} Cl$ and $24.23%$ $^{37} Cl$ . The atomic mass on the periodic table is the weighted average of the isotopic masses, which comes out to $35.45 amu$ .

What is the difference between vapour density and molecular mass?

Vapour density is the ratio of the mass of a certain volume of a gas to the mass of the same volume of hydrogen (under the same T and P). It is dimensionless. Molecular mass is exactly twice the vapour density because hydrogen is diatomic ( $H_{2}$ , mass $2$ ).

Can two compounds have the same molecular formula but different structures?

Yes. They are called isomers. For example, $C_{2} H_{6} O$ is the molecular formula for both ethanol ( $CH_{3} CH_{2} OH$ ) and dimethyl ether ( $CH_{3} OCH_{3}$ ). Their empirical and molecular formulas are identical, but their structures and properties differ.

When does molecular formula equal empirical formula?

When $n = 1$ , i.e. the molecular mass equals the empirical formula mass. Water $H_{2} O$ , ammonia $NH_{3}$ , methane $CH_{4}$ , and carbon dioxide $CO_{2}$ all have $n = 1$ .

Why is Dulong-Petit's law not applicable to non-metals like carbon and boron?

Dulong-Petit assumes the atomic vibrations follow classical predictions at room temperature. Light elements with strong bonds (C in diamond, B, Si, Be) have vibration frequencies so high that quantum effects reduce their specific heat well below the classical value, so the product atomic mass $\times$ specific heat comes out much less than $6.4$ .

How do you find the molecular formula if only percentage composition and vapour density are given?

Three steps. Find the empirical formula from the percentage composition. Compute the empirical formula mass. Compute the molecular mass as $2 \times$ V.D. Then $n = \frac{molecular mass}{empirical formula mass}$ , and molecular formula $= (empirical formula)_{n}$ .

What is the difference between mole, gram atom, and gram molecule?

A mole is a general term meaning $6.022 \times 1 0^{23}$ entities. Gram atom is $1$ mole of atoms of an element (mass $=$ atomic mass in grams). Gram molecule is $1$ mole of molecules of a compound (mass $=$ molecular mass in grams). "Mole" covers both.

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