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UNITS BASICS OF CHEMISTRY PART 3/3

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Page 1: UNITS

UNITSBASICS OF CHEMISTRY

PART 3/3

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g/mol

CHEMISTRY

molar mass

MOLAR MASS?

Molar mass is the mass of

one mole particles/atom/

molecule/ion.

Molar mass is measured

in grams per mole

(g/mol).

If ATOMS?

molar mass = mass of

6.022 X 1023 atoms

which is

GRAM ATOMIC MASS

= GAM

If MOLECULE?

molar mass = mass of

6.022 X 1023 molecules

which is

GRAM MOLECULAR

MASS

= GMM

If GASES?

Gaseous molar volume

= Volume of

6.022 X 1023 molecules

= 22.4L @ STP

(1 atm, 273K)

Note: The molar mass in grams is numerically equal to

atomic/molecular mass in (u).

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g/mol

CHEMISTRY

molar mass

Molar Mass Calculation:

Calculate the molar mass of the

Glucose (C6H12O6)

Molar Mass of H = 1 g.mol-1

Molar Mass of C = 12 g.mol-1

Molar Mass of O = 16 g.mol-1

Molar Mass of C6H12O6

= 6(12) + 12(1) + 6(16)

= 72 + 12 + 96

= 180 g.mol-1

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%

CHEMISTRY

Mass % of

element

If Molecular Formula Of

Compound Is Known, One Can

Calculate The Mass Percentage

Of Element Present In That

Compound. In Contrast To This,

If One Knows The Mass

Percentage Of All Elements

Present In A Compound, Its

Molecular Formula Can

Be Determined.

When Compound

Is Formed From Two Or

More Elements, The

Amount Of Element

Present In A Compound

Is Always Proportional

To Its Definite

Mass.

Percentage Composition Of An

Element

Mass % Of An Element

=๐’Ž๐’‚๐’”๐’” ๐’๐’‡ ๐’•๐’‰๐’† ๐’†๐’๐’†๐’Ž๐’†๐’๐’• ๐’Š๐’ ๐’•๐’‰๐’† ๐’„๐’๐’Ž๐’‘๐’๐’–๐’๐’… ร—๐Ÿ๐ŸŽ๐ŸŽ

๐’Ž๐’๐’๐’‚๐’“ ๐’Ž๐’‚๐’”๐’” ๐’๐’‡ ๐’„๐’๐’Ž๐’‘๐’๐’–๐’๐’…

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%

CHEMISTRY

Mass % of

element

Calculate The Percentage Of Each

Element Present In Ethanol (C2H5OH).

Solution :

Atomic mass of C = 12.0 gram mole-1

Atomic mass of H = 1.0 gram mole-1

Atomic mass of O = 16.0 gram mole-1

Molecular mass of ethanol (C2H5OH)

= 2(12) + 6 (1) +1 (16)

= 24 + 6 + 16

= 46 gram mole-1

Mass % of H =6 x 100

46= 13.03%

Mass % of C =2(12) x 100

46= 52.17%

Mass % of O =1(16) x 100

46= 34.78%

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%w/w

%w/v

CHEMISTRY

percent by

weight

PERCENTAGE BY WEIGHT

%w/w,

%w/v

It Is The

Expression Of

Concentration

Of Solute In

Solution

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%w/w

%w/v

CHEMISTRY

percent by

weight

%w/w?The Weight Of

Substance In Gram

Dissolved In 100 Gram

Solution Is Called

Percentage By Weight

(%w/w). It Stands

For Weight Per

Weight

Formula%w/w =

๐’˜๐’†๐’Š๐’ˆ๐’‰๐’• ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’† ๐‘ฟ ๐Ÿ๐ŸŽ๐ŸŽ

๐’˜๐’†๐’Š๐’ˆ๐’‰๐’• ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’Š๐’๐’ ๐’Š๐’ ๐’ˆ๐’Ž

EXAMPLE

In How Many Grams Of Water 36.5 gram HCl Should

Be Dissolved So That 10% w/w Solution Will Be

Obtained?

Total Weight Of Solution = M g,

Weight of Solute (HCl) = 36.5 gm,

w/w% = 10%

%w/w = ๐Ÿ‘๐Ÿ”.๐Ÿ“ ๐‘ฟ ๐Ÿ๐ŸŽ๐ŸŽ

๐Œ= 10%

Total Weight Of Solution M = ๐Ÿ‘๐Ÿ”.๐Ÿ“ ๐‘ฟ ๐Ÿ๐ŸŽ๐ŸŽ

๐Ÿ๐ŸŽ= 365 gm

Weight Of Water = 365 โ€“ 36.5 = 328.5 gm H2O .

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%w/w

%w/v

CHEMISTRY

percent by

weight

%w/v?

The Weight Of Substance In

Gram Dissolved In 100 Ml

Solution Is Called Percentage

By Weight By Volume

(%w/v).

It Stands For Weight

Per Volume.

Formula%w/v =

๐’˜๐’†๐’Š๐’ˆ๐’‰๐’• ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’† ๐‘ฟ ๐Ÿ๐ŸŽ๐ŸŽ

๐’—๐’๐’๐’–๐’Ž๐’† ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’Š๐’๐’ ๐’Š๐’ ๐‘ณ

Example:How Many Grams of NaOH Will Be Will Be Required To

Prepare 500 ml Solution Containing 5% w/v NaOH ?

Ans.: %w/v = ๐’˜๐’†๐’Š๐’ˆ๐’‰๐’• ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’† ๐‘ฟ ๐Ÿ๐ŸŽ๐ŸŽ

๐Ÿ“๐ŸŽ๐ŸŽ= 5 %

Weight of Solute = ๐Ÿ“ ๐‘ฟ ๐Ÿ“๐ŸŽ๐ŸŽ

๐Ÿ๐ŸŽ๐ŸŽ= 25 gm

Weight of Solute = 25 gm

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EXAMPLE

2 L ethanol solution

have 200 ml ethanol

then find out % v/v.

%v/v = ๐Ÿ๐ŸŽ๐ŸŽ ๐‘ฟ ๐Ÿ๐ŸŽ๐ŸŽ

๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ

= 10%v/v1 4

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%v/v

CHEMISTRY

percent by

volume

%v/vThe Volume Of Substance

In ml Dissolved In 100 ml

Solution Is Called Percentage

By Volume (%v/v).

It Stands For Volume

Per Volume

Formula

%v/v =

๐‘ฝ๐’๐’๐’–๐’Ž๐’† ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’† ๐‘ฟ ๐Ÿ๐ŸŽ๐ŸŽ

๐‘ฝ๐’๐’. ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’Š๐’๐’ ๐’Š๐’ ๐’๐’Š๐’•.

Note: This is used when both chemicals in a

solution are liquid.

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CHEMISTRY

mole fraction

What Is โ€œMole Fractionโ€The ratio of the moles of any component to the total

moles in a solution is the mole fraction.

๐ฆ๐จ๐ฅ๐ž ๐Ÿ๐ซ๐š๐œ๐ญ๐ข๐จ๐ง ๐จ๐Ÿ ๐œ๐จ๐ฆ๐ฉ๐จ๐ง๐ž๐ง๐ญ

=๐’Ž๐’๐’๐’† ๐’๐’‡ ๐’„๐’๐’Ž๐’‘๐’๐’๐’†๐’๐’•

๐’๐’–๐’Ž๐’ƒ๐’†๐’“ ๐’๐’‡ ๐’•๐’๐’•๐’‚๐’ ๐’Ž๐’๐’๐’†๐’” ๐’๐’‡๐’„๐’๐’Ž๐’‘๐’๐’๐’†๐’๐’•๐’” ๐’Š๐’ ๐’‚ ๐’”๐’๐’๐’–๐’•๐’Š๐’๐’

NOTE:

The sum of the mole fraction of all the

components in a solution is always equal to 1.

Nos. of Moles of Solute A = nA

Nos. of Moles of Solvent B = nB

Total Nos. of Moles of Solution = nA + nB

๐‘ด๐’๐’๐’† ๐‘ญ๐’“๐’‚๐’„๐’•๐’Š๐’๐’ ๐’๐’‡ ๐‘บ๐’๐’๐’–๐’•๐’† ๐‘จ =๐’๐‘จ

๐’๐‘จ + ๐’๐‘ฉ

๐‘ด๐’๐’๐’† ๐‘ญ๐’“๐’‚๐’„๐’•๐’Š๐’๐’ ๐’๐’‡ ๐‘บ๐’๐’๐’–๐’•๐’† ๐‘ฉ =๐’๐‘ฉ

๐’๐‘จ + ๐’๐‘ฉ

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CHEMISTRY

mole fraction

Mole Fraction Calculation

Example

Find the Mole Fraction Of Water In A Solution When

13.8g NaCl Is Dissolved In 132 ml (g) Of Water.

Ans.:

Molar Mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole Of NaCl =๐Ÿ๐Ÿ‘.๐Ÿ–

๐Ÿ“๐Ÿ–.๐Ÿ“= 0.2358 mol

Molar Mass Of H2O = 2 + 16 = 18 g/mol

Mole Of Water =๐Ÿ๐Ÿ‘๐Ÿ

๐Ÿ๐Ÿ–= 7.3333

Total moles of Solution = 0.2358 + 7.3333

= 7.5691 mol

Mole Fraction of Water =๐Ÿ•.๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘

๐Ÿ•.๐Ÿ“๐Ÿ”๐Ÿ—๐Ÿ= 0.9688

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M

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molarity

What Is โ€œMolarityโ€?

When One

gram mole Of A

Solute Is Dissolved In

Solvent & Then Making

One litre Of Solution It Is

Called One molar

(1M) solution.

OR

One litre Of Solution

Containing One gram

mole Of A Substance Is

Called 1 molar (1 M)

Solution

Molarity or molar

concentration Is The

Number Of moles Of

Solute In 1 litre Of

Solution

SI Unit

mol/lit

OR

mol/m3

Other

Units

M,

mol/dm3

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molarity

Molarity Formulas

Molarity (M) =๐‘ต๐’. ๐’๐’‡๐’Ž๐’๐’๐’†๐’” ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’†

๐‘ฝ๐’๐’๐’–๐’Ž๐’† ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’Š๐’๐’ ๐’Š๐’ ๐’๐’Š๐’•๐’“๐’†

(here volume is in litre)

Molarity (M) =๐‘ต๐’. ๐’๐’‡๐’Ž๐’๐’๐’†๐’” ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’† ร— ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ

๐‘ฝ๐’๐’๐’–๐’Ž๐’† ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’Š๐’๐’ ๐’Š๐’ ๐’Ž๐’

(here volume is in ml)

Molarity (M) =๐‘ด๐’‚๐’”๐’” ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’† ร— ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ

๐‘ฎ๐‘ด๐‘ดร— ๐‘ฝ๐’๐’๐’–๐’Ž๐’† ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’Š๐’๐’ ๐’Š๐’๐’Ž๐’

(here volume is in ml)

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molarity

Molarity Calculation Example

A Solution Prepared Using 15 g Of Sodium

Sulphate. The Volume Of The Solution Is 125

ml. Calculate The Molarity Of The Given

Solution Of Sodium Sulphate.

Solution:

GMM of Na2SO4 = 2(23) + 32 + 4(16)

= 46 + 32 + 64

= 142 g/mol

Molarity (M) =๐‘ด๐’‚๐’”๐’” ๐’๐’‡ ๐’”๐’๐’๐’–๐’•๐’† ร— ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ

๐‘ฎ๐‘ด๐‘ดร— ๐‘ฝ๐’๐’๐’–๐’Ž๐’† ๐’Š๐’๐’Ž๐’

=๐Ÿ๐Ÿ“ ร— ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ

๐Ÿ๐Ÿ’๐Ÿ ร— ๐Ÿ๐Ÿ๐Ÿ“

= 0.85 M

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molality

What Is โ€œMolalityโ€?

One Mole Of

A Solute, When

Dissolved In One

kilogram Of The Solvent

Is Called One molal (m)

Solution Or molality Of

Solution Is 1 m

(1 molal).

OR

The Number Of

Moles Of Solute

Dissolved In 1

kilogram Of

Solvent.

A Solution

Containing

3 moles Of Solute

per kilogram Of

Solvent For Example,

Is Said To Be

3 molal

or 3 m

SI Unit

mol/kg

Other Units

molal

m

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molality

Molality Formulas

Molality (m) =๐‘๐‘œ. ๐‘œ๐‘“ ๐‘š๐‘œ๐‘™๐‘’๐‘  ๐‘œ๐‘“ ๐‘ ๐‘œ๐‘™๐‘ข๐‘ก๐‘’

๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘ ๐‘œ๐‘™๐‘ฃ๐‘’๐‘›๐‘ก ๐‘–๐‘› ๐‘˜๐‘”.

(here weight is in kg.)

Molality (m) =๐‘๐‘œ. ๐‘œ๐‘“ ๐‘š๐‘œ๐‘™๐‘’๐‘  ๐‘œ๐‘“ ๐‘ ๐‘œ๐‘™๐‘ข๐‘ก๐‘’ ร— 1000

๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘ ๐‘œ๐‘™๐‘ฃ๐‘’๐‘›๐‘ก ๐‘–๐‘› ๐‘”๐‘Ÿ๐‘Ž๐‘š

(here weight is in gram)

Molality (m) =๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘ ๐‘œ๐‘™๐‘ข๐‘ก๐‘’ ร— 1000

๐บ๐‘€๐‘€ ร— ๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘ ๐‘œ๐‘™๐‘ฃ๐‘’๐‘›๐‘ก ๐‘–๐‘› ๐‘”๐‘Ÿ๐‘Ž๐‘š.

(here weight is in gram)

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molality

Molality Calculation Example

What is the molality of a solution containing

5.0 g NaCl dissolved in 25.0 g water?

Solution:

Molar mass of NaCl = 23 + 35.5

= 58.5 g/mol

molality (m) =๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘ ๐‘œ๐‘™๐‘ข๐‘ก๐‘’ ร— 1000

๐บ๐‘€๐‘€ ร— ๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘ ๐‘œ๐‘™๐‘ฃ๐‘’๐‘›๐‘ก ๐‘–๐‘› ๐‘”๐‘Ÿ๐‘Ž๐‘š.

(in this example solute = NaCl, solvent = water)

= 5 ร— 1000

58.5 ร— 25

= 3.42 m

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MOLAR MASS01

MASS % OF ELEMENT02

PERCENT WEIGHT03

PERCENT VOLUME04

MOLALITY07

MOLE FRACTION05

MOLARITY06

UNITSBASICS OF CHEMISTRY

END OF

PART 3/3

SU

MM

ARY

Page 19: UNITS

UNITS01

METRIC SYSTEM02

SI UNITS03

IUPAC04

MASS07

IUPAP05

UNIT PREFIXES06

WEIGHT08

VOLUME09

DENSITY10

TEMPERATURE11

MOLE CONCEPT14

ATOMIC MASS12

MOLECULAR MASS13

SU

MM

ARY

Page 20: UNITS

MOLAR MASS15

MASS % OF ELEMENT16

PERCENTAGE BY WEIGHT17

PERCENTAGE BY VOLUME18

MOLALITY21

MOLE FRACTION19

MOLARITY20

SU

MM

ARY

Page 21: UNITS

THANK YOU VERY MUCH