chm 421 - topic 1 - calculations

8/19/2019 CHM 421 - ToPIC 1 - Calculations

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MoleMolarityNormality

Stoichiometry

Zuraida Khusaimi/CHM421/2015 2


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PhysicalQuantities

Unit nameUnitabbreviation

Mass gram g

Amount ofsubstance

Mole mole

Derived units Def. of Quantity Unit abbreviation

Volume (Length)3 dm3 or L

Density Mass/volume g/L

Concentration Mole/Volume mol/L


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Common Decimal Prefixes Used with SI Units


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Amount of atoms, particles, ions etc. 1 mol = 6.02 x 1023 (atoms, particles, ions etc.)

Mol = mass in g / molar mass= g / g/mol= atoms, particles, ions etc. / mol



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1 mole of particles contains 6.02 x 1023

particles

1 mole of C contains 6.02 x 1023

C atoms 1 mole of H2O contains 6.02 x 10

23 H2Omolecules, 2 x 6.02 x 1023 H atoms, or 6.02 x1023O atoms.

1 mole of NaCl contains 6.02 x 1023 particles,6.02 x 1023 Na+ ions and 6.02 x 1023Cl- ions.



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Number of moles A= mass of A (mg)

molar mass (mg/mmol)

103 milimole equals 1.0 mole1 milimole is 1/1000 of a mole



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The atomic mass of any substance expressed ingrams is the molar mass of that substance.

E.g. the atomic mass of iron, Fe is 55.85 amu Therefore the molar mass of iron, Fe is 55.85

g/mol. Since oxygen occurs naturally as diatomic , O2,

the molar mass of oxygen gas is 2 times 16.00i.e. 32.00 g/mol

Read: atomic wt., atomic mass, molecular wt,formula wt.



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Number og grams in 1 mole Equal to the numerical value of the atomic

mass1 mole of C = 12.0 g1 mole of Mg atoms = 24.3 g1 mole of Cu atoms = 63.5 g



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The molar mass of a substance is the sum of themolar masses of each element.

E.g. what is the molar mass of magnesium

nitrate, Mg(NO3)2? Solution:

The sum of the atomic masses is :24.31 + 2(14.01 + 16.00 + 16.00)

= 24.31 + 2(62.01) = 148.33 amuThe molar mass for Mg(NO3)2 is 148.33 g/mol



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Prozac, C17H18F3NO, is a widely usedantidepressant that inhibits the uptake of

serotonin by the brain. Calculate its molarmass. A) 40.0 g/mol B) 262 g/mol C) 309 g/mol



14/46Zuraida Khusaimi/CHM421/2015 14

• Now we will use the molarmass of a compound to convertbetween grams of a substance

and moles of a particles of asubstance.

6.02 x 1023 particles =1 mol = molar mass

• If we want to convert particlesto mass, we must first convertparticles to moles and then wecan convert moles to mass.


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Concentration is the amount of solute in a givenvolume of solution.

Solvent: A substance that generally a liquidpresent in the larger proportion of the solution.Water is considered a universal solvent for mostsolution.

Solute: The substance present in smallerproportion of a solution. It can be solid, liquid orgas.


Concentration (C) = mass (g)volume (L)


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The number of moles of solute per litre ofsolution, given the symbol, M.

M = no. of moles of solute (mol)volume of solution (L)

= no of milimoles of solute (mmol)volume of solution (mL)



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1. What is the molarity of a solution made bydissolving 2.5 g of NaCl in enough water to

make 125 ml of solution?2. How many mole of a solute present in the

following solutions:

a) 16.3 L solution of 0.113 M . (Ans: 1.8419 mole)

b) 15.66 mL solution of 0.025 M. (Ans: 0.0039 mole)



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4. A chemist dissolves 98.4 g of FeSO4 inenough water to make 2.000 L of solution.

What is the molarity of the solution? (Ans:0.324 M)5. How would you prepare 100.0 mL of 0.25 M

KNO3

solution? (Ans: 2.53 g KNO3

)



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Number of moles of solute that dissolved in1000 g of solvent, m (mol/kg).

Preferred expression of concentration involving colligative properties(boiling point, elevation, freezing point depression, osmotic pressure)

m = no of mol of solute

1 kg of solvent



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Number of gram equivalents of reacting unitper litre of solution, given symbol, N

N = no of equivalentsvolume (L)

No of reacting unit = No. of equivalents



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For 1 mol of H2SO4, it has 2 reacting units ofprotons

Equivalent weight = FW / no. of reacting units= (98.08 g/mol) / (2 eq/mol)= 49.04 g/eq

For 1 mol, 1 L of H2SO4, No. of equivalent = 98.08 /

49.04 = 2Normality = 2 / 1L = 2N

Molarity = mol/vol (L) = 1/1 = 1M



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% conc (% w/w, w/v, v/v)Ppm, ppb for solid and liquid

Density

Specific gravity of solution



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% concentration

Weight/weight % (w/w)

Volume/volume % (v/v) Weight/volume % (w/v)


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Percentage = number of parts in 100

parts,

E.g. Gram of A in 100 grams of sample

% of A = X 100

% Concentration

Number of parts of A

Number of parts of sample


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% (w/w) =

ppt (w/w) =

ppm (w/w) =

ppb (w/w) =

ppt (w/w) =

This scale is useful for solids or solutions.

102mass of solute (g)

mass of sample (g))(10

3mass of solute (g)

mass of sample (g))(10 6

mass of solute (g)

mass of sample (g))(109

mass of solute (g)

mass of sample (g) )(1012

mass of solute (g)

mass of sample (g) )(

parts per hundred

parts per thousand

parts per million

parts per billion

parts per trillion


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% (v/v)

❖ % (v/v) =

❖

ppt (v/v) =

❖ ppm (v/v) =

❖ ppb (v/v) =

❖ ppt (v/v) =

103mass of solute (mL)

mass of sample (mL) )(






mass of sample (mL) )(1012mass of solute (mL)


parts per

hundred

parts per

thousand

parts per

million

parts perbillion

parts per

trillion


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% (w/v) =

ppt (w/v) =

ppm (w/v) =

ppb (w/v) =

ppt (w/v) =


mass of sample (mL))(

10

3mass of solute (g)

mass of sample (mL) )(106

mass of solute (g)






parts per

hundred

parts per

thousand

parts per

million

parts perbillion

parts per

trillion


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Concentrated HCl = 15.0 % (w/w)

Alcoholic beverage. = 60.0 % (v/v)

Coloured indicator for titration

102

10

2

15g HCl

100g solution)(

3.00 mL CH 3CH2OH

50 mL beverage )(

0.040 g phenolphthalein

25.0 mL solution )( 102 = 0.160 % (w/v)


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w/w is used to express the concentration of commercial

aqueous reagents.

For example:

Nitric acid, HNO3 is sold a 70% solution. It means the

reagent contains 70g pure HNO3 per 100g solution.

HCl is sold 37% solution, i.e. For every 100g there is 37g pure

HCl.


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Briefly explain how to prepare the following solutions: 1.0 Lof 10.50 %(w/v) aqueous CH3CH2CH2OH.


(1.0 L = 1000 mL)

% (w/v) = mass(g) x 100volume (mL)

Mass of CH3CH2CH2OH = % x V = 10.5 x 1000

100 100= 105 g of CH3CH2CH2OH

Weigh 105 g of CH3CH2CH2OH and dilute it to the mark withdistilled water in 1.0 L volumetric flask.


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❖ Density - expresses the mass of a substance per unit volume. In SI

units, density is expressed in units of kg/L or g/mL or g/cm3

❖ Density = mass of solute / unit volume (g/mL)

❖ Specific gravity - is the ratio of the mass of a substance to the mass of

an equal volume of water. Unit is dimensionless.

❖ Specific gravity = D of solute / D of water

❖ D H2O = 1.00000 g/mL at 4C

❖ DH2O = 0.99821 g/mL at 20 C


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Not all compounds are in solid form.

Acids are purchased as liquids (stocks solution)

It is usually vital to prepare diluted solutions from these stock solutions

The procedure is: Use pipette to measure moles (via volumes), make upthe volumes using volumetric flask.

For the calculation, we will be using this relation:

McVc = MdVd

c = concentrated solution (stock solution)

d = diluted solution (ending solution / desired solution)


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1. What volume of 0.5 M HCl can be preparedfrom 1 L of 12 M HCl?

M1 = 12 mol/vol V1 = 1 L M2 = 0.5 L

M1V1 = M2V2

V2 = M1V1 / M2

V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L


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1. How many mL of a 14 M stock solution must be used to make 250 mLof a 1.75 M solution?

2. You have 200 mL of 6.0 M HF. What concentration results of this isdiluted to a total volume of 1 L?

3. 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that

the resulting solution is 1.5 M?4. What concentration results from mixing 400 mL of 2.0 M NaCl with 600mL of 3.0 M HCl?

5. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixedwith 2 L of water?

6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixedwith 2 L of 0.2 M NaCl?

7. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixedwith 2 L of water?

8. Water is added to 4 L of 6M antifreeze until it is 1.5 M. What is the totalvolume of the new solution?

9. There are 3 L of 0.2 M HF. 1.7 L of this is poured out. What is theconcentration of the remaining HF?


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1. Calculate the molar concentration of HNO3(63 g/mol) in a solution that has a specific

gravity of 1.42 and is 70% HNO3 (w/w).(Ans: 16M)2. Describe the preparation of 100 mL of 6.0 M

HCl from a concentrated solution that has a

specific gravity of 1.18 and is 37% (w/w) HCl(36.5 g/mol).(Ans: )


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3. Calculate the molarity of the following acid:

1. 70% HClO4, density = 1.67 g/mol

2. 96% H2SO4, density = 1.84 g/mol

chm 421 - topic 1 - calculations

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