Basic chemical calculations

When solving numerical problems,

always ask yourself whether your answer makes sense !!!

How many grams of NaCl were excreted in urine in 24 hours?

20 ml of an average sample was titrated by direct argentometric method and the consumption of the standard solution of AgNO3 (c = 100 mmol/L) was 34.2 mL.

Volume of urine = 1 500 mL/day, M(NaCl) = 58.5 g/mol

ANSWERS IN THE TESTS: 1 500 kg

i.e. 1.5 ton !!!

Correct answer: 15 g

SI prefixes (metric prefixes)

10-1 deci d 101 deca da

10-2 centi c 102 hecto h

10-3 mili m 103 kilo k

10-6 micro 106 mega M

10-9 nano n 109 giga G

10-12 pico p 1012 tera T

10-15 femto f 1015 peta P

10-18 atto a 1018 exa E

10-21 zepto z 1021 zetta Z

10-24 yocto y 1024 yotta Y

Converting units

5 mL = 0.005 L 0.750 L = 750 mL

1 L = 1 dm3 1 mL = 1 cm3 1 L= 1 mm3

0.1 mol/L = 100 mmol/L

50 mg = 0.050 g

Converting units

Erythrocyte volume: 85 fL ( = 85 m3 )

85 fL = 85 x 10-15 L = 85 x 10-15 dm3 =

= 85 x 10-18 m3 = 85 m3

Erythrocyte number: 5 x 106 / mm3 = 5 x 106 / L

= 5 x 1012 / L

Rounding off the results

Numbers obtained by measurement are always INEXACT !

"exact" numbers - in mathematics: 10 = 10.000…..

"measured values"

e.g. by measuring the volume 10 mL - it is NOT

exactly 10.00000… mL

Uncertainties ("errors") always exist in measured quantities!

Significant figures

Is there any difference between 4.0 g and 4.00 g ?

Answer: YES !

4.0 g 2 significant figures

4.00 g 3 significant figures

4.00 g is "more precise" than 4.0 g

E.g. 4.003 4 significant figures

6.023 x 1023 4 significant figures

0.0012 2 significant figures

5000 ? 1,2,3 or 4 significant figures

When rounding off the results, you have to consider

significant figures of given numbers !

The precision of the result is limited by the precision

of the measurements !

Rules

1) Multiplication and division

result – must be with the same number of significant figures as

the measurement with the FEWEST signif. fig.

e.g. 6.221 cm x 5.2 cm = 32.3492 cm2 -->

round off to 32 cm2

2) Addition and subtraction

result – cannot have more digits to the right of the decimal point

than any of the original numbers

e.g. 20.4 1 decimal place

1.322 3 decimal places

83 ZERO decimal places

104.722 round off to 105

Rounding off: digits 5,6,7,8,9 --> round up

digits 0,1,2,3,4 --> round down

CONCLUSION:

Your calculator can give the result like this:

100 / 7 = 14.28571429 !!!

DON‘T give as a result of the calculation a number

with 10 digits, which shows your calculator, round

it off to the "reasonable number" of decimal places

calculation with more steps – round off only

THE FINAL RESULT

You have to be familiar with your calculator !

E.g. 1) 103 = 1000 !!!

10 EXP 3 = 10 x 103 = 10000 !!!

2) 50

2 x 5

50 : 2 x 5 = 125 !!!

= 5 !!!

Uncertainties of quantitative methods

1. PRECISION = how closely individual measurements

agree with one another

2. ACCURACY = how closely measurements agree

with the correct ("true") value

"true" value measured values

Precision and accuracy - shooting on the target

a) good precision

good accuracy

b) good precision

poor accuracy

c) poor precision,

but in average

good accuracy

d) poor precision

poor accuracy

Density ( - relation between mass and volume

- the amount of mass in a unit volume of substance

= m / V Mind the units !

SI unit: kg/m3

other units: g/cm3

kg/dm3

note: cm3 = mL dm3 = L

Examples

water 1 000 kg/m3

1 g/cm3

1 kg/dm3

Au 19 300 kg/m3

19.3 g/cm3

19.3 kg/dm3

Calculate the density of 90% H2SO4, if the mass of 200 mL of the solution is 363 g.

m / V

/ 200 = 1.815 g/cm3

Often we know the volume and we need to calculate the mass

and vice versa!

m = V x V = m /

What is the mass of the solution of KOH,

if the volume is 2.5 L and density 1.29 g/cm3 ?

mV x

m x 1.29 = 3 225 g

units ! 2.5 L = 2 500 mL ( cm3 )

What is the volume of the solution of HNO3 if

the mass is 150 g and the density 1.46 g/cm3 ?

Vm /

V / 1.46 = 102.7 cm3 ( mL )

Amount of substance ( n )- base SI quantity

- is in a close relation to the

NUMBER of ELEMENTARY ENTITIES

(atoms, molecules, electrons, …)

unit: MOLE ( 1 mol )

1 mol = as many objects as the number of atoms

in 12 g of the carbon isotope 126C

1 mol = 6.023 x 1023 elementary entities

= AVOGADRO’s number NA

NA = 6.0221367 x 1023 /mol

number of entities = n x NA

analogy: „counting units“

1 pair = 2 1 dozen = 12 1 gross = 144

1 mol = 6.023 x 1023

Calculate the number of elementary units present in 2.5 mol cations Ca2+.

number of Ca2+n x NA

number of Ca2+xx

x

Calculate the number of protons released

during complete dissociation of 2 mol H3PO4.

H3PO4 --> 3 H+ + PO4 3-

n(H+) = 3 x n(H3PO4) n(H+) = 6 mol

number of H+n(H+) x NA

number of H+x xxx

Calculate the number of C atoms in

0.350 mol of glucose.

C6H12O6

n(C) = 6 x nglukosa n(C) = 2.1 mol

number of Cn(C) x NA

number of Cxxx

Calculate the amount of substance:

2.71 x 1024 molecules of NaCl

n = number of NaCl / NA

n =xxmol

Molar mass ( M )- the mass of 1 mol of a substance

- unit: g / mol - can be calculated with the use of relative atomic masses

(PERIODIC TABLE)

Relative atomic mass Ar

Relative molecular mass Mr expressed in atomic mass units

note:

atomic mass unit = 1/12 of the mass of atom 126C u = 1.6605 x 10-24 g

Ar = matom / u Mr = mmolecule / u

Ar ( 126C ) = 12.00 Ar ( H ) = 1.008

relative molecular mass - no real unit in biochemistry: Dalton ( Da )

e.g. protein 55 kDa

What is a molar mass of glucose ?

C6H12O6

Ar (C) = 12.0

Ar (H) = 1.0

Ar (O) = 16.0

M = 6 x 12 + 12 x 1 + 6 x 16 = 180 g/mol

Often we know the mass and need to calculate the amount of substance

and vice versa!

n = m / Mm = n x M

How many moles of NaCl are present in

100 g of this substance ? M = 58.5 g/mol

n = m / M

n =mol

Calculate the mass of 0.433 mol of calcium nitrate.

M = 164.2 g/mol

m = n x M

m = 0.433 x 164.2 = 71.1 g

How many molecules of glucose are in

5.23 g C6H12O6 ? M = 180 g/mol

n = m / M

n = 5.23 / 180 = 0.029056 mol

number of molecules = n x NA

number of molecules = 0.029056 x 6.023 x 1023

= 1.75 x 1022

Solution composition – "Concentration"

solute = the substance which dissolves

solvent = the liquid which does the dissolving

A solution is prepared by dissolving a solute in a solvent.

Solution composition – "Concentration"

to designate amount of solute disolved in a solution

"number of different ways to express concentration"

Molar concentration (molarity) c mol/L

Mass concentration g/L

Mass fraction wMass percentage % % w/w

Volume fraction

Volume percentage   % v/v

Molar concentration ( c ) (substance concentration, MOLARITY)

- the number of moles of substance in 1 L of solution

c = n / V

unit: mol / L

Mind the units ! volume must be in LITRES

Calculate molar concentration of NaCl solution, if 250 mL contain 0.1 mol NaCl.

cn / V

c mol/L

units ! 250 mL = 0.250 L

What is the substance concentration of a solution,

if it contains 15 g NaOH in 600 mL of solution.

( M(NaOH) = 40.0 g/mol )

n = m / M(NaOH)

cn / V

cm / ( M(NaOH) x V )

c = 15 / ( 40 x 0.6 ) = 0.625 mol/L

How many moles of H+ are present in 2 L of H2SO4

solution, if the concentration is 0.1 mol/L ?

H2SO4 --> 2 H+ + SO4 2-

n(H2SO4) = c x V

n(H2SO4) = 0.1 x 2 = 0.2 mol

n(H+) = 2 x n(H2SO4)

n(H+) = 2 x 0.2 = 0.4 mol

Mass concentration ( )

mass of substance in 1 L of solution

= msolute / V

unit: g / l

What is the mass concentration of a solution,

which contains 7.0 g of KCl in 750 ml ?

mKCl / V

g/l

How many grams of AgNO3 do we need to prepare

7 L of solution of mass concentration 0.5 g/L ?

mAgNO3 = x V

mAgNO3 xg

= c x Mc = / M

Why ? = msolute / V

msolute = n x M

= n x M V

c = n / V

Interconverting substance concentration ( c )

and mass concentration ( m )

What is the mass concentration of the NaOH solution,

if the substance concentration is 0.5 mol/L ?M = 40 g/mol

c x M

xg/L

Calculate the substance concentration of the NaCl

solution, if the mass concentration is 10 g/L ?

M = 58.5 g/mol

c M

cmol/L

mc x V x M

Why? m = n x M

n = c x V

Calculation of the mass necessary for making

a solution of given substance concentration

How many grams of Na2SO4 (M = 142 g/mol) are

necessary for 1 500 mL of solution (c= 0.1 mol/L) ?

mc x V x M

mxx21.3 g

Mass fraction ( w )

ratio between a mass of the dissolved substance (msolute) and the total solution mass (msolution)

w = msolute / msolution

unit: -

Mass percentage: mass fraction x 100 %

( i.e. grams of substance in 100 g of solution )

e.g. w = 0.15 15 % solution

Volume fraction ( )

analogy of mass fraction

= Vsolute / Vsolution

unit: -

Volume percentage: volume fraction x 100 %

The use: ETHANOL in alcoholic drinks !!!

e.g. alc. 11.5 % vol.

"Percent concentration" - summary

"concentration 10 %" can be confusing !

it‘s better to specify it:

% w/w percent by mass (mass percentage)

% v/v percent by volume (volume percentage)

% w/v percent by mass over volume (mass-volume percentage)

w ... weight

v ... volume

Calculate %(w/w) concentration of a solution prepared from 15 g NaOH and 80 mL of water.

w = mNaOH / msolution

w( i.e. 15.8 % )

note: density of water 1.0 g/cm3

How many grams of AgNO3 are necessary to prepare 700 g of 2 % (w/w) solution ?

w = 0.02

w = m / msolution

m = w x msolution

mxg

x w M x c M we need to know the density of the solution!

density must be in g/dm3

w =c =

Interconverting substance concentration (c) and mass fraction (w)

What is the substance concentration of 10 %(w/w) solution of Na2CO3 ? Density of this solution is 1.1 g/cm3.

M = 106 g/mol

x w M

1 100 x 0.1

106

c =

c = = 1.04 mol/L

What is the substance concentration of 10 %(w/w) solution of Na2CO3 ? Density of this solution is 1.1 g/cm3.

M = 106 g/mol

other way of calculation:

10 %(w/w) --> 10 g Na2CO3 in 100 g of solution

10 g Na2CO3 is 10 / 106 = 0.09434 mol

the volume of 100 g of solution is 100 / 1.1 = 90.91 mL

substance conc.: c = n / V

c = 0.09434 / 0.09091 =1.04 mol/L

Mixing of solutions( dilution = particular case of mixing )

These rules must be applied:

1) the mass is the sum of masses of the components:

m1 + m2 = m

( conservation of the mass, NOT THE VOLUME !!! )

2) the mass of the solute present in the new solution

formed by mixing is the sum of masses of the solute

dissolved in the components: m1w1 + m2w2 = mw

Equation: m1w1 + m2w2 = (m1 + m2) w

Calculate the concentration of a solution %(w/w)

prepared by mixing 300 g 70 % and 500 g 20 % H2SO4

m1w1 + m2w2 = (m1 + m2) w

300 x 0.7 + 500 x 0.2 = ( 300 + 500 ) w

310 = 800 w

wthat is 38.75 % )

"Cross rule"

= an easy way to do such calculations

a % ( c - b ) mass portions

c %

b % ( a - c ) mass portions

a , b … original concentrations

c … new concentration

How many g of 60 %(w/w) HNO3 do you need to prepare 1 200 g of 10 %(w/w) solution ?

dilution: 60 % HNO3 + water ( 0 % )

60 % 10 - 0 = 10 portions

10 % the sum is 60 port.

0 % 60 - 10 = 50 portions

1 200 g ….. 60 portions --> 1 portion: 1 200 / 60 = 20 g

We need:

10 portions of 60 % HNO3 …. i.e. 10 x 20 = 200 g of 60 % HNO3 50 portions of water …. i.e. 50 x 20 = 1000 g of water

Chemical equations

a A + b B --> c C + d D

Dalton’s law: ratio of amounts of substance of reactants

can be expressed in small whole numbers

n (A) a

n (B) b

a , b … stoichiometric coeficients

= = stoichiometric factor

a) H2SO4 + 2 NaOH --> Na2SO4 + 2 H2O

n(H2SO4) / n(NaOH) = 1 / 2

b) 2 AgNO3 + K2CrO4 --> 2 KNO3 + Ag2CrO4 n(AgNO3) / n(K2CrO4) = 2 / 1

c) 2 KMnO4 + 5 (COOH)2 + 3 H2SO4 --> K2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O

n(KMnO4) / n( (COOH)2 ) = 2 / 5

MANGANOMETRY !

How many moles of NaOH are necessary for

a full neutralization of 1.5 mol of oxalic acid?

2 NaOH + (COOH)2 --> (COONa)2 + 2 H2O

n(NaOH) / n( (COOH)2 ) = 2 / 1

n(NaOH) = 2 x n( (COOH)2 )

n(NaOH) = 2 x 1.5 = 3 mol

How many g of HCl are necessary for a full neutralization of

10 g Na2CO3 ?

M(HCl) = 36.5 g/mol M(Na2CO3) = 106 g/mol

2 HCl + Na2CO3 --> 2 NaCl + CO2 + H2O

n(HCl) / n(Na2CO3) = 2 / 1

[ m(HCl) / M(HCl) ] / [ m(Na2CO3) / M(Na2CO3) ] = 2

m(HCl) = 2 x [ m(Na2CO3) / M(Na2CO3) ] x M(HCl)

m(HCl) = 2 x ( 10 / 106 ) x 36.5 = 6.89 g

How many g of HCl are necessary for a full neutralization of

10 g Na2CO3 ?

M(HCl) = 36.5 g/mol M(Na2CO3) = 106 g/mol

2 HCl + Na2CO3 --> 2 NaCl + CO2 + H2O

2 x 36.5 g HCl ………………… 106 g Na2CO3

? g HCl ………………… 10 g Na2CO3

? 10

2 x 36.5 106

? = 6.89 g

=

How many grams of gold are in a 160 g piece marked 18 carat ?

w = 18 / 24 = 0.750

w = mpure Au / malloy

mpure Au = malloy x w

mpure Au = 160 x 0.750 = 120 g

note: "purity" of gold: CARAT ( pure gold: 24 carat )

THOUSANDS ( pure gold: 1000 / 1000 )

18-carat gold = 18 / 24 = 750 / 1000 = 0.750 i.e. 75% gold