Basic chemical calculations
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
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
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
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
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
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