Chemical calculations
Department of medical chemistry
Pomeranian Medical University
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Percentage Concentration of solutions
Percentage concentration
Ø  Percent composition by mass (% m/m) or by weight (% w/w) –
gives grams of solute per 100 grams of solution.
Ø  Percent by mass-volume (% m/v) –
gives grams of solute per 100 mL of solution.
Ø  Percent by volume-volume (% v/v) –
gives amount of mililiters of solute per 100mL of solution.
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Concentration of solutions
mr → ms
100 → x
Calculation:
mass percent
ms • 100
x = cp =
ms mass of solute [g]
mr mass of solution [g]
cp percentage concentration by mass [%]
mr
Example
How many grams of glucose is needed to obtain a 2500 g solution
of 12% m/m?
100 g → 12 g
2500 g →
xg
x = 300 g
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Percentage
Concentration
mr g → m s g
100 g → x g
x = cp = ms· 100 / mr
m mass of solute [g]
V volume [cm3]
d density [g/cm3]
d =m /V
Example 18 – student’s book
How many miligrams of bromide was provided if the patient was
given 5 cm3 of medicine containing 0,2% KBr, d=1.0 g/cm3.
mass of medicine (solution) provided to patient
mr = 5 cm3 •1,0 g/cm3 = 5 g
100 g → 0,2 g KBr
5g →
x g KBr
x = 0,01 g KBr
1 mole KBr – 119 g
1 mole Br – 79,9 g
119 g KBr → 79,9 g Br
0,01 g KBr →
x g Br
x = 6,7 • 10-3 g Br = 6,7 mg Br
Percentage
Concentration
mr g → m s g
100 g → x g
x = cp = ms· 100 / mr
Example 4 – student’s book
Calculate the volume of a concentrated H2SO4 solution (96%, d=1.84g/cm3)
needed for making 400 cm3 of a 40% m/v solution.
100 mL
400 mL
→ 40 g H2SO4
→
x g H2SO4
x = 160 g H2SO4
100 g → 96 g H2SO4
x g → 160 g H2SO4
x = 166,67 g 96% H2SO4 solution
V = 166,67 g / 1,84 g/mL = 90,6 mL
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Percentage
Concentration
mr g → m s g
100 g → x g
x = cp = ms· 100 / mr
Example
The blood contains 120% m/v (120mg/100cm3) glucose. Calculate glucose
concentration in mmol/L.
1 mole glucose → 180 g
x mole
→
0,120 g
x = 6,67 • 10-4 mola
6,67 • 10-4 mole → 100 mL
6,67 • 10-3 mole → 1000 mL
c = 6,67 • 10-3 mol/L = 6,67 mmol/L
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Percentage
Concentration
mr g → m s g
100 g → x g
x = cp = ms· 100 / mr
Example
How many milliliters of 95,57% m/m ethyl alcohol (EtOH) solution is
necessary to make 500g of 70% m/m.
A [%, mol/L,etc.] • amount [g,mL] = B [%, mol/L,etc.] • amount [g,mL]
A - concentration of chemical A
B - concentration of chemical B
95,57% • x = 70% • 500g
x = 366g (95,57% EtOH solutions)
500g solution = x g H2O (solvent) + 366g EtOH (solute)
x = 134g H2O
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Molar Concentration of solutions
Molarity (molar concentration, CM )
Molar concentration is the number of moles of solute per liter (1000mL) of solution.
Practically, a mole of the compound is numerically equal to its molecular mass.
It is expressed in: CM [mol/L] or [M]
CM = n / V
M - molecular mass of solute [g/mol]
n – amount of certain solute in mole [mol]
n=m/M
m – amount of certain solute in mass [g]
CM = m / M • V
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Molar Concentration of
solutions
CM = n / V = m / MV
Example 5 – student’s book
Calculate the molar concentration of a 37% m/m HCl solution
with d=1.19g/cm3.
1 mol HCl – 36,45 g
1 mL → 1,19 g
1000 mL → 1190 g
100 g → 37 g HCl
1190 g → x g HCl
x = 440,3 g HCl
1 mol HCl → 36,45 g
x mol HCl → 440,3 g
x = 12,06 mol in 1000 mL → 12,06 mol/mL
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Molar Concentration of
solutions
CM = n / V = m / MV
Example
How much of a 36% m/m HCl solution with d=1.19g/cm3 must be added to
obtain 2L of 0,5 mol/L solution?
1 mol → 36,45 g
0,5 mol → 18,23 g
18,23 g → 1 L
xg → 2L
x = 36,45 g HCl
100 g → 36 g HCl
x g → 36,45 g
x = 101,25 g 36% HCl solution
V = mr/d = 101,25 g / 1,19 g/mL = 85 mL 36% HCl
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Molar Concentration of
solutions
CM = n / V = m / MV
Example 5 – student’s book
0,5 liter of 2mol/L NaOH solution was mixed with 1,5 liters of 6mol/L NaOH
solution. Calculate the concentration of the obtained solution.
1 L → 2 mol
0,5 L → 1 mol
1 L → 6 mol
1,5 L → 9 mol
total moles number = 1 + 9 = 10
total volume = 0,5 + 1,5 = 2 L
c = 10/2 mol/L = 5 mol/L
or
c1 · v1 +c2 · v2 = cr · vr
0,5 L · 2 mol/L + 1,5 L · 6 mol/L = cr · 2 L
cr = 5 mol/L
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