IB1 Chemistry
Quantitative 1b
.
Topic 1: Quantitative chemistry
1.1 The mole concept and Avogadro’s constant
1.1.1 Apply the mole concept to substances.
1.1.2 Determine the number of particles and the amount of substance (in moles).
1.2 Formulas
1.2.1 Define the terms relative atomic mass (Ar) and relative molecular mass (Mr).
1.2.2 Calculate the mass of one mole of a species from its formula.
1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass and molar mass.
1.2.4 Distinguish between the terms empirical formula and molecular formula.
1.2.5 Determine the empirical formula from the percentage composition or from other experimental data.
1.2.6 Determine the molecular formula when given both the empirical formula and experimental data.
1.3 Chemical equations
1.3.1 Deduce chemical equations when all reactants and products are given.
1.3.2 Identify the mole ratio of any two species in a chemical equation.
1.3.3 Apply the state symbols (s), (l), (g) and (aq).
1.4 Mass and gaseous volume relationships in chemical reactions
1.4.1 Calculate theoretical yields from chemical equations.
1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.
1.4.3 Solve problems involving theoretical, experimental and percentage yield.
1.4.4Apply Avogadro’s law to calculate reacting volumes of gases.
1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations.
1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.
1.4.7 Solve problems using the ideal gas equation, PV = nRT.
1.4.8 Analyse graphs relating to the ideal gas equation.
1.5 Solutions
1.5.1 Distinguish between the terms solute, solvent, solution and concentration (g dm–3 and mol dm–3).
1.5.2 Solve problems involving concentration, amount of solute and volume of solution.
Yield
The yield is the amount of product obtained experimentally
Percentage yield=
actual yield
× 100
theoretical yield
Reacts can be:
limiting reagent
to excess
Yield example
0.24±0.01g of magnesium react with excess dilute sulphuric acid
to give a gas and a solution.
The solution is evaporated and the evaporating basin (mass
28.83±0.01g) weighs 28.03±0.01g with the salt.
Calculate the percentage yield.
Yield
Balanced equation for the reaction
Mole ratio
Mass
Molar mass
No. Moles
Uncertainties
Measuring chemical quantities: gases
 in volume units (cm3, dm3, etc.) using a gas syringe
 volume depends on temperature and pressure
Propeties of gases
 Variable volume and shape
 Expand to occupy volume available
 Can be easily compressed
 Exert pressure on whatever surrounds them
 Volume, Pressure, Temperature, and the number of moles
present are interrelated
 Easily diffuse into one another
Mercury barometer
 Defines and measures atmospheric
pressure
 Mercury column rises to 760 mm
average at sea level
 This quantity 1 atmosphere = 100 kPa
Pressure
Standard temperature and pressure (STP)
 Standard Temperature and Pressure (IUPAC)
STP = 0oC or 273.15 K and 100kPa
 Reference for comparing gas quantities
 Can calculate volume at various temperatures and
pressures
Assumptions of the ideal gas model
 the particles are indistinguishable, small, hard spheres
 no energy loss in motion or collision
 Newton's laws apply to collisions
 The average distance between molecules is much larger than the
size of the particles
 The molecules are constantly moving in random directions with
a distribution of speeds
 There are no attractive or repulsive forces between the
molecules or the surroundings except during collisions
 Real gases have attractive forces between particles (van der
Waals forces)
  close to an ideal gas at high Temp and low Pressure
Charles’ Law
Charles’ Law: the volume of a gas is proportional to the Kelvin
temperature at constant pressure
V = kT
V1 = T1
V2 T2
Gay-Lussac’s Law
The pressure and temperature of a gas
are directly proportional at constant
volume.
P = kT
P1 = T1
P2 T2
Boyle’s Law
Boyle’s Law: pressure and
volume of a gas are
inversely proportional at
constant temperature.
PV = constant.
P1V1 = P2V2
Boyle’s Law
Combined gas law
V ∝ 1/p (at constant T)
V ∝ T(at constant p)
combine to give
V ∝ T/p or
pV ∝ T
Avogadro’s Law
Equal volumes of a gas under the same
temperature and pressure contain the same
number of particles.
At constant T and p
V ∝n
Universal Gas Constant
pV
nT
= constant, R
universal gas constant , R= 8.31 Jmol-1K-1
(units also dm3kPamol-1K-1)
Universal Gas Equation
pV = nRT
Where
p = pressure
V = volume
T = Kelvin Temperature
n = number of mole
R = 8.31 J mol-1 K-1
Using the universal gas equation
1.
Calculate the volume of 10g of neon at STP.
2.
Calculate the pressure necessary to compress 1g of
hydrogen into 1Litre at room temperature.
3.
A balloon that contains 2x1023 molecules of air at 20C and
takes up 2 litres.
1.
2.
Calculate the number of moles of air molecules
Calculate the pressure inside the balloon.
Volume units. How many…
1.
cm in 1m?
7.
L in 1m3?
2.
cm2 in 1m2?
8.
dm3 in 1L?
3.
cm3 in 1m3?
9.
cm3 in 1mL?
4.
dm in 1m?
5.
dm2 in 1m2?
6.
dm3 in 1m3?
Calculate the volume of 1 mole of gas at STP
Gases
Molar volume of any gas at
STP
22.4 dm3mol-1
Reacting gas volumes
For a gas at a constant temperature and pressure
the volume is proportional to the number of moles.
mole ratio  volume ratio
Calculate the volume of oxygen that reacts
with 2 dm3 of Hydrogen gas. (const. p & V)
2 H2(g) + O2(g)  2 H2O(g)
2dm3
?
?
Under other conditions use pV=nRT
1.
Balanced Equation
2.
Table
3.
Fill in known and ?
4.
Calculate
6.0 g Carbon burns in Oxygen. Give the volume of
formed Carbon dioxide at 400K and 1 Atm.
C
+
O2 
CO2
m
6.0
(g)
M
12
(gmol-1)
n
0.50
(mol)
Image: http://commons.wikimedia.org/wiki/File:Coal_anthracite.jpg
pV=nRT
Solutions
solute : salt
solvent : water
solution : salt-water mixture
Concentration
Mass percentage = Mass of substance/Mass of solution
Volume percent = volume of solute/ total volume
Mol fraction = Xa = na/(na+nb)
gdm-3
moldm-3
Concentration in moldm-3 often represented by square brackets, eg [HCl]
Concentration in gdm-3
concentration = mass
volume
Solubility
the mass of a
particular solvent
that dissolves in a
solvent at a given
temperature
often in g per 100g H2O
Calculate the mass of salt needed for a
concentration of 10gdm-3 in 50cm3
Concentration in moldm-3 (molarity)
concentration
=
number of moles
volume
Calculate the mass of hydrogen chloride
in 50cm3 of 0.5moldm-3 HCl
Comparing concentrations: which is the
most concentrated- convert to moldm-3
10g of copper sulfate in 25cm3 of water
5g of copper sulfate in 10cm3 of water
0.1mol of copper sulfate in 15cm3 of water
0.01mol of copper sulfate in 5cm3 of water