ib1 chemistry quantitative 1b.. topic 1: quantitative chemistry 1.1 the mole concept and...
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IB1 ChemistryQuantitative 1b
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Topic 1: Quantitative chemistry1.1 The mole concept and Avogadro’s constant1.1.1 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 equations1.3.1 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 reactions1.4.1 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 Solutions1.5.1 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 = constant, RnT
universal gas constant , R= 8.31 Jmol-1K-1
(units also dm3kPamol-1K-1)
Universal Gas Equation
pV = nRT
Where p = pressureV = volumeT = Kelvin Temperaturen = number of moleR = 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. Calculate the number of moles of air molecules
2. Calculate the pressure inside the balloon.
Volume units. How many…
1. cm in 1m?
2. cm2 in 1m2?
3. cm3 in 1m3?
4. dm in 1m?
5. dm2 in 1m2?
6. dm3 in 1m3?
7. L in 1m3?
8. dm3 in 1L?
9. cm3 in 1mL?
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.
Image: http://commons.wikimedia.org/wiki/File:Coal_anthracite.jpg
C + O2 CO2m 6.0 (g) M12 (gmol-1)n 0.50 (mol)
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 temperatureoften 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