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IB1 Chemistry Quantitative 1b - PowerPoint PPT Presentation

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

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IB1 ChemistryQuantitative 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 moldm–3).

1.5.2 Solve problems involving concentration, amount of solute and volume of solution.

The yield is the amount of product obtained experimentally

Percentage yield= actual yield × 100

theoretical yield

Reacts can be:

limiting reagent

to excess

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.

Balanced equation for the reaction

Mole ratio

Mass

Molar mass

No. Moles

• in volumeunits (cm3, dm3, etc.) using a gas syringe

• volumedepends 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

• Defines and measures atmospheric pressure

• Mercury column rises to 760 mm average at sea level

• This quantity 1 atmosphere = 100 kPa

• 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

• 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: the volume of a gas is proportional to the Kelvin temperature at constant pressure

V = kT

V1 = T1

V2 T2

The pressure and temperature of a gas are directly proportional at constant volume.

P = kT

P1 = T1

P2 T2

Boyle’s Law: pressure and volume of a gas are inversely proportional at constant temperature.

PV = constant.

P1V1 = P2V2

V ∝ 1/p (at constant T)

V ∝ T(at constant p)

combine to give

V ∝ T/p or

pV∝ T

Equal volumes of a gas under the same temperature and pressure contain the same number of particles.

At constant T and p

V ∝ n

pV = constant, R

nT

universal gas constant , R= 8.31 Jmol-1K-1

(units also dm3kPamol-1K-1)

pV = nRT

Where p = pressure

V = volume

T = Kelvin Temperature

n = number of mole

R = 8.31 J mol-1 K-1

• Calculate the volume of 10g of neon at STP.

• Calculate the pressure necessary to compress 1g of hydrogen into 1Litre at room temperature.

• A balloon that contains 2x1023 molecules of air at 20C and takes up 2 litres.

• Calculate the number of moles of air molecules

• Calculate the pressure inside the balloon.

• cm in 1m?

• cm2 in 1m2?

• cm3 in 1m3?

• dm in 1m?

• dm2in 1m2?

• dm3in 1m3?

• L in 1m3?

• dm3 in 1L?

• cm3 in 1mL?

Molar volume of any gas at STP

22.4 dm3mol-1

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 ? ?

• Balanced Equation

• Table

• Fill in known and ?

• 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 Carbon =nRT

Solutions Carbon

solute : salt

solvent : water

solution : salt-water mixture

Concentration Carbon

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 Carbon -3

concentration = mass

volume

Solubility Carbon

the mass of a particular solvent that dissolves in a solvent at a given temperature

often in g per 100g H2O

Concentration in 10gdm-moldm-3 (molarity)

concentration = number of moles

volume

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