Chemistry Formulae Relative atomic mass

Example: 79Br and 81Br with r.a.m. of 78.92 and 80.92 respectively. Using the r.a.m of bromine 79.90, calculate the relative isotopic abundances of 79Br and 81Br.

_____________________________________________________________________________________________Mole ConceptDefinition: 1 mole contains 6.021023 (particles: atoms/molecules/ions/electrons) 1 mole contains same number of particles as in 12g 12C (6.021023 atom of C)

_____________________________________________________________________________________________Mole Concept and Molar Volume of Gas

Concentrations of SolutionsDefinition: Molarity - Concentration of a solution stated in moldm-3

_____________________________________________________________________________________________Empirical FormulaDefinition: A formula that gives the simplest/smallest ratio of atoms of each element in compound

Example:A compound consists of 58.77% C, 13.81% H and 27.42% N. Determine the empirical formula of the compound.CHN

mass

No. of mole

Simplest ratio

Molecular formulaDefinition: A formula that shows the actual number of atoms of each element in compoundExample: An unknown compound is found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen with a molecular mass of 60.0 g/mol. What is the molecular formula of the unknown compound?

Answer: C2H4O2_____________________________________________________________________________________________Determining the molecular formula from combustion data

_____________________________________________________________________________________________Oxidation Numbers.

_____________________________________________________________________________________________Percentage yieldTheoretical yield = calculated amount of product obtainedActual yield = actual amount of pure product obtained

ExampleFor the balanced equation shown below, if the reaction of 40.8 grams of C6H6O3 produces a 39.0% yield, how many grams of H2O would be produced? C = 12, H = 1, O = 16C6H6O3+6O26CO2+3H2O

Molar mass of C6H6O3 :12 6 = 721 6 = 6 48 + 6 + 72 = 126 grams/mole16 3 = 48Molar mass of H2O :1 2 = 2 16 1 = 16 2 + 16 = 18 grams/mole

Molar mass of substances numbers of mole :C6H6O3: 126 1 = 126H2O: 18 3 = 54

Theoretical yield0.428 40.8 = 17.5

Actual yield= 6.83g

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Ionization EnergyDefinition: Energy required to remove 1 mole of electron from gaseous particles. First IE:Energy required to remove 1e- form each atom, found in 1 mole of gaseous atoms

Second IE:Energy required to remove 1e- form each +1 gas ion, found in 1 mole of +1 gaseous ions.

_____________________________________________________________________________________________Boyles LawDefinition:Boyles law states that volume occupied by gas is inversely proportional to pressure from gas, when mass (or mole) and temperature are made constant.Formula:PV = constant P = pressure from gas V = volume occupied by gas

Example:A sample of gas occupies 24.0 dm3 at 100kPa. What is its new volume at 300kPa if other conditions are held constant?

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Charles LawDefinition:Charles Law states that volume occupied by gas is directly proportional to Kelvin temperature, when mass (or mole) and pressure are constant.Formula:V=kT T = Kelvin temperature V = volume filled by gas k = constant

Temperature: 25C = (25+273) K = 298K

Example:A sample of neon gas occupies 23.0 dm3 at 100C and was heated to 200C. What is the new volume of the gas?

Avogadros LawDefinition:Avogadros law states that volume occupied by gas is directly proportional to moles of gas present, when pressure and temperature are constant.

1 mol of gas at 0C is 22.4 dm31 mol of gas at 25C is 24.0 dm3

Formula:V=kn n = moles of gas V = volume filled by the gas k = constant

Example:1.00 mole gas occupies 24.0 dm3 at constant T and P. Determine the moles of gas that occupies 400 dm3 at the same pressure (P) and temperature (T)?

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Mixture of Gases into container

Mole Fraction

_____________________________________________________________________________________________Partial PressurePartial pressure = mole fraction total pressure orPartial pressure is defined as pressure from one type of gas in a mixture of gases.

Examplea) Calculate the mole fraction ofi. Nitrogen gas

ii. Helium gas

b) Calculate partial pressure of i. Nitrogen gas

ii. Helium gas

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Daltons LawTotal pressure = sum of partial pressure of all gases Example:A container is filled with the gases oxygen, neon and argon with partial pressure 10.0kPa, 30.0kPa and 15.0kPa. Calculate total pressure.

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Ideal Gas Equation

Example: Liquid X is either ethanol C2H5OH or methanol CH3OH.When 0.0800g of liquid X is heated in a syringe to 127C,the vapour produced a pressure 1.00 x 105 Pa and occupies 81.0 cm3. Given R = 8.31a) Calculate relative molecular mass of X (2m)

b) Calculate Mr of C2H5OH. The identify X. (2m)

Specific Heat Capacity Definition: The amount of heat energy required to raise the temperature of 1g of the substance by 1C or 1KHeat CapacityDefinition:The amount of heat energy required to raise the temperature of a substance by 1C or 1KUnit:Jg-1K-1

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Standard Enthalpy of NeutralisationDefinition:Enthalpy change when acid reacts with an alkali to produce 1 moles of water measured at 25C and 1 atm.

Example:50.0 cm3 HCl 1.00 moldm-3 was mixed with 50.0 cm3 NaOH 1.00 moldm-3 in polystyrene cup.Data obtained: Initial temperature, HCl (aq)25.0C

Initial temperature, NaOH (aq)25.0C

Highest temperature of mixture31.8C

Calculatea) Moles of HCl reacted [ 0.0500 mol ]

b) Moles of NaOH reacted [ 0.0500 mol ]

c) Moles of H2O produced [ 0.0500 mol ]

d) Heat released from the mixture in kJ given specific heat capacity of 4.20Jg-1C-1 [-2.856kJ]

e) Enthalpy of neutralisation [-57.1kJmol-1]

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Standard Enthalpy of FormationDefinition:Enthalpy change when 1 mole of substance is formed from its elements, under standard conditions 25C and 1 atm.Formula:

Example:

Calculate the standard enthalpy change of the reaction Hr =

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Standard Enthalpy of CombustionDefinition:Enthalpy change when 1 mole of substance is burned completely, in excess oxygen gas, measured on standard condition 1 atm and 25C.Formula:

Example:

Calculate the enthalpy change of combustion of ethanol in kJmol-1 [-1260kJmol-1]

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Hess LawDefinition:Enthalpy change of reaction depends only on initial state and final state of chemicalsorEnthalpy change of reaction does not depend on the numbers of steps for that change ______________

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Example:

a) Write equation with state symbol for the standard enthalpy of combustion for

i.Hydrogen gas (1m)

ii.Carbon solid (1m)

iii.Cyclohexane liquid (1m)

b) Write the equation for the standard enthalpy of formation for cyclohexane liquid. (1m)

c) Using Hess Law, calculate the standard enthalpy of formation of cyclohexane liquid (2m)

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Bond EnergyDefinition:Average energy needed to break 1 mole of covalent between 2 atoms, in gaseous state.

Formula:

Example:

Given

Calculate the standard enthalpy change of reaction, [-688kJmol-1]

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Standard Enthalpy Change of Solution Definition:Enthalpy change when 1 mol of a substance is dissolved in water to form a solution of infinite dilution under standard conditions.Example:

Stage 1:The ions are separated to form free gaseous ions.

Stage 2:The free gaseous ions are hydrated with water to form hydrated ions.

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Equilibrium constant aA + bB cC + dD

Example:In a 5.00 dm3 container, 10.0 mole of sulfur trioxide gas was allowed to decomposed at 500C. At a constant pressure at this temperature, 20% of sulfur trioxide molecules decomposed as dissociated as follows.

a) Write and expression of equilibrium constant, according to concentration

b) Determine the initial mole, equilibrium mole and the concentration for each of the substanceSO3SO2O2

Initial mole/mol

Equilibrium mole/mol

Concentration/moldm-3

c) The calculate the value of equilibrium constant, Kc according to concentration and state its unit if any [0.0125moldm-3]

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Equilibrium constant, KP

Example:Calculate the KP value for the following reaction.

When SO2(g) and O2(g), at partial pressures of 1.00 atm and 0.500 atm respectively, are allowed to reach equilibrium at 675K, the total pressure in the container is found to be 1.01 atm when equilibrium is attained. [ KP=2.4105 ]14