Stoichiometry
Aqueous Solution
water that contains dissolved substances
Energy and Stoichiometry Steps
- Balance equation - Convert given quantity to moles - Use energy ratio to determine the energy released
Mass to Mass Steps
- Balance equations - Convert Mass to moles - Molar Ratio - Moles to Mass
Mole-Mole Calculations
- Balance the chemical equation - Molar Ratio (substance needed/substance given) - Multiple given # x needed/given
Converting *Moles to Mass*
- 2 moles (2.0 g/mol = 4.0g of hydrogen gas) + 1 mole (32.0 g/mol) of oxygen gas = 2 moles (18.0 g/mol = 36.0 g of water vapour
Converting *Moles to Volume at STP* - only when reactants are gases
- 2 moles (22.4 L/mol) = 44.8 L of hydrogen gas + 1 mole (22.4 L/mol) = 22.4 L of oxygen gas = 2 moles (22.4 L/mol = 44.8 L of water vapour
Moles to Volume Steps
- Balance - Given quanity to Moles - Molar ratio - Moles to required substance to mass or volume
Solving limiting reactant problems
- Balance the equation - Find the number of moles of each reactant - Using the molar coefficients, determine how much of one reactant is needed to use up the other - If the amount in step 3 is more that what is given of that reactant in Step 2, that reactant is the limiting reactant - Use the molar coefficients and the given amount of the limiting reactant to determine the moles of product - Convert moles of product to mass or volume - Determine the moles of excess reactant remaining after the reaction - Determine the mass of the excess reactant
Moles to Mass Steps
- Balance the equation - If necessary, convert the given amount from mass to moles - Molar ratio given # x sub needed / sub given - If necessary, answer x atomic mass of answer / 1 = mass (g)
Steps to solving stoichiometry problems
1. Balance the chemical equation 2. Convert the given amounts from mass or volume to moles (if necessary 3. Use the molar ratio to calculate the moles of the required substance 4. If needed convert the moles of the required substance to mass or volume )
Dissociation Equations Steps
1. Is the compound ionic or molecular 2. Write the ions that will be formed 3. Indicate coefficients from subscripts 4. Determine appropriate state of compound
Steps in the solution process
1. The solvent particles must separate to make room for solute particles 2. The solute particles or ions must separate from the other solute particles at the surface of crystals 3. The forces of attraction between solute and solvent take hold and the particles "snap" back and move closer
Energy n Stoichiometry Example 3
129 kJ of energy is required to decompose 2 moles of NaHCO3. Calculate the amount of energy required to decopose 2.24 moles of NaHCO3 2NaHCO3 +129 kJ = Na2CO3 + H2O + CO2 *Balance* already balanced *Use the appropriate energy ratio to determine the required amount of energy Energy Ratios: 129 kJ/ 2 mol NaHCO3 OR 2 mol NaHCO3/129 kJ 2.24 x 129 kJ / 2 mol = 144 kJ
Converting *Mole to the number of atoms*
2 moles (2 moles/mol) = 4 atoms of hydrogen gas + 1 mole (2 atoms /mol) = 2 atoms of oxygen gas = 2 moles (2 atoms/mol = 4 atoms of hydrogen gas = 2 moles (1 atom/mol) = 2 atoms of oxygen gas
Interpreting coefficients as Moles
2H2 + O2 = 2H2O - 2 moles of hydrogen gas - 2 moles of water vapour - 1 mole of oxygen gas
Balanced equation
2H2 + O2 = 2H2O - The coefficients represent the *number of molecules of each compound involved in the reaction* - 2 molecules of hydrogen gas - 1 molecule of oxygen gas - 2 molecules of water vapour
Ionic compound
A compound that consists of positive and negative ions
Solvent
A liquid substance capable of dissolving other substances Saltwater // water
Solubility
A measure of how much solute can dissolve in a given solvent at a given temperature. - The amount of solute needed to make a saturated solution
Structure of water
A molecule of water is made up of two hydrogen atoms and one oxygen atom - the electrons between the hydrogen and oxygen atoms in each bond lie more towards the oxygen then they do towards the hydrogen
Colligative property
A property that depends on the number of particles of solute, and not upon their identity 1. vapour pressure lowering 2. freezing point depression 3. boiling point elevation - change is produced because of how much solute is added, regardless of the solute's identity
Pure substances
A sample of matter, either a single element or a single compound, that has definite chemical and physical properties
Unsaturated Solution
A solution that contains less solute than a saturated solution does and that is able to dissolve an additional solute Open rooms in hotel
Saturated Solution
A solution that contains the maximum amount of solute the solution can hold at a given temperature and pressure All booked hotel
Soluble substance
A substance that can be dissolved sugar in water
Insoluble substance
A substance that cannot be dissolved oil in water
Solute
A substance that is dissolved in a solution. Saltwater // salt
Effect of solute on vapour pressure
Adding solute to a solvent lowers the vapour pressure of the solvent - The more solute that is added to a liquid, the greater the decrease in vapour pressure
Example 2 Moles to Moles
Al + O2 = Al2O3 How many moles of aluminum are needed to form 3.70 moles of aluminum oxide? *Balance* 4Al + 3O2 = 2Al2O3 *Molar Ratio* Need: Aluminum Given: 3.70 Aluminum Oxide 4 mol Al/2 mol Al2O3 *Multiple* Mol Al = 3.70 x 4 mol Al / 2 mol Al2O3 = 7.40 mol Al - 7.40 moles of aluminum are needed to form 3.70 moles of aluminum oxide.
electrolyte
An ionic compound whose aqueous solution conducts an electric current
Dissociation Equation Example 1
C12H22O11 1. Molecular compound 2. C12H22O11 -> C12H22O11 3. n/a 4. C12H22O11(s) ->. C12H22O11(aq)
Example 2 Moles to Mass (Mass given)
Calculate the moles of hydrogen gas needed to make 15.0 g of ammonia N2 + 3H2 = 2NH3 *Balance* already balanced *Convert mass to moles* Molar mass of ammonia = NH3 = 17.0 g/mol Mol = 15.0 g x 1 mol / 17.0 (molar mass) = 0.88235 mol *Molar Ratio Need: H2 Given: ammonia mol H2 - 0.88235 x 2 mol H2 / 2 mol NH3 = 1.3235 mol H2 - You need 1.32 moles of hydrogen to make 15.0 g of ammonia
Example 1 of Mass to Mass
Calculate the number of grams of NH3 produced when 5.40 g of H2 reacts with an excess of nitrogen N2 + 3H2 = 2NH3 *Balance* already balanced *Mass to Moles* Molar Mass of hydrogen H2 = 2/0 g/mol Mol - 5.40 x 1 mol H2 / 2.0g H2 = 2.70 mol H2 *Molar Ratio* Need: NH3 Given: H2 Mol NH3 = 2.70 (total) x 2 mol NH3 / 3 mol H2 = 1.80 mol NH3 *Moles to Mass* Molar Mass of NH3 = 17.0 g/mol Mass = 1.80 mol NH3 x 17.0g NH3 / 1 mol NH3 = 30.6 g NH3
Mixtures
Different from pure substances is that a mixture contains more than one kind particle
Polar molecules
Electrons are shared unequally - The more electronegative atom attracts electrons more strongly and gains a slightly negative charge
Energy and Chemical Equations
Endothermic change - A + energy = B Exothermic change A = B + energy
Heat of hydration / energy of hydrati0n
Energy that is released in the final step
Limiting Reactant Example 3
Given 160.5 g of sulfur and 268.8 g of oxygen gas, calculate the mass of SO3 gas produced. How many grams of the excess reactant remains? - 2S + 3O2 -> 2SO3 - S = 32.0 // O2 = 16.0 1. Equation is balanced 2. S = 160.5 x 1/32.0 = 5.0 mol S O2 = 268.8 x 1/32.0 = 8.4 mol O2 3. S = 5.0 mol S x 3 mol O2 / 2 mol S = 7.4 O2 O2 = 8.4 mol O2 x 2 mol S / 3 mol O2 = 5.6 S 4. Sulfur is Limiting // Oxygen is Excess 5. SO3 = 8.4 O2 x 2 mol SO3 / 3 mol O2 = 5.6 mol SO3 6. 5.6 mol SO3 x 80.0 g / 1 mole = 448 g/mol SO3
Limiting Reactant Example 1
Given 3.0 moles of methane and 4.0 moles of oxygen gas, calculate the moles of carbon dioxide gas produced - CH4 + 2O2 -> CO2 + 2H2O 1. Equation is balanced 2. CH4 = 3.0 moles CH4 x 2 moles O2 / 1 moles CH4 = 6 mol O2 OR O2 = 4.0 moles O2 x 1 moles CH4 / 2 moles O2 = 2 mol CH4 3. O2 is Limiting // CH4 is Excess 4. 4.0 O2 = 1 mol CO2/ 2 mol O2 = *2 mol CO2*
Limiting Reactant Example 2
Given 5.0 mol of acetylene and 11.0 mol of oxygen gas, calculate the *volume* of CO2 gas produced - 2C2H2 + 5O2 -> 4CO2 + 2H2O 1. Equation is balanced 2. C2H2 = 5.0 moles C2H2 x 5 mol O2 / 2 mol C2H2 = 12.5 O2 OR O2 = 11.0 moles O2 x 2 mol C2H2 / 5 mol O2 = 4.4 C2H2 3. O2 is Limiting // C2H2 os Excess 4. CO2 = 11.0 O2 x 4 mol CO2 / 5 mol O2 = 8.8 mol CO2 5. 8.8 mol CO2 x 22.4 = 197.12 L CO2
Example 1 Moles to Moles
H2 + N2 = HN3 How many moles of ammonia will be produced from the reaction of 6.0 moles of hydrogen gas? *Balance* 3H2 + 1N2 = 2NH3 *Molar Ratio* Need: Ammonia Given: Hydrogen 2 mol NH3/3 mol H2 *Multiply* mol NH3 = 6.0 x 2 mol NH3/3 mol H2 = 4.0 mol NH3 - 4.0 moles of ammonia will be produced from 6.0 moles of hydrogen gas
Example 2 Mass to Mass
How many grams of acrtylene (C2H2) are produced when 5.00 g of calcium carbide (CaC2) is added to water? CaC2 + H2O = C2H + Ca(OH)2 *Balance* CaC2 + 2H2o = C2H2 + Ca(OH)2 *Mass to Moles* Molar Mass of Calcium Carbide = 64.1 g/mol mol 5.00g x 1 mol / 64.1 g = 0.0780 mol CaC2 *Molar Ratio* Need: C2H2 Given: CaC2 Mol C2H2 = 0.0780 x (1 mol C2H2 / 1 mol CaC2) = 0.0780 mol C2H2 *Mass to Moles* Molar Mass of Acetylene is 26.0 g/mol Mass 0.0780 mol C2H2 x 26.0 g / 1 = 2.03 g C2H2
Solubility Example
If 25.0 g of a solute is the maximum amount of solute that can dissolve in 40.0 g of solvent at a certain temperature, what is the solubility in grams of solute /100 g of solvent 25.0 g solute / 40.0 g solvent x 100 g solvent = 62.5 g solute / 100 g solvent
Solubility Example 2
If 30.1 g of a solute can dissolve in 350.0 g of water at a certain temperature, what is the solubility of the substance in g/100 g water? 30.1 g solute / 350.0 g solvent x 100 g solvent = 8.60 g solute / 100 g water
why do solubility rules change for gases
Increasing temperature decreases the solubility of a gas in a liquid because gases respond to temperature changes to a much larger extent than liquids. Gas particles are situated far apart from another - an increased temperature increases the kinetic energy of the gas particles so that the liquid can no longer hold the gas particles in solution, and they "boil" out of the solution.
Dissociation Equation Example 2
Iron (III) sulfate 1. Ionic compound 2. Fe3 SO4 3. Fe2 (SO4) 3 -> Fe2O3 + SO2 + O2 4. Fe2 (SO4) 3 (s) -> Fe2O3 (aq) + SO2 (aq) + O2 (aq)
Emulsifying Agent
Keeps the suspension of liquids from separating from each other upon standing - PB, shampoo, margarine
Defining the Limiting Reactant
Limiting Reactant: the reactant that is completely consumed - limits the number of products that can be formed Excess Reactant: Remaining or left over
Immiscible
Liquid solutes and solvents that are not soluble in each other oil and water
Stoichiometry
Measuring the reactants and products of a reaction
colloids mixture
Mixture of particles that are smaller than those is suspensions but larger than the particles in solutions - cloudy/mily appearances meaning you cannot see through them - glue, whipped cream, paint
Dipolar molecules
Molecules with a somewhat negative end and a somewhat positive end. All diatomic molecules with a polar covalent bond are dipolar
parts per million (ppm)
Number of parts of a chemical found in 1 million parts of a particular gas, liquid, or solid.
Dissociation Equation Example 3
Potassium permanganate 1. Ionic Compound 2. KMnO4 3. KMnO4 -> K2O + MnO + O2 4. KMnO4 (s) -> K2O + MnO + O2
Dissolving process
Solvent particles pull the solute particles into solution and evenly spread them around
Concentrations
The amount of solute per unit of solutions (solute per solvent), unlike solubility that is the amount of solute per unit of solvent
Nonpolar bonds
The atoms in the bond pull equally - the bonding electrons are equally shared - includes diatomic molecules such as H2
Electronegativity
The attraction of a given atom for the electrons of a covalent bond.
boiling point elevation
The increase in temperature at which the solvent boils required to bring the vapour pressure up to the air pressure The magnitude of the boiling point is proportional to the number of solute particles dissolved in the solvent
Percentage weight by volume (%w/v)
The mass (in grams) of solute for every 100 mL of solution (weight/volume) w/v = grams of solute / 100mL solution * 100%
Percentage weight by weight (% w/w)
The mass (in grams) of solute in 100 g of solution (weight/weight) w/w = grams of solute / 100g solution * 100%
Parts per billion (pbb)
The number of grams of solute for every one billion grams of solution
Molar coefficients
The number of moles that combine or are produced
Regelation
The process of melting under pressure and the subsequent refreezing when the pressure is removed. - skating on ice, making a snow ball
The molar ratio
The ratio of the number of moles of each reactant and product in a balanced chemical equation. 2H2 + O2 = 2H2O 4H2 + 2O2 = 4H2O (doubled)
Percentage volume by volume (%v/v)
The volume (in mL) of solute for every 100 mL of solution (volume/volume) % v/v = mL of solute / 100 mL solution * 100%
Temperature and solubility
There is a decrease in solubility as temperature increases. - The solubility of most gases in liquid solvents is greater at colder temperatures then at warmer temperatures.
moles and mass types
Two types - The amount is given in moles and the amount needed is in grams - The amount given is in grams and the amount needed is in moles
Hydrated ions
ions chemically bonded to water molecules
Example 1 Volume to Mass
What Volume of O2 is need to given completely react with 25.0 g of FeS at STP 4FeS + 7O2 = 2Fe2O3 + 4SO2 *Balance* already balanced *Given quantity to moles* FeS = 87.9 g/mol Mol = 25.0 x (1 / 87.9) = 0.2844 mol FeS *Molar Ratio* Need: O2 Given FeS Mol O2 = 0.2844 mol FeS x 7 mol O2 / 4 mol FeS) = 0.4977 mol O2 *Moles to Volume* Vol O2 = 0.4977 mol x (22.4 L / 1) = 11.1 L O2 - 11.1 L of oxygen is needed to react with 25.0 g of FeS
Example 1 of Moles to Mass (moles given)
What mass of NaCl is formed from the reaction between 3.20 moles of chlorine gas and an excess of sodium, according the equation below Na + Cl2 = NaCl *Balance* 2Na + Cl2 = 2NaCl *Already in moles* 3.20 moles *Molar Ratio* Need: NaCl Given: Cl2 mol NaCl = 3.20 mol C2 x 2 mol NaCl/1 Cl2 = 6.40 mol NaCl *Moles to Mass* NaCl = 58.5 g/mol Mass 6.40 (Total from ratio) x 58.5 (Atomic Mass of NaCl) / 1 = 374 g NaCl - 374 g NaCl can be made from 3,20 moles of chlorine gas
Example 2 Volume to Mass
What mass of ammonia can be produced from 50.0 L of nitrogen gas at STP N2 + H2 = NH3 *Balance* N2 + 3H2 = 2NH3 *Convert given quanity to moles* Mol 50.0 L x 1 / 22.4 L (STP) = 2.232 mol *Molar Ratio* Need: NH3 Given: N2 Mol NH3 = 2.232 mol N2 x (2 mol NH3 / 1 mol N2) = 4.464 mol NH3 *Moles to Mass* Molar Mass of ammonia = 17.0 g/mol Mass NH3 = 4.464 mol x 17.0 g / 1 mol) = 75.9 g NH3 75.9 of ammonia is produced from 50.0 L of nitrogen gas at STP
Energy n Stoichiometry Example 2
What mass of water is produced when 8000. kJ of energy is released by the reaction 4NH3 + 3O2 = 2N2 + 6H2O + 300 kJ - For every 6 moles of water, 300. kJ are released 300 kJ/6mol H2O or 6 mol H20/300 kJ *Balance* already balanced *Use the energy ratio to determine the moles of the required substance Mass H20 = 8000. kJ x (6 mol H20/300. kJ) = 160. mol H20 *Convert the moles of the required substance to mass Molar Mass of water = 18.9 g/mol Mass H2O = (160. mol) x (18.0g/1) = 2880 g H20
Energy n Stoichiometry Example 1
What quanity of heat is produced in the complete combustion of 60.2 g of ethane (C2H6), if the heat of combustion is 1560 kJ/mol of ethane? 2C2H6 + 7O2 = 4Co2 + 6H2O The energy erm 1560 kJ/mol can also be written as an energy ratio: 1560kJ /1mol or 1 mol/1560kJ *Balance* already balanced *Given quantity to Moles* The molar mass of ethane is 30.0 g/ mol Mol = 60.2 g x (1 mol/30.0 g) = 2.007 mol *Use Energy ratio to determine energy release* - 1560 kJ is the amount of energy released for every one mole of ethane Energy = 2.007 mol x (1560 kJ / 1 mol) = 3130 kJ
Amalgam
a combination of diverse elements; a mixture - silver dental fillings
Molecular compound
a compound that is composed of molecules
Solubility Curve
a graphic representation of the variation with changing temperature of the solubility of a given substance in a given solvent
A concentration solution
a measure of the amount of solute in a given amount of solvent or solution The drink that tastes very sweet
Homogeneous mixtures
a mixture that is the same throughout - coffee with sugar
Dilute Solution
a solution that contains a small amount of solute relative to the solvent. The drink that tastes watery
Supersaturated Solution
a solution that contains more dissolved solute than a saturated solution contains under the same conditions Overbooked hotel
thermal pollution
a temperature increase in a body of water that is caused by human activity and that has a harmful effect on water quality and on the ability of that body of water to support life
Polar bonds (hydrogen fluoride)
a type of covalent bond between atoms that differ in electronegativity. the shared electrons are pulled closer to the more electronegative atom. making one slightly negative and the other slightly positive
Matter
anything that has mass and volume (takes up space)
Miscible
describes liquids that dissolve in one another in all proportions alcohol and water
Suspension mixture
heterogeneous mixture with large, often visible solutes that tend to settle out - flour and water
Solutions
homogeneous mixtures
Pressure and solubility
increase pressure, increase solubility - increasing pressure also increases the solubility of a gas in a liquid, because the increased pressure forces the gas into contact with the liquid. This added pressure forces large quantities of gas into solution and allows the liquid to hold onto the gas particles.
Hydrogen Bonds
intermolecular interaction among water molecules - stronger than the normal attractive forces that occur between neighbouring molecules
Solubility Equation
mass of solute / mass or volume of solvent x mass or volume of solvent needed
Alloys
mixtures of two or more metals
Heterogenous mixtures
sand in water, vegetable soup The sand mixes with the water as you swirl the mixture, but eventually settles out on the bottom of the container so that the individual grains of sand are easily distinguished from the water
Heat of solution
the amount of heat energy absorbed or released when a specific amount of solute dissolves in a solvent
freezing point depression
the difference in temperature between the freezing point of a solution and the freezing point of the pure solvent - the greater the amount of solute, the greater the freezing point depression (the lower the freezing point) Example: 100mL solution w/ 10g of salt = FP of -6C 100mL solution w/ 20g of salt = FP of -16C
Electron affinity
the energy change associated with the addition of an electron to a gaseous atom
Ionization energy
the energy required to remove an electron from an atom
Molarity
the number of moles of solute per litre of solution Concentration © = moles of solute / litres of solution
Solvation
the process by which the positive and negative ions of an ionic solid become surrounded by solvent molecules
Dissociation
the separation of positive and negative ions that occurs when an ionic compound dissolves
