General Chemistry Exam 2 Review – Huchital
∙ Lavoisier French scientist discovered the law of conservation of matter. Stating matter can not be created nor destroyed in a reaction.
10 g of reactants yield 10 g of products.
1000 atoms of A (reactants) yields 1000 atoms of A (products).
P4 (s) + Cl2 (g) PCl3 (l) *balance the equation* = P4 (s) + 6Cl (g) 4 PCl3 ∙ Chemical equations show the reactants and products in a reaction.
Example: 4Al (s) + 3 O2 (g) 2 Al2O3 (s)
Atoms balance. Moles do not.
∙ The numbers in front are stoichiometric coefficients.
∙ Stoichiometry: th relationship between amounts of reactants and products. Example: Pentane burns in the presence of oxygen to produce gaseous carbon dioxide and water.
C5H12 + 8O2 5CO2 + 6H2O
∙ Balance elements that appear only once on each side first.
∙ Balance polyatomic ions as a group.
Example: AlCl3 + (NH4)2CO3 Al2(CO3)3 + NH4Cl
Balanced : 2 AlCl3 + 3 (NH4)2CO3 Al2(CO3)3 + 6NH4Cl
∙ Balance free elements last.
∙ Reduce coefficients to smallest whole numbers.
∙ Formulas must be right, or the equation is meaningless.
∙ Subscripts can not be changed.
Examples: NO2 + H20 2HNO3 + NO
3NO2 + H20 2HNO3 +NO
∙ Some reactions are reversible.
N2 (g) + 3H2 (g) 2NH (g)
Under these conditions reactions can be written with double arrows to indicate equilibrium.
Equilibrium is when the amounts of reactants and products no longer change. ∙ Reactions that form a lot of product are – product favored. If you want to learn more check out What is the basic meaning of marketing?
∙ Reactions that for little products are – reactant favored.
∙ Reactions in aqueous solutions:
Ionic compounds dissolve in water to produce ions.
NaCl Na+ + Cl
Co(NO3)2 Co +2 + 2NO 3.
Electrolytes: solutions with ions.
Electrolytes conduct electricity.
Strong electrolyte means the solute 100 % dissociates into ions; product favored. (ex. CuCl2).
Weak electrolyte means there are more molecules than ions; partially dissociate, reactant favored. (ex. Acetic acid).
Non electrolytes = no ions. (ex. Sugar).
What happens when you mix NaCl and Ba(NO3)2?
What happens when you mix NaCl with Pb(NO3)2?
A white solid precipitate is formed and left over. PbCl2 (s) indicates the solid formed. Not all ionic compounds are soluble in water: AgCl and MgF2 can not be electrolytes. Don't forget about the age old question of What is the limit of resolution of microscope?
(aq) = aqueous.
∙ Soluble compounds:
almost all salts of Na+, K+, NH+4. Salts of nitrate (NO3), Chlorate(ClO3), perchlorate (CLO4), acetate (CH3CO2).
Almost all salts of Cl, Br, I. * Exceptions: halides of Ag+, Hg2+, Pb +2. Salts containing F. * Exceptions: fluorides of Mg +2, Ca+2, Sr+2, Ba+2, Pb+2. Salts of sulfate (SO4 2) *Exceptions: sulfates of Ca+2, Sr+2, Ba+2, Pb+2, Ag+. ∙ Insoluble compounds: We also discuss several other topics like Is a linear function continuous?
most salts of carbonate (CO3 2), phosphate (PO4 3), oxalate (C2O4 2), chromate (CrO4 2), sulfide (S2). * exceptions: all salts of NH4+ and the alkali metal cations. Most metal hydroxides and oxides. *Exceptions: alkali metal hydroxides and Ba(OH)2 and Sr(OH)2. If you want to learn more check out What is an example of biological perspective?
∙ Chemical reactions in water:
AX + BY AY +BX
The anions exchange places between cations.
A precipitate form if one of the products is insoluble.
The driving force is the formations of an insoluble solid aka a precipitate. Precipitates are determined form the solubility rules.
Water soluble reactants water insoluble product.
Example: Pb(NO3)2 +KI
Balance: Pb(NO3)2 (aq) + 2KI (aq) 2KNO3 (aq) + PbI2 (s)
Complete ionic equation: Pb +2 (aq) + 2NO3 (aq) + 2K +(aq) +2I (aq) 2K+ (aq) + 2NO3 (aq) + PbI2 (s).
Cancel the spectator ions (the ions that appear on both sides of the equation, to produce the net ionic equation.
Therefore, Pb+2 (aq) +2I (aq) PbI2 (s).
No precipitate = no reaction = no equation.
∙ Acids and Bases:
Acid: any substance that increased the H+ (proton) concentration when dissolved in water.
Base: any substance that increases the OH (aq) concentration in an aqueous solution. A strong acid (strong electrolyte) completely dissociates into its ions. Example: HCl + H20 H3O+ + Cl. We also discuss several other topics like What is the difference between the nervous system and the endocrine system?
(6) inorganic strong acids:
1) HCl hydrochloric acid
2) HBr hydrobromic acid
3) HI – hydroiodic acid
4) HNO3 nitric acid
5) HClO4 perchloric acid
6) H2SO4 sulfuric acid
A weak acid does not completely ionize in solution. Most acids are weak. Since weak acids ionize to such a small extent the molecular species predominates. Weak acids (weak electrolytes): HF hydrofluoric acid, H3PO4 phosphoric acid, H2CO3 carbonic acid, CH3CO2H acetic acid, H2C204 oxalic acid, H2C4H4O6 tartaric acid, H3C6H5O7 citric acid, HC9H7O4 aspirin.
Strong bases dissociate completely in water.
Strong bases (strong electrolytes): LiOH lithium hydroxide, NaOH sodium hydroxide, KOH potassium hydroxide, Ba(OH)2 – barium hydroxide, Sr(OH)2 strontium hydroxide. If you want to learn more check out What are the core aspects of marketing?
Weak bases partially dissociate in water. NH3 – ammonia, codeine, morphine. Monoprotic (1 acidic H); HCl, HNO3.
Diprotic (2 acidic H’s); H2SO4, H2CO3, H2C2O4.
Triprotic (3 acidic H’s); H3PO4 (PHOSPHORIC ACID), H3C6H5O7 (CITRIC ACID)
Bronsted Lowry definition:
Acid: any substance that donates H+ (aq) proton to another species in an aqueous solution.
Base: any substance that accepts an H+ (aq) proton in an aqueous solution. All strong acids completely transfer H+ at equilibrium. Product favored reaction 100% product formed with strong acid.
A weak acid does not completely transfer protons (H+). Reactant favored reaction 12% product formed with weak acid.
A weak base does not completely accept protons (H+). reactant favored reaction 2 % product formed with weak acid.
WATER CAN BE A BASE OR AN ACID depending on other partner. (Amphiprotic)
Example: HCl +H20 H3O + + Cl * water is a base to accept a proton. Example: NH3 + H20 NH4 + + OH *water is acid and donates a proton. Nonmetal oxides can form acids in aqueous solutions.
Metal oxides can form bases in aqueous solutions.
∙ Gas forming reactions:
Metal carbonate or hydrogen carbonate + acid metal salt +CO2 + H20. Metal sulfide + acid metal salt + H2S
Metal sulfide + acid metal salt + SO2 + H2O
Ammonium salt + strong base metal salt + NH3 + H2O.
Metal carbonates react with acids to form the corresponding metal salt, water and carbon dioxide gas.
∙ Oxidation reduction reactions:
Oxidation: adding oxygen
Reduction: removing oxygen.
Electrons are transferred between reactants.
Oxidation = loss of electrons.
Reduction = gain of electrons.
O.I.L.R.I.G: Oxidation Is Loss. Reduction Is Gain.
When one reaction loses electrons, another reactant must gain them.
Oxidation of one substance is always accompanied by the reduction of another. What is oxidized is the reducing agent aka reductant.
What is reduced is the oxidizing agent aka oxidant.
A substance is reduced is responsible for oxidation aka oxidizing agent, gains electrons, decrease in oxidation numbers.
A substance is oxidized is responsible for reduction aka reducing agent, loses electrons, increase in oxidation numbers.
Charges on bonded atoms are assigned by giving the more electronegative atom ALL the shared e.
Such charge assignments are called Oxidation numbers.
∙ Oxidation number rules:
1) Each atom in a pure element has an oxidation number of zero.
2) For monatomic ions, oxidation number = charge on ion.
3) F= always 1 when combined with other elements.
4) Oxygen = 2 in MOST cases.
5) Halogens = 01 except if combined with F or O.
6) H = +1 in MOST cases.
7) Sum of the oxidation numbers for the atoms in a neutral compound = zero. In a polyatomic ion = ion charge.
∙ Mole – Mole calculations:
Iodine and Fluorine react to form iodine trifluoride.
I2+ 3F2 2 IF3
∙ HCl + NaOH H2O + NaCl
1 molecule of HCl will react with 1 molecule of NaOH to produce 1 molecule of water and 1 molecule of salt.
6.02 * 10^23 molecules of HCl will react with 6.02 * 10^23 molecules of NaOH to produce 6.02 *10^23 molecules of water and 6.02 *10^23 molecules of salt. Therefore, 1 mole of HCl will react with 1 mol of NaOH to produce 1 mole of water and 1 mol of salt.
∙ To solve stoichiometric problems, you MUST go through moles using molar masses and mole rations as conversion factors.
Balance the chemical equations first.
Grams moles moles grams.
∙ Reactions with a limiting reactant:
Some cases there is insufficient amount of one reactant to consume others properly. The reactant that is in short supply LIMITS the quantity of product formed. The theoretical yield of products is limited by this Limiting Reactant. Nuts and bolts analogy: you have 32 nuts and 28 bolts. A set has 1 nut and 1 bolt ( N+B NB).
How many sets can you make? 28. You only have 28 bolts.
A set has 2 nuts and 1 bolt, 2N + B NB. How many sets can you make? 16. The bicycle:
1 handle bar, 1 frame, 2 wheels.
How many bicycles can you make with 10 handle bars, 12 frames, 22 wheels? 10. ∙ Reaction yields:
Theoretical yield: is the maximum product yield that can be expected based on the masses of the reactants and the reaction stoichiometry. Assume 100%, nothing lost, no other reactions, enough time.
Actual yield: aka experimental yield, actual amount of product obtained. Percent yield; a percentage. Actual yield/ theoretical yield X 100 % = percent yield. ∙ Solutions
Solution = solute + solvent
Solute: that which is dissolved (lesser amount)
Solvent: that which dissolves (greater amount)
∙ Describing solutions:
Limitations: solubility limits, saturated, unsaturated, supersaturated. Extremes: dilute (small ratio of solute to solvent), concentrated (large ratio) Actual amount: concentration, ratio of quantity of solute to quantity of solvent. Molarity = moles of solute/ volume of solution (L)
How many moles of Na3PO4 are there in 2.5 L of a 0.30 M solution? 2.5L X .3 mol/ 1L = 0.75 mol Na3PO4
∙ Concentration; ratio of quantity of solute to quantity of solvent.
Molarity = moles of solute/ volume of solution (L).
How many moles of Na3PO4 are there in a 2.5 L of 0.30 M solution? – 2.5l X .3 mol/ 1L= 0.75 mol Na3PO4
∙ Diluting solutions:
Moles do not change.
You can find the amount you must take from the concentrated solution to form the more difficult solution.
You can use mL or L.
How would you prepare 250 mL of a 0.500 M HCl solution from commercial concentrated HCl (12.0 M)?
(12.0)(Vc) = (.500M) (250Ml)= Vc = 10.4 mL.
pH scale = logarithmic scale.
Acids have lower pH levels, but higher H30 +
Bases have higher pH levels but lower H3O+.
A solution of pH = 3 is 1000 times more acidic than a solution of pH =6. Calculate pH: pH [H3O]+ ? = log [H3O+]
Quantitative analysis of a substance by complete reaction in a solution with a measured quantity of a reagent of known concentration.
Uses to determine concentration of unknown solution, molar mass of a substance, percentage of mass of an active ingredient in a sample.
Example: to find the HCl in a solution it was titrated with KOH. Data collected: KOH + HCl KCl + H2O
Volume of HCl= 50.0 mL
[KOH]= 0.400 M
Volume of KOH solution = 25.0 Ml
25ml X .400M = 50 mL (x)
X = .200 M HCl