chemistry

writing single replacement precipitation reactions

Check the activity series to see if a reaction occurs! Cl2 (g) K+ Br- Br2 (l) K+ Chlorine gas + potassium bromide--> bromine + potassium Cl- chloride MOLECULAR EQUATION: Cl2 (g) + 2KBr (aq) --> Br2 (l) + 2KCl (aq) Assign phases Write formulas Balance reaction Total ionic equation: break all aqueous compounds into their ions: Cl2 (g) + 2K+ (aq) + 2Br- (aq) --> Br2 (l) + 2K+ (aq) + 2Cl- (aq) Net ionic equation: cancel out all spectator ions and rewrite anything left over: Cl2 (g) + 2Br- (aq) --> Br2 (l) + 2Cl- (aq) note: halogens only replace halogens metals only replace metals

precipitation double replacement reactions

-to fall -Double replacement reactions are partner swap reactions between aqueous ionic compounds, where the cation and anion from each compound switch partners Ex: potassium iodide + lead (II) nitrate potassium nitrate + lead (II) iodide Both reactants will be aqueous (dissolved in water) -If the products are also aqueous, no reaction has occurred! -If at least one product is not aqueous (i.e., solid or liquid), a double replacement chemical reaction has occurred! -Precipitation reactions occur when at least one of the products is an insoluble solid called a precipitate

chemical vs. physical change

A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed

synthesis and exothermic reactions

Synthesis reactions are exothermic because new chemical bonds are formed. 2 K (s) + N2 (g) + 3 O2 (g) 2 KNO3 (s) + energy

synthesis

Synthesis: When a compound is formed from its individual elements. 2 K (s) + N2 (g) + 3 O2 (g) 2 KNO3 (s) Only one product! can only write molecular equations!

no reaction occurs if no precipitate forms

no new compounds or bonds just mixing two mixtures no actual chemical reaction

non aqueous reactions recap

non aqueous reactions are those that do not take place in an aqueous solution. there are three types Synthesis: The exothermic formation of a compound from its elements in their natural forms. 2K(s)+N2 (g)+3O2 (g)2KNO3 (s)+energy Decomposition: The endothermic breakdown of a compound into its elements in their natural forms. energy+2CaO(s)2Ca(s)+O2 (g) Combustion: The exothermic reaction of a hydrocarbon (CxHy) with O2 to give water and carbon dioxide. C3H8 (g)+5O2 (g)4H2O(l)+3CO2 (g)+energy

lab 5-1 precipitation double replacement reactions

purpose: investigate which double replacement reactions will result in the formation of a precipitate and which produce no chemical reaction. allow for the development of some basic solubility rules, which will be used to determine the identity of an unknown cation in solution double replacement reactions: many chemical reactions involve the interactions of ions in aqueous solutions. one such type of reaction is known as a precipitation double replacement reaction, in which aqueous solutions of two ionic compounds are allowed to react. for this type of reaction to occur, both of the original ionic compounds should be soluble (able to dissociate into their ions when dissolved in water), which will be indicated by the phase symbol (aq). once the solutions are mixed, the ions essentially change partners to form two new ionic compounds. the cation from the first compound will pair with the anion from the second compound and vice versa. if one of these products is insoluble (does not dissolve or dissociate in water), it is called a precipitate, which will be indicated in a chemical equation by the phase symbol (s). the presence of a precipitate will cause a previously transparent solution to become murky or cloudy, or a solis will be visible settling out in the bottom of the reaction flask. if both products are soluble, they remain dissociated in their ionic forms, and no chemical reaction occurs. an example of a precipitation double replacement reaction is given below word equation: lead (II) nitrate + potassium iodide --> lead (II) iodide + potassium nitrate molecular equation: Pb(NO3)3 (aq) + 2KI (aq) --> Pbl2 (s) + 2 KNO3 (aq) to get a net ionic equation: dissociate all aqueous compounds into their ions. this equation is called the total ionic equation. then cancel out any spectator ions. spectator ions are ions that are present in the exact form and among of both sides of the chemical equation. the total ionic and net ionic equations for the reaction above would be total ionic equation: Pb2+ (aq) + 2 NO3^1- (aq) + 2 K+ (aq) + 2I^1- (aq) --> Pbl2 (s) + 2K+ (aq) + 2NO3^ 1- (aq) net ionic equation: Pb2+ (aq) + 2 I^1- (aq) --> Pbl2 (s) for a double replacement in which both products are soluble, all ions would become spectator ions, and no reaction would have occurred solubility rules: over time, chemists have been able to analyze these patterns and develop a set of solubility rules which will allow you to predict whether or not mixing a given set of ionic compounds will result in the formation of a precipitate. be able to determine and utilize these solubility rules to solve several different types of problems. Also, use the solubility rules to devise a strategy to determine which cation is present in an unknown solution in the lab, given a set of 9 pipets, which each contain a different aqueous ionic compound. some precipitates are easier to see on a dark background, while others are more visible against a white background. when mixing the solutions, you should place the drops on the border between the black and white sections of the boxes -place all drops of a given solution before switching to the next pipette -to is important to hold pipette vertically when dispensing the drops- ensure consistent volume delivery -never allow the tip of the pipette to come in contact with the drops on the chart- prevent cross-contamination, and ensure accurate results analysis: precipitate formation reactions (precipitates observed, cations and anions that are never precipitates, wiring molecular geometry for 5 chosen precipitates) and determining the unknown cation and how you arrived at the conclusion and writing name and chemical formulas of precipitates and writing molecular equations, total ionic equations and net ionic equations refer back to lab/chart

writing chemical equations

Chemical equations are written, shorthand expressions that describe chemical reactions. For example: Hydrogen combines with oxygen to form water. H2 + O2 --> H2O 2 hydrogen atoms- 2 hydrogen atoms 2 oxygen atoms - 1 oxygen atom this equation is not balanced ^. we must fix the ratios you can never change the substance! you have to change the atoms that come before (left of arrow)- ex) H2 and O2 2 H2 + O2 --> 2H2O 4 hydrogen atoms- 4 hydrogen atoms 2 oxygen atoms-2 oxygen atoms now the reaction is balanced integer coefficients are used to balance chemical equations H2 + 1/2 O2 --> H2O 2 hydrogen atoms- 2 hydrogen atoms 1 oxygen atoms- 1 oxygen atoms For diatomic elements, fractions can be use Hydrogen, oxygen, fluorine, bromine, nitrogen, chlorine, iodine

combustion and exothermic reactions

Combustion reactions are exothermic, as evidenced by heat and explosion. C3H8 (g) + 5 O2 (g) 4 H2O (l) + 3 CO2 (g) + energy

combustion

Combustion: Reaction of a hydrocarbon with oxygen gas to give water and carbon dioxide. C3H8 (g) + 5 O2 (g) 4 H2O (l) + 3 CO2 (g) Always a reaction with O2 to give H2O and CO2! CHO order- carbon, hydrogen, oxygen reactants aren't ions can only write molecular equations

review of previous reaction types

DOUBLE REPLACEMENT (PARTNER SWAP) REACTIONS Precipitation: Two aqueous compounds combine to form at least one solid product. Na2SO4 (aq) + Ba(OH)2 (aq) BaSO4 (s) + 2 NaOH (aq) Acid Base: Combination of an acid and base to form water and an ionic compound (salt). HCl (aq) + KOH (aq) H2O (l) + KCl (aq) SINGLE REPLACEMENT REACTIONS Metal / Hydrogen Replacement: A more active element replaces a metal ion in a compound. 2 Al (s) + 3 CuCl2 (aq) 2 AlCl3 (aq) + 3 Cu (s) Halogen Replacement: A more active halogen replaces another halogen in a compound. F2 (g) + 2 KBr (aq) 2 KF (aq) + Br2 (l)

decomposition and endothermic reactions

Decomposition reactions are endothermic since energy input is required to break bonds. energy + 2 CaO (s) 2 Ca (s) + O2 (g) need to put in energy to break the bonds

decomposition

Decomposition: When a compound decomposes back into its individual elements. 2 CaO (s) 2 Ca (s) + O2 (g) Only one reactant! not ions! breaking a compound into its individual elements and how they exist in their natural form can only write molecular equations

Unit 5 tips

Don't forget phases! Don't forget to balance the equation before you do the total and net ionic equations! Don't forget charges in Total ionic equations and net ionic equations! Don't break gas, liquid, or solid into its ions- leave them exactly the same- break aqueous compounds into their ions! Make sure the ratio is the smallest, most reduced it can be! Make sure to put parentheses around compounds that have two capital letters- ex) OH --> (OH)2 HCl- hydrochloric acid HBr- hydrobromic acid HI- hydroiodic acid HNO3- nitric acid H2SO4- sulfuric acid aluminum- Al3+ zinc- Zn2+ silver- Ag+ ammonium- NH4+ hydroxide- OH- carbonate- CO3(2-) nitrate- NO3(-) sulfate- SO4(2-) phosphate- PO4(3-)

how to represent chemical reactions

Example: 2 NaCl (aq) + Pb(NO3)2 (aq) (reactants) --> PbCl2 (s) + 2 NaNO3 (aq) (products) -The reactants are the compounds mixed together to cause a chemical reaction. -The products are the resulting compounds formed as a result of a chemical reaction. -The phases of each compound are indicated in parentheses: -(s) solid, (l) liquid, (g) gas, (aq) aqueous (soluble / dissolved in water) -Each element or molecule is represented with its one correct chemical formula. -Coefficients are used in front of chemical formulas to indicate how many are needed in order to balance the chemical reaction.

exothermic reactions

In chemical reactions, energy is either released or absorbed. Reactions that release energy to the surroundings are called exothermic reactions. 2 C4H10 (g) + 13 O2 (g) 10 H2O (l) + 8 CO2 (g) + energy energy on product side- exothermic exothermic gives away heat energy

Identifying Oxidation Reduction (Redox) Reactions

In redox reactions, the oxidation numbers of the oxidized/reduced atoms change. Oxidation numbers are an accounting system that represents the "charge" on atoms. Ox. numbers are assigned on a per atom basis. 0 +2 -1 +2 -1 0 Ex: Mg (s) + CuCl2 (aq) MgCl2 (aq) + Cu (s) Mg loses electrons / is oxidized. (0 --> +2) Cu gains electrons / is reduced. (+2 --> 0) 0 0 +3 -1 Ex: 2 Fe (s) + 3 Cl2 (g) 2 FeCl3 (s) Fe loses electrons / is oxidized. (0 --> +3) Cl gains electrons / is reduced. (0 --> -1) guidelines for assigning ox. numbers -Elements in their natural state: 0 -Monatomic ions: same as charge -In compounds: Group I, 2, 17 ions: same as charge -Hydrogen: usually +1 -Oxygen: usually -2 -The sum of all oxidation numbers in a compound or ion must equal its overall charge! Notes: reducing and oxidizing agents are on the reactant side, will only be 2 atoms whose numbers change, add electrons to more positive side Same 2 electrons- same number

precipitation reaction

K+ and NO3^- do not participate in any reaction. they are spectator ions (aq) means aqueous (dissolved in water) new compounds and bonds chemical reaction takes place spectator ion: an ion that exists in the same form on both the reactant and product sides of a chemical reaction

Oxidation Reduction (Redox) Reactions

Oxidation reduction (redox) reactions are reactions in which electrons are transferred from one reactant to another. Example: Zn + Cu2+ Zn2+ + Cu Zn is giving its electrons to Cu2+ Cu gaining negativity Zn is the reducing agent and Cu is the oxidizing agent Each Zn atom loses two electrons and its charge changes from 0 --> +2. Each Cu2+ ion gains two electrons and its charge changes from +2 --> 0. OIL Oxidation is the loss of electrons. Zn is oxidized above. RIG Reduction is the gain of electrons. Cu is reduced above. ^redox reactions include both (OIL and RIG) The atom that is oxidized loses its electrons and gives them to the atom that is reduced. Reducing and oxidizing agents are the molecules giving or taking electrons in the reaction. Whatever is oxidized is the reducing agent and whatever is reduced is the oxidizing agent. -add electrons to more positive side -electrons are negative -electron transfer reaction

acid/base double replacement reactions

Reactions of acids with bases are called acid/base or neutralization reactions The products of acid/base reactions are always water + an ionic salt: acid + base --> water + an ionic salt hydrochloric acid + lithium hydroxide --> water + lithium chloride Acids produce H+ ions in aqueous solution. Bases produce OH ions in aqueous solution. Ex: HCl (aq) H+ (aq) + Cl (aq) Ex: Ca(OH)2 (aq) Ca2+ (aq) + 2 OH (aq) Because a non aqueous product, H2O (l), is formed in acid/base reactions, they always occur (i.e. they are never no reaction). Name of acid and formula: Hydrochloric acid: HCl Hydrobromic acid: HBr Hydroiodic acid: HI Nitric acid: HNO3 Sulfuric acid: H2SO4 Name of base and formula: Sodium hydroxide: NaOH Lithium hydroxide: LiOH Potassium hydroxide: KOH Calcium hydroxide: Ca(OH)2 Barium hydroxide: Ba(OH)2

endothermic

Reactions that require energy from the surroundings are called endothermic reactions. energy + 2 KNO3 (s) 2 KNO2 (g) + O2 (g) energy on reactant side- endothermic

instructions for double replacement reactions

STEP 1: PREDICT THE NAMES OF THE PRODUCTS To figure out the products, do your "partner swap" and then write out the names of the products. Remember to keep the name of the metal (or ammonium) cation first. Step 2: WRITE THE CORRECT FORMULA & PHASE FOR EACH COMPOUND Use the ion charges you memorized in Unit 1 to write the formulas for the compounds. USE CORRECT SUBSCRIPTS! (Remember there is NO SUCH THING as Na2Cl2 or Pb(II)CO3. Ionic compounds have the lowest values of subscripts that allow for charge balancing, and NO roman numerals are used in formulas EVER). Use the solubility rules to determine phases for all compounds! Step 3: BALANCE THE FULL EQUATION Now, you can use coefficients to balance the reaction. DO NOT CHANGE SUBSCRIPTS! Step 4: CHECK SOLUBILITY RULES & BREAK UP ONLY AQUEOUS COMPOUNDS Split only SOLUBLE (aqueous) compounds into ions, and DO NOT split insoluble compounds. In our example, barium phosphate is insoluble. Remember that ionic compounds will split into individual ions (just as you memorized them). Subscripts will be multiplied by the coefficients to find the total number of each ion. Ions such as Na3+, Na2+, and Cl2‐ DO NOT EXIST. PLEASE go back and memorize the ions and their charges from Unit 1. Step 5: CANCEL OUT SPECTATOR IONS & GET TO HEART OF THE REACTION Cancel out all ions that exist in exactly the same number and charge on both sides of the reaction. Re‐ write the equation without these spectator ions. (cross out spectator ions from the total ionic equation to the get the net ionic equation- make sure it's the simplified (smallest) ratio that the ions can be in in that specific reaction) Double‐check to make sure atoms and overall charge are balanced (equal on both sides of reaction). You are all done!!

single replacement reactions (metals/hydrogen)

Single replacement (SR) reactions occur when a neutral element replaces the ion of another element that is already in a compound A + BC --> AC + B (A is element by itself and BC is an ionic compound) Metal / Hydrogen SR reactions involve metallic elements (or hydrogen gas) reacting with an ionic compound or acid SR reactions ONLY OCCUR IF the neutral element (A) is higher in the activity series than the metallic ion (B) in the compound (BC) If this isn't the case, there is NO REACTION activity series: Li K Ba Ca Na Mg Al Zn Cr Fe Cd Co Ni Sn Pb H Cu Hg Ag Pt Au (higher up to be in a "couple"- highest would be Li)

single replacement reactions (halogens)

Single replacement (SR) reactions occur when a neutral element replaces the ion of another element that is already in a compound A2 + BC --> BA + C2 Halogen SR reactions involve diatomic halogens reacting with a halogen containing ionic compound SR reactions ONLY OCCUR IF the lone halogen (A2) is higher in its activity series than the halogen ion (B) in the compound (BC). If this isn't the case, there is NO REACTION. Halogen Activity Series- how it is on the periodic table 9 F 19.0 17 Cl 35.5 35 Br 79.9 53 I 126.9

rules for writing balanced chemical equations + more practice

To write and balance a chemical reaction, you must: -Write the correct formula for each reactant and product. Never change these formulas! -Use integer coefficients in front of the correct formulas so that the same number of each type of atom are on each side of the chemical equation (i.e., so mass is conserved!). Double replacement reactions occur when the ions from two ionic (metal + non metal) compounds swap places: REMEMBER METAL COMES FIRST IN AN IONIC COMPOUND) ex) silver would come before chloride- so it would be silver chloride barium chloride + silver nitrate --> barium nitrate + silver chloride (Ba2+ is barium, Cl- is chloride, Ag+ is silver and NO3- is nitrate) BaCl2 + 2AgNO3 --> Ba(NO3)2 + 2AgCl chloride and nitrate are switched ammonium sulfate + aluminum chloride --> 3(NH4)2 SO4 + 2AlCl3 --> 6NH4Cl + Al2(SO4)3 (NH4+ is ammonium, SO4^2- is sulfate, aluminum is Al3+, chloride is Cl-) (ammonium chloride + aluminum sulfate) iron(III) bromide + sodium phosphate FeBr3 + Na3PO4 --> FePO4 + 3NaBr (iron (III) phosphate + sodium bromide) (iron (III) is Fe3+, bromide is Br-, sodium is Na+, and phosphate is PO4^3-) SIDE NOTE: If it is iron (III) it would be Fe3+, if it was iron (II) it would be Fe2+

solubility rules

We use solubility rules to determine whether an ionic compound will be (aq) or (s) in water. compound is soluble if it contains -group 1 ions (ex alkali metal or hydrogen ions) -ammonium ions -nitrate ions -chloride, bromide,or iodide ions with the exceptions of compounds involving silver mercury (II), or lead (II) ions -sulfate ions (SO4-^2-)with the exceptions of silver, mercury (II), or lead (II), calcium, strontium or barium ions aq= soluble=dissolved compound is insoluble if it contains -hydroxide with exceptions of compounds involving barium ions -carbonate ions -phosphate ions -sulfide ions (S2-)

practice with balancing equations

Zn + 2HCL --> ZnCl2 + H2 3Pb(NO3)2 + 2AlCl3 --> 3PbCl2 + 2Al(NO3)3 C3H8 + 5O2 --> 3CO2 + 4H2O CHO- go in that order if you have a C, H, and O in an equation

writing double replacement acid/base reactions

acid/ base products are always water + salt nitric acid + barium hydroxide --> water + barium nitrate write all formulas: (nitric acid- HNO3- H+ and NO3-) (barium is Ba2+ and hydroxide is OH-) assign phases and balance reaction: 2HNO3 (aq) + Ba(OH)2 (aq) --> 2H2O (l) + Ba(NO3)2 (aq) total ionic equation: break all aqueous compounds into their ions -- do not separate liquid into its ions or give it charges) 2 H+(aq) +2 NO3- (aq) + Ba2+ (aq) + 2OH- (aq) --> 2H2O (l) + Ba2+ (aq) + 2NO3- (aq) net ionic equation: cancel out all spectator ions and rewrite anything left over: 2H+ (aq) + 2OH- (aq) --> 2H2O (l) but can cancel out 2's to make coefficients completely reduced: H+ (aq) + OH- (aq) --> H2O (l) if you do the other problem you end up getting the same exact net ionic equation 95% of acid/base reactions will have H+ (aq) + OH- (aq) --> H2O (l) as their net ionic equation because they are making water

writing single replacement precipitation reactions

check the activity series to see if a reaction occurs! potassium bromide + iodine solid--> NO REACTION-- bromine and iodine are competing, but bromine is already higher up and is already in the compound

writing single replacement precipitation reactions

check the activity series to see if a reaction occurs! (Cu) (Ag+) (NO3-) (Cu2+) copper solid + silver nitrate- silver solid and copper (III) (NO3-) nitrate molecular equation: -assign phases -write formula -balance reaction Cu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu (NO3)2 (aq) total ionic equation: break all aqueous compounds into their ions Cu (s) + 2Ag+ (aq) + 2NO3- (aq) --> 2Ag (s) + Cu2+ (aq) + 2NO3- (aq) net ionic equation: cancel out all spectator ions and rewrite anything left over Cu (s) + 2Ag+ (aq) --> 2Ag (s) + Cu2+ (aq) Cu (s) to Cu2+ (aq) --> Cu metal dissolved 2Ag+ (aq) to 2Ag (s) --> Ag solidified copper solid is a neutral element (no charge) and silver nitrate is an ionic compound (charges) B when gets "kicked" out- its charge is stripped away A when forms compound with C, gains a charge (no charge on solid)

writing single replacement precipitation reactions

check the activity series to see if a reaction occurs! zinc metal + magnesium bromide --> zinc + magnesium competing- magnesium higher up already (one in compound) --> so, no reaction NO REACTION!

writing single replacement precipitation reactions

check the activity series to see if a reaction occurs! HCl Mg hydrochloric acid + magnesium metal --> hydrogen gas + magnesium chloride (something hydroxide for base) (red on periodic table means gas) molecular equation: -assign phases -write formulas -balance reaction 2HCl (aq) + Mg (s) --> H2 (g) + MgCl2 (aq) if the higher one is already in the compound, then there will be no reaction total ionic phases: break all aqueous compounds into their ions 2H+ (aq) + 2Cl- (aq) + Mg (s) --> H2 (g) + Mg2+ (aq) + 2Cl- (aq) net ionic equation: cancel out all spectator ions and rewrite anything left over 2H+ (aq) + Mg (s) --> H2 (g) + Mg2+ (aq) 2H+ (aq) to H2 (g) --> forms bubbles Mg (s) to Mg2+ (aq) --> Mg (s) dissolves

lab 5-2: the metal/ hydrogen activity series

purpose: the purpose of this experiment is to test the activity of a variety of metals and hydrogen, and use the data to generate a partial activity series for the elements tested in a single replacement reaction, an element replaces an ion in a compound. Metallic elements and their ability to replace a different metal cation (or the hydrogen ion) in an aqueous compound. 2Fe (s) + 6HCl (aq) --> 2FeCl3 (aq) + H2 (g) Elemental iron replaces the hydrogen ion in hydrochloric acid, producing iron (III) chloride and hydrogen gas. But, not all single replacement reactions will occur. In order for this type of action to take place, the elemental metal must be "more active" than the metal it is replacing. The activity of a metal is related to the periodic property of metallic character- a metal can replace a metal cation if it is has a greater metallic character. Remember that metallic character is how easily an element will lose electrons in a chemical reaction. In order for the element to replace the ion, it much lose electrons (become oxidized) while the cation gains electrons (is reduced). Thus, the element that is more likely to lose electrons is more active and can replace the less active metal cation, if the elemental metal is less active, there is no reaction Al (s) + MgCl2 (aq) --> no reaction Wrote own method in google doc- lab 5-2 pre-lab assignment Analysis is creating an activity series for the unknown metals (X, Y, Z and the known metals Ag, H, Na, Al) and using the activity series to come up with possible identities for metals X, Y, and Z, and questions about metal activity (what happens to the electron count of the more active metal and what happens to the electron count of the less active metal when a more active metal replaces a less active metal) and doing molecular equations, total and net ionic equations for 5 different single replacement reactions that were "no reaction" refer back to lab/chart

writing double replacement precipitation reactions

silver nitrate + magnesium bromide --> silver bromide + magnesium nitrate (predict products via partner swap) molecular: -correct formulas -phases -balance equation write all formulas: Ag+, NO3^-, Mg2+, Br- assign phases and balance reaction: 2AgNO3 (aq) + MgBr2 (aq) --> 2AgBr (s) + Mg(NO3)2 (aq) total ionic equation: break all aqueous compounds into their ions: -- do not separate solid into its ions or give it charges 2Ag+ (aq) + 2NO3- (aq) + Mg2+ (aq) + 2Br- (aq) --> 2AgBr (s) + Mg2+ (aq) + 2NO3- (aq) net ionic equation: cancel out all spectator ions and rewrite anything left over: rewrite total ionic equation and cancel out spectator ions: 2NO3-, Mg2+, 2NO3- 2Ag+ (aq) + 2Br- (aq) --> 2AgBr (s) but can make all 2's 1's because smallest ratio and all coefficients have to be completely reduced Ag+ (aq) + Br- (aq) --> AgBr (s)

lab 5.3 - cu again- a copper cycle

the purpose is to perform a series of different reactions involving copper, beginning with an ionic compound containing the copper (II) ion an ending with a reaction that will produce copper metal. key to this lab activity are (1) recognizing the evidence that chemical reactions have taken place and (2) verifying that mass is conserved during a chemical reaction hydrogen compounds- certain ionic compounds are "hydroscopic", which means they have a tendency to absorb water molecules from the atmosphere. these water molecules are integrated into their crystal structures, thereby forming what are called "hydrated crystals" or "hydrated salts." the same compounds without water are referred to as "anhydrous crystals" or anhydrous salts" different salts absorb different amounts of water, but the ratio of the molecules of water absorbed to the molecules of the anhydrous (dry) salt usually works out to be a whole number -the formula of a hydrated crystal is written as the formulas of the salt * n H2O, where n represents the whole number of molecules of water absorbed per molecule of anhydrous salt. the * symbol is not a multiplication symbol; it is simply a punctuation mark meaning "and" that links the formula of the salt with how many water molecules are attached to it ex) Ni(NO3)2 * 6 H2O (refer back to lab to see exactly how it is written copper reactions- in this lab, you will convert copper (II) nitrate trihydrate into copper (II) hydroxide, then copper (II) oxide, then copper (II) chloride and finally copper metal. chemical reactions are often accompanied by the formation of a precipitate, evolution of gas, change in color, or pronounced temperature change. you should watch for these signs of a reaction as you proceed with this exercise and record any observations in your data table percent yield- recall that the law of conservation of mass states that in a chemical reaction, atoms are conserved (neither created nor destroyed). ideally, the number of copper atoms that you initially mass into a test tube should equal the number of copper atoms left in the test tube at the end of the experiment. in addition to writing balanced equations for all the reactions you observe, you will be asked to calculate the initial mass of copper you reacted and compare that to the mass of copper remaining at the end of the exercise. the theoretical yield is the mass of copper that you started with, and therefore, should obtain after carrying out these reactions. the actual yield is the mass of the copper actually obtained at the end of the experiment. the percent yield is a comparison of the actual yield and the theoretical yield % yield= (actual yield/ theoretical yield) *100 for the analysis, we are writing word equations, as well as balanced equations, identifying the type of reaction and citing evidence, and writing net ionic equations, calculating the ideal percent composition, and determining mass, calculating actual yield, the percent yield, describing a piece of error, comparing masses and why it changed or stayed the same refer back to lab/chart