Chemistry Q3 Exam
Find the number of mole ratios in a chem. eqn.
#mole ratios= n(n-1) n-# of substances Use this to know how many combos there are and make ratios
Percent yield
%yield= actual/theoretical x100
Ex. Solid silver chromate (Ag2CrO4) forms when potassium chromate (K2CrO4) is added to a solution containing 0.500g silver nitrate (AgNO3). Determine theoretical yield of Ag2CrO4 and calculate %yield of rxn. produces 0.455g Ag2CrO4. K2CrO4(aq)+2AgNO39(aq)-->Ag2CrO4(s)+2KNO3(aq)
*0.455g Ag2CrO4 is the actual yield (experimental yield) Set up dim analysis using 0.500gAgNO3. Multiply 1molAgNO3/169.872gAgNO3 (molar mass). Then multiply 1 mol Ag2CrO4/2 mol AgNO3 (mole ratio). Finally, multiply 331.728gAg2O4/1molAg2CrO4 (molar mass). This answer will be your theoretical yield. Theoretical yield= 0.488gAg2CrO4 %yield= 0.455g/0.488gx100= 93.2%
Ch 11 Stuff to Know via Hermann
*Some reaction eqns are balanced and some are not. You cannot trust that they will be balanced! -Stoichiometry -Mole ratios-> what they are and how to get them -Mole->mole -Mole->mass -Mass-> -Multiple choice that is essentially fill in the blank dimensional analysis. The dim. analysis is set up and you have to fill in the missing ratio -Limiting reactants->make up two short answer questions 1 is just finding the limiting reactant and 1 requires you to go further with it. For one of them, be able to explain in words why the LR you chose is the LR. You also have to find the theoretical/percent yield for one of the LR short answers
Ch 10 Stuff to Know via Hermann
-Convert between moles and atoms/molecules/formula units (avogadro's #) -Calculate molar mass -Convert between mass and moles -Convert between atoms and mass -Empirical and Molecular formulas -% composition -hydrates -Chapter 10 definitions +"Essay" question (about 4 sen.) is about material from chapter 10. We "will be able to write something"
Ch 13 Stuff to Know via Hermann
-Gas laws +Formula sheet is on exam (like test) -Boyle's, Charles', Gay-Lusaac's, and Combined gas laws -22.4 L/1 mol @STP!!! remember this value! -Convert between Celcius and Kelvin -Gas stoichiometry +Be able to do at STP or not at STP -Volume->Volume -Volume->Mass -Mass->Volume -Volume->Mass @STP +Short answer-gas stoichiometry -In order to do this problem, must be able to write rxn eqn and balance-> it's a combustion reaction +Short answer only writing- 4 sen- We "will be able to write something" *The info for this short answer is from Ch 10
R values (will be given)
0.0821 -> Lxatm/molxK 8.314 -> lxkPa/molxK 62.4 -> LxmmHg/molxK
Rounding Rules for emp. formula problems
0.999- rounded up to next whole number 0.0001- round down to nearest whole number 0.5- multiply all subscripts by 2 (mole ratio values) 0.333- multiply all subscripts by 3
Avogadro's number
1 mol=6.022x10^23 representative particles *representative particles: -atoms ex. Cu -molecules ex. H20 -formula units ex. NaCl
Molar mass of CCl2F2
1(12.011)+2(35.453)+2(18.998) = 120.913 1xmolar mass of C + 2xmolar mass of Cl + 2xmolar mass of F
Steps in LR problem (I'm not typing an example problem because i'm lazy. Also sorry for the bad explanation walk through)
1. Determine moles of reactants Use given mass values to set up dim analysis and convert to moles. The values you get are the mol amounts at the start of the reaction. 2. Using mole ratios, calculate moles of product possible for each reactant. Use each mole value from step 1 to set up a dim analysis and multiply using mole ratios. The LR is the reactant that had the least amount of product formed, and the ER is the other reactant. 3. Determine mass of product formed. Set up dim analysis using dim analysis answer from step 2 (smallest answer). Multiply by molar mass/1mol. Your answer will be the mass of product formed 4. Determine mass of excess left ove. Use the LR value, set up a dim analysis and multiply mole ratio. Finally, multiply by molar mass/1mol. This value is the amount of substance that reacted. Subtract this value from the original mass value given in the problem, and your answer is the amount of substance left over.
Mole ratios for CCl2F2
1molC/1molCCl2F2, 2molCl/1molCCl2F2, 2molF/1molCCl2F2 *moles do NOT equal mass mol=amount
Ratios
6.022x10^23 particles/1 mol 1 mol/6.022x10^23 particles
Hydrate
A compound that has a specific number of water molecules bound to its atoms ex. Opal SiO2 with a certain number of H2O molecules attached -formulas: Na2Co3x10H2O -het samples to drive off water (dehydrate); use before and after data to determine amt. of H2O lost (compound is called anhydrous after heating)
Standard Temperature and Pressure (STP)
A temperature of 273 K (0 degrees celcius) and a pressure of 1.00 atm *Gas Stoichiometry problems at STP ALWAYS require 22.4L/1mol or 1mol/22.4L
CCl2F2
C- 1 atom- 1molC -> 1mol of C in every mol of the compound Cl-2 atoms- 2molCl F-2 atoms- 2molF
Ex. Determine the emp. form. for methyl acetate, which has the following chem. analysis: 48.64% carbon, 81.6% hydrogen, 43.20% oxygen
Change each of the percent values to grams (48.64%->48.64 g) Set each value up in an individual dimensional analysis (3 in this case) and multiply each value by 1mol/molar mass You will end up with 3 different mole values. +4.05molC +8.095molH +2.700molO Divide each value by the lowest value you get (Divide each value by 2.700 in this case) you will then get 3 values which will end up being the subscripts for your formula. If they do not end up as whole numbers, use the rounding rules (see next card) to make them whole. + 3 (C) + 6 (H) + 2 (O) Emp. form. C3H6O2
Mass->Moles
Chem eqn given Mass: Known Moles: ? Mole ratio Set up dim analysis with given mass value. Multiply 1mol/molar mass, and then mulitply mol ratio
Setting up a combustion reaction
CxHy+O2(g)-->CO2+H2O balance carbons, then hydrogens, then oxygens
Extra note about molec. formula problems
Do all of the steps from the empirical formula problems and then divide exp. molar mass/molar mass of emp. form.
Ex. Acetylene and Benzene both have a CH as emp. form but drastically different properties. Experimentally determined molar mass of acetylene is 26.04 g/mol and benzene is 78.12 g/mol. What are the molecular formulas for each? Molar mass of emp. form. is 13.019 g/mol
Experimental molar mass of compound/ molar mass of emp. form Acetylene: 26.04 g/mol/13.019g/mol= 2 +2 is the multiplier for emp. form. subscripts 2(CH) -> Molec. form. for acetylene = C2H2 Benzene: 78.12g/mol/13.019g/mol=6 Molec form for benzene= C6H6
Calculate the number of each element in 1.25molC6H12O6.
Find mole ratios for carbon, hydrogen, and oxygen Set up 3 separate dimensional analyses For each separate problem, multiply 1.25molC6H12O6 by the mole ratio for the substance you're looking for ex. Carbon's mole ratio: 6molC/1molC6H12O6 Answers: C= 7.50molC H= 15.0molH O= 7.50molO
Mass->Mass
Given Chem eqn Mass 1: Given Mass 2: ? Mole ratio Set up dim analysis using known mass. Multiply by 1mol/molar mass. Multiply mole ratio. Finally, multiply molar mass/1mol
Volume->Volume
Given chem eqn Volume 1: Given Volume 2: Not given Find volume ratio: unknown L/known Set up dim analysis with known volume. Multiply volume ratio to get answer.
Volume->Mass
Given chem eqn Volume: Known Mass:? Volume ratio Start setting up what looks similar to information from an ideal gas law problem. Write down given volume (V), pressure (P), and temperature (T). Remember that you are looking for mass. Set up a dim analysis with your given volume. Multiply by volume ratio. Now, set up an ideal gas problem (PV=nRT) and solve for moles. *Remember the volume you use in the ideal gas problem is the one you found, no the one given to you in the problem Next, set up a dim analysis with the mole value and multiply molar mass/1 mol to get mass.
Mole->Mass
Given chem eqn. Moles: Given Mass: ? Mole ratio: unknown/known Set up dim analysis with given mole value. Multiply mole ratio, and then multiply by molar mass/1mol
Mole->Mole
Given chem. eqn. Known mol: Given (obviously) Unknown mol: ? Find mole ratio: unknown/known Set up dimensional analysis using given mole value. multiply by mole ratio
Volume->Mass at STP
Given rxn eqn Volume: given Mass: ? volume ratio: unknown/known +22.4L/1mol Set up dim analysis with given volume. Multiply volume ratio. Then multiply 1mol/22.4L. Finally, multiply molar mass/1 mol to get answer.
Percent composition problem example: Calculate the % comp. of phosphoric acid
H3PO4 molar mass: 3(1.008) + 30.974 + 4(15.999) = 97.994 %massH= 3(1.008)/97.994 x100= 3.09%H %massP= 30.974/97.994 x100= 31.61%P %massO= 4(15.999)/97.994 x100= 65.31%O
Ex. What volume of O2 is needed for the complete combustion of 4.00L of propane (C3H8)? C3H8(g)+5O2(g)-->3CO2(g)+4H2O(g)
Known volume: 4.00L C3H8 Unknown volume: L O2 volume ratio: 5L O2/1L C3H8 Set up dim analysis with 4.00L C3H8. Multiply by 5L O2/1L C3H8. Answer: 20.0LO2
Ex. How many moles of CO2 is produced when 10.0 mol C3H8 is burned in a gas grill? C3H8(g)+5O2(g)-->3CO2(g)+4H2O(g)
Known: 10.0molC3H8 Unknown: ? Mole ratio= unknown/known= 3molCO2/1molC3H8 Set up dim analysis with 10.0molC3H8 and multiply 3molCO2/1molC3H8 Answer: 30.0molCO2
Ex. Ammonium nitrate produces dinitrogen oxide gas and water when it decomposes. Determine the mass of water produced from the decomposition of 25.0g of solid ammonium nitrate. NH4NO3(s)-->2H2O(l)+N2O(g)
Mass 1: 25.0g NH4NO3 Mass 2: ? mole ratio: 2molH2O/1molNH4NO3 Set up dim analysis using 25.0gNH4NO3. Multiply 1molNH4NO3/80.043gNH4NO3. Multiply 2molH2O/1molNH4NO3. Finally, multiply 18.015gH2O/1molH2O Answer: 11.3gH2O
Ex. A mass of 2.50g of blue (Cu+2), hydrated copper sulfate is placed in a crucible and heated. After heating, 1.59g of white anhydrous copper sulfate remains. What is the formula and name for the hydrate?
Mass before heating: 2.50g Mass after: 1.59g 2.50g-1.59g= 0.91g H2O lost Set up a dimensional analysis with the mass of the substance (1.59g CuSO4) and another with the mass of water lost (0.91gH20) Multiply each value by 1mol/molar mass +0.0505molH2O +9.962x10^3mol CuSO4 (CuSO4)x(XH2O) X=molH2O/molCuSO4=5 Answer= CuSO4x5H20
Ex. Natural gas in homes is methane (CH4). Calculate the volume that 2.00kg of methane will occupy at STP.
Mass: 2.00kg CH4 Volume: ? Set up dim analysis using 2.00 kg. Convert to grams by multiplying 10^3 g/1kg. Then multiply by 1/molar mass (16.043g). Finally, multiply 22.4 L/1mol. Answer: 2.79x10^3L +Remember 22.4L over 1 mol (or 1 mol over 22.4 L) at STP->This value will NOT be given on exam +Mass->Moles->Volume
Atoms->Mass
Mass: ? Molar Mass: found on periodic table Atoms: Given Set up dimensional analysis with given value Multiply given atom value by 1 mol/6.022x10^23 atoms Then multiply by molar mass/1 mol
Mass->Atoms
Mass: Given Molar Mass: found on periodic table Atoms: ? Set up dimensional analysis with given value Multiply given mass value by 1 mol/molar mass Then multiply by 6.022x10^23 atoms/1 mol
Mass->Moles
Mass: Given Molar mass: found on periodic table Mol: ? Set up dimensional analysis with given value and multiply given mass value by 1 mol/ (molar mass) x g
Mass->Volume
Mass: Known Volume: ? vol ratio Label your P,V (unknown),T, and m (mass) given to prepare for solving the ideal gas problem. Set up a dim analysis using given mass. Multiply by 1 mol/molar mass to get a mol value. Set up an ideal gas law problem and solve for the volume of substance you are given. Finally, use that value and multiply by the vol ratio to get the volume of the substance you're looking for.
Ex. How many moles of O2 are required to make 100g of H2O? 2C8H18(l)+25O2(g)-->16CO2(g)+18H2O(g)
Mass:100gH2O Moles:? Mole ratio: 25molO2/18molH2O Set up dim analysis using 100gH2O. Multiply lmolH2O/18.015gH2O. Then multiply 25molO2/18molH2O Answer: 7.71molO2
Moles->Mass
Mass:? Molar mass of substance: found on periodic table Mol: Given Set up dimensional analysis with given value and multiply given mol value by (molar mass) x g/ 1 mol
Moles->Mass of Compound
Mol of Compound: Given Mass of Compound: ? Molar mass of compound: find using values on periodic table [ex. H2O: 2(1.008)+ 15.999] Set up dimensional analysis with given value Multiply given mol of compound by molar mass of compound/1molcompound
Mass->Moles of Compound
Mol of compound: ? Mass of Compound:Given Molar mass of compound: find using values of the periodic table Set up dimensional analysis with given value multiply given mass of compound by 1molcompound/molar mass of compound
Moles->Particles
Moles: given Atoms:? Set up a dimensional analysis and multiply given mole value by 6.022x10^23 atoms/1mol
Ex. Determine the mass of NaCl produced when 1.50molCl2 reacts with excess Na. 2Na(s)+Cl2(g)-->2NaCl(s)
Moles:1.50 mol Cl2 Mass: ? Mole ratio: unknown/known= 2molNaCl/1molCl2 Set up dim analysis with given mol value. Multiply 2molNaCl/1molCl2 and then multiply 58.443gNaCl/molNaCl Answer: 175gNaCl
Particles->Moles
Moles:? Atoms: Given Set up a dimensional analysis and multiply given atom value by 1 mol/6.022x10^23 atoms
Mass->Volume at STP
Multiply given value by 1mol/16.043g (molar mass). Then multiply 22.4L/1mol
Polyatomic Ions
NH4+ ammonium NO3- nitrate OH- hydroxide MnO4- Permanganate ClO3- chlorate CO3-2 carbonate SO4-2 sulfate O2-2 peroxide PO4-3 phosphate
Gay-Lussac's Law
P1/T1=P2/T2
Combined Gas Law
P1V1/T1=P2V2/T2
Boyle's Law
P1V1=P2V2
Ex. When iron rusts, it undergoes the following rxn: 4Fe(s)+3O2(g)-->2Fe2O3(s) Calculate the volume of O2 (at 280K and 650mmHg) that is required to completely react with 0.520 kg of Fe.
P: 650mmHg V: ? T: 280K mFe: 0.520kg Set up dim analysis using 0.520kgFe. Convert to grams by multiplying 10^3g/1kg . Finally, multiply by 1mol/55.847gFe= 9.31molFe Set up an ideal gas law problem and solve for VFe VFe= (9.31mol Fe)(62.4 LxmmHg/molxK)(280K)/650mmHg= 250.3 LFe Set up a dimensional analysis with 250.3LFe and multiply by 3L O2/4L Fe Answer: 188L O2
Ideal Gas Law
PV=nRT n=# of moles R=ideal gas content
Covalent Nomenclature
Rules for binary molecules: 1. The first element in the formula is always named first using entire name 2. The second element is named using the root and adding -ide 3. Prefixes are used to indicate the number of atoms of each element present 4. Never put mono in front of first word 1-mono 2-di 3-tri 4-tetra 5-penta 6-hexa 7-hepta 8-octa 9-nona 10-deca Rules for binary acids: 1. first word has prefix hydro-, then root of second element, then -ic 2. second word is acid Rules for oxyacids: 1. first take the root of oxyanion and any prefix in the original word then add either -ic or -ous, depending on ending of oxyanion -ate -> ic -ite-> ous
Ionic Nomenclature
Rules: 1. Name the cation first, then the anion 2. For the anion, if it's monoatomic, change ending to -ide 3. If multiple oxidation numbers possible, use Roman numerals 4. For polyatomic ions, simply name the polyatomic ion
mole
SI base unit for amount of substance -mol -defined as the # of carbon atoms in exactly 12g of pure carbon-12
Mass->Particles multi-step problem (so fun)
This should be in your notes if information was given consistently... She made a note saying it "probably wouldn't be on the test" I can't remember if it was so I won't bother putting it on here so review that as you wish
Charles' Law
V1/T1=V2/T2
Volume->Volume->Moles->Mass (not at STP) (mass->moles->volume->volume)
Volume->Volume->Moles->Mass ^ ^ ^ vol. ratio->ideal gas law->molar mass
Volume->Volume->Moles->Mass (at STP) (mass->moles->volume->volume)
Volume->Volume->Moles->Mass ^ ^ ^ Vol. ratio-> 22.4L/1mol->molar mass
Ex. Use the rxn below to calculate the mass of solid ammonium nitrate that must be used to obtain 0.100L of dinitrogen oxide gas at STP. NH4NO3(s)-->N2O(g)+2H2O(g)
Volume: 0.100L N2O Mass? vol ratio: 1 LNH4NO3/1LN2O Set up dim analysis with 0.100L N2O. Multiply 1L NH4NO3/1L N2O. Multiply 22.4 L/1 mol. Finally, multiply 80.043gNH4NO3/1molNH4NO3. Anser: 0.357gNH4NO3
Ex. If 5.00L of N2 reacts completely with H2 at a pressure of 3.00 atm and a temperature of 298K, how much ammonia, in grams, will be produced? N2(g)+3H2(g)-->2NH3(g)
Volume: 5.00L N2 Mass: g NH3 Volume ratio: 2L NH3/1L N2 ---------------- VN2=5.00L N2 P= 3.00atm T= 298K mNH3=? Set up dim analysis with 5.00gN2. Multiply 2LNH3/1L N2 =10.0L (this will be the mass you use for the ideal gas law problem) Set up an ideal gas law problem and solve for nNH3 (moles). nNH3 =(3.00atm)(10.0LNH3)/(0.0821 Lxatm/molxK)(298K)= 1.23molNH3 Set up a dim analysis using 1.23molNH3. Multiply 17.031gNH3/1molNH3 Answer: 20.9gNH3
Mole ratio
a ratio between the numbers of moles of any two substances in a balanced chemical equation
Actual yield
actual amount of product formed 9experimental yield)
Limiting Reactant (LR)
limits the extent of the rxn; determines amount of products formed; rxn stops when run out of LR
Molar mass
mass (g) of one mole of any substance (found on periodic table) 1 amu= 1 g/mol = 1 g/ 1 mol
Theoretical yield
maximum amount of product that could be produced
Mole ratio in word problems (ex. mole->mole, mass->mole, etc.)
mole ratio=moles unknown/moles known
Determine the number of moles of Al in 1.25molAl2O3
mole ratios: 2molAl/1molAl2O3, 3molO/1molAl2O3 Set up dimensional analysis with given value multiply 1.25molAl2O3 by 2molAl/1molAl2O3 =2.5molAl (moles Al2O3 cancel so you are left with molAl) Answer: 2.5molAl/1.25molAl2O3
Percent composition
percent by mass of every element in a compound %mass= molar mass of element/molar mass of compound x100
Excess Reactant (ER)
reactants left over when rxn stops
Molecular formula
specifies the actual number of atoms of each element in one molecule or formula unit of the substance
Empirical Formula +I have an example below but still recommend that you use your notes for further clarification
the formula with the smallest whole-number mole ration +really only applies to covalent compound, but the first example she gives in an ionic compound ex. calcium oxide Ca2O2 -> emp form. = CaO ex. hydrogen peroxide H2O2 -> emp. form.= HO -use % comp. to determine emperical formula -assume mass of compound is 100.00g-> %mass value=mass
Stochiometry
the study of quantitative relationships between amounts of reactants used and products formed by a chemical reaction -based on Law of Conservation of Mass (don't need to know how to do these problems)
Volume ratio
unknown L/known L
Molar Volume
volume that one mol occupies at STP
