mass divided by volume
The ratio of the density of a substance to the density of water
Properties independent of mass. Examples: density, specific heat, equivalent weight
Properties dependent of mass. Example: volume and mass
Volume of a cylinder
V = pi(r)^2h
Mass of water displaced by a rock with 143g in air and 82 g in water
Mass of water displaced = Mass of rock in air – mass of rock in water
Volume of water displaced by a rock
V = mass of water displaced divided by density of water
Density of rock with 143g in air, 82 g in water
= mass of rock in air divided by the volume it displaces in water
If a piece of PVC is suspended in solution than the density of the PVC is
equal to the density of the solution
Which has a greater density: water or ice?
Water (Ice floats on water)
What are the units of density?
g/mL and g/cm^3
Why is density important?
It identifies a substance
What is the conversion factor of cm^3 and mL
1 cm^3 = 1 mL
What is the normal order of density for the states of matter?
Gas is the least dense, liquid is more dense and solid is the most dense.
What property of adipic acid is used in the separation during the adipic acid lab?
What is the formula for Adipic Acid?
What is the unit of specific heat?
there is no unit
What does density tell us?
Density relates information on how the atoms or molecules that make it up are arranged in space.
Substance with larger density is more tightly packed than the substance with smaller one.
What is the structure of adipic acid?
What is adipic acid used for?
Production of nylon; additive to drinks
Solubility of Adipic Acid
Adipic acid’s solubility is temperature dependent. It is more soluble in hot water and slightly soluble in cold water. This property is the principle of recrystallization of adipic acid.
What impurities make the crude adipic acid brown?
Oxidized compounds and fine carbon particles.
What substance absorbs the impurities?
Charcoal. One scoop is added to the beaker.
Why is fluted filter paper used instead of flat filter paper?
It has a larger surface area for contact with the solution, thus leading to a more rapid rate of filtration.
Why was the adipic acid heated?
To dissolve is completely in hot water.
Why was the adipic acid cooled?
So the acid will crystallize
How do you calculate percent yield of adipic acid?
Weight of pure adipic acid/Weight of crude adipic acid x 100
What is the difference in melting point between Crude and Pure adipic acid?
Pure adipic acid will melt at a temperature close to the literature value. It will have a narrow melting point range. The purer the compound is, the smaller is melting point depression (Literature value – experimental value) and the smaller is melting point range. The crude will have a large melting point range and the melting point will be less than the pure on. The impurity interrupts the normal intermolecular force.
How much pure adipic acid was recovered if there was a crude sample of adipic acid that weighed 20.421 g and a yield of pure adipic acid of 73.5%?
Pure/Crude x 100 = 73.5%
20.421g(0.735) = 15.009 g
If the adipic acid was 15 degrees below the literature value, was his adipic acid pure or impure?
What could cause the adipic acid to still be impure?
The impurities were not all filter out.
The solubility vs. temperature curve:
What is the waste product that is emitted from industrial adipic acid synthesis?
N2O (Nitrous oxide)
What property makes it possible for adipic acid to be purified by recrystallization method?
The solubility of adipic acid and its ability to recrystallize. REVIEW ME
Pour the boiling solution into the fluted filter paper, you should be careful to pour the liquid
on the side
the adipic acid should be poured quickly so that
the adipic acid does not begin to crystallize
When impurities are present in the adipic acid, the melting point is expected to be
lower than the pure compound
Compared to the pure compound of adipic acid, the range of temperatures from when is starts melting to when it is completely melted will be
Amount of crude adipic acid used when there is a yield of pure adipic acid of 25% and 7.2 grams of pure adipic acid recovered?
7.2 g / .25 = 28.8 g
The crystals obtained from the adipic acid lab should have been
large because it was allowed to cool and was not mixed. The crystals should have been colorless.
What are some physical properties of pure compounds?
Melting point, crystal structure and color
what compounds or elements are present in the substance
how much of a particular compound or element is present in a sample
procedures that depend on weighing
Which substances would contaminate the Ba(OH)2 sample during the conductivity and gravimetric determination experiment?
Strontium, mercury (I), and lead (II) because they could co-precipate with the Ba(OH)2
Given a 10.0 mL sample of the unknown Ba(OH)2 solution, calculate the concentration of the solution if it took 12.08 mL of a 0.0980 M H2SO4 to reach the equivalence point.
!2.08 mL H2SO4 (0.0980 M H2SO4) = 10.0 mL Ba(OH)2 (x Ba(OH)2)
1.1834 = 10.0 (x)
x = 0.118 M Ba(OH)2
Suppose that in a 10.0 mL sample of Ba(OH)2 the mass of the crucible with the precipitate was 17.550 g and the mass of the empty crucible (tare weight) was 17.410 g. Calculate the concentration of the Ba(OH)2 solution from this data.
17.550-17.410 = .140 g BaSO4/MW BaSO4 = .140g/233.4 g/mol BaSO4 = 5.998 x 10^-4 mol
5.998 x 10^-4 mol/ 0.010 L = .06 M Ba(OH)2
If the barium sulfate in the crucible was not dry then the concentration calculated for the barium hydroxide solution would be
larger than its actual concentration
Balanced equation of the conductimetric titration and gravimetric determination experiment of BaSO4
Ba^2+(aq) + 2OH^-(aq) + 2H^+(aq) + SO4^2-(aq) –> BaSO4(s) + 2H2O(l)
State the five requirements for gravimetric analysis to work:
1. The compound formed must be of pure and known stoichiometry
2. The precipitation reaction must be virtually compete (99.9%)
3. The precipitation reaction should be specific for the substance being determined, interference by others should be minimal.
4. The solid which is precipitated should be in the form of large and well formed crystals so that the filtration can be easily finished.
5. The molecular weight of the solid formed should be large so that enough precipitation can be generated when the percent of the substance being determined is low.
How many moles of sulfate anions will generate one mol of BaSO4?
If you get N grams of BaSO4 in the end of the reaction, there should be how many grams of sulfate in the unknown powder?
(N*(233g/mol-137g/mol)/233 = N*96/233
During the conductimetric titration why do you need to rinse the reagent reservoir with H2SO4 solution before transferring 40 m: of the H2SO4 to the reagent reservoir?
To get rid of the H2O that could interfere
During the conductimetric titration of BaSO4, the graph resembles a
V shape with volume on the x-axis and conductivity on the y-axis.
The equivalence point on the volume vs. conductivity of BaSO4 graph is
the volume of BaSO4 where conductivity reaches a minimum
Given a 20 mL sample of an unknown Ba(OH)2 solution, calculate the concentration of the solution if it took 22.45 mL of 0.389 M H2SO4 to reach the equivalent point.
22.45 mL (0.389 M H2SO4)/20 mL = .44 M Ba(OH)2
aggregating into a mass
What is the purpose of flocculating during the gravimetric determination of BaSO4?
To ensure complete precipitation formation
A student performs the gravimetric determination of BaSO4 experiment. Volume of original Ba(OH)2 solution is 5 mL, mass of dry BaSO4 is 0.2334 g. What is the concentration of original Ba(OH)2 solution?
MV = mol = .2334 g/233.4g/mol = .001 mol BaSO4 = M(.005 L)
.001 mol/.005 L = .02 M Ba(OH)2
Na2SO4 + Ba(NO3)2 –> 2NaNO3 + BaSO4
Sn2+ –> Sn3+ + e-
Fe3+ + e- –> Fe2+
CH4 + 2O2 –> CO2 + 2H2O
AB + CD –> AC + BD
HNO3 + NH3 –> NH4NO3
AB + C –> AC + B
CaCO3 –> CaO + CO2
Define Conductivity of a solution
The ability of the substance to conduct electricity.
How do size and charge of ions affect the conductivity of a solution?
Smaller ions are better conductors than larger ones of the same charge. Higher charged ions are more conductive but less efficient because they are solvated–surrounded by water molecules.
Which is more conductive? Li+ or Cs+
Which is more conductive? Mg2+ or Al3+
Given a 10mL solution of NaOH at known concentration, would would happen to the conductivity of the solution as 20 mL of an equimolar solution of HCL is added drop wise?
The conductivity would increase once the H ions rain out of OH to pair with. The graph of drops vs. conductivity would be v-shaped.
What would happen to the observed temperature in an endothermic reaction? Why?
The observed temperature would decrease because the reaction absorbs heat.
HCl + H2O –> H3O+ + Cl-
H+ + OH- –> H2O
Acid-Base, Double Displacement
Cu2+ + 4NH3 –> Cu(NH3)4 2+
CO3 2+ + 2H+ –> H2O + CO2
Ba 2+ + SO4 2—> BaSO4
Precipitation, Double Displacement
2Fe^3+ + Sn^2+ –> 2Fe^2+ + Sn^4+
Mg + 2H+ –> Mg2+ + H2
Oxidation Reduction, Gas Evolution, Single-displacement
In the Types of Reactions lab, what two measurements are we obtaining other than visible changes
Conductivity and temperature
a reaction that evolves heat (Temp increases)
a reaction that absorbs heat (temp decreases)
Which has the creates conductivity: H+ or Na+?
Which has the greatest conductivity: Fe3+ or Mg2+?
If a reactant is oxidized, it
If a reactant is reduced
it gains electrons
HCl + NH3 –> NH4Cl
2H2 + O2 –> 2H2O
FeCl3 + 3AgNo3 –> Fe(NO3)3 + 3AgCl
Precipitation; double displacement
Fe2+ –> Fe3+ + e-
MgCO3 –> MgO + CO2
HCl + NaCO3 –> NaCl + which gas?
2Na + 2H2O –> 2NaOH + H2
redox reaction; gas evolution
Na is oxidized; H2O is reduced
What is an acid?
An acid is a proton donor
What is a base?
A base is a proton acceptor
The negative log of the hydrogen ion concentration
[H+] = 1×10^-4 so [OH] =
1 x 10^-10
[H+] = 10^-9.5 = 3.16 x10^-10
[OH-] = 10^-4.5 = 3.16 x10^-5
Litmus paper turns
red in presence of an acid and blue in presence of a base
pH equals 9.3 [OH] =
10^-4.7 = 2×10^-5
In the proton donor model, an acid is a proton
donor and base is a proton acceptor
Which is the most accurate way to determine acids and bases?
gives a number representation of the pH of a substance and is the most accurate way to determine acids and bases
When gas is dissolved in water to form the H2CO3 present in soda? How does the pH of the resulting soda compare to the pH of pure water?
H2O + CO2 –> H2CO3
CO2 is dissolved in water. the pH is less than water
Which two tests are created in the acid base lab?
grape juice and red cabbage
What happens if the red cabbage is boiled to long?
the pigments will be denatured, rendering it useless
at a constant temperature, a fixed amount of gas occupies a volume inversely proportional to pressure exerted on it.
To provide a linear relationship of Boyle’s Law
P vs. 1/V
When using Charles’ Law, where is the experimentally determined value for absolute zero?
Where the line intersects the y-axis
at a constant pressure, volume and temperature or a gas are proportional
Charles’ Law of Temp(K) vs. Volume (L)
positive diagonal line starting at the origin
In the procedure for Boyle’s Law, the probe used is
What is the purpose of using the 60 mL syringe setup in the Gas Law 1 Lab?
The volume will change as the pressure changes.
unite of pressure in Gas Laws 1 for LabPro
If there is no significant resistance when depressing the plunger in Gas Laws 1:
there might be a leak
What is the real volume in the syringe when you set the plunger to 30 mL in Gas Laws 1?
what specific statistical measure will you get form the data of Pressure vs. 1/V linear fit
coefficient constant–close to 1
Probe used for Charles’ law?
Why is the pressure below the glycerin plug assumed to be constant at atmospheric pressure?
amount is negligible
In Charles’ Law experiment, as water with pipet in it is heated
volume will increase
lowest attainable temperature OK and -273 degrees C
Assumption in Boyle’s Law
Compression of a gas heats it up so the gas in the syringe is actually hotter than outside of it, however this difference is negligible
Denver vs. Miami both at 85 F. Balloon would be bigger where?
Less pressure in Denver, so balloon would be bigger.
If volume increased to infinity, pressure would go to
% error =
actual value – experimental value/actual value x 100%
the weight of the substance that will react with or produce one mole of hydrogen
What would happen if Mg ribbon were left out
a layer of magnesium oxide would form, reacting with the
Why is volume of dry gas being calculated?
so the ideal gas law can be used, and Pwv is taken out
105.33mmH2O = how many torr
105.33/13.66 = 7.74 torr
Pw in torr of a water column in a gas buret with height of 26.2cm?
Pw = 262mm/13.6 = 19.265 torr
How would some of the equivalent weight be affected if some of the H2 gas produced during the reaction of Mg(s) and acid escaped from the gas buret?
The equivalence weight would be higher because it would seem like the same amount of Mg would result in less H2
What is the reaction taking place in gas laws 2
Mg(s) + 2HCl(aq) –> MgCl2(aq) + H2(g)
How much magnesium is necessary to react with 1 g of H atoms MW=24.3g/mol
24.3 g/mol(1 mol Mg/1mol H2)(1 mol H2/2 mol H) = 12.15g
2Mg(s) + O2(g) –> MgO
MgO + 2HCL –>
How can you tell if the solution is complete in gas laws II?
it will stop bubbling
pure H2(g) without water vapor
What would be the effect on this experiment if a thin layer of MgO were present on the surface of the Mg ribbon? Would this produce a higher or lower result?
1 torr = _____ mmHg
13.6mmH2O = _____ torr
moles solute/kg solvent
moles solute/L solvent
If 45g of C12H22O11 is dissolved in 0.65L H2O what is the molality?
45 g/ (342 g/mol) = .13158 mol/.65 kg = .202 molal
If 45g of C12H22O11 is dissolved in 0.65L H2O what is the molarity?
45 g/ (342 g/mol) = .13158 mol/.65 L = .202 M
Kf of solution with .202 molal if freezing point changed from 0C to -4.3C
changeT = Kfm
Kf = changeT/m = (4.3)/.202 = 21.29C/m
Why and how does the introduction of a solute into a solution change the freezing point?
The impurities interfere with the normal molecular interactions. More energy must be taken out of the system.
Where is the freezing point seen on the graph of temp vs. time of the Freezing Point depression lab?
The freezing point is the flat portion of the curve.
A mixture of C2H6O2 and water is mixed to lower the boiling point and raise the melting point. How could could it get before a 25%mass solution of antifreeze in water would freeze?
changeT = Kfm = 1.86C/m(m)
m = (25 g/62 g/mol)/.075 kg H2O = .403 mol/.075 kg = 5.37
1.86(5.37) = 10C
Given 1 L of water which contains 50 g of NaCl, will the freezing point be higher or lower compared to 1 L of water containing 70 g of NaCl?
Solute of Freezing point depression lab
Solvent is Lauric Acid
36g KCl(74.6g/mol) to a beaker of 0.75 L of a substance (d=1.5g/mL) and allow it to dissolve. What is the molality of the solution? if the normal melting point of the substance is 0.4C, calculate the Kf of the solution if the freezing point of the solution (with the KCl) was -8.2C.
m = vd = 1.5(750) = 1125 g
molality = (36/74.6)/1.125 kg = .439 molal
Kf = .4+8.2/.429 = 20.05C/m
Formula for a weak acid
HX(aq) + H2O <-> H3O+(aq) + X-
where X- is the conjugate base
Formula for a weak acid reacting with the strong base NaOH
HX(aq) + NaOH(s) –> NaX(aq) + H2O(l)
If a weak acid is titrated with a strong base, will the equivalence point in the titration occur at above or below pH 7.0
Ka of a weak acid .65 M and pH = 3.68
pH = 3.68
[H+] = 10^-3.68 = 2.09×10^04 M
Ka = [H+][X-]/[HX] = (2.09×10^-4)^2/0.65 = 6.72×10^-8
What is the difference between a strong acid and a weak acid?
A strong acid will go to completion
2 ways to determine Ka of a weak acid being titrated at a given temperature?
1. Calculating Ka with a known concentration of a weak acid.
2. Calculating Ka using the half-neutralization pt.
What is the acid and base being used in the determination of Ka lab?
Weak acid: acetic acid HC2H3O2
Strong base: NaOH
Equation of reaction occurring during Ka of a weak acid lab?
HC2H3O2(acid) + H2O(base) <--> H3O+(conjugate acid) + C2H3O2-(conjugate base)
It 24 mL of a 0.3 M NaOH solution to titrate 20 mL of a weak acid to its equivalence point. This weak acid had an original pOH of 10.56. What is the Ka of this weak acid?
0.45 M solution of an unknown weak acid. pH = 2.75. pH1/2= 5.15. Use both ways to determine Ka of the unknown acid.
HX <-> H+ + X-
[H+] = 10^-2.75 = 1.78×10^-3
Ka = [1.78×10^-3]^2/[.45] = 7.03×10^-6
1/2pH= 5.15 = pKa = 10^-5.15 = 7.08×10^-6
Molcular geometry, #lone pairs, # bonding paris, bond angles, steric number number of bonds: CO2, H2O, CH4, PF5, SF6
CO2: linear, 0, 4, 180, 2, 2 double
H2O: bent, 4, 1, 109.5, 4, 2 single
CH4: Tetrahedral, 0, 4, 109.5, 4, 4 single
PF5: Trigonal Bypyramidal, 0, 5, 90,120, 5, 5 single
SF6: Octahedral, 0, 6, 90, 6, 6 single
To determine the molarity of a weak acid
=[base](volume of the base/volume of the acid)
If the base has a .1M concentration and a volume of 18 mL, then what is the concentration of the acid with a volume of 20mL?
.1(18/20) = .9 M acid
Ka if the initial pH is 3 and [HA] = .9M
Kf = (10^-3)^2/.9
Ka if volume of base at equivalent is 18mL and half-equivalent is 9mL with pH 5.4
Ka = 10^-5.4 = 3.98 x10^-6
Pg of H2 if Pb equals 753.1 torr, Pwv = 18.65 torr and the height of water is 250.0 mm.
Pg = Pb – Pwv – Pw
(753.1torr/760) – (18.65/760) – (250/13.6)/760 = .942 atm
Volume of H2 if collected at STP
(Pg = .942 atm and T=294 K and Vf = 30.0 mL)
PgVf/T = PstpVstp/Tstp
Vstp = PgVfTstp/TPstp = .942atm(.030L)(273K)/(294K)(1atm) = .02624 L at STP
# of equivalents of H2 is 11.2L at STP = 1 equivalent and .02624 L at STP
.02624/11.2 = .002343 equivalents
# of equivalents of Mg needed to produce the above (H2 at STP = .002343)
.002343 equivalents of Mg
Equivalent weight of Mg if there are 0.0282 g of Mg
# g of Mg/# of Equivalents = .0282/.002343 = 12.04 g/Equivalent
Valence of Mg
= atomic weight/equivalent weight = 24.35g/mol/12.04 g/equivalent = 2.02 = 2
% error in equivalent weight of Mg
24.3/2 = 12.16 – 12.04 = .09 1.0-.09 = .91
In the freezing point depression lab, the heat of fusion
was the flat portion of the graph, where the temperature did not change because the energy lost during the freezing process is used in packing the Lauric acid molecules into a dense, solid state.
as we remove heat from a liquid, we can temporarily cool it below its freezing point without forming a solid . eventually, stirring and and agitation will cause the liquid to freeze, and the temperature will return to that of the freezing point.
[H+] = Ka[HA]/[A-]
pH = -logKa + log[A-]/[HA] = pKa + log[A-]/[HA]
At the half-neutralization point:
pH1/2 = pKa = -logKa
In the gravimetric/conductimetric lab: If you have water in your precipitate the weight of BaSO4 will _______ so your concentration will be _____ than it actually is
go up so your concentration will be higher than it actually is
In the gravimetric/conductimetric lab: If Lead chloride got into your sample you would get ______ precipitate and use ______ H2SO4 titrant which means _______ equivalence volume.
MORE precipitate and use more H2SO4 titrant which means higher equivalence volume.
In the gravimetric/conductimetric lab: Why did you put the crucible and your solid in the oven? What happens if you do not do this?
To dry it out
In the molecular modeling lab: Hartree-Fock for
and Semi-emprical AM-1 for
Hartree-Fock for H2O, NH3. CH4
Semi-emprical AM-1 for PF5, SF4, ClF3, SF6, and NO3-, and SO42-