AP CHEMISTRY LAB REPORT FORMAT
Lab reports are to follow the guidelines listed below. Furthermore, lab reports are to be entirely handwritten, unless told to
type it, on graph paper (regular lined or blank paper will NOT be accepted). Failure to do so will result in a penalty.
1) The following are to be given at the top of the report:
a) student’s name
b) period number
c) date of the lab
d) title of the lab
2) Purpose: State the purpose of the lab—what are you trying to determine.
3) Materials: Make a list of the equipment and chemicals used in the lab. Give the size of glassware and the name of
any software used. For chemicals, give the name and the formula. Also include the concentration for solutions.Do
this in a column format:
Equipment
100 mL graduated cylinder
400 mL beaker
balance
LoggerPro
temperature probe
Chemicals
zinc, Zn
hydrochloric acid, HCl (1.0 M)
4) Data table:
a) The format of the data table will be continued being supplied.
b) If there are no lines on your paper, supply lines by USING A STRAIGHTEDGE.
c) Include unit labels, either at the top of the columns or with the data description in the first column.
d) Enter numbers neatly! And with the correct number of significant digits.
e) Include the unknown number if applicable.
5) Calculations:
a) Need ALL calculations for the first trial, regardless of how trivial the mathematics is. This includes any averages
or other calculations requiring data from all your data.
b) Tell what the calculations are for.
c) Provide an equation for all calculations.
d) Do the calculation—show the algebra.
e) Follow significant digits.
f) Do this in a single column.
6) Questions:
a) Write out the questions followed by the answers.
7) Conclusion:
a) Must answer to what the purpose of the lab is.
b) Explain how you arrived at your result(s).
c) If there is an “accepted” value or reason for the results, compare your results with the accepted value.
d) Explain any relevant events in the lab that influenced the results in a negative manner—errors. Keep in mind,
“measured wrong” is NOT generally acceptable as an excuse.
8) Graphs:
a) Attach to the end of the report.
b) Graphs need to be titled and axes labeled correctly.
c) Any extensions to the graph (e.g. interpolation, best fit lines with slopes, etc.) need to be on the graph handed in.
Determining Molar Mass by Freezing Point Depression
Student’s Name
Period Number
Date
Purpose: To determine the molar mass of sulfur by the freezing point depression of a naphthalene-sulfur solution.
Material:
Equipment
test tube
balance
1000 mL beaker
burner
ring stand and ring
wire gauze
LoggerPro software
temperature probe
Chemicals
naphthalene, C10H8
sulfur, S
distilled water
DATA TABLE
kf of naphthalene
freezing point of naphthalene
mass of empty test tube (g)
mass of test tube + naphthalene (g)
mass test tube + naphthalene + sulfur (g)
freezing point of solution (from graph) (°C)
ΔTf (°C)
mass of naphthalene (g)
mass of sulfur (g)
molality of solution (m)
molar mass of sulfur (g/mol)
average molar mass of sulfur (g/mol)
molecular formula of sulfur
6.9 °C/m
80.5 °C
31.55
41.80
42.32
78.13
2.4
10.25
0.52
0.35
140
31.49
41.53
42.04
78.16
2.3
10.04
0.51
0.33
150
140
S4
CALCULATIONS:
∆Tf:
mass naphthalene:
mass sulfur:
molality of solution:
molar mass sulfur:
∆Tf = FPnaphthalene − FPsoln
∆Tf = 80.5°C – 78.13°C = 2.37°C = 2.4°C
mass naphthalene = mass empty test tube – mass test tube and naphthalene
mass naphthalene = 41.80 g – 31.55 g = 10.25 g
mass sulfur = mass test tube, naphthalene and sulfur – mass test tube and naphthalene
mass sulfur = 42.32 g – 41.80 g = 0.52 g
∆Tf = Kfm
m=
∆Tf
2.4°C
=
= 0.35 m
Kf
6.9 °C
m
m=
g sulfur
(MM sulfur)(kg naphthalene)
MM sulfur =
average molar mass sulfur:
g sulfur
0.52 g
g
=
= 144.95 mol = 140
(0.35 m)(0.01025 kg)
(m)(kg naphthalene)
average molar mass=
average molar mass=
sum of MM for all trials
number of trials
g
g
140 mol
+ 150 mol
g
g
= 145 mol = 140 mol
2
g
mol
molecular formula:
g
140 mol
average molar mass
=
g = 4.4
molar mass S
32.07 mol
S4
QUESTIONS:
1) Define the term “colligative property”.
A colligative property is one that is dependent only on the amount of solute particles in a solution.
2) For a given a given number of moles of solute, why do ionic substances have a larger effect on the freezing and
boiling points of solvents than do nonionic substances?
Ionic compounds break down into their ions when dissolved whereas nonionic substances dissolve as single
molecules. Therefore, ionic compounds have a higher “i” value (more solute particles per unit formula) than nonionic
substances (which has i = 1) and will affect the freezing point of the solvent more.
3) Explain why the temperature doesn’t level off with the solution as it would with a pure substance.
As a solution freezes, only the solvent freezes. Thus, any remaining unfrozen solvent will become more concentrated
since the amount of solute in solution remains the same. Therefore, as the concentration of the unfrozen solution
increases its freezing point decreases resulting in a curved line on the graph.
CONCLUSION:
From the graph, the freezing points of my solutions were 78.13°C and 78.16°C. From the freezing point of pure
naphthalene, 80.5°C, the ∆Tf were 2.4°C and 2.3°C respectively, resulting in concentrations of 0.35 m and 0.33 m. From
this the molar mass of the sulfur were determined to be on the average of 140 g/mol, resulting in a molecular formula of
S4.
Since sulfur is generally in the form of S or S8, it is obvious there is a flaw in the experiment. One possible reason for this
is that not all the sulfur made it into the solvent—some of it stuck to the side of the test tube. This would make the molar
mass of the sulfur less than it should have been. Another possible reason is that I didn’t determine the freezing point of
the naphthalene alone and therefore would not have “calibrated” the temperature probe. I most likely would have had a
more accurate ∆T if I had done so since thermometers are better in determining change in temperature than in finding a
stand-alone temperature as I did in this lab.