Report Sheet
SOLUTION
Laboratory Orientation
Part 2 – Measurements and Handling Data in the Lab
CHEM 110
* Remember that the scale on the measuring instrument can be read to a fraction of the
LAST NAME
smallest
FIRST upon
NAME the smallest scale division and
LOCKER
scale division. That fraction will depend
the#
judgement of the person reading the scale. In the case of the 100 mL graduated cylinder, the
most likely precision for most people would be half of the smallest scale division (i.e.,
0.5 mL)
DATE
LAB ±
SECTION
although one fifth (± 0.2 mL) could also be possible. It’s unlikely that the scale could be read to
± 0.1 mL though. In the case of the 25 mL graduated cylinder, it’s unlikely that the scale could
A. Liquid
Measuring
(Personal
Locker)
be read
any moreDevices
precisely than
to one fifth
of the smallest scale division (± 0.1 mL).
Calibration
Smallest scale
division


Device
Precision of
calibrated scale
100 mL graduated cylinder TD 
TC 
1
_________
mL
25 mL graduated cylinder
TC 
0.5 mL
_________
*
0.1* mL
±_________
10 mL
_________
1
±_________
mL
25 mL
_________
5
±_________
mL

TD 

150 mL beaker
TD 
400 mL beaker
TD 

TC 

TC 
0.5 mL
±_________
1. Based upon the scale precision you reported above for a 100 mL graduated cylinder,
calculate the percent uncertainty for the given volume readings. Show a sample
calculation, as required in the box below. In each of the 5 cases below, report your results
to the proper number of significant figures:
Volume Reading
100.0 mL
80.0 mL
60.0 mL
40.0 mL
20.0 mL
Percent Uncertainty
0.5
0.6
0.8
1
3
%
%
Show how this result was calculated:
0.5 mL
×
100
100.0 mL
%
%
%
Your were asked to report these answers to the proper
amount of sig figs because they are not being
converted back to absolute uncertainties. See the
section on Absolute and Percent Uncertainties in your
lab manual for more information.
2. Which is the most suitable device to make a single measurement of 26.5 mL of water?
a)
b)
c)
d)
100 mL graduated cylinder
400 mL beaker
150 mL beaker
25 mL graduated cylinder
IFS/06/08(rev05/09,05/11)
ORT_1
2
B. Liquid Measuring Devices
(Fume Hood at the end of your bench)
Go to the fume hood at the end of your bench, where several pieces of glassware have
been partially filled with solutions. Holding each piece of glassware, so that the meniscus is
level with your eye, record and report the reading of the liquid with the appropriate
degree of precision:
Note: Readings may vary
but sig figs should stay the
Scale
increments
of
50
mL.
Fume Hood 1 
Fume Hood 2 
Fume Hood 3 
Readable to only the nearest 10 mL.
same
So,
e.g.,
a
reading
of
380
mL
would
Device
Reading
Significant Figures
be reported as 3.8 x 10 mL (2 sf).
~350-400
A 600 mL beaker
2
mL
Scale increments of 10 mL. Readable to the nearest 1 mL at best.
~35
B 50 mL beaker
mL
2
Scale increments of 10 mL (less widely spaced than B). Readable to the nearest 5 mL at best.
C 1000 mL graduated cylinder
~850
mL
3
Scale increments of 1 mL. Readable to the nearest 0.5 mL at best.
D 100 mL graduated cylinder
~80-90
mL
3
Scale increments of 0.2 mL. Readable to the nearest 0.1 mL at best.
~8.5
E 10 mL graduated cylinder
mL
2
SHOW THIS PAGE TO YOUR INSTRUCTOR BEFORE LEAVING THE LAB
FOR INSTRUCTOR USE ONLY:
Parts A and B: Complete 
Incomplete 
Student Contract submitted:
Yes 
No 
Student Questionnaire submitted:
Yes 
No 
Comments:
Instructor’s
initials
3
C. Practice Problems: Handling Numbers and Dimensional Analysis
If you’re not able to complete this part in the available time, the following questions can be
answered at home. Be sure to check your answers on our website and submit the
completed sheet at the beginning of the next lab session.
1. Examine the millilitre scale on the glassware below and notice the meniscus. Read the
level of liquid, determine the uncertainty and report the number of significant figures in
the reading:
Reading
31.5
35
Uncertainty ± 0.5
30
Sig figs
3
25
or
mL
Convert this reading to units of
litres. Show calculation with
conversion factor(s):
0.2
(maybe
even
0.1)
mL
A fraction of the smallest
scale division, the value
of which depends upon
the scale and the person
reading it.
31.5 mL ×
1L
1000 mL
0.0315 L
2. A chemist reported a measured volume of a solution as 400 mL ± 10 mL. What is the
value of this measurement in scientific notation?
a)
b)
c)
d)
4 × 102 mL
4.0 × 10-2 mL
4.0 × 102 mL
4.00× 102 mL
3. Express the following numbers as decimals:
a) 2.300 × 10-3
0.002300
b) 8.08 × 10-6
0.00000808
4. Express the results of the following calculations in scientific notation to the correct
number of significant figures:
a) (3.6 × 10-3) × (2.40 × 10-6)
8.6 x 10-9
b) 30.0 ÷ (2.75 × 10-7)
1.09 x 108
c) (2.01 + 3.0) ÷ 0.05482
9.1 x 101
L
4
5. How many significant figures are there in each of the following measurements?
a) 0.0002 L
1
b) 430 g
2 or 3
c) 0.020 kg
2
d) 23.20 mL
4
It’s ambiguous: if the measurement
was made to the nearest gram, it’s
3 sf; if the measurement was made
to the nearest 10 g, it’s 2 sf.
If you missed any of these, please
review Significant Figures in your
Laboratory Guide
6. A builder is to construct a swimming pool on the roof of a building. The dimensions of
the pool are 25.0 m x 10.0 m x 1.0 m. What mass, in units of tonnes, of water must the
building’s structure be able to support when the pool is filled with water? Assume the
density of water = 1.00 g/cm3. See Section 5 in your Laboratory Guide for conversion
factors which will be needed to solve this problem.
Adopt the Strategy for Problem Solving – see your lab manual.
•
What has to be solved?
o START WITH pool dimensions (metres)  END WITH mass of water (tonnes).
•
Gather Information:
o Given
 pool dimensions 25.0 m x 10.0 m x 1.0 m
 density of water: 1.0 g/cm3
o Obtain
 conversion factors…
1 t = 103 kg
1 kg = 103 g
1 m = 100 cm
∴ 1 m3 = 106 cm3 (derived from 1 m = 100 cm)
•
Solve problem using dimensional analysis
First find volume of pool in cubic metres:
25.0 m × 10.0 m × 1.0 m = 250 m3 (no rounding – intermediate result)
Now use DA to find your way from cubic metres of water to tonnes of water:
250 m
3
×
106cm3
1 m3
×
1.00 g
×
cm3
1 kg
103 g
×
Answer:
1t
103 kg
= 250 t
2.5 × 102 t
(2 sf)