Table 4 contains data in regards to the volume and mass of both a predetermined amount of water
(20
mL, 40 mL), as well as a 50 mL buret. Such additionally displays the the percent error
of the 50 mL buret in measuring the volume of water in each trial.
Sample Calculations for Tables 1-4:
Using data from Table #1, Trial #1.
Mass of Water = Mass of Full Container with Water - Mass of Empty Dry Container
19.828 g
= 48.573 g
- 28.745 g
Calculated Glassware Volume = Mass of Water / Density of Distilled Water
19.872 mL = 19.828 g / 0.997773 g/mL
% Error for Volume Measured = |Experimental Value - Accepted Value|
x
100
Accepted Value
0.6 % = (|20 mL
- 19.872 mL| / 19.872 mL) x 100
Task 2:
Table 5: Measurements for an Average Volume and Mass of a Drop of Water
Container Used: 10 mL Graduated Cylinder
Distilled water has a density of: 0.997773 g/mL
Temperature of: 22.1
℃
Trial
#
Experimentally
Measured
Number of
Drops
Mass of
Empty Dry
Container
(g)
Mass of Full
Container
with Water
(g)
Mass of
One Drop
of Water
(g)
Calculated
Volume of
One Drop
(mL)
Calculated
Number of
Drops/mL
(drops/mL)
1
40
29.942
31.269
0.03
0.03
33

2
60
29.942
31.928
0.03
0.03
33
3
80
29.942
32.537
0.03
0.03
33
Average:
0.03
0.03
33
Table 5 contains data in regards to the average volume and mass of a single drop of water.
Additionally, through the use of the mass of a full container with water, the mass of a dry
container, and the mass of a single drop of water, the table displays the number of drops/mL
present in the 10 mL graduated cylinder for each trial.
Sample Calculations for Table 5:
Using data from Table #5, Trail #1.
Mass of One Drop of Water = (Mass of Full Container with Water - Mass of Empty Dry
Container) / Number of Drops
0.03 g
= (31.269 g
- 29.942 g) / 40 drops
Volume of One Drop of Water = Mass of One Drop of Water / Density of Water
0.03 mL = 0.03 g / 0.997773 g/mL
Number of Drops/mL = 1 Drop / Volume of 1 Drop
33 drops/mL = 1 Drop / 0.03 mL
Task 3:
Mass of Sand
0.100 g
1.000 g
10.000 g
Trial 1
Trial 2
Trial 1
Trial 2
Trial 1
Trial 2
Initial mass (g)
0.100
0.107
0.999
1.015
10.006
10.002
Transferred mass
0.096
0.102
0.994
1.004
10.005
10.000

(g)
% difference
4.081%
4.784%
0.5021%
1.089%
0.00999%
0.01998%
This table shows the semi-accurate sand measurements made with the balance.
Two trials were conducted for transferring 0.100 g, 1.000 g, and 10.000 g of sand from one
weighing boat to another. The percent difference was calculated for each trial. The percent
difference can be found by subtracting the transferred mass from the initial mass and that will be
divided by half of the initial and transferred mass combined. That total will be multiplied by 100
to give you the correct percentage (CalculatorSoup).
Percent Difference:
From 0.100 g, Trial 1:
Task 4:
Volume and Concentration Calculations for Solutions
Initial
Concentration of
Stock Dye
Solution
Stock Dye
Solution Volume
Diluted Water
Volume
Total Volume
Concentration of
Diluted Solution
0.30 M
2 mL
8 mL
10 mL
1.5 M
0.75 M
5 mL
5 mL
10 mL
1.5 M
1.35 M
9 mL
1 mL
10 mL
1.5 M
This table is indicative of the amount (in milliliters) of stock dye solution and distilled water
needed to create a 10 mL diluted solution. Additionally, it indicates the molarity of the initial

stock dye and diluted solutions.
Example Calculation for 0.30 M Stock Dye Solution:
M V =M V
₁
₁
₂
₂
(
1.5
)
V
₁
=(
0.30
M
)(
10
mL
)=
2
mL
V
= 2 mL of stock dye solution necessary
₁
Discussion:
The purpose of this investigation encompassed the idea of gaining familiarity with the
laboratory prior to running other, more challenging experiments. Upon the completion of this
investigation, familiarity with measuring liquid volumes, using common laboratory glassware,