# Table 4 contains data in regards to the volume and

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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 ( 1.5 ) V =( 0.30 M )( 10 mL )= 2 mL V = 2 mL of stock dye solution necessary 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,
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