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lab6report

# lab6report - 1.0782g EDTA x(1mol EDTA/372.24g EDTA...

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Carla Paulo Exercise 6 Determination of Mg by EDTA Purpose: To determine the amount of Mg in a Magnesium Oxide compound by titrating it with EDTA. Unknown #: 6C Reaction: Mg 2+ + H 2 Y 2- MgY 2- + 2H + Safety Factors: Gloves and goggles must be worn during the experiment. Solutions must be prepared with deionized water. Raw data: Initial weight of vial (g) Final weight of vial (g) Total mass (g) EDTA 32.9032 31.8250 1.0782 Unknown 28.1628 27.3649 0.7979 Titration data: Trial 1 Trial 2 Trial 3 Trial 4 Final EDTA reading(mL) 29.75 29.70 29.40 29.55 Initial EDTA reading(mL) 3.60 3.50 3.70 3.65 Volume EDTA used (mL) 26.15 26.20 25.70 25.90 Sample Calculations and Results: 1. Molarity of EDTA in the solution prepared:

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Unformatted text preview: 1.0782g EDTA x (1mol EDTA/372.24g EDTA) x (1/0.200L) = 0.0145 M 2. Mass of MgO in unknown: Trial 1 Calculation: 0.02615L EDTA x (0.0145mol EDTA/ 1L) x (1mol MgO/ 1mol EDTA) x (40.31g MgO/ 1mol MgO) x (5 portions) = 0.07642g MgO in the Unknown Trial Mass MgO in the Unknown (g) 1 0.07642 2 0.07656 3 0.07511 4 0.07569 3. %MgO in the Unknown Sample Trial 1 Calculation: (0.07642g MgO/0.79790g sample) x 100 = 9.5776% Trial %MgO in sample 1 9.5776 2 9.5952 3 9.4135 4 9.4862 Average %MgO: 9.5181% Standard Deviation: 0.0845 Conclusion: The average % MgO in the unknown sample 6C is 9.5181 ± 0.0845%...
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lab6report - 1.0782g EDTA x(1mol EDTA/372.24g EDTA...

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