# lab 4z - Subtract the amount in excess by the amount...

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Results and Discussion In order to determine the calcium content in the milk, we must first calculate the molarity of the EDTA used in the experiment. In order to do so, we must use the equation M EDTA = M(Ca 2+ ) x V(Ca 2+ ) / V EDTA . M(Ca 2+ ) = .03 [mol / L] V(Ca 2+ ) = 2.50 ml .03 x 2.50 / 10.00 = M EDTA V EDTA = 10.00 ml M EDTA = .75 [mol / L] Next, the moles of EDTA started with were found by multiplying the molarity of EDTA from above by the volume of EDTA used. The equation is Moles EDTA = M EDTA x V EDTA . M EDTA = .028 [mol / L] Start Moles EDTA = .028 x .01 V EDTA = .01L Start Moles EDTA = 2.8 x 10 -4 mol Next, we must calculate the excess amount of EDTA. The necessary operations are to multiply the molarity of the calcium by the average volume of calcium to give the number of moles of calcium. Then the amount of calcium is the same amount of the excess EDTA.
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Unformatted text preview: Subtract the amount in excess by the amount started with to find the amount of EDTA used M(Ca 2+ ) = .03 [mol / L] V avg (Ca 2+ ) = (8.19 + 7.38 +7.42) / 3 = .00766 L Moles(Ca 2+ ) = .03 x .00766 Moles(Ca 2+ ) = 2.3 x 10-4 mol Excess Moles EDTA = 2.3 x 10-4 mol Moles EDTA = 2.8 x 10-4- 2.3 x 10-4 Moles EDTA = .00005 mol The next step is to find the amount, in milligrams, of calcium that are in the calcium. Multiplying the number of moles by the molar mass will give us the answer in the grams. In order to convert to milligrams, we must multiply the mass by 1000. Moles(Ca 2+ ) = .00005 mol Mass(Ca 2+ ) = (.00005) x (40.078) Molar Mass (Ca) = 40.078 g Mass(Ca 2+ ) = .002 g = 2 mg Since there are 2 mg of calcium in 2.5 ml of Dean’s 2% milk, then there must be .8 mg in 1.0 ml of Dean’s 2% milk....
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