Lab Experiment #603
Resolving a Two-Component Mixture
Binary Mixture: a mixture containing only two components. Such a mixture can be separated into its components using a variety of methods, both physical and chemical.
Physical separation methods are based on differences between the physical properties of components.
Decantation: a physical method of separation where the solid is allowed to settle at the bottom of the container. The liquid, called the supernatant liquid (or for purposes of brevity, supernate)carefully poured off into another container without disturbing the solid.
Filtration: a physical method of separation involving pouring the mixture into a pourous material, whereas the solid (residue) is unable to pass through and the liquid (filtrate) passes through easily.
Evaporation: physical method of separation during which a solution is heated to vaporize and thus remove the solvent. The solid remaining after total evaporation of solvent is the solute.
Extraction: best used when one component of a binary mixture is soluble in a particular solvent and the other component is not. We add binary mixture to solvent, dissolving one component and then seperate the mixture by means of filtration.
Worked example #1:
- Determine the percent recovery of silver bromide (AgBr) and potassium bromide (KBr) when separated from a sample of a binary mixture weighing 2.18g. Water was added to the mixture with stirring in order to extract the KBr. The remaining solid, AgBr, was collected on a piece of filter paper weighing 0.88g. When dried the mass of the paper plus the dry AgBr was 1.82g. The filtrate was collected in a beaker weighing 69.15g. After evaporation and cooling the beaker plus the residue weighed 70.33g.
#1. Calculate the mass of AgBr recovered
#2. Calculate the mass of KBr recovered.
#3. Calculate the percent AgBr in the mixture.
#4. Calculate the percent KBr in the mixture.
#5. Calculate the percent recovery of the mixture components.
#1. To calculate the amount of mass of the silver bromide we recovered we must subtract the mass of the filter paper from the mass of dried filter paper + AgBr, hence: 1.82g - 0.88g = 0.94g
So we recovered 0.94g of AgBr.
#2. The calculate the mass of KBr we recovered, we must subtract the beaker weight from the weight of beaker + residue to isolate the weight of the residue, hence: 70.33g -69.15g = 1.18g
So we recovered 1.18g of KBr.
#3. Our intial mixture weighed 2.18g, so to calculate the percent of AgBr in the mixture, we divide the amount of AgBr we recovered (from calculation #1) and divide it by initial sample and multiply it by 100 (to gain amount in %), %AgBr = gAgBr recovered/intial sample X 100, hence:
0.94g/2.18g=0.43, 0.43X100 = 43%
So our mixture had 43% AgBr in it.
#4. Once again, we take the recovered amount of KBr and divide it by our initial sample weight to determine the percentage of KBr that was in the initial sample. We use the same formula as we did with AgBr. 1.18g/2.18g = 0.54, 0.54X100= 54%
Therefore, we determine that our initial sample had 54% KBr in it.
#5 To calculate our overall percent of recovery of mixture components, we add the amounts we recovered for AgBr and KBr and then divide them by the weight of our initial sample, then multiply the result by 100 to get the results in terms of percentage.
Amount of AgBr recovered: 0.94
Amount of KBr recovered: 1.18
Weight of initial sample: 2.18
Therefore: 0.94+1.18 = 2.12
2.12/2.18 = 0.97
0.97(100)= 97%
So overall we recovered 97% of our components, NOT BAD!!!