PRE QuESTIONS E-LAB Why does the precipitation of the Cu out of solution go more

Question

PRE QuESTIONS E-LAB Why does the precipitation of the Cu out of solution go more quickly if the Cuis removed from the Al wire periodically? 2. What is the relationship between the number of moles of an element in acompound and its subscript in the empirical formula? 3. A student performs an experiment and finds that for a compound of iron and oxygen, Fe,o, a sample weighing 1.083 g contains 0.757 gFe. SHow ALL CALCULATION SETUPS ple? (Hint: you know the mass of the Fe in the sample) a. How many moles of Fe are in the sam b. How many grams ofoare in the sample? (Hint: the sample only contains Fe and O) c. How many moles of oare in the sam you know the mass of the oin the sample) d. What is the ratio of moles Fe to moles O in the sample? (Determine the mole ratio by dividing smallest number of moles inno all of her subsances) e. What is the empirical formula of iron oxide in smallest whole numbers? 4. A student doing this experiment collects the following data: Mass of Crucible t Sample: 15.937 g Mass of Empty Crucible: 14.912 g Mass of Crucible Sample: 15.720 g (after heating) 0.382 g Mass of Cu Collected: From this data, determine the mass of the original hydrated sample, the mass of the dehydrated sample, the mass of water removed by heating and the mass of chlorine in the sample.

Explanation / Answer

Pre-lab

1. Removing the product Cu changes the equilibrium status of the reaction and according to LeChatellier's principle the shift in equilibrium would be towards the product with more formation of Cu until it reaches equilibrium state again.

2. number of moles of an element is equal to multiple of subscript of element.

3. for the given data,

a. moles Fe = 0.757/55.845 = 0.0135 mol

b. grams O = 1.083 - 0.757 = 0.326 g

c. moles O = 0.326/16 = 0.0204 mol

d. mole ratio (Fe/O) = 0.662

e. Empirical formula = Fe2O3

4. From the data

mass of original hydrated sample = 15.937 - 14.912 = 1.025 g

mass of dehydrated sample = 15.720 - 14.912 = 0.808 g

mass of water = 1.025 - 0.808 = 0.217 g

mass of Cl = 0.808 - 0.382 = 0.426 g

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