Question

determination	1	2
mass of Zn strip, g	25.469	23.887
mass of empty evaporating dish, g	23.274	20.740
mass of dried unreated Zn, g	24.678	23.372
mass of evaportating dish plus Cu, g (first heating)	24.741	24.499
second heating	24.066	21.240
third heating	23.980	21.240

mass of copper chloride in 25.0 ml of solution, g = 2.018g

PLEASE SHOW ALL WORK FOR THESE PROBLEMS AS WELL AS BRIEFLY EXPLAINING WHEN NECESSARY PLEASE AND THANK YOU!

1. (A) Is the calculated empirical formula the same for both determinations? Should the formula be the same for both? Why?

(B) Write the experimentally determined empiral formula for Cu(x)Cl(y) into the chemical equation below. Then balance the equation.

____________ (aq, blue) + Zn (s, silvery) ---> Cu(s,brown) +ZnCl2(aq, colorless)

(C) Based on the data from each determination, calculate the mass of Zn consumed by the reaction described by the chemical equation in part B.

(D) Calcualte the number of moles of Zn consumed for each determination.

(E) Compare the molar ration of Cu to Zn in your balanced chemical equation in (B) with the molar ratio determined using your experimental results. Briefly account for any differences between the ratios.

2. (A) Calculate the mass of Cu produced in both determinations.

determination 1:_______ determination 2:________

(B) Calculate the mass of chlorine (Cl) in your original Cu(x)Cl(y) sample.

determination 1:_______ determination 2:________

which of the reactants was the limiting factor? Which reactant was in excess?

(C) Calculate the number of moles of Cu present in your Cu(x)Cl(y) sample and calculate the number of moles of Cl present in the Cu(x)Cl(y) sample.

mole of Cu determination 1:_______ determination 2:________

moles of Cl determination 1:_______ determination 2:________

(D) Determine the molar ratio of Cu to Cl in Cu(x)Cl(y) then use the ration to write the empirical formula for copper chloride.

determination 1:_______ determination 2:________

(E) Determine the pecent of Cu and the percent of Cl in copper(II) chloride using the formula CuCl2

(F) Using the mass of Zn reacted in each determination calculate the theoretical yield of Cu for each using the below equation

theoretical yield of Cu = (mass of Zn reacted) x 1molZn/65.38g Zn x 1mol Cu/1mol Zn x 63.55g Cu/1mol Cu

Finally determine the percent yield from each of the determinations for Cu using the following equation

percent yield of Cu= actual mass of Cu produced/ theoretical yield of Cu X 100

If the yeild is not 100% explain why it is not.

PLEASE SHOW ALL WORK FOR THESE PROBLEMS AS WELL AS BRIEFLY EXPLAINING WHEN NECESSARY PLEASE AND THANK YOU!

Answer 1

Experiment 1 )

Wt. of copper ............ % ......No. of.moles ................................whole No. ratio

0.706 gms.................34.99......34.99 / 63.5 =...0.55...............................1

wt. of chlorine...............% .........No .of moles ..........................whole number ratio

1.312 gms................65.01........65.01 /35.5 =.1.831...............................3

Experiment 2 )

Wt of copper ..................% .................No. of moles ---------------whole no. ratio   

0.70 gms......................34.69....................0.55...................................1

wt. of chlorine...................%................No.of moles.......................whole number ratio

1.318 gms.......................65.31................ 1.84...................................3

a Calculated empirical formula is same in both the experiments.

The formula should be same , because law of constant composition ' holds good

b) Cu Cl 2+ Zn (s., silvery) = Cu (s. brown ) + ZnCl2 (colorless)

mass of zinc consumed ...........Expt 1 .........................................................Expt. 2

..........................................25.469 - 24.678 = 0.791gms...........................23.887 -23.372 = 0.515 gms

no. of mols of Zn..................0.791 / 65.37 = 1.2 x 10^-2------------------------------0.515 / 65.37 = 7.87 x 10^-3

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Glad to help . Please post the remaining part ( ie. E ) and also the other question separately for detailed answers.