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Determining empirical formula for hydrated cobalt chloride compound. RESULTS, DATA...

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stella01 | Student, College Freshman | eNoter

Posted May 5, 2011 at 5:38 PM via web

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Determining empirical formula for hydrated cobalt chloride compound.

RESULTS, DATA CHART, QUESTION LISTED BELOW

Results:
Trial 1
Mass of test tube=10.41g
Mass of test tube&hydrated cobalt chloride=12.43g
Mass of test tube and contents after heating=11.52g

Trial 2
Mass of test tube=10.42g
Mass of test tube and hydrated cobalt chloride=13.06g
Mass of test tube & contents after heating=11.86g

Data Chart:so far I hav gotten this
Trial #1 & Trial #2

Mass of test tube & hydrated cobalt chloride
Trial #1=12.43g   Trial#2=13.06g
Mass of test tube
Trial #1=10.41g    Trial#2=10.42g
Mass of hydrated cobalt chloride
Trial #1=2.02g   Trial#2=2.64g
Mass of test tube and cobalt chloride after heating
Trial #1=11.52g   Trial#2=11.86g
Mass of Cobalt Chloride
Trial #1=1.11g   Trial#2=1.44g
Mass of H2O Trial #1=??   Trial#2=??
Mole of cobalt chloride(NOT hydrated)
Trial #1=??   Trial#2=??
Mole of H20 Trial #1=??   Trial#2=??
Mole ratio-H20/Cobalt chloride
Trial #1=??   Trial#2=??
Mass of H2O predicted from mole ratio
Trial #1=??   Trial#2=??
Percentage error Trial #1=??   Trial#2=??
NEED HELP WITH THE QUESTION(??)MARKS

Questions(where possible show work)
1Find initial &final mass of cobalt chloride compound
2mass of water that was driven off?
3# of mole of dry CoCl2 there were
4# of mole of H2O driven off?
5Find mole ratio of mole of H2O to mole of CoCl2 &formula for hydrated compound in the form of CoCl2*xH20
6Find actual % by mass of water in CoCl2*6H2O.This compound has 1Co+2Cl+12H+6Ototal to find molar mass
7Mass of H2O shud hav driven off wid no error?AND % error for each trial

4 Answers | Add Yours

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msscience2004 | High School Teacher | eNoter

Posted May 28, 2011 at 1:10 PM (Answer #1)

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Mass of H2O

You simply deduce the mass of hydrated chloride to the mass of cobalt chloride which is after heating process.

#1= 2.02- 1.11= 0.91g

#2= 2.64-1.44= 1.2g

mole of unhydrated CoCl2

You need to determine first the molar mass of CoCl2

CoCl2= 129.83 g/mol

#1.) 1.11 g CoCl2 ( 1mol/ 129.83 g CoCl2) =0.0085 mol CoCl2

#2.) 1.44 ( 1mol/ 129.83 g CoCl2)= 0.0110 mol CoCl2

mole of H2O

from the mass of water we can get the mole by using the molar mass of water which is

H2O= 18. 016 g/mol

#1.) 0.91 g H2O ( mol/ 18.016g) = 0.0505 mol H2O

#2) 1.2 g ( mol/ 18.016g) = 0.0666 mol H2O

mole ratio of H2O to CoCl2

you simply use the mole of water divided by the moles of CoCl2

#1.) 0.0505 mol H2O/0.0085 mol CoCl2= 5.94 or 6

formula of hydrated CoCl2= CoCl2 6H20

#2.) 0.0666 mol H2O/0.0110 mol CoCl2=6.05

formula of hydrated CoCl2= CoCl2 6H20

therefore there are 6 moles of water per mole of CoCl2 based from the mole ratio acquired which is 6.

finding the actual % of water from the CoCl2 6H20

you will use the actual mass of the sample. use a littel stoichiometry to derive the mass of water. here it is how:

you need to determine first the molar mass of CoCl2 6H20

#1.) 2.02 g CoCl2 6H2O ( mol CoCl26H2O/ 237.926 g) = 0.0084 mol

0.0084 mol CoCl2 6 H2O ( 6 mol H2O/ 1 mol CoCl2) (18.016 gH2O/ 1mol H2O) = 0.908 g H2O

#2.) 2.64 g CoCl2 6H2O ( mol CoCl26H2O/ 237.926 g) =0.0110 mol

0.0110 molCoCl2 6 H2O ( 6 mol H2O/ 1 mol CoCl2) (18.016 gH2O/ 1mol H2O) = 1.189 g H2O

determining your percent error from the theoretical yield and actual yield

% error =   | your result - accepted value | x  100 %
accepted value

#1.) % error= 0.91-0.908/ 0.908 X 100 = 0.22 %

#.2) % error= 1.2- 1.189/1.189 X 100= 0.92 %

 

 

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stella01 | Student, College Freshman | eNoter

Posted June 1, 2011 at 9:24 AM (Answer #2)

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thanks for the help but I dont understand it like which answer is for which if you can indicate the values for each of the question so I can scroll up to see your shown work. Thanks =)

0.91g the initial mass of cobalt chloride and 1.2g the final mass?

whats the mass of water that was driven off?

number of mole of dry CoCl2 there were

number of mole of H2O driven off?

5Find mole ratio of mole of H2O to mole of CoCl2 &formula for hydrated compound in the form of CoCl2*xH20

6Find actual % by mass of water in CoCl2*6H2O.This compound has 1Co+2Cl+12H+6Ototal to find molar mass

7Mass of H2O shud hav driven off wid no error?AND % error for each trial

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stella01 | Student, College Freshman | eNoter

Posted June 5, 2011 at 12:33 PM (Answer #3)

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Sorry, I looked into it in depth and I understood the chart. Thanks

If you could answer these questions that would be great.

1 Determine the mole ratio of moles of H2O to moles of CoCl2 and the formula for the hydrated compound in the form of CoCl2* XH2O

2 Determine the actual percent by mass of the water in CoCl2* 6H2O.
this compound has 1Co + 2Cl + 12H + 6O to total in order to find the molar mass

3 Determine the mass of water that should have been driven off in each trial assuming no sources of error.

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ndnordic | High School Teacher | (Level 2) Associate Educator

Posted June 26, 2011 at 11:33 PM (Answer #4)

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Start out by writing the decomposition reaction that is taking place.

CoCl2(xHOH) -->  CoCl2    +  x HOH

You are trying to determine the value of x os you can find the formula of the hydrate.

Trial 1:  You started with 12.43-10.41 = 2.02 g of hydrate.  You ended with 11.52-10.41 = 1.11 g of CoCl2.  The difference is 0.91 g of water.

Molecular mass of CoCl2 = 129.83 and molecular mass of water is 198.

Trial 1 you produced 1.11/129.83 = 0.0085 moles of CoCl2 and 0.91/18 = 0.05055 moles of water.  Mole ratio is 0.05055/.0085 = 5.95:1.

Trial 2.  You started with 13.06-10.42 = 2.64 g of hydrate. You produced 11.86-10.42 = 1.44 g of CoCl2 + 1.2 g of water.

1.44/129.83 = 0.0111 moles of CoCl2 and 1.2/18 = 0.0666 moles of water.  mole ratio is 0.0666=0.0111 = 6.00 : 1.

So the correct formula for the hydrate is 6 moles of water per mole of CoCl2.  CoCl2-6HOH and the molar mass is 237.83.

In trial 1 you started with 12.43 - 10.41 = 2.02 g of hydrate.  2.02/237.83 = 0.0085 mol.  You should produce 0.912 g and you did so percent error = 0.

In trial 2 you started with 13.06- 10.42 = 2.64 g of hyrate. 2.64/237.83 = 0.0111 moles. You should produce 1.2 g of water and you did so percent error = 0 again.

 

 

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