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A dry-cleaning solvent (molecular weight = 146.99 g/mol) that contains C, H, and Cl is suspected to be a?

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A dry-cleaning solvent (molecular weight = 146.99 g/mol) that contains C, H, and Cl is suspected to be a?

Postby landers » Wed Sep 14, 2011 1:21 pm

a cancer-causing agent. When a 0.367 g sample was studied by combustion analysis, 0.659 g of CO2 and 0.0892 g of H2O formed.
How many moles of each of the following were in the original sample?
_____ moles of C
_____ moles of H
_____ moles of Cl

I got 1.50 x 10^(-2) moles of C, 1.00 x 10^(-2) moles of H, and 5.01 x 10^(-3) moles of Cl. My moles of hydrogen is wrong. Can someone help me?
landers
 
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Joined: Fri Apr 01, 2011 11:06 pm

A dry-cleaning solvent (molecular weight = 146.99 g/mol) that contains C, H, and Cl is suspected to be a?

Postby celestine » Wed Sep 14, 2011 1:33 pm

here is your solution and the molecualr formula ..

# moles = mass / molar mass and molar mass of CO2 = 44 g/ mole so
0.659 g of CO2 has 0.659 /44 = 0.0150 moles of CO2
there is 1 mole of C in CO2 and all the C from the compound becomes CO2 so moles of C in the compound = 0.0150 moles
mass of C = 0.0150 x 12 = 0.1797 g

molar mass of H2O = 18 g/ mole
0.089 g of H2O has 0.0892 / 18 = 0.0050 moles of H2O
there are 2 moles of H in H2O so moles of H in the compound = 0.0099 moles
mass of H = 0.0099 x 1.0079 = 0.0100 g

mass of H + C = 0.1897 g
mass of sample = 0.3670 g
mass of Cl by difference = 0.1773 g
moles of Cl = 0.0050 moles

molar ratio of C : H : Cl = 0.0150 : 0.0099 : 0.0050
smallest number 0.0050
divide the ratio by the smallest number we get
molar ratio of C : H : Cl = 3.00 1.98 1

empirical formula is C3H2Cl
this has a formula weight of ( 3x12+2+35.5) or 73.5 g

which is 1/2 the molecular mass

so the molecular formula is 2 x the empirical formula or

C6H4Cl2 dichloro benzene a cancer-causing agent ..
celestine
 
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Joined: Fri Apr 01, 2011 3:26 am


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