Asked by Karisa
                A certain compound contains 8.74% Carbon, 13.8% Fluorine, 77.4% Chlorine by weight. A possible molecular formula for this compound might be?
I think the answer is CF2Cl26
multipy mass of each element by the percent:
c: 12.01*.0874=1.049674
F: 19*.138=2.622
Cl: 35.45*.774=27.4383
Then divide each by 1.05 because you divide by the smallest number and get C-1 F-2.049 Cl-26.1317
            
        I think the answer is CF2Cl26
multipy mass of each element by the percent:
c: 12.01*.0874=1.049674
F: 19*.138=2.622
Cl: 35.45*.774=27.4383
Then divide each by 1.05 because you divide by the smallest number and get C-1 F-2.049 Cl-26.1317
Answers
                    Answered by
            DrBob222
            
    NOPE.
You divide, not multiply.
mols C = 8.74/12.01 = ?
mols F = 13.8/19 = ?
mols Cl = 77.4/35,45 = ?
Then find the ratio as you did. I get CFCl2.
    
You divide, not multiply.
mols C = 8.74/12.01 = ?
mols F = 13.8/19 = ?
mols Cl = 77.4/35,45 = ?
Then find the ratio as you did. I get CFCl2.
                    Answered by
            Karisa
            
    So:
C=.7277
F=.7263
Cl=2.183
then
C=1
F=0.99
Cl=2.9998
so wouldn't Cl round up to 3?
CFCl3
    
C=.7277
F=.7263
Cl=2.183
then
C=1
F=0.99
Cl=2.9998
so wouldn't Cl round up to 3?
CFCl3
                    Answered by
            DrBob222
            
    Yes it does. Rounding to the nearest whole number you have C1, F1, Cl3 and I made a typo. So the empirical formula is CFCl3. I'm glad  you caught that.
    
                    Answered by
            Karisa
            
    Thank you!
    
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