An unbranced alkene A reaction with hydrogen formed a compound B. On analysis, B had the composition by mass of C, 82%, H , 17.2% with a relative molecular mass of 58.
Calculate the empirical formula and molecular formula of B
For empirical formula is it CH2 or CH3? Not sure how to calculate molecular formula though..
C H
% 82 17.8
/AR 6.83 17.8
/small no. 1 2.5
Therefore empirical formula is C2H5.
To get the molecular formala
58 / ( 2X 12+ 5 X1 ) = 2
Therefore, molecular formula is 2 (C2H5) = C4H10
B is butane so A should be butene
2. What bonds are made when butane burns?
Since H2O and CO2 are produced, so covalent bonds? lol.
Originally posted by anpanman:2. What bonds are made when butane burns?
Since H2O and CO2 are produced, so covalent bonds? lol.
Got so weird qn de ah. Should be covalent bah even if i see no sense in this question...
Originally posted by dkcx:C H
% 82 17.8
/AR 6.83 17.8
/small no. 1 2.5
Therefore empirical formula is C2H5.
To get the molecular formala
58 / ( 2X 12+ 5 X1 ) = 2
Therefore, molecular formula is 2 (C2H5) = C4H10
B is butane so A should be butene
Oh, so you have to multiply to get a whole number. HAHA. I thought you have to round up or round down. Thanks. By the way, have you had any idea what 3 other substances present in chimney gas besides CO2, SO2 and oxides of nitrogen? I think CO is one of them but what about other substances?
Thanks
Originally posted by anpanman:Oh, so you have to multiply to get a whole number. HAHA. I thought you have to round up or round down. Thanks. By the way, have you had any idea what 3 other substances present in chimney gas besides CO2, SO2 and oxides of nitrogen? I think CO is one of them but what about other substances?
Thanks
Should have CO and maybe unburnt C as well which is what makes the smoke black. Not too sure what else, try googling.