really hard exam question please help me!! Watch

usernamenew
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Isooctane reacts with an excess of chlorine to form mixture of chlorinated compounds, one of these compunds has 24.6%Carbon 2.56Hydrogen 72.8%Chlorine, calculate molecular formula

so I calculated the empirical and got 1:1.25:1 and the markscheme says multiply that by 4 to give the empricial formula C4H5CL4
but I don't understand why u multiply by 2 to give the molecular formula C8H10CLl8?why ?? how do u get the molecular formula? normally they give the rmm of the molecule and then u have to calculate the empirical mass which for C4H5C4 is 195?
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charco
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(Original post by usernamenew)
Isooctane reacts with an excess of chlorine to form mixture of chlorinated compounds, one of these compunds has 24.6%Carbon 2.56Hydrogen 72.8%Chlorine, calculate molecular formula

so I calculated the empirical and got 1:1.25:1 and the markscheme says multiply that by 4 to give the empricial formula C4H5CL4
but I don't understand why u multiply by 2 to give the molecular formula C8H10CLl8?why ?? how do u get the molecular formula? normally they give the rmm of the molecule and then u have to calculate the empirical mass which for C4H5C4 is 195?
You have not given us all of the required information (like the actual question)
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usernamenew
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(Original post by charco)
You have not given us all of the required information (like the actual question)
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charco
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Isooctane has the formula C8H18
Monochloroisooctane has the formula C8H17Cl

There are two isotopic products one with a Mr of 148 and the other with Mr = 150

The ratio of the two products is given as 1.5 to 1.0 so the average Mr = [(3 x 148)+(2 x 150)]/5 = 148.8

24.6% carbon
2.56% hydrogen
(100 - 24.6-2.56)% chlorine

2.05 C; 2.56 H; 2.05 Cl

C; 1.25H; Cl

Multiply through by 4 to remove fraction.

C4H5Cl4

BUT you are told that it's a chlorinated octane, i.e it has eight carbon atoms, so you must MTB 2 to get eight C atoms

Molecular formula C8H10Cl8
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usernamenew
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(Original post by charco)
Isooctane has the formula C8H18
Monochloroisooctane has the formula C8H17Cl

There are two isotopic products one with a Mr of 148 and the other with Mr = 150

The ratio of the two products is given as 1.5 to 1.0 so the average Mr = [(3 x 148)+(2 x 150)]/5 = 148.8

24.6% carbon
2.56% hydrogen
(100 - 24.6-2.56)% chlorine

2.05 C; 2.56 H; 2.05 Cl

C; 1.25H; Cl

Multiply through by 4 to remove fraction.

C4H5Cl4

BUT you are told that it's a chlorinated octane, i.e it has eight carbon atoms, so you must MTB 2 to get eight C atoms

Molecular formula C8H10Cl8
thank u so much for explaining
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