CHEM1 15 May 2012 AQA OFFICIAL POST EXAM

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Wasn't the molecular formula ratio 1:2? so it was BCl2???

That would be the emperical formula. But they gave you the total molecular mass of the compound which was doublt the molecular mass of BCl2, so it's B2Cl4.

Reply 61

CelineG

Original post by firoza59

what did u guys get for atom economy question??

It was 8.98 % but I rounded it up to 9%!
It was the molecular mass of the desired product divided by the molecular mass of the reactants times 100!

Reply 62

DisturbingKand0R

Original post by firoza59

and PH3 what that trignol planner??

trigonal pryramidal, like ammonia, it had a lone pair

Reply 63

purplebells

What did everyone put for the reagent for converting the alkene into an alkane? It wasn't on the spec so I guessed H2SO4.

Reply 64

AishaTara

can we have an unofficial mark scheme?

Reply 65

DisturbingKand0R

Original post by CelineG

It was 8.98 % but I rounded it up to 9%!
It was the molecular mass of the desired product divided by the molecular mass of the reactants times 100!

same i rounded it up to 9% they never said if it should be to 3sf or 1dp

Reply 66

DisturbingKand0R

Original post by purplebells

What did everyone put for the reagent for converting the alkene into an alkane? It wasn't on the spec so I guessed H2SO4.

H2 (hydrogen) as it was hydrogenation

Reply 67

An Obese Turtle

Original post by Scienceisgood

Got the same as me.
I tried it with 20 and 50 though. Got 86.4
Tried it with another two sets of numbers, got 86.4 again.

So, I think 86.4 is a safe bet.

For the mass, did anyone get 2.18?
Also, for the last Q, I got B2Cl7. =l

Last Q was B2Cl4

The mass you had to take into account the water used up in the reaction and take that amount away from the 500cm3 - almost everyone apart from me at my 6th form didnt get that one

Hopefully grade boundaries are now :smile:

Reply 68

Poolwizard

Original post by purplebells

What did everyone put for the reagent for converting the alkene into an alkane? It wasn't on the spec so I guessed H2SO4.

i put that as well

Reply 69

DisturbingKand0R

Original post by CelineG

It was 8.98 % but I rounded it up to 9%!
It was the molecular mass of the desired product divided by the molecular mass of the reactants times 100!

personally i prefer dividing by total Mr of PRODUCTS, (conservation of mass/energy)
as you have done half the work already

Reply 70

Poolwizard

Original post by DisturbingKand0R

NaBO2 is right, the 'ate' means it has oxygen in it, and i just wiki.ed it

but did you balance it? i tried for about 5 minutes but didn't work.

Reply 71

DisturbingKand0R

Original post by Poolwizard

but did you balance it? i tried for about 5 minutes but didn't work.

yeah it did for me , there was 2H2O as well

Reply 72

thetransitionary

Not sure if I did Pv=nRT correct as i didn't divide, ended up with 14g, most people in my class seem to have to that, people who did divide ended up with about 2g and some others got 28g.

I got the concentration question to be about 2.something mol dm^-3

Also for relative mass I just did the [(85*2.5)+(87*1)] / 3.5 to get 85.6

How were we supposed know Sodium Borate? I took a guess at NaBO2

Also put H2 for turning X into Y

Unofficial mark scheme would be handy, overall I didn't think the paper was too bad

Reply 73

Lily.K

Original post by DisturbingKand0R

but there was more of the 87 isotope, therefore the number has to be close to 87
so probably around 86.4

That's the wrong way round, 85 was 2.5 times the abundance of 87

Reply 74

Poolwizard

Original post by lochbeau

Was B2Cl4 for definite I think, I think I done 35.5/86.3? and got 0.4, then done 10.8/13.7? and got 0.8, so 1:2 ratio, which would have meant empirical formula was BCl2, that was half of 163.6, so just doubled up and got B2Cl4. And let us hope 86.4 is right :tongue: