aqa a level chem amount of substance question help
Hi everyone, I'm having trouble understanding how to solve this question and looking at the mark scheme confuses me even more lol. If anyone could explain how to tackle this question as well as the reasoning behind it, it would be great thanks.
Compound L (Mr = 88.0) contains carbon, hydrogen and oxygen only.
A 6.56 × 10–4 mol sample of L burns completely in air to form 2.62 × 10–3 mol of
water and 2.62 × 10–3 mol of carbon dioxide.
Deduce the formula of L.
Show your working.
M1 (2.62x10^-3)/(6.56x10^-4) = 4 or (6.56x10^-4)/(2.62x10^-3) = 0.25
M2 Hence 4CO2 + 4H2O
M3 So 4C and 8H in L
M4 Hence 2O so C4H8O2
C3H4O3 scores 1 if no other mark scored
Alternative method
M1
nH in L = 5.24 × 10-3
Hence mass H = 5.24 × 10-3 g
M2
n C in L = 2.62 × 10-3
Hence mass C = 2.62 × 10-3 × 12
= 3.144 × 10-2 g
M3
Mass L = 6.56 × 10-4 × 88
= 0.057728 g
Mass O = 0.057728 - ( 5.24 × 10-3 + 3.144 × 10-2)
= 0.021048 g
M4
nO = 0.021048 / 16
= 1.3155 × 10-3
MF = (88/44) x C2H4O
= C4H8O2
C3H4O3 scores 1 if no other mark scored
Compound L (Mr = 88.0) contains carbon, hydrogen and oxygen only.
A 6.56 × 10–4 mol sample of L burns completely in air to form 2.62 × 10–3 mol of
water and 2.62 × 10–3 mol of carbon dioxide.
Deduce the formula of L.
Show your working.
M1 (2.62x10^-3)/(6.56x10^-4) = 4 or (6.56x10^-4)/(2.62x10^-3) = 0.25
M2 Hence 4CO2 + 4H2O
M3 So 4C and 8H in L
M4 Hence 2O so C4H8O2
C3H4O3 scores 1 if no other mark scored
Alternative method
M1
nH in L = 5.24 × 10-3
Hence mass H = 5.24 × 10-3 g
M2
n C in L = 2.62 × 10-3
Hence mass C = 2.62 × 10-3 × 12
= 3.144 × 10-2 g
M3
Mass L = 6.56 × 10-4 × 88
= 0.057728 g
Mass O = 0.057728 - ( 5.24 × 10-3 + 3.144 × 10-2)
= 0.021048 g
M4
nO = 0.021048 / 16
= 1.3155 × 10-3
MF = (88/44) x C2H4O
= C4H8O2
C3H4O3 scores 1 if no other mark scored
Reply 1
I’d recommend writing out like an equation to visualise it easier (L -> ?CO2 + ?H2O where ? is the ratio number)
First you’d find the ratio difference by dividing them by each other (which is 4 in this case) so now yk that ? = 4
Now you can assume there are 4 C and 8 H (because in 4H2 it would be multiplied together) as the equation needs to be balanced (not 12 O because you have to consider that O2 is being taken in during combustion)
so because the mr of L is 88 and mr of C is 12 so 12x4 =48 and mr of H is 1 so 8x1 =8therefore 48 + 8 =56 and 88 - 56 = 32.
Since the only element left over is oxygen and the mr of oxygen is 16 , 32/2 =16 so there would be 2 oxygens
therefore the formula of L would be C4H4O2
hope this helps!!!😛
First you’d find the ratio difference by dividing them by each other (which is 4 in this case) so now yk that ? = 4
Now you can assume there are 4 C and 8 H (because in 4H2 it would be multiplied together) as the equation needs to be balanced (not 12 O because you have to consider that O2 is being taken in during combustion)
so because the mr of L is 88 and mr of C is 12 so 12x4 =48 and mr of H is 1 so 8x1 =8therefore 48 + 8 =56 and 88 - 56 = 32.
Since the only element left over is oxygen and the mr of oxygen is 16 , 32/2 =16 so there would be 2 oxygens
therefore the formula of L would be C4H4O2
hope this helps!!!😛
Reply 2
makkuro
OP
3
Original post
by r378x_I’d recommend writing out like an equation to visualise it easier (L -> ?CO2 + ?H2O where ? is the ratio number)
First you’d find the ratio difference by dividing them by each other (which is 4 in this case) so now yk that ? = 4
Now you can assume there are 4 C and 8 H (because in 4H2 it would be multiplied together) as the equation needs to be balanced (not 12 O because you have to consider that O2 is being taken in during combustion)
so because the mr of L is 88 and mr of C is 12 so 12x4 =48 and mr of H is 1 so 8x1 =8therefore 48 + 8 =56 and 88 - 56 = 32.
Since the only element left over is oxygen and the mr of oxygen is 16 , 32/2 =16 so there would be 2 oxygens
therefore the formula of L would be C4H4O2
hope this helps!!!😛
First you’d find the ratio difference by dividing them by each other (which is 4 in this case) so now yk that ? = 4
Now you can assume there are 4 C and 8 H (because in 4H2 it would be multiplied together) as the equation needs to be balanced (not 12 O because you have to consider that O2 is being taken in during combustion)
so because the mr of L is 88 and mr of C is 12 so 12x4 =48 and mr of H is 1 so 8x1 =8therefore 48 + 8 =56 and 88 - 56 = 32.
Since the only element left over is oxygen and the mr of oxygen is 16 , 32/2 =16 so there would be 2 oxygens
therefore the formula of L would be C4H4O2
hope this helps!!!😛
Hi thanks for the reply, your explanation really helps! I'm only confused on how you get the ratio difference, why do you divide the moles of H20/CO2 by the mol sample of L?
Reply 3
Original post
by makkuroHi thanks for the reply, your explanation really helps! I'm only confused on how you get the ratio difference, why do you divide the moles of H20/CO2 by the mol sample of L?
Because that gives the mole ratio of H2O : CO2: L in the form n : m : 1
This makes it fairly straightforward to tell how many carbons and hydrogens there are per molecule of L (i.e one molecule of L should contain 2n hydrogens and m carbons).
(edited 2 months ago)
Reply 4
Hi yep it’s basically how ‘TypicalNerd’ explained. Dividing them by each other allows you to compare how many C’s and H’s in L there are compared to how many there are in CO2 and H2O
Reply 5
Original post
by makkuroHi thanks for the reply, your explanation really helps! I'm only confused on how you get the ratio difference, why do you divide the moles of H20/CO2 by the mol sample of L?
By diving u get the mole ratio of CO2 : L & H2O🤔 I.e. 4:1
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