Question

Upon graduation, in a misguided attempt to pay off your student loans more quickly you become involved in the trade of illicit narcotics. Your usual supplier gets eliminated by the competition and you are forced to procure your product from a new supplier. You purchase what your new supplier claims to be codeine (C18H23NO3) of >99% purity but a simple (and ill-advised) taste test leaves you suspicious that it has been thinned with sucrose (C12H22O11). You look up the heat of combustion of codeine (2327 kcal/g) and that of sucrose (2222 kJ/mole). You then take 1.0g of your product and find that it releases 2095 kcal upon combustion. What is the purity of your sample, assuming it contains only codeine and sucrose?

Answer 1

Given the sample contain only codeine and sucrose with an unknown ratio.

Now the heat of combustion of codeine is 2327 kcal/g

i.e 4 C₁₈H₂₃NO₃ + 89 O₂ ------->  72 CO₂ + 46 H₂O + 2 N₂ and ΔH = 2327 kcal/g

Similarly the heat of combustion of sucrose is 2222 kJ/mole

i.e C₁₂H₂₂O₁₁+12 O₂ ------------> 12 CO₂ + 11 H₂O and ΔH = 2222 kJ/mole

= 531.07 kcal / moles

Now we know mass of 1 mole sucrose = 342.3 g/mol

So we can say, from burning of 342.3 g of sucrose, we get 531.07 kcal of heat

So the heat of combustion of 1 g of sucrose = 531.07 kcal / 342.3 g

= 1.55 kcal/g

• Now given when we take 1 g of sample, the total heat released = 2095 kcal
• Now lets assume in our sample, we have x g of codeine and y g of sucrose

So heat released from x g of codeine = x g * 2327 kcal/g

   = 2327x kcal

And heat released from y g of sucrose = y g * 1.55 kcal/g

   = 1.55y kcal

And total heat released from this 1g sample = 2095 kcal

That means 2327x kcal + 1.55y kcal = 2095 kcal

Or 2327x + 1.55y = 2095 ------------------------------------ (1)

Again since total mass of the sample = 1g

We can write x + y = 1 ---------------------------------------- (2)

Now solving the 2 equations-

• 2327x + 1.55y = 2095 ------------------------------------ (1)
• x + y = 1 ---------------------------------- (2)

Or multiplying 1.55 in eqn-2 and substracting it from eqn-1

• 2327x + 1.55y = 2095 ------------------------------------ (1)
• 1.55 x + 1.55 y = 1.55 ---------------------------------- (2)

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2,325.45 x = 2,093.45‬

So x =   2,093.45‬ / 2,325.45

= 0.9

So   y = 1 - 0.9 = 0.1 g

That means if we have a sample of only codeine of 1g, the purity would be = 100%

Since we have actual codeine of 0.9 g, the purity would be = 0.9/1 * 100

= 90%