Question

A 1.149 g sample contains only vitamin C (C6H8O6) and sucralose (C12H19Cl3O8). When the sample is dissolved in water to a total volume of 41.1 mL, the osmotic pressure of the solution is 3.18 atm at 287 K. What is the mass percent of vitamin C and sucralose in the sample?
a) mass percent vitamin C:%?
b) mass percent sucralose:%?

Answer 1

Answer:

See explanation below

Explanation:

in this case, the expression for osmotic pressure is the following:

π = MRT

We have the temperature of 287 K, we have R which is the constant and is 0.08206 L atm / mol K, and the pressure of 3.18 atm.

We do not have molarity but that does not matter, we can express that in moles and volume:

π = nRT/V

From here, we solve for n, which would be the total moles of the sample that contains the vitamin and sucralose. So let's solve first for the moles:

n = πV/RT

n = 3.18 * (0.0411) / 0.08206 * 287

n = 0.555 moles

Now, we do know that these total moles, is the sum of the moles between the vitamin C and the sucralose:

n = nvit + nsucr

the vitamin C I will call it X and the sucralose would be Y.

Now, in order to calculate the %mass we need to know the mass of the vitamin and the sucralose. To calculate mass:

m = n * MM

the molecular mass of vitamin C and sucralose are 176 g/mol and 397.35 g/mol. So, to calculate the mass of each compound, we need to express one of them in function of another. In other words:

n = x + y

Solve for x:

y = n - x (1)

The moles for each compound:

x = mx/MMx (2)

y = my/MMy   -----> my = y * MMy  (3)

Replacing (2) in (1):

y = n - (mx/MMx)  (4)

Now, we do the same thing but we the mass:

1.149 = mx + my

my = 1.149 - mx (5)

Replacing (3) in (5):

y * MMy = 1.149 - mx

y = 1.149 - mx/MMy  (6)

Finally, we replace (6) in (4) and solve for mx:

1.149 - mx/MMy = n - mx/MMx

(1.149 - mx/397.35) = (0.555 - mx/176)

176(1.149 - mx) = 397.35(0.555 - mx)

87.324 - 176mx = 220.53 - 397.35mx

397.35mx - 176mx = 220.53 - 87.324

221.35mx = 133.21

mx = 133.21/221.35

mx = 0.602 g

Now for the mass of sucralose:

my = 1.149 - 0.602

my = 0.547 g

Finally the %mass of each:

%x = 0.602/1.149 * 100 = 52.39%

%y = 100 - 52.39 = 47.61 %

Answer 2

The mass percentage of vitamin C is 74.5% and the mass percentage of sucralose is 25.5%.

Osmotic pressure (π) = iCRT

Where;

π = Osmotic pressure = 3.18 atm

i = Van't Hoff factor = 1 (molecular solution)

C = concentration (in mol/L)

R = gas constant = 0.082 LatmK-1mol-1

T = temperature =  287 K

But C = number of moles(n) /Volume (V)

volume = 41.1 mL or 0.0411 L

Substituting into the given equation;

π = n/V × RT

Make n the subject of the formula;

n = πV/RT

n = 3.18 atm ×  0.0411 L/0.082 LatmK-1mol-1 × 287 K

n = 0.0056 moles

Mass of vitamin C = a

Mass of sucralose = b

Mass of sample = 1.149 g

Hence;

a + b = 1.149

b = 1.149 - a

Molar mass of vitamin C = 176 g/mol

Molar mass of sucralose = 398 g/mol

Total number of moles = number of moles of vitamin C + number of moles of sucralose

 0.0056  moles = a/176  + (1.149 - a)/398

70048 ( 0.0056 ) = 398a + 176 (1.149 - a)

392.269 = 398a + 202.224 - 176a

392.269 - 202.224 = 398a  - 176a

190.045 = 222a

a = 190.045/222

a = 0.856 g

But;

b = 1.149 - a

Where a =  0.856 g

b = 1.149 - 0.856

b = 0.293 g

Mass percent vitamin C =  0.856 g/1.149 g × 100/1

= 74.5%

Mass percent sucralose = 0.293/1.149 g × 100/1

= 25.5%

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