A 1.167 g sample contains only vitamin C (C6H8O6) and sucralose (C12H19Cl3O8). When the sample is dissolved in water to a total volume of 35.9 mL, 35.9 mL, the osmotic pressure of the solution is 3.67 atm 3.67 atm at 285 K. 285 K. What is the mass percent of vitamin C and sucralose in the sample?

Mass % Vitamin C: 73.3%

Mass % sucralose: 26.7%

Explanation:

The mass of Vitamin C (a) + Sucralose (b) is 1.167g. Thus, it is possible to write:

a + b = 1.167g (1)

The osmotic pressure formula in solution is:

π = i*M*R*T

Where π is osmotic pressure (3.67atm), i is Van't Hoff factor (1 for Vitamin C and Sucralose), M is molarity of solution (moles / 0.0359L), R is gas constant (0.082atmL/molK) and T is temperature (285K). Replacing:

3.67atm = 1 * moles / 0.0359L * 0.082atmL/molK * 285K

0.00564 = moles of solution (Moles of Vitamin C + Sucralose)

These moles can be written as:

0.00564 mol = a * (1mol / 176.12g) + b * (1mol / 397.64g) (2)

Where 176.12 is molar mass of vitamin C and 397.64 is molar mass of sucralose

Replacing (1) in (2)

0.00564 = (1.167 - b) / 176.12 + b / 397.64

0.000986 = 0.003163 b

0.312g = b. → Mass of sucralose.

Mass of Vitamin C → 1.167g - 0.312g = 0.855g

Thus, Mass percent is:

Mass % Vitamin C: 0.855g / 1.167g * 100 = 73.3%

Mass % sucralose: 0.312g / 1.167g * 100 = 26.7%

Mass of sucrose and sucralose = 1.167 g

Molar mass of sucrose = (C6H8O6)

= (12*6) + (8*1) + (6*16)

= 176 g/mol

Molar mass of = C12H19Cl3O8

= (12*12) + (19*1) + (35.5*3) + (16*8)

= 397.5 g/mol

Pressure, P = 3.67 atm

Volume, V = 35.9 ml

Temperature, T = 285 K

Using PV = nRT

nt = (3.67 * 35.9 * 10^-3) / (285 * 0.08205)

= 0.00563 moles.

Molar mass = mass/number of moles

let x be the mass of sucralose

nt = n1 + n2

0.00563 = (1.167 - x) / 176 + x/397.5

0.00563 = 0.00663 - 0.0057x + 0.00252x

0.001 = 0.00318x

x = 0.3145 g

Mass % = mass of sample/total mass * 100

Sucralose mass % = 0.3145/1.167 * 100

= 26.95%

Sucrose mass % = (1.167 - 0.3145) / 1.167 * 100

= 73.05%