Question

Saccharin, an artificial sweetner, has the formula C7H5NO3S. Suppose you
have a sample of a saccharin-containing sweetner with a mass of 0.2140 g.
After decomposition to free the sulphur and convert it to the SO4
2-ion, the
sulphate ion is trapped as water-insoluble BaSO4. The quantity of BaSO4
obtained is 0.2070 g. What is the mass percent of saccharin in the sample of
sweetner?

Answer 1

Assuming there were no other sulfur containing compounds in the sample, the same amount of sulfur, in BaSO4, is the same as in C7H5NO3S.

so the moles (n) of BaSO4:

n=m/M=(0.207 g)/(233.43 g/mol)= 0.0008867754787302 moles

since there is only one sulfur atom in the sweetener also, they're in a 1:1 ratio.

find the mass of C7H5NO3S:

m= n*M= (0.0008867754787302 moles)(183.18 g/mol)= 0.16243953 g

so in the 0.2140 g sample, there was 0.16243953 g of saccharin.

i'm sure you can finish the question

Answer 2

thanx