Question

1. Saccharin is an artificial sweetener that was discovered in 1879 when a chemist spilled some of his product on his sandwich, and when he ate his sandwich discovered it was particularly sweet. It is synthesized according to the following reaction and the product is isolated after addition of HCI. -NH₂ + NaOH + KMnO4 NH OS=0 a) What safety rule did our chemist violate when discovering saccharin? (1 pt) b) Redraw the structure of the carbon-containing starting material, explicitly showing all carbon and hydrogen atoms and lone pairs. [3 pts] c) Determine the chemical formula and calculate the molecular weight for the starting material in (b). [3 pt]

Answer 1

a)

The Chemist bought food(sandwich) into the laboratory.

b)

The structure of our organic starting material showing all the atoms and lone pair is as follows.

NH2 C C C H c H I-

Note that N atom has 5 outermost valence electrons and O atoms have 6 outermost valence electrons. Since N has made 3 single bonds, it has a lone pair. Similarly, O atoms are forming double bonds and hence have 2 lone pairs on them.

c)

While writing the chemical formula, we calculate the number of times each element appears in the structure above

C: 7

H: 9

N: 1

O:2

S:1

Hence, the chemical formula is .

C7H9N02S

Now, we can calculate the molecular weight of the compound by summing up the atomic mass of each element that appears multiplied by the number of times each element appears.

Hence,

M.W. = 7 x m(C) + 9 x m(H) + m(N) + 2 x m(0) + m(S) =MW = 7x 12 g/mol+9x 1.0 g/mol+14 g/mol+2 x 16 g/mol+32 g/mol M.W. = 171 g/mol

Hence, the molecular weight of the compound is 171 g/mol approximately.

d)

The mass, molecular weight and number of moles of a compound are related as follows:

mass number of moles = mass M.W. 50.150 mol = 171 g/mol mass = 0.150 mol x 171 g/mol = 25.65 g

Hence, the mass of the starting material that contains 0.150 mol is 25.65 g approximately.

e)

Concentration of NaOH being used = 0.400 M = 0.400 mol/L.

Hence, 1 L of the solution contains 0.400 mol of NaOH.

We need to take 0.150 mol of NaOH. Hence, the volume of solution that contains 0.150 mol is

FX 0.150 mol = 0.375 L = 375 mL 0.400 mol

Hence, we should use 375 mL of the 0.400 M NaOH solution.

f)

The number of equivalent of KMnO4 that is being used = 1.60 equiv.

We know that 1 equiv of NaOH = 0.150 mol

Hence, number of moles of KMnO4 that equals 1.60 equivs is

0.150 mol x 1.60 equiv = 0.240 mol 1 equiv.

Hence, 0.240 mol of KMnO4 is needed.

g)

The molecular weight of KMnO4 can be calculated as follows

M.W = m(K) + m(Mn) + 4 x m(O) MW. = 39 g/mol + 55 g/mol + 4 x 16 g/mol = 158 g/mol

Hence, the mass of KMnO4 needed to be weighed is

mass = molar mass x number of moles = 158 g/mol x 0.240 mol = 37.92 g

Hence, we need to weigh 37.92 g of KMnO4 approximately.

h)

We know that 1 mol of organic starting material will result in 1 mol of sachharin theoretically.

Given that molecular weight of sachharin is 183.18 g/mol.

The theoretical yield of sachharin is 0.150 mol if 0.150 mol of organic starting material is taken.

Hence, theoretical yield in grams is

183.18 9 0 150 mol = 27.477 9 1 mol

Now, the actual yield is reported as 19.5 g.

Hence, the % yield of the reaction can be calculated as

19.5 9 actual yield theoretical vield x 100 = 57.477 X 100 x 100 71.0%