percent composition question

Question

One hundred grams of a dye compound is equivalent to 0.1871 moles. Its percent composition is shown below:

ElementPercent	Composition  Atomic Mass
C	35.95%	12.011 amu
H	1.69%	1.008 amu
N	10.49%	14.007 amu
Na	12.92%	22.990 amu
O	26.97%	15.999 amu
S	11.98%	32.06 amu

 

A. What is its empirical formula? (Enter 1 if the corresponding subscript is 1.)

C_{Answer 1}H_{Answer 2}N_{Answer 3}Na_{Answer 4}O_{Answer 5}S_{Answer 6}

 

B. What is the empirical mass of the compound? Answer 7 g/mol

C. What is its molecular formula? (Enter 1 if the corresponding subscript is 1.)

Answer 1

Part A:

First we need to convert mass percentages to mole percentages. To do this we have to divide them by atomic mass of each element:. This means basically dividing the 2nd column in the table by the 3rd column to result in the following:

C: 2.995

H: 1.68

N: 0.75

Na: 0.56

O: 1.68

S: 0.374

Now divide these mole fractions by the smallest number which is 0.374 for S, so we can get better whole numbers as follows:

C: 8

H: 4.5

N: 2

Na: 1.5

O: 4.5

S: 1

Since we still don't quite have whole numbers, we would now multiply by 2 which we would give us the following:

C: 16

H: 9

N: 4

Na: 3

O: 9

S: 2

So the empirical formula would be: C₁₆H₉N₄Na₃O₉S₂

Part B:

For the empirical mass, we simply add the masses of all elements in this molecule:

16 x 12 + 9 x 1 + 4 x 14 + 3 x 23 + 9 x 16 + 2 x 32 = 534 g / mol

Part C:

Now we need the actual molar mass of this compound using the info given in the problem statement:

100g / 0.1871 mol = 534.5 g/mol

Now divide the actual molar mass by the empirical mass: 5234.5 / 534 = 1

Therefore, our molecular formula is the same as our empirical formula: C₁₆H₉N₄Na₃O₉S₂