Question

A 2.52 g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess O2 to yield 3.52 g of CO2 and 0.839 g of H2O. Another sample of the same compound, of mass 4.14 g, yielded 1.75 g of SO3. A reaction to determine the nitrogen content was carried out on a third sample with a mass of 5.66 g and yielded 1.88 HNO3 (assume that all of the nitrogen in HNO3 came from the compound).

Calculate the empirical formula of the compound.Express your answer as a chemical formula. Enter the elements in the order: C, H, S, N, O.

Answer 1

The answer to your question is: C₄H₄NO₂S

Step-by-step explanation:

Data

2.53 g CvHwNxOySv ⇒ 3.52g CO2 + 0.839g H2O

4.14 g CvHwNxOySv ⇒ 1.75 g SO3

5.66g CvHwNxOySv ⇒ 1.88 g HNO3

Empirical formula

For Carbon

MW CO2 = 44g

44g of CO2 ------------------- 12 g of C

3.52 of CO2 ----------------- x

x = (3.52 x 12)/44

x = 0.96 g of C

12 g ------------------ 1 mol

0.96g ---------------- x

x = (0.96 x 1) / 12

x = 0.08 mol of C

Percent of Carbon in the sample

2.53 g of sample -------------- 100%

0.96 g of Carbon ------------ x

x = (0.96 x 100) / 2.53

x = 37.94 % of Carbon in the sample

For Hydrogen

MW H2O = 18 g

18g of H2O -------------- 2 g of H2

0.839 g ------------ - x

x = 0.093 g of H2

1 mol of H2 ------------- 1 g of H2

x -------------- 0.093 g of H2

x = 0.093 mol of H2

Percent of H2 in the sample

2.53 g of sample -------------- 100%

0.093 --------------- x

x = 3.68 % of Hydrogen in the sample

For Sulfur

MW of SO3 = 80 g

80g of SO3 ---------------- 32 g

1.75 g ---------------- x

x = 0.7 g of S

32g S ----------------- 1 mol

0.7g ------------------- x

x = 0.022 mol of S

Percent of S in the sample

4.14 g ------------------ 100%

0.7 g ------------------ x

x = 16.9%

For Nitrogen

MW of HNO3 = 63g

63g of HNO3 ---------------- 14 g of N

1.88 g ------------------ x

x = 0.42 g of N

14 g ----------------- 1 mol of N

0,42 g of N ------ x

x = 0.03 mol of N

Percent of N in the sample

5.66 g --------------- 100%

0.42 g ----------------- x

x = 7.4%

Percent of Oxygen = 100 - (37.94 + 3.68 + 16.9 + 7.4)

Percent of oxygen = 34.08

Grams of Oxygen 2.53 g of sample ------------- 100%

x ------------ 34.08

x = 0.86 g of Oxygen

16 g of O ----------------- 1 mol

0.86 g of O --------------- x

x = 0.054 mol of Oxygen

Divide by the lowest # of mol

C = 0.08/ 0.022 = 3.6 ≈ 4

H = 0.093/ 0.022 = 4.2 ≈ 4

N = 0.022/ 0.022 = 1

O = 0.053/ 0.022 = 2.4 ≈ 2

S = 0.03/ 0.022 = 1.4 ≈ 1

Empirical formula

C₄H₄NO₂S