the analysis of an organic compound showed that it contained 0.175 mol of C, 0.1

Question

the analysis of an organic compound showed that it contained 0.175 mol of C, 0.140 mol of H, and 0.0350 mol of N. its molar mass is 156 g/mol. respectively, how many atoms of carbon are there in the empirical formula and in the molecular formula of the compound? the analysis of an organic compound showed that it contained 0.175 mol of C, 0.140 mol of H, and 0.0350 mol of N. its molar mass is 156 g/mol. respectively, how many atoms of carbon are there in the empirical formula and in the molecular formula of the compound? the analysis of an organic compound showed that it contained 0.175 mol of C, 0.140 mol of H, and 0.0350 mol of N. its molar mass is 156 g/mol. respectively, how many atoms of carbon are there in the empirical formula and in the molecular formula of the compound?

Explanation / Answer

The determination of molecular formula of compound is as shown below.

Given

moles of C = 0.175

moles of H = 0.140

moles of N = 0.035

calculate mole ratio

C = 0.175/0.035 = 5

H = 0.140/0.035 = 4

N = 0.035/0.035 = 1

Thus,

Empirical formula of the compound = C5H4N

Empirical formula mass = 5 x 12 + 4 x 1 + 1 x 14 = 78

ratio of molar mass/empirical formula mass = 156/78 = 2

Thus,

Molecular formula of the compound = C10H8N2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.