A 2.50 mg sample of a compound containing carbon, hydrogen, and oxygen underwent

Question

A 2.50 mg sample of a compound containing carbon, hydrogen, and oxygen underwent combustion producing 5.69 mg of CO_2 and 2.32 mg of H_2O. What is the empirical formula of the compound? (Type your answer using the format CxHyOz for the compound C_xH_yO_z) A compound is composed of elements C, H, and N. A 2.625-mg sample of the unknown compound was combusted, producing 7.121 mg of CO_2 and 2.042 mg of H_2O. What is the empirical formula for this compound? (Type your answer using the format CxHyNz for the compound C_xH_yN_z)_______________If the compound has a molar mass of 160 plusminus 5 g/mol, what is its molecular formula?____________________

Explanation / Answer

(a) moles of C = moles of CO2 = 5.69 mg/44 g/mol = 0.13 mmol

mass of C = 0.13 x 12 = 1.56 mg

moles of H = 2 x moles of H2O = 2 x 2.50 mg/18 g/mol = 0.28 mmol

mass of H = 0.28 x 1 = 0.28 mg

mass of oxygen = 2.32 - (1.56 + 0.28) = 0.48 mg

moles of O = 0.48 mg/16 g/mol = 0.03 mmol

Divide moles by smallest factor

C = 0.13/0.03 = 4

H = 0.28/0.03 = 10

O = 0.03/0.03 = 1

Empirical formula = C4H10O

(b) moles of C = moles of CO2 = 7.121 mg/44 g/mol = 0.1618 mmol

mass of C = 0.162 x 12 = 1.942 mg

moles of H = 2 x moles of H2O = 2 x 2.042 mg/18 g/mol = 0.2269 mol

mass of H = 0.227 x 1 = 0.227 mg

mass of O = mass of sample - (mass of C + H) = 0.456 mg

moles of O = 0.456 mg/16 g/mol = 0.028 mmol

Divide by smallest factor

C = 0.162/0.028 = 6

H = 0.227/0.0285 = 8

O = 0.028/0.028 = 1

Empirical formula = C6H8O

Molecular formula factor = 160/(12 x 6 + 8 + 16) = 1.66

Molecular formula = C10H13O2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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