A. A 3.06 mg sample of a compound containing carbon, hydrogen, and oxygen underw

Question

A. A 3.06 mg sample of a compound containing carbon, hydrogen, and oxygen underwent combustion producing 6.95 mg of CO2 and 2.84 mg of H2O. What is the empirical formula of the compound? (Type your answer using the format CxHyOz for the compound CxHyOz)

B. A compound is composed of elements C, H, and N. A 3.675-mg sample of the unknown compound was combusted, producing 9.969 mg of CO2 and 2.858 mg of H2O. What is the empirical formula for this compound? (Type your answer using the format CxHyNz for the compound CxHyNz)

C.If the compound has a molar mass of 160 ± 5 g/mol, what is its molecular formula?

Explanation / Answer

A) answer : C5H5O2

(6.95 mg CO2) / (44.00964 g CO2/mol) x (1 mol C / 1 mol CO2) x (12.01078 g C/mol) = 1.8967 mg C

(2.84 mg H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) x (1.007947 g H/mol) = 0.15890 mg H

Use the Law of Conservation of Mass:

3.06 mg total - 1.8967 mg C - 0.15890 mg H = 1.0044 mg O

(1.8967 mg C) / (12.01078 g C/mol) = 0.15792 mmol C

(0.15890 mg H) / (1.007947 g H/mol) = 0.15765 mmol H

(1.0044 mg O) / (15.99943 g O/mol) = 0.062777 mmol O
Divide by the smallest number of moles:

(0.15792 mmol C) / 0.062777 mmol = 2.52

(0.15765 mmol H) / 0.062777 mmol = 2.51

(0.062777 mmol O) / 0.062777 mmol = 1.00

To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C5H5O2

B) answer : C5H7N

same above procedure.

C)   C5H7N empirical formula mass = 5 x 12 + 7 x 1 + 14 = 83 g /mol

   molar mass = 165

n = 165/83 = 2 (nearly)

molecular formula = n x empirical formula

                              = 2 x C5H7N

                              = C10H14N2

molecular formula = C10H14N2

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