When a sample of a compound in the vitamin D family was burned in a combustion analysis, 5.983 mg of the compound gave 18.490 mg of CO2 and 6.232 mg of H2O. This compound was found to have a molecular mass of 399. What is the molecular formula of this compound? Put your answer in form of CxHyOz.

We calculate it as follows:

Moles CO2 = 0.01849 g / 44 = 0.000420
Mass C = 0.000420 x 12 = 0.00504 g
Moles H = 2 x 0.006232 / 18 = 0.000692
Mass H = 0.000692 g
Mass O = 0.005982 - ( 0.00504 + 0.000692) = 0.00025
Moles O = 0.00025 / 16 = 0.0000156
C 0.000420
H 0.000692
O 0.0000156

divide each by the smallest value, giving you the chemical formula as:

C27H44O

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Annainn Annainn · Answer 2 · 2019-06-16T07:15:03+02:00

Answer:

Molecular formula C33H27O3

Explanation:

Given:

Mass of CO2 produced = 18.490 mg

Mass of H2O produced = 6.232 mg

Mass of sample = 5.983 mg

To determine:

Molecular formula of the compound

Calculation:

Step 1: Calculate the mass of carbon in the sample

Molar Mass of CO2 = 44 g/mol

Moles of CO2 = [tex]\frac{Mass}{Molar mass } = \frac{18.490}{44} =0.4202[/tex]

The carbon content in the sample corresponds to the amount of CO2 produced

Moles of C in the sample = 0.4202

Mass of C = moles * atomic mass of C

= 0.4202 moles * 12 g/mol = 5.0424 g

Step 2 : Calculate the mass of hydrogen in the sample

Molar Mass of H2O = 18 g/mol

Moles of H2O = [tex]\frac{Mass}{Molar mass } = \frac{6.232}{18} =0.3462[/tex]

The hydrogen content in the sample corresponds to the amount of H2O produced

Moles of H in the sample = 0.3462

Mass of H = moles * atomic mass of H

= 0.3462 moles * 1 g/mol = 0.3462 g

Step 3: Mass of O in the sample

Mass of O = Total mass - Mass of C - Mass of H

= 5.983 - 5.0424-0.3462=0.5944 g

Step 4: Moles of O

Moles of O = mass of O/atomic mass = 0.5944/16 = 0.03715

Step 5: Find the empirical formula

Moles of C = 0.4202

Moles of H = 0.3462

Moles of O = 0.03715

Ratio

C = [tex]\frac{0.4202}{0.03715} =11.31\\[/tex]

H = [tex]\frac{0.3462}{0.03715} =9.32\\[/tex]

O = [tex]\frac{0.03715}{0.03715} =1.00\\[/tex]

Empirical formula = C11H9O

Step 6: Find n

Empirical formula mass = 12(11) + 1(9) + 16 = 157

n = [tex]n = \frac{Molar\ Mass}{Empirical\ Mass} = \frac{399}{157} =2.54[/tex]

n = 3.0

Step 7: Find molecular formula

(C11H9O)3 = C33H27O3