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Musya8 [376]
3 years ago
7

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.
Chemistry
2 answers:
enyata [817]3 years ago
6 0

Answer:

Molecular formula C33H27O3

Explanation:

<u>Given:</u>

Mass of CO2 produced = 18.490 mg

Mass of H2O produced = 6.232 mg

Mass of sample = 5.983 mg

<u>To determine:</u>

Molecular formula of the compound

<u>Calculation:</u>

<u>Step 1</u>: Calculate the mass of carbon in the sample

Molar Mass of CO2 = 44 g/mol

Moles of CO2 = \frac{Mass}{Molar mass } = \frac{18.490}{44} =0.4202

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

<u>Step 2 </u>: Calculate the mass of hydrogen in the sample

Molar Mass of H2O = 18 g/mol

Moles of H2O = \frac{Mass}{Molar mass } = \frac{6.232}{18} =0.3462

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

<u>Step 3</u>: 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

<u>Step 4</u>: 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 = \frac{0.4202}{0.03715} =11.31\\

H = \frac{0.3462}{0.03715} =9.32\\

O = \frac{0.03715}{0.03715} =1.00\\

Empirical formula = C11H9O

<u>Step 6</u>: Find n

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

n = n = \frac{Molar\ Mass}{Empirical\ Mass} = \frac{399}{157} =2.54

n = 3.0

<u>Step 7</u>: Find molecular formula

(C11H9O)3 = C33H27O3

Jet001 [13]3 years ago
5 0
We calculate it as follows:

Moles CO2 = 0.01849 g / 44 = 0.000420 
<span>Mass C = 0.000420 x 12 = 0.00504 g </span>
<span>Moles H = 2 x 0.006232 / 18 = 0.000692 </span>
<span>Mass H = 0.000692 g </span>
<span>Mass O = 0.005982 - ( 0.00504 + 0.000692) = 0.00025 </span>
<span>Moles O = 0.00025 / 16 = 0.0000156 </span>
<span>C 0.000420
H 0.000692
O 0.0000156 
</span>
<span>divide each by the smallest value, giving you the chemical formula as:
</span><span>
C27H44O</span>
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