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Nat2105 [25]
3 years ago
15

g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empir

ical formula of the compound?
Chemistry
1 answer:
sammy [17]3 years ago
7 0

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

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Kryger [21]

The empirical formula : C₁₂H₄F₇

The molecular formula : C₂₄H₈F₁₄

<h3>Further explanation</h3>

mol C (MW=12 g/mol)

\tt \dfrac{18.24}{12}=1.52

mol H(MW=1 g/mol) :

\tt \dfrac{0.51}{1}=0.51

mol F(MW=19 g/mol)

\tt \dfrac{16.91}{19}=0.89

mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7

Empirical formula : C₁₂H₄F₇

(Empirical formula)n=molecular formula

( C₁₂H₄F₇)n=562 g/mol

(12.12+4.1+7.19)n=562

(281)n=562⇒ n =2

Molecular formula : C₂₄H₈F₁₄

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