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VLD [36.1K]
4 years ago
14

A compound is found to have a percent composition (by mass) of 85.63% C and 14.37% H. The molar mass of the compound was found t

o be 42.0 g/mol. Which of the following is the empirical formula and the molecular formula, respectively a. C2H6 and C3H9 b. CH2 and C3H6 c. CH and C3H3 d. CH3 and C2H6 e. C2H3 and C4H6
Chemistry
1 answer:
AysviL [449]4 years ago
6 0

Answer:

Empirical formular = CH₂

Molecular formular = C₃H₆

Explanation:

Percent Mass of C = 85.63%

Percent Mass of H = 14.37%

Molar mass = 42.0 g/mol

Steps in Calculating the Empirical Formular;

1 - Divide the percent mass of the Elements with their respective atomic masses. Carbon = 12 and Hydrogen = 1

Carbon = 85.63 / 12 = 7.14

Hydrogen = 14.37 / 1 = 14.37

2 - Divide the values by the smallest one. In this case Carbon has the lowest value of 7.14

Carbon = 7.14 / 7.14 = 1

Hydrogen = 14.37 / 7.14 =  2.013

3. Approximate the values to the nearest whole number.

Hydrogen ≈ 2

Hence the empirical formular is given as; CH₂

Molecular formular is given as;

[12 + 2(1)]n = 42

14n = 42

n = 3

Molecular formular = C₃H₆

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Explanation:

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If electronegativity difference between two atoms is greater than 1.8 then bond formed is an ionic bond.

The electronegativity value of a nitrogen atom is 3.04. Hence, the electronegativity difference of N_{2} molecule is as follows.

Electronegativity difference = electronegativity value of N - electronegativity value of N\\= 3.04 - 3.04\\= 0As the electronegativity difference is 0 which is less than 0.4.

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Explanation :

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4 years ago
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Answer:

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