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kiruha [24]
3 years ago
15

Which of the following pairs does not share the same empirical formula? C2H6 and C6H18 C3H6O3 and C2H4O2 C3H6 and CH4 CH2O and C

3H6O3
Chemistry
2 answers:
DerKrebs [107]3 years ago
7 0

Answer:

c) C₃H₆ and CH₄

Explanation:

The empirical formula is obtained from a molecular formula by dividing it with a whole number with which all elements are divisible completely. Empirical formula is the simplest ratio of all the elements present in a molecule.

Let us consider all the examples

a) C₂H₆ and C₆H₁₈

Empirical formula of C₂H₆ = CH₃[by dividing with two]

Empirical formula of C₆H₁₈= CH₃ [by dividing with six]

b) C₃H₆O₃and C₂H₄O₂

Empirical formula of C₃H₆O₃ = CH₂O[by dividing with three]

Empirical formula of C₂H₄O₂= CH₂O [by dividing with two]

c) C₃H₆ and CH₄

Empirical formula of C₃H₆ = CH₂[by dividing with three]

Empirical formula of CH₄= CH₄ [already simplest ratio]

So these are not same.

d) CH₂O and C₃H₆O₃

Empirical formula of C₃H₆O₃ = CH₂O[by dividing with three]

Empirical formula of CH₂O= CH₂O [already simplest ratio]

denpristay [2]3 years ago
3 0
C3H6 and CH4

The empirical number is found by looking at the ratio of the numbers. For example, H20 and H402 have the same empirical formula because the ratio remained similar. 2 H to 1 O. 
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Determine the molecular formula for the unknown if the molecular mass is 60.0 amu and the empirical formula is ch2o.
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Answer:

The molecular formule for this unknow molecule is C2H4O2

Explanation:

The empirical formula is CH2O  ( or better said CnH2nOn)

This means there are 3 elements in the formula of this molecule

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We can also notice that the amount of hydrogen should 2x the amount of carbon ( also 2x the amount of oxygen).

The mass of the empirical formule = 12g/ mole + 2* 1 g/mole + 16 g/mole = 30 g/mole

To know what number is n in CnH2nOn we should divide the molecular mass by the empirical mass:

60 g/mole / 30g/mole = 2

this means n = 2

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We can control this to calculate the molecular mass:

2*12 + 4* 1 + 2*16 = 24 + 4 + 32 = 60 g/mole

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To summarize, this mechanism takes places in two separate steps. The mechanism is attached below.

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The mechanism is attached as well.

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