Question

What's the empirical and molecular weight of C14H22N4O8? What's the molecular formula of C14H22N4O8?

Answer 1

374u

187u

C₁₄H₂₂N₄O₈

Explanation:

To find the molecular weight of the compound C₁₄H₂₂N₄O₈ we simply sum that atomic masses of the given elements in the compound.

 The empirical weight is determined by using the simplest ratio of the elements involved in the compound;

Molecular weight of C₁₄H₂₂N₄O₈;

atomic mass of C = 12g/mol

                           H = 1g/mol

                            N = 14g/mol

                            O = 16g/mol

 Molecular weight = 14(12) + 22(1) + 4(14) + 8(16)

                                = 168 + 22 + 56 + 128

                                 = 374u

Empirical weight:

  Empirical formula:

                        C₁₄    H₂₂      N₄     O₈

                         14  :    22 :    4  :     8

   divide by 2:

                          7   :    11    :    2  :    4

     empirical formula  C₇H₁₁N₂O₄

     empirical weight = $\frac{molecular weight}{2}$ = $\frac{374}{2}$ = 187u

The molecular formula is the actual combination of atoms in a compound. so the molecular formula of the compound is C₁₄H₂₂N₄O₈

learn more:

Molecular mass brainly.com/question/5546238

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