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2 years ago

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

Chemistry

1 answer:

Inessa05 [86]2 years ago

8 0

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

#learnwithBrainly

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The reaction;
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To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

First Formula would be $ZnF_{2}$

Second empirical formula is between $Zn^{2+}(Positive ion) and O^{2-}(Negative ion)$

Second Formula would be $Zn_{2}O_{2}$

Note : When the subscript are same they get cancel out, so $Zn_{2}O_{2}$ would be written as $ZnO$

Third empirical formula is between $Ni^{4+}(Positive ion) and F^{-}(Negative ion)$

Third Formula would be : $NiF_{4}$

Forth empirical formula is between $Ni^{4+}(Positive ion)and O^{2-}(negative ion)$

Forth Formula would be : $Ni_{2}O_{4}$ or $NiO_{2}$

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