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Fiesta28 [93]
3 years ago
11

During a qualitative analysis experiment a student found his unknown salt to contain mn2 and po43- ions. write the chemical form

ula for their unknown salt. do not include phases with your answer.

Chemistry
1 answer:
MrMuchimi3 years ago
3 0

The ions in the unknown salt are:

Mn^{2+} and PO_{4}^{3-}

The chemical formula of the salt is determined as shown in the picture.

The chemical formula of the salt is Mn_3(PO_{4})_{2}.


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The 1H-NMR of a compound with molecular formula C6H12O2 consists of four signals: 1.1 (triplet, integrating to 3 Hydrogens), 1.2
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Answer:

Isopropyl propionate  

Explanation:

1. Information from formula

The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.

2. Information from the spectrum

(a) Triplet-quartet

A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group

(b) Septet-doublet

A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group

(c) The rest of the molecule

The ethyl and isopropyl groups together add up to C₇H₁₂.

The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.

The compound is either

CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.

(d) Well, which is it?

The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.

The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.  

The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.

We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.

3. Summary

My peak assignments are shown in the diagram below.

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