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Inessa [10]
3 years ago
10

A reaction was set up to convert an alcohol to a carbonyl Samples of the reaction mixture were checked by infrared spectrometry

every couple of minutes. Use the infrared spectra provided to determine if the reaction mixture contains: pure alcohol, a mixture of alcohol and carbonyl, or pure carbonyl pure alcohol O pure carbonyl O mixture of alcohol and carbonyl 1000 500 O O O pure alcohol pure carbonyl mixture of alcohol and carbonyl 70

Chemistry
1 answer:
QveST [7]3 years ago
6 0

Answer:

look at peaks:

sharp peak at ~ 1700 = carbonyl

broad peak at ~ 3300 = alcohol

both = mixture

Explanation:

In the first spectrum, the broad peak occurring in the range of 3300-3700 cm-1 corresponds to OH bond stretch frequency. Carbonyl C=O bond stretch has a frequency range of 1670-1820 cm-1. The absence of any peaks in this region in the given spectrum rules out the presence of carbonyl compound in the reaction mixture. So, the reaction mixture contains only pure alcohol.

The second spectrum shows a peak around 1700 cm-1 corresponding to the carbonyl compound along with the broad peak around 3300 cm-1 corresponding to alcohol. So, the reaction mixture contains both alcohol and carbonyl compounds.

A table containing different functional groups and their corresponding Infra Red frequencies is given below for further references

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The answer for the following problem is mentioned below.

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Given:

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We know;

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = 120 + 3(95)

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} )  = 120 +285 = 405 grams

<em>We also know;</em>

No of molecules at STP conditions(N_{A}) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

\frac{m}{M} =\frac{N} }{}N÷N_{A}

\frac{405}{125.3} =\frac{N}{6.023*10^23}

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<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules</em></u>

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