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ankoles [38]
3 years ago
15

Write the molecular formula for the compound that exhibits a molecular ion at M+ = 112.0499. Assume that C, H, N, and O might be

present, and use the exact masses below: Exact mass of carbon = 12.000 Exact mass of hydrogen = 1.0078 Exact mass of nitrogen = 14.003 Exact mass of oxygen = 15.995 (The order of atoms should be carbon, then hydrogen, then the others in alphabetical order. If there is more than one answer, just give one. ) Molecular formula:
Chemistry
1 answer:
IceJOKER [234]3 years ago
7 0

Answer:

C₅H₈N₂O

Explanation:

The molecular formula denotes the various forms of atoms contained in a molecule at a particular fixed proportion.

The molecular ion M⁺ = 112.0499

and the exact mass values are given as follows:

C = 12.000

H = 1.0078

N = 14.003

O = 15.995

By assumption:

C = 12.000 × 5 = 60.0000

H = 1.0078 × 8 =    1.0078

N = 14.003 × 2 =  28.0060

<u>O = 15.995 × 1 =   15.9950    </u>

<u>                         = 112.0634     </u>

This is approximtely equal to 112.0499.

As such the Molecular formula for the compound = C₅H₈N₂O

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A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
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Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

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