Molecular Formula Problems Directions: Show all work for this problem. You may w

Question

Molecular Formula Problems Directions: Show all work for this problem. You may want to write your answers on a piece of notebook paper. Answer all questions by showing all mathematical work and/or interpreting all major peaks in the IR and all peaks in the 'H-NMR spectra. You may wish to do complete your assign- ment on another sheet of paper. 1. A researcher analyzed an unknown solid, extracted from the bark of spruce trees, to determine its percentage composition. An 11.32 mg sam- ple was burned in a combustion apparatus. The carbon dioxide (24.87 mg) and water (5.82 mg) were collected and weighed. a. From the results of this analysis, calculate the percentage composition of the unknown solid. b. Determine the empirical formula of the unknown solid.

Explanation / Answer

For CO2: moles = 24.87 mg/44.01g/mol = 0.02487g / 44.01g/mol = 5.65 x 10^-4 mol
For H2O: moles = 5.82 mg/18.02g/mol = 0.00582g / 18.02g/mol = 3.23x 10^-4 mol

For each mole of CO2, one mole of carbon exists. So, the amount of carbon is 5.65 x 10^-4 mol
For each mole of H2O, two moles of hydrogen exist. So, the amount of hydrogen is 2 * 3.23 x 10^-4 mol = 6.46 x 10^-4 mol

If we translate these mole values back into masses using the molar mass of the pure element, we find that:
Carbon: mass = molar mass * moles = (12.01g/mol)(5.65 x 10^-4 mol) = 0.00674 g = 6.74 mg
Hydrogen: mass = molar mass * moles = (1.01g/mol)(6.46 x 10^-4mol) = 6.46 x 10^-4 g = 0.646 mg
mass of oxygen = total mass - mass of carbon - mass of hydrogen = 11.32mg - 6.74 mg - 0.646 mg = 3.934 mg

So, this particular compound should have 6.74 mg carbon, 0.646 mg hydrogen, and 3.934 mg oxygen, for a combined mass of 11.32 mg. To determine the percent composition, take each mass and divide it by the total mass:
Carbon: 6.74 mg / 11.32 mg = 59.54%
Hydrogen: 0.646mg / 11.32 mg = 5.70%
Oxygen: 2.023 mg / 11.32 mg = 34.7%

Thus, the percent composition of this compound is 59.54% carbon, 5.7% hydrogen, and 34.7% oxygen

1) Assume 100 g of the compound is present. This changes the percents to grams:

C 59.54 mg = 0.059g
H 5.7 mg = 0.0057g
O 34.7 mg = 0.0347g

2) Convert the masses to moles:

C 0.059 g / 12 = 0.0049
H 0.0057 g / 1 = 0.0057
O 0.0347 g / 16 = 0.0021

3) Divide by the lowest, seeking the smallest whole-number ratio:

C 0.0049 / 0.0021 = 2.33
H 0.0057 / 0.0021 = 2.71
O 0.0021 / 0.0021 = 1

4) Write the empirical formula:

C3H3O

5) Determine the molecular formula:

"EFW" 36+3+16 = 55

MF = MFW/EFW = 450/55 = 8

Molecular formula = (C3H3O)8 = C24 H24 O8

For the formula C24 H24 O8 the degree of unsaturation is, degree of unsaturation = 1/2 (2 c+ 2H -O) =13

It may have 3 aromatic rings

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