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2- A 0.60 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24 grams of carbon, 0.040 grams of hydrogen, with the rest being oxygen. If the molecular weight of the substance is 90 grams/n, what is the molecular formula

Sagot :

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Answer:

C₃H₆O₃

Explanation:

To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).

With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:

Moles C -Molar mass:12.0g/mol-

0.24g * (1mol/12.0g) = 0.020 moles C

Moles H = Mass H because molar mass = 1g/mol:

0.040 moles H

Moles O -Molar mass: 16g/mol-

Mass O: 0.60g - 0.24g - 0.040g = 0.32g O

0.32g O * (1mol/16g) = 0.020 moles O

Ratio of atoms (Dividing in moles of C: Lower number of moles):

C = 0.020 moles C / 0.020 moles C = 1

H = 0.040 moles H / 0.020 moles C = 2

O = 0.020 moles O / 0.020 moles C = 1

Empirical formula:

CH₂O.

Molar mass CH2O:

12g/mol + 2*1g/mol + 16g/mol = 30g/mol

As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:

C₃H₆O₃

Oseni

The molecular formula of the organic acid would be C3H6O3

Molecular formula

Molecular formula = [empirical formula]n

Where n = molar mass/mass of empirical formula

Empirical formula

C = 0.24/12 = 0.02

H = 0.040/1 = 0.04

O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02

Divide by the smallest

C = 1

H = 2

O = 1

Empirical formula = CH2O

Empirical formula mass = 12 + 2 + 16 = 30

n = 90/30 = 3

Molecular formula = [CH2O]3

                               = C3H6O3

More on molecular formula can be found here: https://brainly.com/question/1247523

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