What are the molecular weights of the following pharmaceuticals? (b) C22H27F3O4S (fluticasone, anti-inflammatory)

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Identify the number of each type of atom in the molecular formula C22H27F3O4S for fluticasone. This includes 22 carbon (C) atoms, 27 hydrogen (H) atoms, 3 fluorine (F) atoms, 4 oxygen (O) atoms, and 1 sulfur (S) atom.

Look up the atomic masses for each type of atom from the periodic table. Typically, the atomic masses are approximately: Carbon = 12.01 amu, Hydrogen = 1.008 amu, Fluorine = 19.00 amu, Oxygen = 16.00 amu, Sulfur = 32.07 amu.

Multiply the atomic mass of each element by the number of atoms of that element in the molecule. For example, for carbon, multiply 12.01 amu by 22.

Add up all the products from the previous step to get the total molecular weight of the molecule.

Ensure the units are consistent and typically expressed in atomic mass units (amu) or grams per mole (g/mol).

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Molecular Weight Calculation

Molecular weight, or molar mass, is the sum of the atomic weights of all atoms in a molecule. It is calculated by multiplying the number of each type of atom by its atomic weight (from the periodic table) and summing these values. For example, in C22H27F3O4S, you would calculate the contributions from carbon, hydrogen, fluorine, oxygen, and sulfur to find the total molecular weight.

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Molecular Formula Calculation Example

Atomic Weights

Atomic weights are the average masses of an element's isotopes, measured in atomic mass units (amu). These values are essential for calculating molecular weights, as they provide the necessary data for each element in a compound. For instance, carbon has an atomic weight of approximately 12.01 amu, while oxygen is about 16.00 amu, and these values are used in the molecular weight calculation.

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Atom Structure

Chemical Formula Interpretation

A chemical formula, like C22H27F3O4S, indicates the types and numbers of atoms in a molecule. The subscripts denote the number of each atom present; for example, 'C22' means there are 22 carbon atoms. Understanding how to interpret these formulas is crucial for determining the molecular weight and for grasping the composition of the compound.

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