How many milligrams of fluoride ions are present in a 2.2 mg tablet of NaF?

To determine how many milligrams of fluoride ions are present in a 2.2 mg tablet of sodium fluoride (NaF), one must understand the composition of sodium fluoride and the molar masses of its components.

Sodium fluoride is made up of sodium ions (Na+) and fluoride ions (F-). The molar mass of NaF is approximately 41.99 grams per mole, with the fluoride ion being 19 grams per mole. When sodium fluoride is dissolved, it dissociates to release one fluoride ion for each molecule of NaF.

Calculating the weight of fluoride in a 2.2 mg tablet of NaF involves finding the proportion of fluoride in sodium fluoride. The fluoride ion constitutes about 19 grams per 41.99 grams of NaF. Therefore, the percentage of fluoride in sodium fluoride can be calculated as follows:

Fluoride percentage = (Molar mass of F / Molar mass of NaF) = (19 g/mol / 41.99 g/mol)

Carrying out the calculation yields approximately 0.453, which signifies that 45.3% of the weight of NaF is fluoride. So, when applied to the 2.2 mg tablet:

Amount of fluoride =