A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na2PO3F) in 100 mL. (a) What mass of fluorine atoms in mg was present? (b) How many fluorine atoms were present?

a. 100 mg

b. 3.18×10²¹ atoms of F

Explanation:

Let's determine the moles of Na₂PO₃F that represents 0.76 g of compound

0.76 g . 1mol / 143.95 g = 0.00528 moles

1 mol of Na₂PO₃F has 2 moles of Na, 1 mol of P, 3 moles of O, 1 mol of F

Then, 0.00528 moles of Na₂PO₃F must have 0.00528 moles of F

Look that ratio is 1:1

Let's convert the moles to mass (g), and then from g to mg

0.00528 mol . 18.99 g/1mol = 0.100 g

0.1 g . 1000 mg/1g = 100 mg

Let's calculate the number of atoms:

1 mol has 6.02×10²³ atoms

Then, 0.00528 moles may have (0.00528 . 6.02×10²³) /1 = 3.18×10²¹ atoms

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