# How many molecules are in 8.0 g of ozone, O3?

• n = m / M

n = 8 / 48

n = 1/6

1/6 x 6.02×10^23 (avagadro’s number)

= 1.003333333×10^23

• You should review concepts of molecular weight (mass), converting grams of substances to moles of substances, and Avogadro’s number concepts to solve this. (1 mole of anything = 6.02 x 10^23 of those things)

(Forget significant figures for simplicity, please)

Convert Grams of ozone to moles of ozone, so you need the MW of ozone. 3 oxygen atoms at 16.0 grams/mole = MW of 48.0 g/mol

8.0 g O3 x (1 mole O3 / 48.0 g O3) = 0.167 mol O3

0.167 mol O3 x (6.02 x 10^23 molecules of anything / 1 mole of O3) =

1.00 x 10^23 molecules of Ozone in 8 g of Ozone

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Think about it: One mole of ozone weighs 48.0 grams. That would be 6.02 x 10^23 molecules. You only have 8.0 grams O3. So you SHOULD expect less than Avogadro’s number of molecules. and you DO. A LOT less…

1.00x 10^23 molecules vs. 6.02 x10^23 molecules…. you have 6 times less approximately, both on molar and gram quantities.

48 g O3 / 8 g O3 = 6

Source(s): Any general chemistry text, and practice
• 8/48 times Avagadro’s number. You know that number I suppose if you are in Chemistry or can look it up. You do the math.

• pV=nRT keeping temperature and quantity consistent, the rigidity of the gas modifications basically with the no. of moles of gas. on the grounds that 2 moles of ozone could variety 3 moles of O2, and the preliminary rigidity of O3 is a million.2atm, on condition that rigidity is quickly proportional to the no. of moles of gas, rigidity of O2 could be (3/2)(a million.2)=a million.8atm