Can You Calculate the Mass of a Water Molecule Using Avogadro's Number?

[Kamataat]

Is there any way to find the mass of a water molecule when only Avogadro's number is given?

Thanks!

- Kamataat

 

[Smith4046]

Think I got it!

The mass of one mole of water = 18 grams.

Avogadro's = 6.02x10^23 molecules/mole


18gms/mole divided by 6.02x10^23 molecules/mole should work, if my chemistry isn't too rusty. The mole units will cancel each other, leaving you with units of gms/molecule.

Should get something like 3x10^-23 gm/molecule. Actually a little less, since I rounded all numbers off.

Please correct me (anyone) if I am incorrect!

Good luck.

Steve

 

[Kamataat]

Yeah, that's how I'd do it, but I was asking because of the stupid exercise book. It gives all sorts of extra info in the back, in case it's required to solve a problem, but it doesn't give any molar masses. So I thought that maybe there's a clever way to solve this w/o the molar mass.

All I know is Avogadro's number, the definition of a mole (# of atoms in 0.012kg of $C^{12}$) and Clapeyron's equation.

Thanks anyway!

- Kamataat

 

[moose]

The easiest way to figure out the molar mass is to look at a pereodic table of elements...Off the top of my head, two hydrogen(1.01) and one oxygen(16), you just add those and get the molar mass.

Am I missing something...?

 

[bross7]

No, that definitely should be what you do. Most problems just don't give molar masses for things like water, carbon dioxide, etc. in some chemistry books because the value is given so many times over that it becomes one of those redundant things.

Given only avogadro's number, you definitely need to have the molar mass to find the number of molecules.

 

[whozum]

The average molar mass for an element is given on any periodic table, and certain isotopes can be looked up. Water is 2 hydrogens (1.01 each) and one oxygen (16) which gives 16 + 2(1.01) = 18.02 g/mol.

$$\left(\frac{18.02 g}{1 mol}\right) \left(\frac{1 mol}{6.02 \times 10^{23} molecules}\right) = \left(\frac{ x grams}{ 1 molecule }\right)$$

 

[theCandyman]

Smith basically used this then multiplied by one mole then divided by one mole.

 

[Kamataat]

Thanks everyone for making this clear for me!

- Kamataat