# What is the mass of 3.34xx10^23 "water molecules"?

Feb 18, 2016

$6.022140857 \times {10}^{23}$ water molecules have a mass of $\left(15.999 + 2 \times 1.008\right) \cdot g$ $\approx$ $18.0 \cdot g$

#### Explanation:

The number $6.022140857 \times {10}^{23}$ is ${N}_{A}$, Avogadro's number. If I have precisely this number of atoms, or molecules, I have, by definition, A MOLE of such ATOMS or MOLECULES. You have $3.34 \times {10}^{23}$ individual water molecules; you have slightly over $\frac{1}{2}$ $\times$ ${N}_{A}$, i.e. $9 - 10$ $g$.

Feb 18, 2016

the mass of $3.34 \times {10}^{23}$ water molecules is 9.911 grams

#### Explanation:

Okay so firstly, you want to be converting the number of molecules of water you have into number of moles
As you probably know
Number of Particles = Number of Moles x Avogadro's constant
thus, no of moles = number of particles/Avogadro's constant
$= 3.34 \times {10}^{23} / 6.02 \times {10}^{23}$
$= 0.55$
Thus 0.55 Moles
Now number of moles = mass/molar mass
We have to calculate the mass, thus
Mass = Number of Moles x Molar Mass
The molar mass of ${H}_{2} O$ is
H - 2(1.01) = 2.02
O- 16
18.02
Thus, the molar mass is 18.02 g/mol
Therefore, now we have the required values to find the mass
Mass = $0.55 \times 18.02$
Mass = $9.911$
Thus, the mass of $3.34 \times {10}^{23}$ water molecules is 9.911 grams