Quantized Charge Problem, Why is this right?

[PFStudent]

Homework Statement

26. Calculate the number of coulombs of positive charge in 250 $cm^3$ of (neutral) water. (Hint: A hydrogen atom contains one proton; an oxygen atom contains eight protons.)

Homework Equations

$$q = n_{e}e, n_{e} = \pm1, \pm 2, \pm 3,...,$$

e $\equiv$ elementary charge

$$e = 1.60217646{\textcolor[rgb]{1.00,1.00,1.00}{.}}x{\textcolor[rgb]{1.00,1.00,1.00}{.}}10^{-19}$$

The Attempt at a Solution

$$q = n_{e}e, n_{e} = \pm1, \pm 2, \pm 3,...,$$

$n_{p} \equiv$ number of protons
$q_{p} \equiv$ charge on a single proton

$$q_{p} = +e$$

$$q = \left(n_{p}\right)\left(q_{p}\right)$$

Z $\equiv$ Atomic Number (Number of Protons)
m $\equiv$ mass
M $\equiv$ Molar Mass ([kg]/[mols])

$$\rho_{w} = \frac{m_{w}}{V_{w}}$$

$$n_{p} = \frac{m_{w}}{m_{H_{2}O}} \cdot \frac{Z_{H_{2}O }}{1}$$

$$n_{p} = \frac{m_{w}}{\left(2m_{H}+1m_{O}\right)}} \cdot \frac{\left(2Z_{H}+1Z_{O}\right)}{1}$$

$$q = \left(\frac{\left(V_{w}\rho_{w}\right)}{2m_{H}+1m_{O}} \cdot \frac{2Z_{H}+1Z_{O}}{1}\right)q_{p}$$

$$q = \left(\frac{V_{w}\rho_{w}}{2\left(\frac{M_{H}}{N_{A}}\right)+1\left(\frac{M_{O}}{N_{A}}\right)} \cdot \frac{2Z_{H}+1Z_{O}}{1}\right)q_{p}$$

$$q = \left(\frac{N_{A}V_{w}\rho_{w}}{2M_{H}+1M_{O}} \cdot \frac{2Z_{H}+1Z_{O}}{1}\right)q_{p}$$

The above equation yields the correct solution, however my question is why is this right as opposed to the following?

$$q = \left(\frac{m_{w}}{m_{p}}\right)q_{p}$$

$$q = \left(\frac{V_{w}\rho_{w}}{m_{p}}\right)q_{p}$$

Why is the previous equation wrong?

Any help is appreciated.

Thanks,

-PFStudent

 

[Kurdt]

The last equation assumes that the mass of the water is made entirely from protons which of course is not true.