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.

 

[m2003]

As a scientist, it is important to understand the concept of quantized charge and how it applies to this problem. In this case, we are dealing with the charge on protons, which is a fundamental unit of charge and is quantized in nature. This means that it can only exist in discrete, integer multiples of a certain value (in this case, the elementary charge e). Therefore, when calculating the charge in a certain volume of water, we must take into account the number of protons present, which is a whole number, and not just the mass or volume of the water. This is why the first equation, which takes into account the number of protons, is the correct one.

The second equation, which only considers the mass or volume of the water, does not take into account the quantized nature of charge and therefore does not yield the correct solution. In order to accurately calculate the charge, we must consider the number of protons, which is a discrete value.

In conclusion, the first equation is correct because it takes into account the quantized nature of charge and the discrete value of the number of protons in the water sample.