Propose a mathematical equation to calculate the charge of an ion from the number of protons and electrons in an ion confirm that your equation works using two positive ion examples and two negative ion examples

The mathematical equation to calculate the charge of an ion is

We can calculate the charge (q) of an ion using the following expression.

where,

Let's see it work using two positive ion examples.

H⁺ (q = 1+) has 1 proton and 0 electrons.

Be²⁺ (q = 2+) has 4 protons and 2 electrons.

Now, let's see it work using two negative ion examples.

F⁻ (q = 1-) has 9 protons and 10 electrons.

O²⁻ (q = 2-) has 8 protons and 10 electrons.

The mathematical equation to calculate the charge of an ion is

You can learn more about ions here: link

Here we have to get the equation to determine the charge of an ion in terms of protons and electrons.

The general formula to determine the charge of an ion is

number of protons - number of electrons = positive (+ve) value represent the positive ion.

number of protons - number of electrons = negative (-ve) value represent the negative ion.

The protons carry positive charge and the electrons carry negative charge. The ion in which the positive charge is more then the negative charge that will be positive ion

i.e. number of protons - number of electrons = positive (+ve) value represent the positive ion.

number of protons - number of electrons = negative (-ve) value represent the negative ion.

Examples:

Positive ion:

1. H⁺ in which number of protons 1 and number of electrons is 0.

Thus 1-0 = 1 (+ve) value.

2. Mg²⁺ in which number of protons 12 and number of electrons are 10.

Thus 12 - 10 = 2 (+ve).

Negative ion:

1. F⁻ in which number of protons are 9 and number of electrons are 10.

Thus 9 - 10 = -1 (-ve).

2. O²⁻ in which number of protons are 8 and number of electrons are 10.

Thus 8 - 10 = -2 (-ve)

Thus the general formula for the determining the charge of an ion is determined.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g