# How to find out the no. of protons, neutrons and electrons in the species CO and NO+, and lithium fluoride? Is there a specfic formula to calculate? cos I only understand how to calculate the no. of subatomic particles in an element.

Here’s how you could do that.

To find the number of protons, neutrons, and electrons in a molecule or formula unit you basically add up the number of protons, neutrons, and electrons each individual atom that’s part of that compound brings to the table.

I’ll use carbon monoxide, ##”CO”##, and the nitrosonium ion, ##”NO”^(+)##, as examples and leave lithium fluoride, ##”LiF”##, to you as practice.

So, let’s start with carbon monoxide. One molecule of carbon monoxide contains

• one atom of carbon
• one atom of oxygen

Start with the carbon atom. Your tool of choice here will be the . You an find carbon located in period 2, group 14. The element has an equal to ##6##, which means that is has ##6## protons located in its nucleus.

Now, to find the number of neutrons in a carbon atom, you need to first determine its , which as you know tells you how many protons and neutrons can be found in the nucleus.

Notice that carbon’s atomic weight is listed in as being equal to ##”12.011 u”##. The of the most abundant carbon isotope will be whole number that’s closest to the value of the atomic weight.

Simply put, you round ##12.011## to the closest integer to get ##12##. This means that the most abundant isotope of carbon has

• ##6## protons in its nucleus
• ##12 – 6 = 6## neutrons in its nucleus

A neutral atom will always have equal numbers of protons in its nucleus and electrons surrounding its nucleus. This means that the neutral carbon atom will have a total of ##6## electrons surrounding its nucleus.

Now do the same for the oxygen atom. Oxygen has an equal to ##8##, so it will have ##8## protons in its nucleus.

Round its atomic weight, ##”15.9994 u”##, to the nearest integer to get the mass number of the most abundant isotope of oxygen. In this case, you have

##15.9994 ~~ 16##

Therefore, you can say that the most abundant isotope of oxygen will have

• ##8## protons in its nucleus
• ##16 – 8 = 8## neutrons in its nucleus

Once again, a neutral atom will have equal numbers of electrons and of protons. This means that the oxygen atom will have a total of ##8## electrons.

So, the carbon monoxide molecule will have

##”Protons: ” overbrace(6)^(color(red)(“from C”)) + overbrace(8)^(color(blue)(“from O”)) = 14##

##”Neutrons: ” overbrace(6)^(color(red)(“from C”)) + overbrace(8)^(color(blue)(“from O”)) = 14##

##”Electrons: ” overbrace(6)^(color(red)(“from C”)) + overbrace(8)^(color(blue)(“from O”)) = 14##

The exact same approach can be used for the nitrosonium ion, with the important difference that you’re now dealing with an ion, so you need to keep track of that .

So, for nitrogen you will get

• ##7## protons
• ##7## neutrons
• ##7## electrons

You already know the numbers for oxygen. This means that a neutral ##”CO”## molecule, not an ion, will have

##”Protons: ” overbrace(7)^(color(red)(“from N”)) + overbrace(8)^(color(blue)(“from O”)) = 15##

##”Neutrons: ” overbrace(7)^(color(red)(“from N”)) + overbrace(8)^(color(blue)(“from O”)) = 15##

##”Electrons: ” overbrace(7)^(color(red)(“from N”)) + overbrace(8)^(color(blue)(“from O”)) = 15##

However, since ##”NO”^(+)## carries a ##1+## charge, it will be one electron short compared with the neutral molecule. This means that its actual number of electrons will be

##”Electrons: ” overbrace(7)^(color(red)(“from N”)) + overbrace(8)^(color(blue)(“from O”)) – overbrace(1)^(color(purple)(“from charge”)) = 14##

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