How Do You Calculate the Empirical Formula of Magnesium Nitride?

[thomas49th]

Hi. I am having problems with finding the empirical formulae

What is the emprical formula of magnesium nitride if 0.72g of magnesium combines with 0.28g nitrogen?


So what would my first step be? Find magesuim... which would be 0.44g?
Can you take me through the steps?

Thx

 

[hage567]

Hi. I am having problems with finding the empirical formulae

What is the emprical formula of magnesium nitride if 0.72g of magnesium combines with 0.28g nitrogen?


So what would my first step be? Find magesuim... which would be 0.44g?
Can you take me through the steps?

Thx

I would start by converting your Mg and N into moles. The basic idea is to look at the ratios of one substance to another to find the emiprical formula.

I don't know why you subtracted up there. Mass must be conserved, so if you start out with 0.72g and 0.28g, how much must you have after the reaction?

 

[neelneel311]

First calculate the number of moles of both of them by dividing each by their atomic weights.
Moles are obtained

Now divide each mole by the smallest mole obtained i.e. 0.02 mole of nitrogen, ratio obtained will be 1.5 of magnesium and 1 of nitrogen

as 1.5 is not a whole number multiply it by 2 to get a whole number and you will get as 3 and 2 respectivley. thus from this you can get that 3 moles of magnesium reacts with 2 moles of nitrogen to give magnesium nitride.

I hope this will help you out.
Regards

 

[thomas49th]

Hahaha thanks :D But a little late, I did my GCSE chemistry over 3 years ago and got my A - never said thank you hage567. Infact, I'm starting university tomorrow!

Thanks
Thomas :)

 

[DeeNos]

Hello,

Thank you for reaching out. I would be happy to assist you with calculating the empirical formula of magnesium nitride.

First, let's start by understanding what empirical formula means. An empirical formula is the simplest ratio of whole number subscripts in a compound. This means that we need to find the ratio of magnesium and nitrogen in the compound.

To start, we need to convert the given masses of magnesium and nitrogen into moles. We can do this by dividing the mass of each element by its respective molar mass. The molar mass of magnesium is 24.31 g/mol and the molar mass of nitrogen is 14.01 g/mol.

So, for magnesium, we have 0.72g/24.31 g/mol = 0.0296 mol. Similarly, for nitrogen, we have 0.28g/14.01 g/mol = 0.0199 mol.

Next, we need to find the smallest whole number ratio between these two values. To do this, we divide both values by the smaller one, which is 0.0199 mol in this case. This gives us a ratio of 0.0296 mol/0.0199 mol = 1.49. We can round this to the nearest whole number, which is 1.

This means that for every 1 mole of nitrogen, we have 1.49 moles of magnesium. To get whole numbers, we multiply both values by 2, giving us a ratio of 2 moles of nitrogen to 2.98 moles of magnesium.

Finally, we write the empirical formula using these whole number ratios. In this case, it would be Mg2N3.

I hope this helps guide you through the steps of calculating empirical formulae. Let me know if you have any further questions. Good luck!