# Does mass equal to moles times molar mass take into account coefficients?

I have an equation $$\ce{ 2NaN3 -> 2Na + 3N2}$$ I've calculated there are $\pu{4.4 moles}$ of $\ce{N2}$ (using the data given in the question). I then did $\large{\frac{4.4}{3 \cdot 2}}$ to get number of moles required of $\ce{NaN3}$ (although if this is wrong then correct me). And now I want to calculate the mass of $\ce{NaN3}$ used (knowing there are $\pu{3 moles}$).

Where I have $\pu{ 3 moles}$ of $\ce{2NaN3}$ and I want to calculate mass do I use the $\mathrm{M_r }$of $\ce{M_r(NaN3)\cdot 2}$ or just a single mole of it.

Eg:
$\text{mass = moles} \cdot \mathrm{M_r} = 3 \cdot 65$ or $\text{mass = }~3 \cdot 130$?

• Apparently you are confusing relative molecular mass $M_\mathrm r$ and molar mass $M$. Relative molecular mass $M_\mathrm r$ is a dimensionless quantity. – user7951 Mar 21 '18 at 18:18
• Furthermore, the quantity ‘amount of substance’ $n$ shall not be called ‘moles’; i.e. don’t confuse the quantity and the unit. – user7951 Mar 21 '18 at 18:20
• What part of my question confuses them? Is mass = moles * M? – Ben Hughes Mar 21 '18 at 18:21

## 1 Answer

What do you mean by $\pu{3 moles}$ of $\ce{ 2NaN3}$? The coefficient (the 2) is the number of moles of that molecule that you have.

If you're looking at an equation with $\ce{ 2NaN3}$ written in it, and you want that 3 times, then you're going to have $\pu{6 moles}$ of $\ce{NaN3}$, so $\text{mass}~= ~6 \cdot 65$ .

• I have an equation 2NaN3 --> 2Na + 3N2. I've calculated there are 4.4 moles of N2 (using the data given in the question). I then did 4.4/3 * 2 to get number of moles required of NaN3 (although if this is wrong then correct me). And now I want to calculate the mass of NaN3 used (knowing there is 3 moles). Hope this has clarified. – Ben Hughes Mar 21 '18 at 18:24
• I've edited my question to put this in to help other try and understand where I am getting confused – Ben Hughes Mar 21 '18 at 18:27
• Ah I see. The coefficients in the chemical equation simply show the ratio between the number of moles of each compound that are involved in the reaction. So for every 2 moles of NaN3 you form 2 moles of Na and 3 moles of N2. This is distinct from the actual number of moles you have of each chemical. You have calculated that if 4.4 moles of N2 are formed, then 2/3 as many moles of NaN3 must have reacted (ie 3 moles). Therefore mass = 3 * 65. – Matt Whitelock Mar 21 '18 at 18:33