Question

What number of atoms of nitrogen are present in 5.40 g of each of the following?

a) glycine: atoms N

b) magnesium nitride: atoms N

c) calcium nitrate: atoms N

d) dinitrogen tetroxide: atoms N

Answer 1

a) glycine formula is: C₂H₅NO₂

_{$\frac{5.4g C2H5NO2}{1}*\frac{1 mol C2H5NO2}{75.06 gC2H5NO2}*\frac{1 mol N}{1 molC2H5NO2}*\frac{6.022x10^{23}atoms N}{1 mol N}=4.3323x10^{22}$}

There are 4.3323X10²² atoms of N in 5.4 g of glycine.

b) The formula of magnesium nitrid is: Mg₃N₂

_{$\frac{5.4g Mg3N2}{1}*\frac{1 mol Mg3N2}{100.95 gMg3N2}*\frac{2 mol N}{1 molMg3N2}*\frac{6.022x10^{23}atoms N}{1 mol N}=6.4425x10^{22}$}

There are 6.4425X10²² atoms of N in 5.4 g of magnesium nitrid.

c)The formula of calcium nitrate is: Ca(NO₃)₂

_{$\frac{5.4g Ca(NO3)2}{1}*\frac{1 mol Ca(NO3)2}{164.09 gCa(NO3)2}*\frac{2 mol N}{1 molCa(NO3)2}*\frac{6.022x10^{23}atoms N}{1 mol N}=3.9635x10^{22}$}

There are 3.9635X10²² atoms of N in 5.4 g of calcium nitrate.

d)The formula of dinitrogen tetroxide is: N₂O₄

$\frac{5.4g N2O4}{1}*\frac{1 mol N2O4}{92.01 gN2O4}*\frac{2 mol N}{1 molN2O4}*\frac{6.022x10^{23}atoms N}{1 mol N}=7.069x10^{22}$

There are 7.069X10²² atoms of N in 5.4 g of dinitrogen tetroxide