How many atoms of nitrogen are present in 5.00g of C2H5O2N?

? atoms N = 5.00g C2H5O2N x 1 mole C2H5O2N/75.08g C2H5O2N x 1 mol N/1 mol C2H5O2N x (6.022x10^23 atoms)/1 mol N = 4.01 x 10^22 atoms N

To find 75.08g:

2C 2 x 12.01g = 24.02g

5H 5 x 1.01g = 5.05g

2O 2 x 16.00g = 32.00g

1N 1 x 14.01g = 14.01g

Add the four numbers together and you get: 1 mole of C2H5O2N = 75.08 grams

You use this as a conversion factor in the problem above.