The molecular formula of the artificial sweetener aspartame is C14H18N2O5. How many millgrams of nitrogen are present in 3.15 grams of aspartame?

Question

Chemistry mass 

Answer 1

Hello Matt

 

The process for solving this question is:  g aspartame → mol aspartame → mol nitrogen → g nitrogen → mg nitrogen

 

Molar mass of aspartame and nitrogen are 294 g/mol and 14 g/mol, respectively.

 

For mol aspartame → mol nitrogen, there are 2 moles of nitrogen present in mole of aspartame as per its chemical formula.

 

3.15 g aspartame x 1 mol / 294 g aspartame x 2 mol N / 1 mol aspartame x 14 g N / 1 mol N x 1000 mg / 1 g = 300 mg