What is the empirical formula of a compound that contains 6.10 g of hydrogen and 28 g of nitrogen?

NH₃

Explanation:

mass H = 6.10 grams

mass N = 28.00 grams

mass cpd = (6.10 + 28.00) grams = 34.10 grams

%H/100wt = (6.10/34.10) 100% = 17.9% w/w

%N/100wt = (28.00/34.1) 100% = 82.1% w/w

%/100wt = > grams/100wt = > moles = > ratio = > reduce = > emp ratio

%H/100wt = 17.9% w/w = > 17.9g = > (17.9/1) moles = 17.9 moles H

%N/100wt = 82.1% w/w = > 82.1g = > (82.1/14) moles = 5.9 moles N

Ratio N:H = > 17.9 : 5.9

Reduce mole ratio (divide by smaller mole value) = > 17.9/5.9 : 5.9/5.9

=> 3HY:1H empirical ratio = > empirical formula NH₃ (ammonia)