Answer to Question #138919 in General Chemistry for Ada

Question

The most common experimental technique to perform elemental analysis is combustion analysis, where a sample is burned in a large excess of oxygen and the combustion products are trapped in a variety of ways. A 99.99% pure, 0.4882 g sample containing only carbon, hydrogen, and nitrogen is subjected to combustion analysis, resulting in the formation of 1.394 g CO2, 0.3307 g H2O, and 0.1573 g NO. What is the empirical formula of the sample

Answer 1

moles = mass / molar mass

We calculate the mass and number of moles of C and H first 

1.394 g of CO2 has 1.394/44 = 0.03168 moles of CO2

there is 1 mole of C in CO2 and all the C from the compound becomes CO2

so moles of C in the compound = 0.03168 moles

mass of C = 0.03168 x 12 = 0.38016 g

molar mass of H2O = 18 g/mole

0.3307 g of H2O has 0.3307/18 = 0.01837 moles of H2O

there are 2 moles of H in H2O so moles of H in the compound = 0.03674 moles

mass of H = 0.03674x1 = 0.03674 g

now we do the calculation for N

moles of NO = 0.1573/30 = 0.00524 moles

moles of N in the sample = 0.00524

mass of N = 0.00524x14 = 0.07336 g

we calculate the O by difference in 0.4882 g 

mass C + H + N = 0.38016 + 0.03674 + 0.07336 = 0.49

So there is no O in the compound. 

We now have all the number of moles, now we can do the empirical formula as the ratio of the elements 

molar ratio of C : H : N = 0.03168 : 0.03674 : 0.00524

Divide by the smallest number to simplify the ratio and we get

C : H : N = 6:7:1

Empirical formula is C6H7N