A sample of a substance(containing only C, H, and N) is burned in oxygen. 1.691

Question

A sample of a substance(containing only C, H, and N) is burned in oxygen.
1.691 g of CO2, 3.462×10-1 g of H2O and 3.459 g of NO are thesole products of combustion.
a) What is the empirical formula of the compound? b) What is the mass of the initial sample burned? The correct answer for a) is CHN3, but I dont know how tosolve it. How do you get Moles of C and H and N? Form there, Ithink I can get the answer. The corrwct answer for b) is2.115g If someone could help me ASAP I would greatly appreciateit! Thanks A sample of a substance(containing only C, H, and N) is burned in oxygen.
1.691 g of CO2, 3.462×10-1 g of H2O and 3.459 g of NO are thesole products of combustion.
a) What is the empirical formula of the compound? b) What is the mass of the initial sample burned? The correct answer for a) is CHN3, but I dont know how tosolve it. How do you get Moles of C and H and N? Form there, Ithink I can get the answer. The corrwct answer for b) is2.115g If someone could help me ASAP I would greatly appreciateit! Thanks b) What is the mass of the initial sample burned? The correct answer for a) is CHN3, but I dont know how tosolve it. How do you get Moles of C and H and N? Form there, Ithink I can get the answer. The corrwct answer for b) is2.115g If someone could help me ASAP I would greatly appreciateit! Thanks

Explanation / Answer

mass of C = 1.691 g *(1 mol CO2 / 44 g) *( 12 g C / 1 molCO2) = 0.4611 g C
mass of H = 0.3462 g *( 1mol H2O/ 18 g) *( 2 g H/ 1 mol H2O)= 0.0384 g H
mass of N = 3.459 g *( 1mol NO/ 30 g) *( 14 g N / 1mol NO ) = 1.6142 g N

moles C = 0.4611 g C *( 1mol / 12 g) = 0.0384 mol C
moles H = 0.0384 g H *(1 mol / 1 g) = 0.0384 mol H
moles N = 1.6142 g H *(1 mol / 14 g) = 0.1153 mol N
Ratio of C : H :N is 0.0384 : 0.0384 : 0.1153
                             1 : 1 : 3

=> C H N 3   

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