Question

An unknown powder has been found at a crime scene. It is your job to identify the chemical formula for the substance and its common or generic name by searching the internet. Preliminary tests reveal that the possible elements that can be in the material are carbon, hydrogen, oxygen, and nitrogen. Complete combustion of a 1.6 g sample of the unknown powder with excess O2 produced 3.6665 g of CO2, 1.1601 g of H2O, and 0.1061 g N2. The molar mass of the substance was found to be about 211 g/mol.

a) C10H20O5N
b) C9H18O4N2
c) C8H16O3N3
d) C7H14O2N4

Answer 1

The chemical formula of the unknown powder is identified through combustion analysis by calculating the moles of carbon, hydrogen, and nitrogen. The moles are calculated using the mass of CO2, H2O, and N2 produced. The empirical formula is then determined and compared to the options provided, with option a) C10H20O5N being the closest match.

Step-by-step explanation:

The task involves identifying the chemical formula of an unknown powder through combustion analysis. The mass of CO2, H2O, and N2 produced from burning a 1.6 g sample indicates the amounts of carbon, hydrogen, and nitrogen in the compound, respectively.

To find the number of moles of carbon, we use the mass of CO2 produced: 3.6665 g CO2 x (1 mol CO2/44.01 g CO2) x (1 mol C/1 mol CO2) = 0.0833 moles of carbon. Similarly, for hydrogen, we have 1.1601 g H2O x (1 mol H2O/18.015 g H2O) x (2 mol H/1 mol H2O) = 0.1287 moles of hydrogen. For nitrogen, 0.1061 g N2 x (1 mol N2/28.02 g N2) x (2 mol N/1 mol N2) = 0.0076 moles of nitrogen. By multiplying with the atomic masses (C = 12.01 g/mol, H = 1.008 g/mol, N = 14.01 g/mol), we get the mass of each element and can find the empirical formula which matches the molar mass of the substance (about 211 g/mol).

After determining the number of moles of each element, the ratios are found to be approximately C10:H20:N1, which is closest to option a) C10H20O5N. Since the molar mass is given as 211 g/mol, the number of oxygen atoms can be found by subtracting the mass contribution of C, H, and N from 211 g/mol. We already know the total moles of C, H, and N which are 0.0833 mol, 0.1287 mol, and 0.0076 mol, respectively. By multiplying these figures by their molar masses, we can determine the mass contribution and subtract from the total molar mass to obtain the mass and subsequently the moles of oxygen.

On calculating the moles of oxygen, you align the atomic ratios to the nearest whole number, which gives the empirical formula. Comparing the sum of the atomic weights in the empirical formula to the given molar mass confirms the molecular formula of the compound.