(I)
2.241 g CO2 x (1 mol CO2/molar mass CO2) x (atomic mass C/1 mol C) = g carbon.
(II)
1.2359 g sample x 0.2884 = g nitrogen.
(III)
1.2359 g sample - g C - g N -g H = grams O.
(IV)
Convert g C to percent C.
Convert g N to percent N.
Convert g H to percent H.
Convert g O to percent O.
Take a 100 g sample which will give you the grams of C, H, O, and N equal to the percents of each.
Convert g C, H, O, and N to mols by dividing by the atomic mass of C, H, O, and N respectively.
Convert to the ratio of small whole numbers. The easiest way to do this is to divide the smallest number of mols by itself (which gives it a value of 1.000). Divide all the other mole values by the same small number, then take the ratio of those whole numbers.
Post your work if you get stuck. Check my thinking. Check my work.
18. Answer the following question that relate to the analysis of chemical compounds.
(a) A compound containing the elements C, H, N, and O is analyzed. When a 1.2359g sample is burned in excess oxygen, 2.241g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.048g of H.
(i) Determine the mass, in grams, of C in the 1.2359g sampleof the compound.
(ii) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84%. Determine the mass, in grams, of N, in the original 1.2359g sample of the compound.
(iii) Determine the mass, in grams, of O in the original 1.2359g sample of the compound.
(iv) Determine the empirical formula of the compound.
3 answers
thanks so much for your help!
yes