A compund containing the elements C, H, N, and O is analyzed.When a 1.2359g samo

Question

A compund containing the elements C, H, N, and O is analyzed.When a 1.2359g samole is burned in excess oxygem, 2.241 g ofCO2(g) is formed. The combustion analysis also showedthat th sample contained 0.0648 g of H. i) what is the mass, in grams, of C in the 1.2359 gsample of the compund? ii) When the compund is analyzed for N content only, the masspercent of N is found to be 28.84 percent. What is the mass, ingrams, of N in the original 1.2359 g sample of the compound? iii) What is the mass, in grams, of O in the original 1.2359 gsamole of the compound. iv) What is the empirical formula of the compound? A compund containing the elements C, H, N, and O is analyzed.When a 1.2359g samole is burned in excess oxygem, 2.241 g ofCO2(g) is formed. The combustion analysis also showedthat th sample contained 0.0648 g of H. i) what is the mass, in grams, of C in the 1.2359 gsample of the compund? ii) When the compund is analyzed for N content only, the masspercent of N is found to be 28.84 percent. What is the mass, ingrams, of N in the original 1.2359 g sample of the compound? iii) What is the mass, in grams, of O in the original 1.2359 gsamole of the compound. iv) What is the empirical formula of the compound?

Explanation / Answer

(i) Moles of CO2 formed = mass / molar mass                                     =2.241 g / 44.01 g/mol                                     =0.0509mol So moles of carbon = 0.0509 mol Mass of carbon = moles * atomic mass                         = 0.0509 mole * 12.01 g/mol                         =0.6113 g (ii) Mass % = 28.84 % (28.84 / 100 ) * 1.2359 g = 0.3564 g This is the mass of N in the original sample (iii) Mass of O = mass of sample - ( mass of C + mass of N +mass of H)                  = 1.2359 g - (0.6113 g + 0.3564 + 0.0648 g)                   = 0.2034 g (iv) Moles of N = 0.3564 g / 14.01 g/mol                    =0.0254 mol Moles of O = 0.2034 g/ 16.00 g/mol                    =0.0127 mol Moles of H = 0.0648 / 1.008 g/mol                   = 0.0642 mol                            C              N               H              O Moles            0.0509      0.0254         0.0642      0.0127 Dividing all by 0.0127                         4               2                  5                  1 So the empirical formual is C4 N2H5 O

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