Chemistry 103 Finals NAME SHOW SET-UP AND SOLUTION 1. Determine the molecular fo

Question

Chemistry 103 Finals NAME SHOW SET-UP AND SOLUTION 1. Determine the molecular formula of a compound that contains “ and has a molecular weight of 72 2. A compound contains only carbon, hydrogen, and nitrogen After combustion at a0.500 sample, the mass of the first trap (collecting H ,01 increses b 068 rap (leting CO, ncreises by 097 g. Whet i the emerkal ftrmula of eis t and he mass of the compound? 3, 3. What is the molecular formula of a subitance that comains only O n0 of carbon and o of hydrogen and has a molecular welght of 54? 4. Given the following equation, caleulate the mass of O eeded to react NO 2NO

Explanation / Answer

Empirical formula = C4H8O

Empirical formula = 4*(12) + 8 *(1) + 1*(16) = 72

number empirical units n = molecular mass/empirical mass = 72/72 = 1

molecular formula = n * empirical formula = 1 * C4H8O

molecular formula = C4H8O

2)

Mass of the compound = 0.5 g

Mass of H2O = 0.698 g

mass of CO2 = 0.977g

Percentage of H = 2/18 *0.698/0.5*100 = 15.51%

Percentage of C = 12/44*0.977.0.5*100 = 53.29%

Percentage of N = 100-(15.51+53.29) = 31.2%

Empirical formula = C2H7N

3)

mass of C = 0.801 g

mass of H = 0.101 g

Percentage of C = 0.801/(0.801+0.101) *100 = 88.88%

Percentage of H = 0.101/(0.801+0.101)*100 = 11.2%

Empirical formula = C2H3

Empirical mass = 2*12+3*1 = 27

number of empirical units = molr mass / empirical mass = 54/27 = 2

molecular formula = 2 * (C2H3) = C4H6

Element Percentage percentage ratio number of moles simple whole number C 66.6 66.6/12 = 5.55 5.55/1.3875 = 4 4 H 11.2 11.2/1 = 11.2 11.1/1.3875 = 8 8 O 22.2 22.2/16 = 1.3875 1.3875/1.3875 = 1 1

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