%HW3 X Molecular wei...Taylor Swift- Ghttps:/ qa/post Zconvert moles e How many

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%HW3 X Molecular wei...Taylor Swift- Ghttps:/ qa/post Zconvert moles e How many at . G determine the Minbox (16,876) G molar mass of + , ii e www.webassign.net/web/Student/Assignment-Responses/submit2dep=12881136 a molar mass of sodium carbonate 12. + -11 points ZumChem8 3.CW 12. My Notes Question: 1 Para-cresol, a substance used as a disinfectant and in the manufacture of several herbicides, is a molecule that contains the elements carbon, hydrogen, and oxygen. Complete combustion of a 0.345 g sample of p-cresol prodced 0.983 g carbon dioxide and 0.230 g of water Find the empiical formula for p-cresol. Input the elements in your formula in the order shown above. Enter a subscript by surrounding it with underscore"_"characters. For example H2O would be entered as H 2_O If parentheses are needed in the formula, be sure to include them in your answer. Submit Show Hints 8:12 PM 1/29/2016 Search the web and Windows

Explanation / Answer

Let the formula of the para-cresol be CxHYOz

Now, CxHYOz + {x+(y/4)-z}O2 -------> xCO2 + (y/2)H2O

Molar mass of CxHYOz = 12x + y + 16z

Thus, moles of CxHYOz in 0.345 g of it = mass/molar mass = 0.345/(12x + y + 16z)

Now, molar mass of CO2 = 44 g/mole

Molar mass of H2O= 18 g/mole

Thus, moles of CO2 produced = mass/molar mass = 0.983/44 = 0.0223

Thus, moles of C = moles of CO2 = 0.0223

Mass of C = 0.0223*12 = 0.2676 g

moles of O = 2*moles of CO2 = 0.0446

Mass of O = 0.0446*16 = 0.7136 g

moles of H2O produced = mass/molar mass = 0.23/18 = 0.0128

Thus, moles of H = 2*0.0128 = 0.0256

mass of H = 0.0256*1 = 0.0256 g

moles of O = 0.0128

mass of O = 0.0128*16 = 0.2048 g

Total mass of H, C & O = 0.0256 + 0.2676 + 0.2048 + 0.7136 = 1.2116 g

Thus, mass of O2 used in combustion = 1.2116 - 0.345 = 0.8666 g

Mass of O in CxHYOz = 0.345 - 0.2676 - 0.0256 = 0.0518 g

Moles of O in CxHYOz = mass/molar mass = 0.0518/16 = 0.00324

Thus, C , H & O are present in the molar ratio of x:y:z = 0.0223:0.0256:0.00324 = 7:8:1

Thus, the empirical formula is :-

C7H8O

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