You perform a combustion analysis on a 255 mg sample of a substance that contain

Question

You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg of CO_2 is produced, along with 306 mg of H_2O. a. If the substance contains only C, H, and O, what is the empirical formula? b. If the molar mass of the compound is 180g/mol, what is the molecular formula of the compound? Potassium superoxide, KO2, is employed in a self-breathing apparatus used by emergency personnel as a source of oxygen. The reaction is: 4KO_2 (s) + H_2O (l) a 4KOH (s) + 2O_2 (g) If a self-contained breathing apparatus is charged with 750 g of KO_2 and then is used to produce 188 g of oxygen, was all of the KO_2 consumed in this reaction? If the KO_2 wasn't all consumed, how much is left over and what mass of additional O_2 could be produced?

Explanation / Answer

Question 1.

m = 255 mg = 0.255 g of sample

m = 0.561 g of CO2

m = 0.306 g of H2O

a)

find empirical formula

first, calculate moles:

mol of CO2 = mass/MW = 561/44 = 12.75 mmol

mol of H2O = mass/MW = 306/18 = 17 mmol

mol of H = 2*17 = 34 mmol

therefore

mmol of C = 12.75; mass of C = 12.75*12 = 153 mg

mmol of H = 34; mass of H = 34*1 = 34 mg

mass of O = (255) - (153 + 34) = 68 mg of O

mol = mass/MW = 68/16 = 4.25 mmol

ratio:

C:O = 12.75/4.25   = 3

H:C = 34/12.75 = 2.666

H:O = 34/4.25 = 8

Assume, 1 Oxygen, so C = 3, H = 8

empirical formula is C3H8O

b)

if MW = 180 g/mol

MW empirical = 12*3 + 8*1 + 16 = 60

rati = 180/60 = 3

then

C3H8O --> C9H24O3

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