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Question

?OWLv21Orine teachin × do?locator-assignment-take&takeAssignmentSessiont.ocator; assignment-tak 1 CO f pts s1 pts i. 1 pts at 1 pts onst pts When 1.489 grams of a hydrocarton, Gah, were burned im a combustion analysis apparatus, 4.907 grams of CO, and 1.339 grams of H:O were produced In a separate experiment, the molar mass of the compound was found to be 40.06 g mol Determine the empirical formula and the molecular formula of the hydrocarbon Ester the elements in the order presented in the uestion. 1 pts empirical formula molecular fomula 7 more group attempts remaining ", , to." '" "?? "brt sc ", delete home 567890-=?backspace

Explanation / Answer

1)
No of mol of co2 = 4.907/44 = 0.111 mol

No of mol of c = 0.111 mol

No of mol of H2O = 1.339/18 = 0.0744 mol

no of mol of H = 2*0.0744 = 0.1488 mol

simplest ratio :

C = 0.111/0.111 = 1

H = 0.1488/0.111 = 1.34

simplest whole number ratio : 3:4

C = 3
H = 4

empirical formula: C3H4

empirical formula mass = 12*3+4*1 = 40 g/mol

n = 40.06/40 = 1

molecular formula = C3H4

2)

No of mol of co2 = 12.97/44 = 0.295 mol

No of mol of c = 0.295 mol

No of mol of H2O = 5.309/18 = 0.295 mol

no of mol of H = 2*0.295 = 0.59 mol

mass of O present in sample = 7.275 - (0.295*12+0.59*1) = 3.145 g

no of mol of O present in sample = 3.145/16 = 0.196 mol

simplest ratio :

C = 0.295/0.196 = 1.5

H = 0.59/0.196 = 3

O = 0.196/0.196 = 1

simplest whole number ratio : 3:6:2

C = 3

H = 6

O = 2

empirical formula: C3H6o2

empirical formula mass = 12*3+6*1+2*16 = 74 g/mol

n = 74.08/74 = 1

molecular formula = C3H6o2

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