Jawaban:

c. C9H18O3

Penjelasan:

Mis: massa senyawa 100 gram

maka massa C = 62,079 gram

massa H = 10,34 gram

massa O = 27,59 gram

Hitung perbandingan mol masing" unsur: mol = massa / Mr

mol C : mol H : mol O

=> ( 62,079 / 12 ) : ( 10,34 / 1 ) : ( 27,59 / 16 )

=> 5,17 : 10,34 : 1,72

=> 3 : 6 : 1

Rumus empiris: C3H6O

T = 80° C = 353° K

P = 37 cm Hg = 0,49 atm

1 atm = 76 cm Hg

V = 743 ml = 0,743 L

R = 0,082 L atm / mol K => tetapan gas ideal

Rumus gas ideal:

P V = n R T

0,49 x 0,743 = n x 0,082 x 353

0,36 = n x 28,95

n = 0,01 mol

massa senyawa X = n x Mr

1,45 = 0,01 x Mr

Mr = 145

Mencari rumus molekul:

Rumus empiris x n = Mr senyawa

( C3H6O ) n = 145

( 3 x Ar C + 6 x Ar H + 1 x Ar O ) n = 145

( 36 + 6 + 16 ) n = 121

58 n = 145 => n = 2,5 ~ 3

Rumus molekul ( C3H6O ) 3 => C9H18O3