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Pembakaran sempurna 22 gram suatu senyawa organik ternyata menghasilkan 66 gram CO2​, dan 36 gram H2​O. Tentukan RE dan RM jika Mr​ senyawa diketahui 132!  (Ar​ C=12, O=16, H=1)

Pertanyaan

Pembakaran sempurna 22 gram suatu senyawa organik ternyata menghasilkan 66 gram C O subscript 2, dan 36 gram H subscript 2 O. Tentukan RE dan RM jika italic M subscript italic r senyawa diketahui 132!space 

left parenthesis A subscript r space C equals 12 comma space O equals 16 comma space H equals 1 right parenthesis

I. Solichah

Master Teacher

Jawaban terverifikasi

Jawaban

rumus empiris dan rumus molekul senyawa tersebut berturut-turut adalah Error converting from MathML to accessible text.space  

Pembahasan

Menentukan massa masing-masing unsur penyusun senyawa

massa space C equals fraction numerator A subscript r space C cross times valensi space C over denominator M subscript r space C O subscript 2 end fraction cross times massa space C O subscript 2 massa space C equals fraction numerator 12 space gram space mol to the power of negative sign 1 end exponent cross times 1 over denominator 44 space gram space mol to the power of negative sign 1 end exponent end fraction cross times 66 space gram space mol to the power of negative sign 1 end exponent massa space C equals 18 space gram  massa space H equals fraction numerator A subscript r space H cross times valensi space H over denominator M subscript r space H subscript 2 O end fraction cross times massa space H subscript 2 O massa space H equals fraction numerator 1 space gram space mol to the power of negative sign 1 end exponent cross times 2 over denominator 18 space gram space mol to the power of negative sign 1 end exponent end fraction cross times 36 space gram space mol to the power of negative sign 1 end exponent massa space H equals 4 space gram  massa space O double bond massa space C subscript x H subscript y O subscript z minus sign left parenthesis massa space C and massa space H right parenthesis massa space O equals 22 space gram minus sign left parenthesis 18 space gram plus 4 space gram right parenthesis massa space O equals 22 space gram minus sign 22 space gram massa space O equals 0 space gram 
 

Menghitung rumus empiris dengan menggunakan perbandingan mol C H dan O

table attributes columnalign right center left columnspacing 0px end attributes row cell koefisien space C space colon space koefisien space H end cell equals cell mol space C space colon space mol space H space end cell row cell koefisien space C space colon space koefisien space H end cell equals cell fraction numerator massa space C over denominator A subscript r space C end fraction space colon space fraction numerator massa space H over denominator A subscript r space H end fraction end cell row cell koefisien space C space colon space koefisien space H end cell equals cell fraction numerator 18 space gram over denominator 12 space gram space mol to the power of negative sign 1 end exponent end fraction space colon space fraction numerator 4 space gram over denominator 1 space gram space mol to the power of negative sign 1 end exponent end fraction end cell row cell koefisien space C space colon space koefisien space H end cell equals cell 1 comma 5 space colon space 4 end cell row cell koefisien space C space colon space koefisien space H end cell equals cell 3 space colon space 8 end cell row cell Rumus space Empiris end cell equals cell open parentheses C subscript 3 H subscript 8 close parentheses subscript n end cell end table 
 

Menentukan rumus molekul

table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript r space open parentheses C subscript 3 H subscript 8 close parentheses subscript n end cell equals cell 132 space gram space mol to the power of negative sign 1 end exponent end cell row cell left parenthesis left parenthesis 3 cross times A subscript r space C right parenthesis plus left parenthesis 8 cross times A subscript r space H right parenthesis right parenthesis subscript n end cell equals cell 132 space gram space mol to the power of negative sign 1 end exponent left parenthesis left parenthesis 3 cross times 12 space gram space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 8 cross times 1 space gram space mol to the power of negative sign 1 end exponent right parenthesis right parenthesis subscript n equals 132 space gram space mol to the power of negative sign 1 end exponent end cell row cell left parenthesis 36 space gram space mol to the power of negative sign 1 end exponent plus 8 space gram space mol to the power of negative sign 1 end exponent right parenthesis subscript n end cell equals cell 132 space gram space mol to the power of negative sign 1 end exponent 44 n equals 132 n equals 3 end cell end table Rumus molekulnya adalah open parentheses C subscript 3 H subscript 8 close parentheses subscript 3 equals C subscript 9 end subscript H subscript 24 end subscript 

Jadi rumus empiris dan rumus molekul senyawa tersebut berturut-turut adalah Error converting from MathML to accessible text.space  

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