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Analisa terhadap suatu senyawa organik menunjukkan bahwa senyawa tersebut mengandung begin mathsize 14px style 52 percent sign space C comma space 13 percent sign H comma space dan space 35 percent sign space O end style, rumus molekul senyawa tersebut adalah ...

begin mathsize 14px style left parenthesis A subscript r space C equals 12 space gram space mol to the power of negative sign 1 end exponent comma space H equals 1 space gram space mol to the power of negative sign 1 end exponent comma space O equals 16 space gram space mol to the power of negative sign 1 end exponent comma space M subscript r space Senyawa equals 46 space gram space mol to the power of negative sign 1 end exponent right parenthesis end stylespace space  

  1. begin mathsize 14px style C subscript 2 H subscript 4 O end stylespace 

  2. begin mathsize 14px style C subscript 2 H subscript 6 O end stylespace 

  3. begin mathsize 14px style C subscript 3 H subscript 6 O end stylespace 

  4. begin mathsize 14px style C subscript 3 H subscript 6 O subscript 2 end stylespace 

  5. begin mathsize 14px style C subscript 4 H subscript 8 O subscript 2 end stylespace 

L. Avicenna

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.space 

Pembahasan

Mencari Rumus Empiris

begin mathsize 14px style jumlah space mol space C colon space jumlah space mol space H colon space jumlah space mol space O equals fraction numerator percent sign space C over denominator A subscript r space C end fraction colon fraction numerator percent sign space H over denominator A subscript r space H end fraction colon fraction numerator percent sign space O over denominator A subscript r space O end fraction jumlah space mol space C colon space jumlah space mol space H colon space jumlah space mol space O equals fraction numerator 52 percent sign over denominator 12 end fraction colon fraction numerator 13 percent sign over denominator 1 end fraction colon fraction numerator 35 percent sign over denominator 16 end fraction jumlah space mol space C colon space jumlah space mol space H colon space jumlah space mol space O equals 4 comma 3 colon 13 colon 2 comma 18 jumlah space mol space C colon space jumlah space mol space H colon space jumlah space mol space O equals 2 colon 6 colon 1 end style 

Jadi rumus empirisnya adalah begin mathsize 14px style C subscript 2 H subscript 6 O end style.
 

Mencari Rumus Molekul

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript r space Senyawa end cell equals cell open parentheses RE close parentheses cross times n end cell row cell 46 space g space mol to the power of negative sign 1 end exponent end cell equals cell open parentheses C subscript 2 H subscript 6 O close parentheses cross times n end cell row cell 46 space g space mol to the power of negative sign 1 end exponent end cell equals cell left curly bracket left parenthesis 2 cross times 12 space g space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 6 cross times 1 space g space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 16 space g space mol to the power of negative sign 1 end exponent right parenthesis right curly bracket cross times n end cell row cell 46 space g space mol to the power of negative sign 1 end exponent end cell equals cell left curly bracket 24 space g space mol to the power of negative sign 1 end exponent plus 6 space g space mol to the power of negative sign 1 end exponent plus 16 space g space mol to the power of negative sign 1 end exponent right curly bracket cross times n end cell row cell 46 space g space mol to the power of negative sign 1 end exponent end cell equals cell 46 space g space mol to the power of negative sign 1 end exponent cross times n end cell row n equals 1 end table end style 

Jadi Rumus Molekulnya adalah begin mathsize 14px style C subscript bold 2 H subscript bold 6 O end style. 

Oleh karena itu, jawaban yang benar adalah B.space 

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