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Senyawa yang mengandung jumlah mol paling banyak adalah...

Pertanyaan

Senyawa yang mengandung jumlah mol paling banyak adalah...undefined 

  1. 10,0 g begin mathsize 14px style C subscript 2 H subscript 6 end style (Mr = 30)undefined 

  2. 11,0 g begin mathsize 14px style C O subscript 2 end style (Mr = 44)undefined 

  3. 12,0 g begin mathsize 14px style N O subscript 2 end style (Mr = 46)undefined 

  4. 17,0 g begin mathsize 14px style Cl subscript 2 end style (Mr = 71)undefined 

  5. 20,0 g begin mathsize 14px style C subscript 6 H subscript 6 end style (Mr = 78)undefined 

Pembahasan Soal:

Bagan di bawah ini merupakan rangkuman dari konsep mol.


 


Berdasarkan bagan di atas, untuk mencari jumlah mol ketika massa zat telah diketahui, dapat menggunakan rumus:


begin mathsize 14px style mol space zat equals fraction numerator massa space zat over denominator Mr space zat end fraction end style 

 

  • mol begin mathsize 14px style C subscript 2 H subscript 6 end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n space C subscript 2 H subscript 6 end cell equals cell massa over Mr end cell row blank equals cell fraction numerator 10 space gram over denominator 30 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 0 comma 33 space mol end cell end table end style 

     
  • mol begin mathsize 14px style C O subscript 2 end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n space C O subscript 2 end cell equals cell massa over Mr end cell row blank equals cell fraction numerator 11 space gram over denominator 44 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 0 comma 25 space mol end cell end table end style 
     
  • mol begin mathsize 14px style N O subscript 2 end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n space N O subscript 2 end cell equals cell massa over Mr end cell row blank equals cell fraction numerator 12 space gram over denominator 46 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 0 comma 26 space mol end cell end table end style 

     
  • mol begin mathsize 14px style Cl subscript 2 end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n space Cl subscript 2 end cell equals cell massa over Mr end cell row blank equals cell fraction numerator 17 space gram over denominator 71 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 0 comma 24 space mol end cell end table end style 

     
  • mol begin mathsize 14px style C subscript 6 H subscript 6 end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n space C subscript 6 H subscript 6 end cell equals cell massa over Mr end cell row blank equals cell fraction numerator 20 space gram over denominator 78 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 0 comma 26 space mol end cell end table end style 


Jadi, jawaban yang benar adalah A.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Nur

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 30 Maret 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Tentukan rumus empiris dan rumus molekul dari senyawa yang terdiri dari:

Pembahasan Soal:

Rumus empiris dapat dihitung dengan cara sebagai berikut:undefined


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell C space colon space H end cell equals cell fraction numerator massa space C over denominator A subscript r space C end fraction colon fraction numerator massa space H over denominator A subscript r space H end fraction end cell row cell C space colon space H end cell equals cell fraction numerator 0 comma 9 over denominator 12 end fraction colon fraction numerator 0 comma 1 over denominator 1 end fraction end cell row cell C space colon space H end cell equals cell 0 comma 075 space colon space 0 comma 1 end cell row cell C space colon thin space H end cell equals cell 3 space colon space 4 end cell end table end style 


Maka rumus empirisnya adalah begin mathsize 14px style C subscript 3 H subscript 4 end style. Sedangkan rumus molekulnya dapat dicari dengan cara berikut.


begin mathsize 14px style M subscript r space open parentheses C subscript 3 H subscript 4 close parentheses subscript n equals 40 end style  
begin mathsize 14px style left parenthesis 3 middle dot A subscript r space C plus 4 middle dot A subscript r space H right parenthesis subscript n equals 40 end style 
begin mathsize 14px style left parenthesis 3 middle dot 12 plus 4 middle dot 1 right parenthesis subscript n equals 40 end style 
begin mathsize 14px style left parenthesis 36 plus 4 right parenthesis subscript n equals 40 end style 
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 40 right parenthesis subscript n end cell equals 40 row n equals 1 end table end style 
 

Rumus molekulnya yaitu begin mathsize 14px style open parentheses C subscript 3 H subscript 4 close parentheses subscript 1 equals C subscript 3 end subscript H subscript 4 end subscript end style.undefined

Jadi, jawaban yang tepat adalah seperti penjelasan di atas.undefined

0

Roboguru

Tentukan rumus empiris dan rumus molekul dari senyawa yang terdiri dari:

Pembahasan Soal:

Rumus empiris dapat dihitung dengan cara sebagai berikut:undefined


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell C space colon thin space H space colon space O end cell equals cell fraction numerator massa space C over denominator A subscript r space C end fraction colon fraction numerator massa space H over denominator A subscript r space H end fraction colon fraction numerator massa space O over denominator A subscript r space O end fraction end cell row cell C space colon space H space colon space O end cell equals cell fraction numerator 7 comma 92 over denominator 12 end fraction colon fraction numerator 1 comma 32 over denominator H end fraction colon fraction numerator 7 comma 04 over denominator 16 end fraction end cell row cell C space colon space H space colon space O end cell equals cell 0 comma 66 space colon space 1 comma 32 space colon space 0 comma 44 end cell row cell C space colon space H space colon O end cell equals cell 3 space colon space 6 space colon space 2 end cell end table end style 


Maka rumus empirisnya adalah begin mathsize 14px style C subscript 3 H subscript 6 O subscript 2 end style. Sedangkan rumus molekulnya dapat dicari dengan cara berikut.undefined


begin mathsize 14px style M subscript r space open parentheses C subscript 3 H subscript 6 O subscript 2 close parentheses subscript n equals 74 end style 
begin mathsize 14px style left parenthesis 3 middle dot A subscript r space C plus 6 middle dot A subscript r space H and A subscript r space O right parenthesis subscript n equals 74 end style 
begin mathsize 14px style left parenthesis 3 middle dot 12 plus 6 middle dot 1 plus 2 middle dot 16 right parenthesis subscript n equals 74 end style
begin mathsize 14px style left parenthesis 36 plus 6 plus 32 right parenthesis subscript n equals 74 end style 
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 74 right parenthesis subscript n end cell equals 74 row n equals 1 end table end style 


Rumus molekulnya yaitu begin mathsize 14px style open parentheses C subscript 3 H subscript 6 O subscript 2 close parentheses subscript 1 equals C subscript 3 end subscript H subscript 6 end subscript O subscript 2 end subscript end style.undefined

Jadi, jawaban yang tepat adalah seperti penjelasan di atas.

0

Roboguru

Lengkapi tabel berikut ini!

Pembahasan Soal:

Langkah 1: Melengkapi tabel NO

1. Mol NO (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript NO end cell equals cell fraction numerator jumlah space partikel subscript NO over denominator L end fraction end cell row blank equals cell fraction numerator 6 comma 02 cross times 10 to the power of 23 space over denominator 6 comma 022 cross times 10 to the power of 23 end fraction end cell row blank equals cell 1 space mol end cell end table


2. Mr NO 

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript NO end cell equals cell Ar subscript N and Ar subscript O end cell row blank equals cell 14 plus 16 end cell row blank equals cell 30 space begin inline style bevelled gram over mol end style end cell end table


3. Massa NO (m)

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript NO end cell equals cell n subscript NO cross times Mr subscript NO end cell row blank equals cell 1 space mol space cross times 30 space begin inline style bevelled gram over mol end style end cell row blank equals cell 30 space gram end cell end table


4. Volume NO (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript NO end cell equals cell n subscript NO cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 1 space mol space cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 22 comma 4 space L end cell end table


5. Volume NO (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript NO end cell equals cell n subscript NO cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 1 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 24 space L end cell end table


Langkah 2: Melangkapi tabel H subscript bold 2 

1. Jumlah partikel H subscript bold 2 (x)

    table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times L end cell row blank equals cell 0 comma 1 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 6 comma 022 cross times 10 to the power of 22 space partikel end cell row blank blank blank end table

2. Mr H subscript bold 2

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript H subscript 2 end subscript end cell equals cell 2 cross times Ar subscript H end cell row blank equals cell 2 cross times 1 space begin inline style bevelled gram over mol end style end cell row blank equals cell 2 space begin inline style bevelled gram over mol end style end cell end table
 

3. Massa H subscript bold 2

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times Mr subscript H subscript 2 end subscript end cell row blank equals cell 0 comma 1 space mol space cross times 2 space begin inline style bevelled gram over mol end style end cell row blank equals cell 0 comma 2 space gram end cell end table


4. Volume H subscript bold 2 (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 1 space mol space cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 2 comma 24 space L end cell end table


5. Volume H subscript bold 2 (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 1 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 2 comma 4 space L end cell end table


Langkah 3: Melangkapi tabel N H subscript bold 3

1. mol N H subscript bold 3 (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript N H subscript 3 end subscript end cell equals cell fraction numerator V subscript STP subscript N H subscript 3 end subscript end subscript over denominator 22 comma 4 space bevelled L over mol end fraction end cell row blank equals cell fraction numerator 4 comma 48 space L over denominator 22 comma 4 space bevelled L over mol end fraction end cell row blank equals cell 0 comma 2 space mol end cell end table
 

2. Jumlah partikel N H subscript bold 3 (x)
     table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times L end cell row blank equals cell 0 comma 2 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 1 comma 2 cross times 10 to the power of 23 space partikel end cell end table


3. Mr N H subscript bold 3

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript N H subscript 3 end subscript end cell equals cell Ar subscript N plus left parenthesis 3 cross times Ar subscript H right parenthesis end cell row blank equals cell 14 space begin inline style bevelled gram over mol end style plus left parenthesis 3 cross times 1 space begin inline style bevelled gram over mol right parenthesis end style end cell row blank equals cell 17 space begin inline style bevelled gram over mol end style end cell end table
 

4. Massa N H subscript bold 3

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times Mr subscript N H subscript 3 end subscript end cell row blank equals cell 0 comma 2 space mol space cross times 17 space begin inline style bevelled gram over mol end style end cell row blank equals cell 3 comma 4 space gram end cell end table


5. Volume N H subscript bold 3 (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 2 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 4 comma 8 space L end cell end table


Langkah 4: Melangkapi tabel C H subscript 4

1. mol C H subscript 4 (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C H subscript 4 end subscript end cell equals cell fraction numerator V subscript RTP subscript C H subscript 4 end subscript end subscript over denominator 24 space bevelled L over mol end fraction end cell row blank equals cell fraction numerator 12 comma 3 space L over denominator 24 space bevelled L over mol end fraction end cell row blank equals cell 0 comma 5125 space mol end cell end table 
 

2. Jumlah partikel C H subscript 4 (x)
     table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times L end cell row blank equals cell 0 comma 512 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 3 comma 08 cross times 10 to the power of 23 space partikel end cell end table


3. Mr C H subscript 4

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript C H subscript 4 end subscript end cell equals cell Ar subscript C plus left parenthesis 4 cross times Ar subscript H right parenthesis end cell row blank equals cell 12 space begin inline style bevelled gram over mol end style plus left parenthesis 4 cross times 1 space begin inline style bevelled gram over mol right parenthesis end style end cell row blank equals cell 16 space begin inline style bevelled gram over mol end style end cell end table
 

4. Massa C H subscript 4

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times Mr subscript C H subscript 4 end subscript end cell row blank equals cell 0 comma 512 space mol space cross times 16 space begin inline style bevelled gram over mol end style end cell row blank equals cell 8 comma 2 space gram end cell end table 


5. Volume C H subscript 4 (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 512 space mol cross times 22.4 space begin inline style bevelled L over mol end style end cell row blank equals cell 11 comma 48 space L end cell end table  


Dengan demikian, tabel lengkapnya adalah

0

Roboguru

Tentukan rumus empiris dan rumus molekul dari senyawa yang terdiri dari:

Pembahasan Soal:

Rumus empiris dapat dihitung dengan cara sebagai berikut:undefined


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K space colon thin space Mn space colon space O end cell equals cell fraction numerator massa space K over denominator A subscript r space K end fraction colon fraction numerator massa space Mn over denominator A subscript r space Mn end fraction colon fraction numerator massa space O over denominator A subscript r space O end fraction end cell row cell K space colon space Mn space colon space O end cell equals cell fraction numerator 1 comma 56 over denominator 39 end fraction colon fraction numerator 2 comma 2 over denominator 55 end fraction colon fraction numerator 2 comma 56 over denominator 16 end fraction end cell row cell K space colon space Mn space colon space O end cell equals cell 0 comma 04 space colon space 0 comma 04 space colon space 0 comma 16 end cell row cell K space colon space Mn space colon space O end cell equals cell 1 space colon space 1 space colon space 4 end cell end table end style 


Maka rumus empirisnya adalah begin mathsize 14px style K Mn O subscript 4 end style. Sedangkan rumus molekulnya dapat dicari dengan cara berikut.undefined


begin mathsize 14px style M subscript r space open parentheses K Mn O subscript 4 close parentheses subscript n equals 158 end style 
begin mathsize 14px style left parenthesis A subscript r space K and A subscript r space Mn plus 4 middle dot A subscript r space O right parenthesis subscript n equals 158 end style 
begin mathsize 14px style left parenthesis 39 plus 55 plus 4 middle dot 16 right parenthesis subscript n equals 158 end style 
begin mathsize 14px style left parenthesis 39 plus 55 plus 64 right parenthesis subscript n equals 158 end style 
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 158 right parenthesis subscript n end cell equals 158 row n equals 1 end table end style 


Rumus molekulnya yaitu undefined.undefined

Jadi, jawaban yang tepat adalah seperti pada penjelasan di atas.undefined

0

Roboguru

Tentukan rumus empiris dan rumus molekul dari senyawa yang terdiri dari:

Pembahasan Soal:

Rumus empiris dapat dihitung dengan cara sebagai berikut:undefined


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Fe space colon space O end cell equals cell fraction numerator massa space Fe over denominator A subscript r space Fe end fraction colon fraction numerator massa space O over denominator A subscript r space O end fraction end cell row cell Fe space colon thin space O end cell equals cell fraction numerator 0 comma 336 over denominator 56 end fraction colon fraction numerator 0 comma 144 over denominator 16 end fraction end cell row cell Fe space colon thin space O end cell equals cell 0 comma 006 space colon space 0 comma 009 end cell row cell Fe space colon space O end cell equals cell 2 space colon space 3 end cell end table end style 


Maka rumus empirisnya adalah begin mathsize 14px style Fe subscript 2 O subscript 3 end style. Sedangkan rumus molekulnya dapat dicari dengan cara berikut.undefined


begin mathsize 14px style M subscript r space open parentheses Fe subscript 2 O subscript 3 close parentheses subscript n equals 160 end style 
begin mathsize 14px style left parenthesis A subscript r space Fe plus 3 middle dot A subscript r space O right parenthesis subscript n equals 160 end style 
begin mathsize 14px style left parenthesis 2 middle dot 56 plus 3 middle dot 16 right parenthesis subscript n equals 160 end style  
begin mathsize 14px style left parenthesis 112 plus 48 right parenthesis subscript n equals 160 end style 
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 160 right parenthesis subscript n end cell equals 160 row n equals 1 end table end style 


Rumus molekulnya yaitu begin mathsize 14px style open parentheses Fe subscript 2 O subscript 3 close parentheses subscript 1 equals Fe subscript 2 end subscript O subscript 3 end subscript end style.undefined

Jadi, jawaban yang tepat adalah seperti penjelasan di atas.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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