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Berapa massa nitrogen pada: 12,6 gram  dan 36 gram ?

Pertanyaan

Berapa massa nitrogen pada:

  1. 12,6 gram H N O subscript 3 dan
  2. 36 gram C O open parentheses N H subscript 2 close parentheses subscript 2?

Pembahasan Soal:

Massa unsur dalam senyawa dapat ditentukan dengan rumus berikut:

massa space X equals fraction numerator indeks cross times A subscript r space X over denominator M subscript r space senyawa end fraction cross times massa space senyawa 


Massa N dalam 12,6 g H N O subscript 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell massa space N end cell equals cell fraction numerator indeks space N cross times A subscript r space N over denominator M subscript r space H N O subscript 3 end fraction cross times massa space H N O subscript 3 end cell row blank equals cell fraction numerator 1 cross times 14 over denominator 63 end fraction cross times 12 comma 6 end cell row blank equals cell 2 comma 8 space g end cell end table 


Massa N dalam 36 g C O open parentheses N H subscript 2 close parentheses subscript 2 

table attributes columnalign right center left columnspacing 0px end attributes row cell massa space N end cell equals cell fraction numerator indeks space N cross times A subscript r space N over denominator M subscript r space C O open parentheses N H subscript 2 close parentheses subscript 2 end fraction cross times massa space C O open parentheses N H subscript 2 close parentheses subscript 2 equals fraction numerator 2 cross times 14 over denominator 60 end fraction cross times 36 end cell row blank equals cell 16 comma 8 space g end cell end table 


Jadi, massa N dalam 12,6 gram H N O subscript bold 3 adalah 2,8 gram, sedangkan dalam 36 gram C O bold open parentheses N H subscript bold 2 bold close parentheses subscript bold 2 adalah 16,8 gram.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

K. Nurul

Mahasiswa/Alumni Universitas Lambung Mangkurat

Terakhir diupdate 02 Mei 2021

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Pertanyaan yang serupa

Tentukan persentase kadar masing-masing unsur dalam senyawa berikut. e.

Pembahasan Soal:

Diketahui italic A subscript r C = 12, O = 16, H = 1, sehingga nilai italic M subscript r space C subscript 12 H subscript 22 O subscript 11:

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space C subscript 12 H subscript 22 O subscript 11 end cell equals cell left parenthesis 12 cross times italic A subscript r space C right parenthesis plus left parenthesis 22 cross times italic A subscript r space H right parenthesis plus left parenthesis 11 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 12 cross times 12 right parenthesis plus left parenthesis 22 cross times 1 right parenthesis plus left parenthesis 11 cross times 16 right parenthesis end cell row blank equals cell 144 plus 22 plus 176 end cell row blank equals cell 342 space g space mol to the power of negative sign 1 end exponent end cell end table 

Menentukan persentase kadar masing-masing unsur:

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign C end cell equals cell fraction numerator 12 cross times italic A subscript r space C over denominator italic M subscript r space C subscript 12 H subscript 22 O subscript 11 end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 12 cross times 12 over denominator 342 end fraction cross times 100 percent sign end cell row blank equals cell 144 over 342 cross times 100 percent sign end cell row blank equals cell 42 comma 11 percent sign end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign H end cell equals cell fraction numerator 22 cross times italic A subscript r space H over denominator italic M subscript r space C subscript 12 H subscript 22 O subscript 11 end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 22 cross times 1 over denominator 342 end fraction cross times 100 percent sign end cell row blank equals cell 22 over 342 cross times 100 percent sign end cell row blank equals cell 6 comma 43 percent sign end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign O end cell equals cell fraction numerator 11 cross times italic A subscript r space O over denominator italic M subscript r space C subscript 12 H subscript 22 O subscript 11 end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 11 cross times 16 over denominator 342 end fraction cross times 100 percent sign end cell row blank equals cell 176 over 342 cross times 100 percent sign end cell row blank equals cell 51 comma 46 percent sign end cell end table 

Jadi, % C, % H dan % O dalam senyawa berturut-turut adalah 42,11%, 6,43% dan 51,46%.

0

Roboguru

Tentukan massa pupuk ZA yang harus digunakan apabila kita ingin mendapatkan 28 gram nitrogen dari pupuk Za,  dengan kemurnian 66%.

Pembahasan Soal:

Senyawa begin mathsize 14px style open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript end style memiliki massa molekul 126 maka untuk mendapatkan 28 gram nitrogen maka dapat digunakan perbandingan massa atom dengan molekul.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Massa subscript N end cell equals cell 14 over 126 cross times Massa space ZA end cell row 28 equals cell 14 over 126 cross times Massa space ZA end cell row cell Massa space ZA end cell equals cell 126 over 14 cross times 28 end cell row blank equals cell 252 space gram end cell end table end style


Pupuk ZA hanya memiliki kemurnian 66% maka massa pupuk ZA yang sebenernya adalah


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 66 over 100 cross times Massa space ZA space kotor end cell equals 252 row cell Massa space ZA space kotor end cell equals cell 252 cross times 100 over 66 end cell row blank equals cell 381.8 space gram end cell end table end style 


Jadi, massa pupuk ZA yang diperlukan adalah 381,8 gram.space 

0

Roboguru

Tentukan persentase kadar masing-masing unsur dalam senyawa berikut. d.

Pembahasan Soal:

Diketahui italic A subscript r Fe = 56, O = 16, N = 14, sehingga nilai italic M subscript r space Fe open parentheses N O subscript 3 close parentheses subscript 3:

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space Fe open parentheses N O subscript 3 close parentheses subscript 3 end cell equals cell left parenthesis 1 cross times italic A subscript r space Fe right parenthesis plus left parenthesis 3 cross times italic A subscript r space N right parenthesis plus left parenthesis 9 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 1 cross times 56 right parenthesis plus left parenthesis 3 cross times 14 right parenthesis plus left parenthesis 9 cross times 16 right parenthesis end cell row blank equals cell 56 plus 42 plus 144 end cell row blank equals cell 242 space g space mol to the power of negative sign 1 end exponent end cell end table 

Menentukan persentase kadar masing-masing unsur:

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign Fe end cell equals cell fraction numerator 1 cross times italic A subscript r space Fe over denominator italic M subscript r space Fe open parentheses N O subscript 3 close parentheses subscript 3 end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 1 cross times 56 over denominator 242 end fraction cross times 100 percent sign end cell row blank equals cell 56 over 242 cross times 100 percent sign end cell row blank equals cell 23 comma 14 percent sign end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign N end cell equals cell fraction numerator 3 cross times italic A subscript r space N over denominator italic M subscript r space Fe open parentheses N O subscript 3 close parentheses subscript 3 end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 3 cross times 14 over denominator 242 end fraction cross times 100 percent sign end cell row blank equals cell 42 over 242 cross times 100 percent sign end cell row blank equals cell 17 comma 36 percent sign end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign O end cell equals cell fraction numerator 9 cross times italic A subscript r space O over denominator italic M subscript r space Fe open parentheses N O subscript 3 close parentheses subscript 3 end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 9 cross times 16 over denominator 242 end fraction cross times 100 percent sign end cell row blank equals cell 144 over 242 cross times 100 percent sign end cell row blank equals cell 59 comma 5 percent sign end cell end table 

Jadi, % Fe, % N dan % O dalam senyawa berturut-turut adalah 23,14%, 17,36% dan 59,5%.

0

Roboguru

Tentukan molalitas dari 500 mL larutan yang mengandung 5% massa , jika diketahui massa jenis larutan 1,2 gram/mL!

Pembahasan Soal:

Molalitas (m) adalah mol zat terlarut tiap kg pelarutnya. Massa larutan dapat kita cari dari massa jenis dan volume yang telah diketahui.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row rho equals cell massa over V end cell row cell 1 comma 2 space g forward slash mL end cell equals cell fraction numerator massa over denominator 500 space mL end fraction end cell row massa equals cell 500 space gram end cell end table end style 

Massa larutan (keseluruhan zat) adalah 500 gram, jika terdapat 5% begin mathsize 14px style H subscript 2 S O subscript 4 end style, artinya:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell massa space H subscript 2 S O subscript 4 end cell equals cell 5 over 100 cross times 500 space gram end cell row blank equals cell 25 space gram end cell end table end style 

maka, massa pelarut adalah 500 gram - 25 gram = 475 gram.

Kemudian dicari mol begin mathsize 14px style H subscript 2 S O subscript 4 end style, massa atom relatif dapat dilihat pada tabel periodik unsur.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Mr space H subscript 2 S O subscript 4 end cell equals cell 2 A subscript r space end subscript H and A subscript r space S and 4 A subscript r space O end cell row blank equals cell 2 left parenthesis 1 right parenthesis plus 32 plus 4 left parenthesis 16 right parenthesis end cell row blank equals 98 end table end style 

Sehingga molalitasnya adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row m equals cell massa over M subscript r cross times 1000 over p end cell row blank equals cell 25 over 98 cross times 1000 over 475 end cell row blank equals cell 0 comma 54 space m end cell end table end style 

Jadi, molalitas larutan tersebut adalah 0,54 m.

0

Roboguru

Campuran, 23 gram alkohol  dengan 180 gram air, menghasilkan 200 ml larutan campuran . Tentukanlah:     %b/b bpj/ppm m M

Pembahasan Soal:

1. begin mathsize 14px style X subscript t end style = fraksi mol zat terlarut

Fraksi mol merupakan mol bagian, langkah pertama adalah mencari mol zat terlarut dan mol pelarut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript t end cell equals cell massa over M subscript r end cell row blank equals cell fraction numerator 23 space g over denominator 46 space g forward slash mol end fraction end cell row blank equals cell 0 comma 5 space mol end cell row cell n subscript p end cell equals cell massa over M subscript r end cell row blank equals cell 180 over 18 end cell row blank equals cell 10 space mol end cell end table end style 

Maka fraksi mol zat terlarut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell X subscript t end cell equals cell fraction numerator n subscript t over denominator n subscript t and n subscript p end fraction end cell row blank equals cell fraction numerator 0 comma 5 over denominator 0 comma 5 plus 10 end fraction end cell row blank equals cell fraction numerator 0 comma 5 over denominator 10 comma 5 end fraction end cell row blank equals cell 0 comma 0476 end cell end table end style 

2. begin mathsize 14px style X subscript p end style = fraksi mol pelarut

fraksi mol pelarut dapat dicari menggunakan rumus:

begin mathsize 14px style X subscript p equals fraction numerator n subscript p over denominator n subscript t and n subscript p end fraction end style 

atau karena di awal sudah diperoleh nilai begin mathsize 14px style X subscript t end style, dapat menggunakan persamaan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell X subscript t and X subscript p end cell equals 1 row cell X subscript p end cell equals cell 1 minus sign 0 comma 0476 end cell row blank equals cell 0 comma 9524 end cell end table end style 

3. %b/b = persen massa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign b forward slash b end cell equals cell fraction numerator massa space zat space terlarut over denominator massa space larutan end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 23 space gram over denominator left parenthesis 23 plus 180 right parenthesis gram end fraction cross times 100 percent sign end cell row blank equals cell 11 comma 33 percent sign end cell end table end style 

4. bpj/ppm = bagian per juta

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row bpj equals cell fraction numerator massa space zat space terlarut over denominator massa space larutan end fraction cross times 10 to the power of 6 end cell row blank equals cell fraction numerator 23 space gram over denominator left parenthesis 23 plus 180 right parenthesis space gram end fraction cross times 10 to the power of 6 end cell row blank equals cell 1 comma 133 cross times 10 to the power of 5 space bpj end cell end table end style 

5. m = molalitas: mol tiap kg pelarut

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row m equals cell mol over P end cell row blank equals cell fraction numerator 0 comma 5 space mol over denominator 0 comma 18 space kg end fraction end cell row blank equals cell 2 comma 78 space m end cell end table end style 

6. M = molaritas: mol tiap liter larutan

(anggap massa jenis larutan = 1 g/mL)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row M equals cell mol over V end cell row blank equals cell fraction numerator 0 comma 5 space mol over denominator 0 comma 18 space L end fraction end cell row blank equals cell 2 comma 78 space M end cell end table end style 

Jadi, fraksi mol terlarutnya = 0,0476; fraksi mol pelarut = 0,9524: persen massa = 11,33%; bagian per juta = begin mathsize 14px style bottom enclose bold 1 bold comma bold 133 bold cross times bold 10 to the power of bold 5 end enclose end style; molalitas = 2,78 m; molaritas = 2,78 M.

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