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Senyawa berikut yang kadar nitrogennya terbesar adalah ... ( H = 1, C = 12, N = 14, O = 16, P = 31, dan S= 32)

Pertanyaan

Senyawa berikut yang kadar nitrogennya terbesar adalah ... (italic A subscript r H = 1, C = 12, N = 14, O = 16, P = 31, dan S= 32)

  1. N H subscript 3 

  2. C O open parentheses N H subscript 2 close parentheses subscript 2 14

  3. open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript 

  4. open parentheses N H subscript 4 close parentheses subscript 3 P O subscript 4 end subscript 

  5. N subscript 2 H subscript 4 

Pembahasan Soal:

A. N H subscript 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space N H subscript 3 end cell equals cell left parenthesis 1 cross times italic A subscript r space N right parenthesis plus left parenthesis 3 cross times italic A subscript r space H right parenthesis end cell row blank equals cell left parenthesis 1 cross times 14 right parenthesis plus left parenthesis 3 cross times 1 right parenthesis end cell row blank equals cell 14 plus 3 end cell row blank equals cell 17 space g space mol to the power of negative sign 1 end exponent 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 italic A subscript r space N over denominator italic M subscript r space N H subscript 3 end fraction cross times 100 percent sign end cell row blank equals cell 14 over 17 cross times 100 percent sign end cell row blank equals cell 82 comma 35 percent sign end cell end table 

B. C O open parentheses N H subscript 2 close parentheses subscript 2 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space C O open parentheses N H subscript 2 close parentheses subscript 2 end cell equals cell left parenthesis 1 cross times italic A subscript r space C right parenthesis plus left parenthesis 1 cross times italic A subscript r space O right parenthesis plus left parenthesis 2 cross times italic A subscript r space N right parenthesis plus left parenthesis 4 cross times italic A subscript r space H right parenthesis end cell row blank equals cell left parenthesis 1 cross times 12 right parenthesis plus left parenthesis 1 cross times 16 right parenthesis plus left parenthesis 2 cross times 14 right parenthesis plus left parenthesis 4 cross times 1 right parenthesis end cell row blank equals cell 12 plus 16 plus 28 plus 4 end cell row blank equals cell 60 space g space mol to the power of negative sign 1 end exponent 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 2 cross times italic A subscript r space N over denominator italic M subscript r space C O open parentheses N H subscript 2 close parentheses subscript 2 end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 2 cross times 14 over denominator 60 end fraction cross times 100 percent sign end cell row blank equals cell 46 comma 6 percent sign end cell end table 

C. open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript end cell equals cell left parenthesis 2 cross times italic A subscript r space N right parenthesis plus left parenthesis 8 cross times italic A subscript r space H right parenthesis plus left parenthesis 1 cross times italic A subscript r space S right parenthesis plus left parenthesis 4 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 2 cross times 14 right parenthesis plus left parenthesis 8 cross times 1 right parenthesis plus left parenthesis 1 cross times 32 right parenthesis plus left parenthesis 4 cross times 16 right parenthesis end cell row blank equals cell 14 plus 8 plus 32 plus 64 end cell row blank equals cell 118 space g space mol to the power of negative sign 1 end exponent 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 2 cross times italic A subscript r space N over denominator italic M subscript r space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 2 cross times 14 over denominator 118 end fraction cross times 100 percent sign end cell row blank equals cell 23 comma 73 percent sign end cell end table 

D. open parentheses N H subscript 4 close parentheses subscript 3 P O subscript 4 end subscript 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space open parentheses N H subscript 4 close parentheses subscript 3 P O subscript 4 end cell equals cell left parenthesis 3 cross times italic A subscript r space N right parenthesis plus left parenthesis 12 cross times italic A subscript r space H right parenthesis plus left parenthesis 1 cross times italic A subscript r space P right parenthesis plus left parenthesis 4 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 3 cross times 14 right parenthesis plus left parenthesis 12 cross times 1 right parenthesis plus left parenthesis 1 cross times 31 right parenthesis plus left parenthesis 4 cross times 16 right parenthesis end cell row blank equals cell 42 plus 12 plus 31 plus 64 end cell row blank equals cell 149 space g space mol to the power of negative sign 1 end exponent 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 open parentheses N H subscript 4 close parentheses subscript 3 P O subscript 4 end subscript end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 3 cross times 14 over denominator 149 end fraction cross times 100 percent sign end cell row blank equals cell 28 comma 19 percent sign end cell end table 

E. N subscript 2 H subscript 4 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space N subscript 2 H subscript 4 end cell equals cell left parenthesis 2 cross times italic A subscript r space N right parenthesis plus left parenthesis 4 cross times italic A subscript r space H right parenthesis end cell row blank equals cell left parenthesis 2 cross times 14 right parenthesis plus left parenthesis 4 cross times 1 right parenthesis end cell row blank equals cell 28 plus 4 end cell row blank equals cell 32 space g space mol to the power of negative sign 1 end exponent end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell bold percent sign N end cell equals cell fraction numerator 2 cross times italic A subscript r space N over denominator italic M subscript r space N subscript 2 H subscript 4 end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 2 cross times 14 over denominator 32 end fraction cross times 100 percent sign end cell row blank equals cell bold 87 bold comma bold 5 bold percent sign end cell end table 

Senyawa yang kadar nitrogennya terbesar adalah N subscript 2 H subscript 4.

Jadi, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Chandra

Mahasiswa/Alumni Universitas Negeri Jakarta

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

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Pembahasan Soal:

Rumus kimia menunjukkan jumlah atom-atom penyusun suatu zat. Oleh karena massa atom suatu unsur sudah tertentu, maka dari rumus kimia tersebut dapat juga ditentukan persentase atau komposisi masing-masing unsur dalam suatu zat. Dalam senyawa Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3, tersusun dari 3 jenis unsur, yaitu besi (Fe), belerang (S) dan oksigen (O). Massa masing-masing unsur dalam senyawa Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 adalah sebagai berikut.space


table attributes columnalign right center left columnspacing 0px end attributes row cell massa space Fe end cell equals cell fraction numerator jumlah space atom space Fe cross times A subscript r space Fe over denominator M subscript r space Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end fraction cross times massa space Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end cell row cell massa space Fe end cell equals cell fraction numerator 2 cross times 56 over denominator 400 end fraction cross times 20 end cell row cell massa space Fe end cell equals cell 5 comma 6 space gram end cell row blank blank blank row cell massa space S end cell equals cell fraction numerator jumlah space atom space S cross times A subscript r space S over denominator M subscript r space Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end fraction cross times massa space Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end cell row cell massa space S end cell equals cell fraction numerator 3 cross times 32 over denominator 400 end fraction cross times 20 end cell row cell massa space S end cell equals cell 4 comma 8 space gram end cell row blank blank blank row cell massa space O end cell equals cell massa space Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 minus sign left parenthesis massa space Fe and massa space S right parenthesis end cell row cell massa space O end cell equals cell 20 space gram minus sign left parenthesis 5 comma 6 plus 4 comma 8 right parenthesis space gram end cell row cell massa space O end cell equals cell 20 space gram minus sign 10 comma 4 space gram end cell row cell massa space O end cell equals cell 9 comma 6 space gram end cell end table  


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

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Suatu senyawa mengandung 23,3% unsur ; 25,3% ; dan 51,4%  . Tentukan rumus empiris senyawa tersebut!

Pembahasan Soal:

Rumus empiris senyawa dapat dilihat dari perbandingan mol masing-masing unsur. Untuk menentukannya diperlukan data massa atom relatif masing-masing atom (dapat dilihat di tabel periodik unsur).

begin mathsize 14px style A subscript r space Co equals 59 comma space Mo equals 96 comma space Cl equals 35 comma 5 end style  

Perbandingan massa masing-masing atom kita ambil dari persen massanya,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Co space colon space Mo space colon space Cl end cell equals cell fraction numerator percent sign space Co over denominator A subscript r space Co end fraction space colon space fraction numerator percent sign space Mo over denominator A subscript r space Mo end fraction space colon space fraction numerator percent sign Cl over denominator A subscript r space Cl end fraction end cell row blank equals cell fraction numerator 23 comma 3 over denominator 59 end fraction space colon space fraction numerator 25 comma 3 over denominator 96 end fraction space colon space fraction numerator 51 comma 4 over denominator 35 comma 5 end fraction end cell row blank equals cell 0 comma 395 space colon space 0 comma 26 space colon space 1 comma 45 end cell row blank equals cell 1 comma 5 space colon space 1 space colon space 5 comma 6 end cell row blank equals cell 3 space colon space 2 space colon space 11 end cell end table end style 

Jadi, rumus empiris senyawa tersebut adalah begin mathsize 14px style bottom enclose Co subscript bold 3 Mo subscript bold 2 Cl subscript bold 11 end enclose end style.

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Pembahasan Soal:

m subscript Zn double bond x n subscript Zn equals x over 65  m subscript Al double bond y n subscript Al equals y over 27  n subscript O subscript 2 end subscript equals fraction numerator 1 comma 28 over denominator 32 end fraction equals 0 comma 04 space mol top enclose V space equals fraction numerator 1 over denominator 0 comma 04 end fraction equals 25 space L forward slash mol  n subscript H subscript 2 end subscript equals 20 over 25 equals 0 comma 8 space mol 

open parentheses x over 65 close parentheses minus sign 0 comma 2 equals 0 space space space space space space space space space x over 65 equals 0 comma 2 space space space space space space space space space space space space x equals 13 space g  left parenthesis y over 27 right parenthesis minus sign 0 comma 4 equals 0 space space space space space space space space y over 27 equals 0 comma 4 space space space space space space space space space space y equals 10 comma 8 space g  

Dengan demikian, maka jawaban yang tepat adalah massa Zn adalah 13 g dan Al adalah 10,8 g.

Roboguru

Hitunglah persentase air dalam senyawa hidrat berikut: kalsium fluorofosfat dihidrat, !

Pembahasan Soal:

Persentase air dalam senyawa Kalsium flourofosfat dihidrat dapat ditentukan sebagai berikut:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign H subscript 2 O end cell equals cell fraction numerator i middle dot Mr space H subscript 2 O over denominator Mr space Ca F P O subscript 3 middle dot 2 H subscript 2 O end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 2 middle dot 18 over denominator 174 end fraction cross times 100 percent sign end cell row blank equals cell 36 over 174 cross times 100 percent sign end cell row blank equals cell 20 comma 6 percent sign end cell end table
 

Dengan demikian maka persentase air dalam senyawa tersebut adalah 20,6%.space

Roboguru

Berapa kadar dalam 100 gram ? (Ar: = 12, = 16, = 40)

Pembahasan Soal:

Rumus kimia suatu senyawa menyatakan jenis dan jumlah atom dalam senyawa tersebut, sehingga perbandingan massa dan kadar unsur dalam senyawa dapat ditentukan dari rumus senyawanya. Kadar zat atau kadar unsur umumnya dinyatakan dalam persen massa (% massa). Misalkan kadar unsur A dalam senyawa begin mathsize 14px style A subscript x B subscript y end style adalah:

begin mathsize 14px style massa space A equals fraction numerator x middle dot Ar space A over denominator Mr space A subscript x B subscript y end fraction cross times massa space A subscript x B subscript y atau space percent sign space A equals fraction numerator x middle dot Ar space A over denominator Mr space A subscript x B subscript y end fraction cross times 100 percent sign end style 

Untuk menghitung kadar begin mathsize 14px style Ca end style dalam 100 gram begin mathsize 14px style Ca C O subscript 3 end style, dengan cara :

1) Tentukan Mr begin mathsize 14px style Ca C O subscript 3 end style

begin mathsize 14px style Mr space Ca C O subscript 3 equals 1 cross times Ar space Ca plus 1 cross times Ar space C plus 3 cross times Ar space O space space space space space space space space space space space space space space space space equals 1 cross times 40 plus 1 cross times 12 plus 3 cross times 16 space space space space space space space space space space space space space space space space equals 100 end style

2) Menentukan massa begin mathsize 14px style Ca end style dalam 100 gram begin mathsize 14px style Ca C O subscript 3 end style

begin mathsize 14px style massa space Ca space dalam space Ca C O subscript 3 equals fraction numerator 1 cross times Ar space Ca over denominator Mr space Ca C O subscript 3 end fraction cross times massa space Ca C O subscript 3 massa space Ca space dalam space Ca C O subscript 3 equals fraction numerator 1 cross times 40 over denominator 100 end fraction cross times 100 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 40 space gram end style

3) Menentukan kadar begin mathsize 14px style Ca end style dalam 100 gram begin mathsize 14px style Ca C O subscript 3 end style

begin mathsize 14px style percent sign Ca space dalam space Ca C O subscript 3 equals fraction numerator massa space Ca over denominator massa space Ca C O subscript 3 end fraction cross times 100 percent sign percent sign Ca space dalam space Ca C O subscript 3 equals 40 over 100 cross times 100 percent sign equals 40 percent sign end style 

Maka, kadar begin mathsize 14px style Ca end style dalam 100 gram begin mathsize 14px style Ca C O subscript 3 end style adalah 40%.

Roboguru

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