{"id":275416,"date":"2023-06-05T10:34:09","date_gmt":"2023-06-05T03:34:09","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=275416"},"modified":"2023-06-05T10:34:20","modified_gmt":"2023-06-05T03:34:20","slug":"stoikiometri-kimia-kelas-10","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/mapel\/kimia\/stoikiometri-kimia-kelas-10\/","title":{"rendered":"Materi Stoikiometri Rumus, Persamaan dan Contoh Soal"},"content":{"rendered":"

\"Stoikiometri<\/p>\n

Hai Quipperian, bagaimana kabarnya? Semoga selalu sehat dan tetap semangat, ya!<\/span><\/p>\n

Siapa di antara Quipperian yang pernah memupuk tanaman? Saat memupuk tanaman, konsentrasi pengenceran pupuk harus didasarkan pada perbandingan tertentu. Konsentrasi pupuk tidak boleh terlalu pekat, maupun terlalu encer. Mengapa demikian? Bukankah semakin pekat pupuk, tanaman akan mendapatkan unsur hara yang semakin banyak? <\/span><\/p>\n

Jika konsentrasi pupuk semakin banyak, proses metabolisme di dalam tubuh tanaman akan terganggu karena komposisi unsur kimia di dalam pupuk sudah ditentukan melalui stoikiometri. Apa itu stoikiometri? Temukan jawabannya di pembahasan Quipper Blog kali ini.<\/span><\/p>\n

Sebelum lanjut ke bagian stoikiometri, Quipperian harus tahu apa itu rumus molekul dan rumus empiris.<\/span><\/p>\n

Pengertian Rumus Molekul dan Rumus Empiris<\/strong><\/h2>\n

\"\"<\/p>\n

Seperti pada pembahasan di awal, pupuk merupakan senyawa penting yang berperan dalam menyuburkan tanaman. Pupuk yang biasa digunakan adalah urea. Pupuk urea memiliki rumus molekul CH<\/span>4<\/span><\/sub>N<\/span>2<\/span><\/sub>O. Rumus molekul adalah rumus yang menunjukkan jumlah atom setiap unsur dalam suatu senyawa. Ternyata, rumus molekul inilah yang biasa digunakan untuk lambang suatu senyawa. Membahas rumus molekul tentu tidak bisa lepas dari rumus empiris. Apa itu rumus empiris? Rumus empiris adalah rumus yang memiliki perbandingan atom-atom paling sederhana dalam suatu senyawa.<\/span><\/p>\n

Menentukan Rumus Empiris<\/strong><\/h2>\n

\"\"<\/p>\n

Untuk menentukan rumus empiris, ada beberapa langkah yang harus Quipperian lakukan, yaitu sebagai berikut.<\/span><\/p>\n

    \n
  1. Rumus empiris bisa ditentukan melalui rumus molekul, yaitu dengan menyederhanakan jumlah atom penyusun rumus molekul agar diperoleh bilangan bulat paling kecil.<\/span><\/li>\n
  2. Rumus empiris juga bisa ditentukan melalui perbandingan mol unsur-unsur penyusun senyawa, dengan syarat massanya diketahui.<\/span><\/li>\n<\/ol>\n

    Untuk meningkatkan pemahaman Quipperian, simak contoh soal berikut.<\/span><\/p>\n

    Contoh soal 1<\/strong><\/h3>\n

    Tentukan rumus empiris senyawa C<\/span>6<\/span><\/sub>H<\/span>12<\/span><\/sub>O<\/span>6<\/span><\/sub>!<\/span><\/p>\n

    Pembahasan:<\/strong><\/p>\n

    Rumus empiris diperoleh dengan menyederhanakan rumus molekul, seperti berikut ini.<\/span><\/p>\n

    C<\/span>6<\/span><\/sub>H<\/span>12<\/span><\/sub>O<\/span>6<\/span><\/sub> = (CH<\/span>2<\/span><\/sub>O)<\/span>6<\/span><\/sub><\/p>\n

    Dengan demikian, rumus empiris dari C6<\/sub>H12<\/sub>O6<\/sub><\/span> adalah CH<\/span>2<\/span><\/sub>O.<\/span><\/p>\n

    Menentukan Rumus Molekul<\/strong><\/h2>\n

    \"\"<\/p>\n

    Rumus molekul merupakan rumus yang jumlah atomnya kelipatan dari rumus empiris. Namun demikian, ada beberapa senyawa yang memiliki rumus molekul dan empiris sama. Untuk menentukan rumus molekul, gunakan persamaan berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Keterangan:<\/span><\/p>\n

    n<\/span><\/em> merupakan kelipatan rumus empiris terhadap rumus molekul.<\/span><\/p>\n

    Untuk meningkatkan pemahaman Quipperian tentang rumus molekul, perhatikan contoh soal berikut.<\/span><\/p>\n

    Contoh soal 2<\/strong><\/h3>\n

    Suatu senyawa memiliki rumus empiris CH<\/span>2<\/span><\/sub>O. Jika <\/span>Mr<\/span><\/em> senyawa tersebut 150, tentukan rumus molekul senyawa tersebut! (<\/span>Ar <\/span><\/em>H = 1, C = 12, dan O = 16)<\/span><\/p>\n

    Pembahasan:<\/strong><\/p>\n

    Sebelum mencari rumus molekul senyawa tersebut, tentukan dahulu massa molekul relatifnya (<\/span>Mr<\/span><\/em>)!<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Selanjutnya, Quipperian harus menentukan nilai <\/span>n<\/span><\/em> menggunakan persamaan berikut.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Jika nilai <\/span>n<\/span><\/em> sudah diketahui, Quipperian bisa mencari rumus molekulnya seperti cara berikut.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Jadi, rumus molekul senyawa tersebut adalah C<\/span>5<\/span><\/sub>H<\/span>10<\/span><\/sub>O<\/span>5<\/span><\/sub>.<\/span><\/p>\n

    Melengkapi Persamaan Reaksi Kimia<\/strong><\/h2>\n

    \"\"<\/p>\n

    Reaksi kimia merupakan reaksi yang berlangsung secara spontan. Jika suatu senyawa yang mengandung ion negatif direaksikan dengan senyawa lain yang mengandung ion positif, maka ion-ion tersebut bisa saling berikatan. Ikatan antarion tersebut bisa Quipperian lihat pada contoh berikut.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Pengertian Stoikiometri<\/strong><\/h2>\n

    \"\"<\/p>\n

    Stoikiometri adalah ilmu kimia yang mempelajari tentang kuantitas suatu zat, meliputi massa, jumlah mol, volume, dan jumlah partikel. Saat mempelajari stoikiometri, Quipperian akan selalu bertemu dengan koefisien. Apa itu koefisien? Koefisien merupakan bentuk perbandingan mol, volume, atau jumlah partikel. Adapun rumus-rumus yang biasa digunakan dalam menyelesaikan stoikiometri adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Pengertian Pereaksi Pembatas<\/strong><\/h2>\n

    \"Buku<\/p>\n

    Membahas reaksi kimia, sama artinya membahas zat pereaksi dan produk hasil reaksi. Zat pereaksi biasanya ditulis di ruas kiri, sedangkan produk ditulis di ruas kanan. Zat pereaksi tersebut akan berubah menjadi produk jika memenuhi perbandingan koefisien tertentu. perbandingan koefisien biasanya mengacu pada jumlah mol yang terlibat selama reaksi berlangsung. Penentuan mol produk didasarkan pada perbandingan koefisien mol zat pereaksi. Nah, jika mol pereaksinya berasal dari dua zat yang bereaksi, maka perlu dicari acuan yang tepat. Acuan itu disebut sebagai pereaksi pembatas. Mol zat yang digunakan sebagai pereaksi pembatas ini akan habis bereaksi.<\/span><\/p>\n

    Penentuan Pereaksi Pembatas<\/strong><\/h3>\n

    Untuk menentukan pereaksi pembatas, Quipperian bisa menggunakan persamaan berikut.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Senyawa Hidrat (Air Kristal)<\/strong><\/h2>\n

    \"\"<\/p>\n

    Kristal adalah zat padat homogen yang dibatasi oleh bidang-bidang datar dan memiliki bentuk tertentu. ada beberapa senyawa yang bisa ditemukan dalam bentuk kristal, contohnya garam Inggris (MgSO<\/span>4<\/span><\/sub>.7H<\/span>2<\/span><\/sub>O). Kristal dalam senyawa tersebut merupakan hasil perubahan wujud air menjadi padatan.<\/span><\/p>\n

    Tampaknya, belajar Kimia tidak akan sah jika belum latihan soal ya Quipperian. Oleh karena itu, Quipper Blog telah merangkum beberapa contoh soal untuk Quipperian. <\/span>Check this out<\/span><\/em>!<\/span><\/p>\n

    Contoh soal 3<\/strong><\/h3>\n

    Perhatikan persamaan reaksi kimia berikut ini.<\/span><\/p>\n

    \"\" <\/span><\/p>\n

    Jika gas SO<\/span>2<\/span><\/sub> yang direaksikan sebanyak 12,8 gram, tentukan volume gas O<\/span>2 <\/span><\/sub>diukur pada keadaan standar (STP) dan massa SO<\/span>3<\/span><\/sub> yang terbentuk!<\/span><\/p>\n

    Pembahasan:<\/strong><\/p>\n

    1. Untuk menentukan volume gas O<\/span>2<\/span><\/sub>, Quipperian harus mencari massa molekul relatif gas O<\/span>2<\/span><\/sub> dan SO<\/span>2<\/span><\/sub> tersebut.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Kemudian, tentukan mol gas SO<\/span>2<\/span><\/sub> dan O<\/span>2<\/span><\/sub>.<\/span><\/p>\n

    \"\" <\/span><\/p>\n

    Volume O<\/span>2<\/span><\/sub> pada keadaan STP dirumuskan sebagai berikut.<\/span><\/p>\n

    \"\" <\/span><\/p>\n

    2. Untuk mencari massa SO<\/span>3<\/span><\/sub> yang terbentuk, Quipperian juga harus mencari massa molekul relatif dan mol gas SO<\/span>3<\/span><\/sub>.<\/span><\/p>\n

    \"\"<\/p>\n

    Massa gas SO<\/span>3<\/span><\/sub> yang terbentuk dirumuskan sebagai berikut.<\/span><\/p>\n

    \"\"
    \n<\/span><\/p>\n

    Jadi, volume gas O<\/span>2 <\/span><\/sub>diukur pada keadaan standar (STP) dan massa SO<\/span>3<\/span><\/sub> yang terbentuk berturut-turut adalah 2,24 L dan 16 gram.<\/span><\/p>\n

    Bagaimana Quipperian, mudah bukan? Ayo lanjut ke contoh soal berikutnya.<\/span><\/p>\n

    Contoh soal 4<\/strong><\/h3>\n

    Sebanyak 22 gram C<\/span>3<\/span><\/sub>H<\/span>8<\/span><\/sub> direaksikan dengan 48 gram O<\/span>2<\/span><\/sub> berdasarkan persamaan reaksi berikut.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Tentukan zat yang merupakan pereaksi pembatas dan massa zat yang tersisa!<\/span><\/p>\n

    Pembahasan:<\/strong><\/p>\n

    1. Untuk menentukan zat pereaksi pembatasnya, Quipperian harus mencari massa molekul relatif dan mol C<\/span>3<\/span><\/sub>H<\/span>8<\/span><\/sub> dan O<\/span>2<\/span><\/sub>.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Setelah massa molekul relatif dan mol dari C<\/span>3<\/span><\/sub>H<\/span>8<\/span><\/sub> dan O<\/span>2<\/span><\/sub> sudah diketahui, Quipperian harus menyetarakan persamaan reaksinya seperti berikut ini.<\/span><\/p>\n

    \"\"<\/p>\n

    Pereaksi pembatasnya bisa ditentukan dengan persamaan berikut.<\/span><\/p>\n

    \"\"<\/span><\/p>\n

    Zat yang berperan sebagai pereaksi pembatas adalah zat yang hasil baginya palingg kecil. Oleh karena hasil bagi O<\/span>2<\/span><\/sub> lebih kecil daripada C<\/span>3<\/span><\/sub>H<\/span>8<\/span><\/sub>, maka zat yang berperan sebagai pereaksi pembatas adalah O<\/span>2<\/span><\/sub>.<\/span><\/p>\n

    2. Pada reaksi di atas, zat yang tersisa adalah C3<\/sub>H8<\/sub><\/span>, sehingga Quipperian harus mencari massa C3<\/sub>H8<\/sub><\/span> yang tersisa. Sebelum mencari massa C3<\/sub>H8<\/sub><\/span> yang tersisa, Quipperian harus tahu berapa jumlah mol C3<\/sub>H8<\/sub><\/span> yang tersisa.<\/span><\/p>\n

    \"\" <\/span><\/p>\n

    Jadi, zat yang merupakan pereaksi pembatas dan massa zat yang tersisa berturut-turut adalah O<\/span>2 <\/span><\/sub>dan 8,8 gram C3<\/sub>H8<\/sub><\/span>.<\/span><\/p>\n

    Apakah Quipperian sudah semakin paham dengan materi stoikiometri? Cukup mudah bukan untuk menyelesaikan soal-soal terkait stoikiometri? Belajar stoikiometri sama artinya belajar Kimia secara matematis. Namun demikian, stoikiometri bisa dengan mudah dikuasai jika Quipperian rajin mengerjakan soal. Nah, soal-soal tersebut bisa Quipperian lihat di Quipper Video. Tidak hanya latihan soal beserta pembahasannya, namun juga pembahasan tutor Quipper Video tentang materi Kimia yang lain. <\/span>So<\/span><\/em>, tunggu apa lagi. Ambil <\/span>handphone<\/span><\/em>mu dan gabung bersama Quipper Video.<\/span><\/p>\n

    Penulis: Eka Viandari<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

    Hai Quipperian, bagaimana kabarnya? Semoga selalu sehat dan tetap semangat, ya! Siapa di antara Quipperian yang pernah memupuk tanaman? Saat memupuk tanaman, konsentrasi pengenceran…<\/p>\n","protected":false},"author":99400369,"featured_media":275444,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_crdt_document":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[679384879,679384888],"tags":[],"ppma_author":[679386827,679386839],"class_list":["post-275416","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-kimia","category-mapel"],"acf":[],"yoast_head":"\nMateri Stoikiometri Rumus, Persamaan dan Contoh Soal - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Stoikiometri adalah ilmu kimia yang mempelajari tentang kuantitas suatu zat. Penasaran dengan materi lengkapnya? 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