Stoikiometri (7) | Menentukan Rumus Kimia Senyawa Hidrat | Kimia kelas 10

Kimatika
27 Apr 202108:37

Summary

TLDRThis chemistry video for 10th graders delves into the topic of stoichiometry, focusing on determining the chemical formula of hydrates. Hydrates are compounds that bind water molecules as part of their crystal structure. The tutorial explains the process of finding the formula of a hydrate, using examples like magnesium sulfate heptahydrate and copper sulfate pentahydrate. It guides through the steps of heating the hydrate to release water molecules, calculating the mass of water released, determining moles of the anhydrous compound and water, and finally deriving the chemical formula of the hydrate based on mole ratios.

Takeaways

  • 🔍 The video discusses the topic of stoichiometry in chemistry, focusing on how to determine the chemical formula of hydrates.
  • 💧 Hydrates are compounds that bind a certain number of water molecules (H2O) as part of their crystal structure.
  • 🔥 When heated or dissolved, hydrates release their water molecules, leaving behind the anhydrous crystal or salt.
  • 📘 Examples of hydrates include calcium sulfate dihydrate (CaSO4·2H2O) and magnesium sulfate heptahydrate (MgSO4·7H2O).
  • 🧪 The process of determining the chemical formula of a hydrate involves calculating the number of water molecules in the crystal.
  • 📊 The video provides a step-by-step method to determine the formula, using the example of magnesium sulfate hydrate (MgSO4·xH2O).
  • ⚖️ The method involves heating the hydrate to evaporate the water, then calculating the mass of the remaining anhydrous compound.
  • 📐 The molar mass of the anhydrous compound and water are used to determine the moles of each component in the hydrate.
  • 🔢 The molar ratio of the anhydrous compound to water is found, which corresponds to the stoichiometric coefficients in the chemical formula.
  • 📝 The video demonstrates the calculation using the example of copper(II) sulfate hydrate, determining its formula as CuSO4·5H2O based on the mass loss upon heating.
  • 👨‍🏫 The video concludes with a reminder that understanding these concepts is important for grasping the principles of chemistry.

Q & A

  • What is the topic of the video script?

    -The topic of the video script is about stoichiometry, specifically learning how to determine the chemical formula of hydrated compounds in chemistry for 10th-grade students.

  • What are hydrated compounds?

    -Hydrated compounds are compounds that bind a certain number of water molecules (H2O) as part of their crystal structure.

  • What happens to hydrated compounds when they are heated or dissolved?

    -When hydrated compounds are heated or dissolved, the water molecules are released, leaving behind the anhydrous crystal or salt.

  • Can you give an example of a hydrated compound mentioned in the script?

    -Examples of hydrated compounds mentioned in the script are calcium sulfate dihydrate (CaSO4·2H2O) and magnesium sulfate heptahydrate (MgSO4·7H2O).

  • What is the basic principle behind determining the chemical formula of a hydrated compound?

    -The basic principle is to determine the number of water molecules in the crystal or H2O within the hydrated compound.

  • How is the chemical formula of sodium sulfate decahydrate determined in the script?

    -The script explains that the chemical formula of sodium sulfate decahydrate is determined by knowing the formula of the anhydrous salt (Na2SO4) and then finding the number of water molecules (x) in the hydrated form.

  • What is the molar mass of MgSO4 and H2O used in the example calculation?

    -The molar mass of MgSO4 is 120 g/mol, and the molar mass of H2O is 18 g/mol.

  • How is the mass of released water determined in the example of MgSO4 xH2O?

    -The mass of released water is determined by the difference in mass before and after heating the hydrated compound, which is 38g - 20g = 18g.

  • What is the mole ratio of MgSO4 to H2O in the hydrated compound based on the example?

    -The mole ratio of MgSO4 to H2O is determined to be 1:6 based on the mass and molar mass calculations.

  • What is the final chemical formula of the hydrated compound in the example of MgSO4?

    -The final chemical formula of the hydrated compound is determined to be MgSO4·6H2O.

  • How is the chemical formula of copper(II) sulfate pentahydrate determined in the script?

    -The chemical formula of copper(II) sulfate pentahydrate is determined by calculating the mass of released water and then finding the mole ratio of CuSO4 to H2O, which results in the formula CuSO4·5H2O.

Outlines

00:00

🔍 Determining the Chemical Formula of Hydrates

This paragraph introduces the concept of hydrates in chemistry, which are compounds that bind a certain number of water molecules as part of their crystalline structure. It explains that when heated or dissolved, the water molecules are released, leaving behind an anhydrous compound or salt. Examples of hydrates given are calcium sulfate dihydrate (CaSO4·2H2O) and magnesium sulfate heptahydrate (MgSO4·7H2O). The main focus is on determining the chemical formula of hydrates, which involves finding the number of water molecules in the crystal. The process is illustrated with an example of magnesium sulfate hydrate, where the mass of the dehydrated crystal and the molar masses of the components are used to calculate the number of water molecules, resulting in the formula MgSO4·6H2O.

05:03

📚 Calculation of Hydrate Formula Using Empirical Data

The second paragraph delves into another example of determining the chemical formula of a hydrate, specifically copper(II) sulfate hydrate. It starts by hypothesizing the formula of the hydrate as CuSO4·xH2O and uses the known mass percentage of anhydrous copper sulfate to deduce the mass of water released upon heating. The molar masses of CuSO4 and H2O are calculated to find the moles of each component in the 100-gram sample. By comparing the moles, a mole ratio of CuSO4 to H2O is established, which is then translated into the coefficient x in the hydrate formula, resulting in CuSO4·5H2O. This step-by-step approach demonstrates the application of stoichiometry to determine the composition of hydrates.

Mindmap

Keywords

💡Stoichiometry

Stoichiometry is the quantitative relationship between the amounts of reactants and products in a chemical reaction. It is fundamental to understanding how chemical reactions occur in specific proportions. In the video, stoichiometry is applied to determine the chemical formula of hydrates, which is essential to understanding the composition of these compounds.

💡Hydrate

A hydrate is a chemical compound that includes water molecules within its crystal structure. These water molecules can be released upon heating or dissolving the compound. In the script, examples of hydrates include calcium sulfate dihydrate and magnesium sulfate heptahydrate, which are used to illustrate the process of determining the number of water molecules in a hydrate's chemical formula.

💡Chemical Formula

A chemical formula represents the composition of a substance, indicating the elements present and the number of atoms of each element in a molecule. The video focuses on determining the chemical formula of hydrates by calculating the number of water molecules associated with the compound, as seen in the examples provided.

💡Crystal Water

Crystal water refers to the water molecules that are part of a hydrate's crystal lattice. It is distinct from the water of hydration, which is water surrounding a compound but not part of its crystal structure. The script discusses the release of crystal water when a hydrate is heated.

💡Anhydrous

An anhydrous compound is one that does not contain any water molecules within its structure. In the context of the video, anhydrous refers to the state of a compound after the removal of water molecules, such as when a hydrate is heated to become an anhydrous form, as demonstrated in the example of magnesium sulfate heptahydrate.

💡Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is used to convert between mass and the number of moles in a chemical calculation. In the script, molar masses of MgSO4 and H2O are used to calculate the number of moles of each substance in the hydrate.

💡Mole Ratio

The mole ratio is the ratio of the number of moles of different substances in a chemical reaction or compound. It is used to determine the coefficients in a balanced chemical equation. In the video, the mole ratio is used to find the relationship between the moles of the anhydrous compound and the water molecules in the hydrate.

💡Dehydration

Dehydration in chemistry refers to the removal of water from a compound, often through heating. The script describes the process of dehydration as it applies to hydrates, where the crystal water is released upon heating, leaving behind the anhydrous form of the compound.

💡Law of Conservation of Mass

The Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction. It is used to ensure that the mass of the reactants equals the mass of the products. In the script, this law is applied to calculate the mass of water released from a hydrate upon heating.

💡Empirical Formula

An empirical formula represents the simplest whole-number ratio of atoms or ions in a compound. In the context of the video, determining the empirical formula of a hydrate involves finding the ratio of the anhydrous compound to the water molecules in the hydrate, as demonstrated with the examples of MgSO4 and CuSO4 hydrates.

Highlights

Introduction to the topic of stoichiometry in chemistry for 10th graders.

Explanation of hydrates as compounds binding water molecules as part of their crystal structure.

The process of determining the chemical formula of a hydrate involves calculating the number of water molecules in the crystal.

Example of hydrates given: calcium sulfate dihydrate and magnesium sulfate heptahydrate.

The method to determine the chemical formula of a hydrate when the formula of the anhydrous salt is known.

A step-by-step approach to determining the formula of a hydrate through an example problem.

The concept of heating a hydrate to release water molecules, leaving behind the anhydrous salt.

Calculation of the mass of water released from magnesium sulfate hydrate upon heating.

Use of the law of conservation of mass to equate the mass of substances before and after the reaction.

Determination of moles of the hydrate and moles of water molecules using their molar masses.

Establishing the mole ratio between the anhydrous salt and water molecules.

Conversion of the mole ratio to the coefficient ratio in the chemical formula.

Final determination of the chemical formula of magnesium sulfate hydrate as MgSO4·6H2O.

Introduction of a second example involving copper(II) sulfate hydrate.

Calculation of the remaining mass percentage of copper(II) sulfate after heating.

Determination of the molar mass of copper(II) sulfate and water for further calculations.

Calculation of moles of copper(II) sulfate and water released upon heating.

Establishing the mole ratio and subsequently the coefficient ratio for the chemical formula of the hydrate.

Conclusion with the chemical formula of copper(II) sulfate hydrate as CuSO4·5H2O.

Closing remarks and thanks to the viewers, with a sign-off in Arabic.

Transcripts

play00:00

Halo Assalamualaikum Halo adik-adik

play00:02

ketemu lagi dengan Kakak di channel

play00:04

kinematika di video kali ini kita akan

play00:07

belajar materi kimia kelas 10 yaitu

play00:10

tentang stoikiometri dimana pada bagian

play00:13

ini yang akan kita bahas adalah cara

play00:16

menentukan rumus kimia senyawa hidrat

play00:20

senyawa hidrat adalah senyawa yang

play00:23

mengikat sejumlah molekul air atau H2O

play00:26

sebagai bagian dari struktur kristalnya

play00:29

nah apabila senyawa ini dipanaskan atau

play00:32

dilarutkan maka molekul airnya akan

play00:35

terlepas menyisakan senyawa kristal

play00:38

padat atau garam anhidrat contoh senyawa

play00:43

hidrat adalah kalsium sulfat dihidrat

play00:45

atau caso4 2 h2o dan magnesium sulfat

play00:50

heptahidrat atau MgSO4 7 H2O sekarang

play00:56

kita lanjut cara menentukan rumus kimia

play00:59

senyawa

play01:00

Hai pada dasarnya penentuan rumus hidrat

play01:03

merupakan penentuan jumlah molekul air

play01:05

kristal atau h-2 o dalam senyawa hidrat

play01:09

jadi untuk rumus kimia kristal garam

play01:12

padatnya sudah diketahui ya misal kita

play01:16

disuruh menentukan rumus kimia dari

play01:18

natrium sulfat berhidrat Nah untuk

play01:22

natrium sulfat nya berarti sudah

play01:23

diketahui rumus kimianya ya yaitu na2s

play01:28

o4 Nah untuk air kristalnya atau jumlah

play01:32

molekul H2O nya inilah yang kita cari

play01:35

kita misalkan dengan x jadi esnya ini

play01:38

yang kita tentukan Bagaimana

play01:41

langkah-langkahnya akan langsung dibahas

play01:43

pada contoh soal ya

play01:46

Hai sebanyak 38 gram kristal MgSO4 xh2o

play01:50

dipanaskan hingga semua air kristalnya

play01:53

menguap masa kristal yang tersisa adalah

play01:56

20gram rumus kristal hidrat tersebut

play01:59

adalah diketahui Mr MgSO4 120 dan mrh2o

play02:05

adalah 18 jadi kita disuruh menentukan

play02:09

rumus senyawa hidrat nya ya

play02:11

langkah-langkahnya adalah yang pertama

play02:13

kita buat dulu reaksi pemanasan dari

play02:16

senyawa hidrat yaitu MgSO4 xh2o jika

play02:25

dipanaskan maka molekul airnya ini akan

play02:29

terlepas menjadi MgSO4 anhidrat tanpa

play02:34

molekul air dan xh2o langkah berikutnya

play02:40

kita Tentukan massa air yang dilepaskan

play02:44

caranya di

play02:46

tahui 38 gram kristal MgSO4 xh2o jadi

play02:51

senyawa hidrat nya sebanyak 38 G setelah

play02:58

dipanaskan airnya lepas kristalnya

play03:01

menjadi 20 gram

play03:06

Oh berarti jumlah molekul air yang lepas

play03:09

sebanyak Tigapuluh delapan dikurang 20

play03:12

gram atau = 18 G sesuai dengan Hukum

play03:18

lavoisier ya massa zat sebelum dan

play03:21

sesudah reaksi sama kemudian langkah

play03:24

berikutnya Kita tentukan mol dari

play03:27

senyawa hidrat dan Mal dari molekul air

play03:31

caranya sama-sama dibagi dengan mr-nya

play03:35

jadi untuk MgSO4 20gram dibagi dengan

play03:39

mr-nya yaitu 120 punthuk H2O mr-nya

play03:43

adalah 18 jadi untuk MgSO4 maunya adalah

play03:49

20 dibagi 120/160 Mal dan untuk H2O

play03:55

maunya adalah

play03:59

Hai ingat perbandingan mol sama dengan

play04:02

perbandingan koefisien jadi kita buat

play04:05

dulu perbandingan maunya di sini MgSO4

play04:08

1/6 disini H2O adalah satu jadi

play04:11

sama-sama kita kali dengan enam jadinya

play04:14

MgSO4

play04:17

Hai H2O adalah 6 jadi perbandingan

play04:20

mallnya 1 banding 6 karena perbandingan

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mol sama dengan perbandingan koefisien

play04:25

langsung kita pindahkan saja ke

play04:27

fashionnya ke atas disini berarti 11

play04:32

biasanya tidak perlu ditulis ya Kemudian

play04:34

untuk H2O disini adalah 6 berarti x-nya

play04:39

adalah 6 jadi rumus senyawa hidrat nya

play04:43

adalah MgSO4 6 H2O Oke bisa dipahami ya

play04:49

sekarang kita lanjut ke contoh yang

play04:51

kedua tembaga 2 sulfat berhidrat

play04:55

dipanaskan ternyata berat padatan yang

play04:58

tersisa 64 persen Tentukan rumus kimia

play05:02

senyawa berhidrat tersebut diketahui Arc

play05:06

63,5 s32 H1 dan Aero = 16 kita buat

play05:14

reaksinya dulu tembaga 2 sulfat

play05:17

Hai berhidrat berarti rumusnya adalah Cu

play05:21

so4 xh2o dipanaskan menghasilkan Cu so4

play05:31

ditambah xh2o masa CuSO4 xh2o nya belum

play05:39

diketahui yang diketahui adalah CuSO4

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anhidrat atau tanpa molekul air yaitu

play05:45

sebanyak enam puluh empat persen jadi

play05:48

kita misalkan CuSO4 xh2o nya adalah 100

play05:53

gram

play05:57

Hai karena cewek so4 xh2o nya 100gram

play06:00

CuSO4 nya 64 persen berarti 64 persen

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dari 100 gram yaitu 64 gram Nah sekarang

play06:11

kita Tentukan massa dari H2O yang

play06:14

dilepaskan yaitu 100 dikurang

play06:20

laundry atau sama dengan 36 G keje last

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ya sekarang kita lanjut mencari mol dari

play06:29

CuSO4 dan Mal dari H2O yaitu sama-sama

play06:34

dibagi dengan mr-nya nah Disini mr-nya

play06:38

belum diketahui yang diketahui baru

play06:40

airnya berarti kita harus cari dulu Mr

play06:43

dari CuSO4 dan H2O untuk Mr CuSO4

play06:51

berarti Arc uh yaitu 63,5 ditambah

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dengan arts yaitu 32 ditambah dengan 24

play07:02

kaliaro Aero 16 berarti empat dikali 16

play07:05

adalah 64 atau sama dengan 159,190

play07:12

Hai Sedangkan untuk mrh2o berarti airnya

play07:18

shiha dua kali yah dua kali 12 ditambah

play07:23

dengan airnya Oh yaitu 16 atau = 18 jadi

play07:28

untuk mencari Mal CuSO4 64 G dibagi

play07:33

dengan mr-nya yaitu 159.com 5 = 0,4 mol

play07:41

Sedangkan untuk mencari mol H2O berarti

play07:44

36 G dibagi dengan email-nya H2O yaitu

play07:47

18 atau sama dengan dua Mal berikutnya

play07:54

kita buat perbandingan maunya ya berarti

play07:56

0,4 dibagi 0,4 jadinya satu banding dua

play08:02

Mal dibagi 0,4 adalah lima jadi

play08:05

perbandingan maunya adalah satu banding

play08:07

lima perbandingan mol sama dengan

play08:10

perbandingan koefisien jadi kita

play08:12

Khan ke fashionnya sehingga diperoleh

play08:17

nilai x yaitu = 5 jadi rumus senyawa

play08:22

hidrat nya adalah CuSO4 5 H2O Oke semoga

play08:29

bisa dipahami ya Nah Sekian dulu untuk

play08:31

video kali ini terima kasih

play08:33

wassalamualaikum warahmatullahi

play08:35

wabarakatuh

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