Stoikiometri (7) | Menentukan Rumus Kimia Senyawa Hidrat | Kimia kelas 10
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
🔍 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.
📚 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
💡Hydrate
💡Chemical Formula
💡Crystal Water
💡Anhydrous
💡Molar Mass
💡Mole Ratio
💡Dehydration
💡Law of Conservation of Mass
💡Empirical Formula
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
Halo Assalamualaikum Halo adik-adik
ketemu lagi dengan Kakak di channel
kinematika di video kali ini kita akan
belajar materi kimia kelas 10 yaitu
tentang stoikiometri dimana pada bagian
ini yang akan kita bahas adalah cara
menentukan rumus kimia senyawa hidrat
senyawa hidrat adalah senyawa yang
mengikat sejumlah molekul air atau H2O
sebagai bagian dari struktur kristalnya
nah apabila senyawa ini dipanaskan atau
dilarutkan maka molekul airnya akan
terlepas menyisakan senyawa kristal
padat atau garam anhidrat contoh senyawa
hidrat adalah kalsium sulfat dihidrat
atau caso4 2 h2o dan magnesium sulfat
heptahidrat atau MgSO4 7 H2O sekarang
kita lanjut cara menentukan rumus kimia
senyawa
Hai pada dasarnya penentuan rumus hidrat
merupakan penentuan jumlah molekul air
kristal atau h-2 o dalam senyawa hidrat
jadi untuk rumus kimia kristal garam
padatnya sudah diketahui ya misal kita
disuruh menentukan rumus kimia dari
natrium sulfat berhidrat Nah untuk
natrium sulfat nya berarti sudah
diketahui rumus kimianya ya yaitu na2s
o4 Nah untuk air kristalnya atau jumlah
molekul H2O nya inilah yang kita cari
kita misalkan dengan x jadi esnya ini
yang kita tentukan Bagaimana
langkah-langkahnya akan langsung dibahas
pada contoh soal ya
Hai sebanyak 38 gram kristal MgSO4 xh2o
dipanaskan hingga semua air kristalnya
menguap masa kristal yang tersisa adalah
20gram rumus kristal hidrat tersebut
adalah diketahui Mr MgSO4 120 dan mrh2o
adalah 18 jadi kita disuruh menentukan
rumus senyawa hidrat nya ya
langkah-langkahnya adalah yang pertama
kita buat dulu reaksi pemanasan dari
senyawa hidrat yaitu MgSO4 xh2o jika
dipanaskan maka molekul airnya ini akan
terlepas menjadi MgSO4 anhidrat tanpa
molekul air dan xh2o langkah berikutnya
kita Tentukan massa air yang dilepaskan
caranya di
tahui 38 gram kristal MgSO4 xh2o jadi
senyawa hidrat nya sebanyak 38 G setelah
dipanaskan airnya lepas kristalnya
menjadi 20 gram
Oh berarti jumlah molekul air yang lepas
sebanyak Tigapuluh delapan dikurang 20
gram atau = 18 G sesuai dengan Hukum
lavoisier ya massa zat sebelum dan
sesudah reaksi sama kemudian langkah
berikutnya Kita tentukan mol dari
senyawa hidrat dan Mal dari molekul air
caranya sama-sama dibagi dengan mr-nya
jadi untuk MgSO4 20gram dibagi dengan
mr-nya yaitu 120 punthuk H2O mr-nya
adalah 18 jadi untuk MgSO4 maunya adalah
20 dibagi 120/160 Mal dan untuk H2O
maunya adalah
Hai ingat perbandingan mol sama dengan
perbandingan koefisien jadi kita buat
dulu perbandingan maunya di sini MgSO4
1/6 disini H2O adalah satu jadi
sama-sama kita kali dengan enam jadinya
MgSO4
Hai H2O adalah 6 jadi perbandingan
mallnya 1 banding 6 karena perbandingan
mol sama dengan perbandingan koefisien
langsung kita pindahkan saja ke
fashionnya ke atas disini berarti 11
biasanya tidak perlu ditulis ya Kemudian
untuk H2O disini adalah 6 berarti x-nya
adalah 6 jadi rumus senyawa hidrat nya
adalah MgSO4 6 H2O Oke bisa dipahami ya
sekarang kita lanjut ke contoh yang
kedua tembaga 2 sulfat berhidrat
dipanaskan ternyata berat padatan yang
tersisa 64 persen Tentukan rumus kimia
senyawa berhidrat tersebut diketahui Arc
63,5 s32 H1 dan Aero = 16 kita buat
reaksinya dulu tembaga 2 sulfat
Hai berhidrat berarti rumusnya adalah Cu
so4 xh2o dipanaskan menghasilkan Cu so4
ditambah xh2o masa CuSO4 xh2o nya belum
diketahui yang diketahui adalah CuSO4
anhidrat atau tanpa molekul air yaitu
sebanyak enam puluh empat persen jadi
kita misalkan CuSO4 xh2o nya adalah 100
gram
Hai karena cewek so4 xh2o nya 100gram
CuSO4 nya 64 persen berarti 64 persen
dari 100 gram yaitu 64 gram Nah sekarang
kita Tentukan massa dari H2O yang
dilepaskan yaitu 100 dikurang
laundry atau sama dengan 36 G keje last
ya sekarang kita lanjut mencari mol dari
CuSO4 dan Mal dari H2O yaitu sama-sama
dibagi dengan mr-nya nah Disini mr-nya
belum diketahui yang diketahui baru
airnya berarti kita harus cari dulu Mr
dari CuSO4 dan H2O untuk Mr CuSO4
berarti Arc uh yaitu 63,5 ditambah
dengan arts yaitu 32 ditambah dengan 24
kaliaro Aero 16 berarti empat dikali 16
adalah 64 atau sama dengan 159,190
Hai Sedangkan untuk mrh2o berarti airnya
shiha dua kali yah dua kali 12 ditambah
dengan airnya Oh yaitu 16 atau = 18 jadi
untuk mencari Mal CuSO4 64 G dibagi
dengan mr-nya yaitu 159.com 5 = 0,4 mol
Sedangkan untuk mencari mol H2O berarti
36 G dibagi dengan email-nya H2O yaitu
18 atau sama dengan dua Mal berikutnya
kita buat perbandingan maunya ya berarti
0,4 dibagi 0,4 jadinya satu banding dua
Mal dibagi 0,4 adalah lima jadi
perbandingan maunya adalah satu banding
lima perbandingan mol sama dengan
perbandingan koefisien jadi kita
Khan ke fashionnya sehingga diperoleh
nilai x yaitu = 5 jadi rumus senyawa
hidrat nya adalah CuSO4 5 H2O Oke semoga
bisa dipahami ya Nah Sekian dulu untuk
video kali ini terima kasih
wassalamualaikum warahmatullahi
wabarakatuh
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