Name: finding empirical and molecular formulas
Uploaded: 2020-09-03T13:38:26.000Z
Duration: 2 min 37 s
Channel: JoVE

K12 Standards

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North Carolina K-12 Science Standard Course of Study

9-12

Chemistry

Matter and its Interactions

PS.Chm.3.3

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Experimental Determination of Chemical Formula

Kimyasal bileşikler genellikle deneysel veya moleküler bir formül kullanılarak açıklanır. Bu formüller, ilgili elementlerin farklı atomlarının miktarı hakkında bilgi sağlar. Peki bu formüller nasıl oluşturulur?Bileşiklerin ayrışması gibi deneysel analiz, bileşikteki kurucu elementlerin göreli kütlelerini hesaplamak için kullanılır. Bu göreli kütleler daha sonra bir kimyasal bileşiğin formülünü belirlemek için her bir elementin mol sayısını hesaplamada kullanılır. Örneğin, bir bileşik numunesinin deneysel olarak 43, 64 gram fosfor ve 56, 36 gram oksijen içerdiği belirlenir.Molar kütleleri dönüştürme faktörleri olarak kullanarak, deneysel verilerden elde edilen bağıl kütleler fosfor için 1, 41 mol ve oksijen için 3, 52 mol'e dönüştürülür. Bu mol değerleri, elementlere geçici alt simgeler olarak atandıklarında, bileşiğin en basit formülünü verir. Mol değerlerinin en küçük mol değerine bölünmesi, bileşikteki elementlerin gerçek oranıyla doğrudan ilişkili olan yaklaşık 2, 5 mol oksijen ile 1 mol fosfor mol oranlarını verir.Bölümlerden biri hala ondalıklı ise tüm sayılar alt simgelerin en küçük tam sayı oranını veren en küçük sayma sayısıyla çarpılarak P2O5 ampirik formülü elde edilir. Bileşiklerin moleküler formülü, deneysel formüllerinden ve molar kütlelerinden veya moleküler ağırlıklarından belirlenebilir. Örneğin, P2O5'in ampirik formülüne sahip kimyasal bileşiğin, mol başına 283, 89 gramlık bir molar kütleye sahip olduğu deneysel olarak ölçülür.Moleküler formülü, ampirik formülünün tam sayı katı iken molar kütlesi, ampirik formül kütlesinin tam sayı katıdır. Molar kütle ve ampirik formül kütlesinin oranı formül birimlerinin sayısını verir. Ampirik formülün, formül birimlerinin sayısıyla çarpılması, moleküler formülü verir.Bu nedenle, P4O10 moleküler formülüne sahip bileşik tetrafosfor dekaoksit olarak tanımlanır veya daha yaygın olarak ampirik formül adıyla fosfor pentoksit olarak bilinir.

kimyasal formülün deneysel olarak belirlenmesi

Bir bileşiğin elemental içeriği kimyasal kimliğini tanımlar ve kimyasal formüller bu elemental içeriği temsil etmenin en özlü yoludur. Bir bileşiğin formülü bilinmediğinde, kurucu elemanlarının kütlesini ölçmek genellikle formülü deneysel olarak belirlemede ilk adımdır.
 
 <h4>Ampirik Formüllerin Belirlenmesi</h4>
 
 Bir bileşiğin kimyasal formülünü belirlemede en yaygın yaklaşım, önce kurucu elementlerin kütlelerini ölçmektir. Bununla birlikte, kimyasal formüller, maddedeki atom kütlelerini değil, nispi sayıları temsil eder. Bu nedenle, kütle içeren deneysel olarak türetilmiş herhangi bir veri, bileşikteki karşılık gelen atom sayısını elde etmek için kullanılmalıdır. Bu ise her bir elementin kütlesini mol sayısına dönüştürmek için molar kütleler kullanılarak gerçekleştirilir. Bu molar miktarlar, maddenin ampirik formülünü elde etmek için kullanılabilecek tam sayı oranlarını hesaplamak için kullanılır.
 
 1,71 gram karbon ve 0,287 gram hidrojen içerdiği belirlenen bir bileşik örneğini düşünün. Karşılık gelen atom sayıları 0,142 mol karbon ve 0,284 mol hidrojendir. Böylece bu bileşik C0,142H0,284 şeklinde formülize edilebilir. Kural olarak formüller, her bir alt simgeyi en küçük alt simgeye (0,142) bölerek elde edilebilen tam sayı alt simgelerini içerir. Bu nedenle, bu bileşik için ampirik formül CH 2 şeklinde yazılabilir. "1" alt simgeleri yazılmaz, ancak başka bir sayı yoksa 1 olduğu varsayılır. Bu bileşiğin moleküler formülü olabilir veya olmayabilir; bununla birlikte, bu tayini yapmak için ek bilgilere ihtiyaç vardır.
 
 İkinci bir örnek olarak, bir bileşik numunesinin 5,31 gram klor ve 8,40 gram oksijen içerdiği belirlenir. Aynı yaklaşım, ClO3,5 şeklindeki geçici ampirik formülü verir. Bu durumda, en küçük alt simgeye bölünmek, ampirik formülde hala bir ondalık bırakır. Bunu bir tam sayıya dönüştürmek için, alt simgelerin her biri iki ile çarpılır, aynı atom oranını koruyun ve son ampirik formül olarak Cl2O7 elde edilir.
 
 <h4>Ampirik Formüllerin Yüzde Bileşiminden Türetilmesi</h4>
 
 Bir bileşiğin yüzde bileşiminin bilindiği durumlarda, bu bilgi bileşikte bulunan elementlerin kütlelerini hesaplamak için kullanılır. Yüzdeler için ölçek 100 olduğundan, 100 gram ağırlığındaki bir numunede bulunan elementlerin kütlesini hesaplamak uygundur. Elde edilen kütleler ampirik formülü elde etmek için kullanılır.
 
 Örneğin, bir gaz halindeki bileşiğin %27,29 C ve %72,71 O içerdiğini varsayalım. Bu nedenle kütle yüzdeleri fraksiyonlar olarak ifade edilir:
 
 <img alt=""Eq1"" src=""https://www.jovecore.com/CDNSource/lessons/1018/1018_Equations-01v2.png"" style=""height: 35px;"" />
 
 <img alt=""Eq2"" src=""https://www.jovecore.com/CDNSource/lessons/1018/1018_Equations-02v2.png"" style=""height: 35px;"" />
 
 Karbon kütlesi, 27,29 g, 2,272 mol karbona karşılık gelir ve oksijen kütlesi, 72,71 g, 4,544 mol oksijene karşılık gelir. Bu nedenle, temsili formül C2,272O4,544 olacaktır. Her bir alt simgeyi 2,272 ile bölmek ampirik formülü verir: CO2.
 
 <h4>Moleküler Formüllerin Türetilmesi</h4>
 
 Bir kovalent bileşiğin tek bir molekülünü oluşturan atomların mutlak sayılarını belirlemek, hem ampirik formülü hem de moleküler kütlesi veya molar kütlesi hakkında bilgi gerektirir. Bu miktarlar çeşitli ölçüm teknikleri ile deneysel olarak belirlenebilir. Örneğin moleküler kütle, genellikle bileşiğin kütle spektrumundan türetilir.
 
 Moleküler formüller, bileşiğin molar kütlesini veya moleküler kütlesini ampirik formül kütlesi ile karşılaştırarak elde edilir. Adından da anlaşılacağı gibi, ampirik bir formül kütlesi, ampirik bir formülde temsil edilen tüm atomların ortalama atom kütlelerinin toplamıdır. Bir maddenin bilinen molar kütlesi ampirik formül kütlesi ile bölünürse, molekül başına ampirik formül birimlerinin sayısını verir (n).
 
 <img alt=""Eq3"" src=""https://www.jovecore.com/CDNSource/lessons/1018/1018_Equations-03v2.png"" style=""height: 40px;"" />
 
 Moleküler formül daha sonra ampirik formüldeki her bir alt simge, genel ampirik formül AxBy ile gösterildiği gibi n ile çarpılarak elde edilir:
 
 <img alt=""Eq4"" src=""https://www.jovecore.com/CDNSource/lessons/1018/1018_Equations-04v2.png"" style=""height: 17px;"" />
 
 Örneğin, bir kovalent bileşiğin ampirik formülü CH2O olarak belirlenir ve ampirik formül kütlesi yaklaşık 30 amu'dur. Bileşiğin moleküler kütlesi 180 amu olarak belirlenirse, bu durum, bu bileşiğin moleküllerinin ampirik formülde temsil edilen atom sayısının altı katını içerdiğini gösterir.
 
 <img alt=""Eq5"" src=""https://www.jovecore.com/CDNSource/lessons/1018/1018_Equations-05v2.png"" style=""height: 35px;"" />
 
 Bu bileşiğin molekülleri daha sonra ampirik formüldekilerden altı kat daha büyük olan alt simgeler içeren bir moleküler formülle temsil edilir: (CH2O)6 = C6H12O6.
 
 Bu metin bu kaynaktan uyurlanmıştır: <a href=""https://openstax.org/books/chemistry-2e/pages/3-2-determining-empirical-and-molecular-formulas"">Openstax, Chemistry 2e, Section 3.2: Determining Empirical and Molecular Formulas.</a>

Kimyasal Formülün Deneysel Olarak Belirlenmesi

Bir bileşiğin temel yapısı, kimyasal kimliğini tanımlar ve kimyasal formüller, bu temel yapıyı temsil etmenin en özlü yoludur. ...

Education

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Molecules, Compounds, and Chemical Equations

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KEY TERMS AND CONCEPTS

TEMEL TERİMLER VE KAVRAMLAR

finding empirical and molecular formulas

Chemical formulas show the elemental makeup of a compound. When a formula is unknown, scientists often start by measuring the masses of the elements in the sample. Those mass data are then used to find the simplest whole-number ratio of atoms, which gives the empirical formula.

To get from mass to atom ratio, the masses of the elements are first converted to moles using molar masses. Chemical formulas describe numbers of atoms, not masses, so this step is necessary. In one example, 1.71 grams of carbon and 0.287 grams of hydrogen become 0.142 moles of carbon and 0.284 moles of hydrogen. Dividing by the smallest value gives the empirical formula CH2, with the 1 omitted by convention.

Sometimes the numbers do not become whole right away. A sample containing 5.31 grams of chlorine and 8.40 grams of oxygen gives a tentative ratio of ClO3.5. Multiplying both subscripts by two keeps the same atom ratio and gives the whole-number empirical formula Cl2O7.

Percent composition can also be used to find an empirical formula. Because percentages are based on 100, it is convenient to imagine a 100-gram sample. For a gas that is 27.29% carbon and 72.71% oxygen, those masses correspond to 2.272 moles of carbon and 4.544 moles of oxygen. Dividing by the smaller amount gives the empirical formula CO2.

A molecular formula gives the actual number of atoms in one molecule of a covalent compound. To find it, both the empirical formula and the molecular mass or molar mass are needed. The empirical formula mass is the sum of the average atomic masses in the empirical formula, and dividing the molar mass by that value gives the number of empirical units per molecule, n. If the empirical formula is CH2O and the molecular mass is 180 amu, then n equals 6, so the molecular formula is C6H12O6.

Finding Empirical and Molecular Formulas

molecules and compounds

Atomic Elements, Molecules, and Bonds

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Molecular Model Types in Chemistry

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Chemical formula mass and percent composition help describe the makeup of compounds. A chemical formula shows which atoms are present and how many of each atom appear in a substance. For covalent compounds, the formula shows the atoms in one molecule. That is why formula mass can also be called molecular mass. Chloroform, CHCl3, is one example. Its formula shows one carbon atom, one hydrogen atom, and three chlorine atoms. The molecular mass is the sum of their average atomic masses. Ionic...

experimental determination of chemical formula

Chemical formulas show the elemental makeup of a compound. When a formula is unknown, scientists often start by measuring the masses of the elements in the sample. Those mass data are then used to find the simplest whole-number ratio of atoms, which gives the empirical formula. To get from mass to atom ratio, the masses of the elements are first converted to moles using molar masses. Chemical formulas describe numbers of atoms, not masses, so this step is necessary. In one example, 1.71 grams...

chemical equations

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Chemical equations show the identities and relative amounts of substances in a chemical reaction. The substances that react are called reactants, and their formulas are written on the left side of the equation. The substances formed are called products, and their formulas are written on the right side. A plus sign (+) separates formulas on the same side, and an arrow (→) shows the change from reactants to products. The numbers in a chemical equation are shown with coefficients. A coefficient...

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Learn how to use mass, percent composition, and molar mass to determine empirical formulas and molecular formulas from experimental data.

Finding Empirical and Molecular Formulas | JoVE K12

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PS.Chm.3.1

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Analyze the structure of atoms and isotopes.

Finding Empirical and Molecular Formulas

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