Show
Solution Part 1: Defining Empirical and molecular formulaEmpirical formula:An empirical formula is a formula in which the atoms in a molecule are present in their lowest ratio.An empirical formula gives the simplest ratio of atoms in a molecule or compound.Molecular formula:The molecular formula is the formula in which the atoms in a molecule is present in fixed ratio.The molecular formula gives the exact number of atoms in a molecule or compound.Relation between molecular formula and empirical formula:n=MolecularformulaEmpiricalformula Where, n is the whole number multiple that is 1,2,3,4….so on.Part 2: Determining the Empirical formula of C2H2The molecular formula of the compound is C2H2.In this compound, the value of ‘n’ is 2 that means the subscript of Carbon and Hydrogen are divided by the whole number 2 and we get an empirical formula for C2H2. 2= C2H2EmpiricalformulaEmpiricalformula=C2H22Empiricalformula=CHPart 3: Determining the Empirical formula of C6H6The molecular formula of the compound is C6H 6.In this compound, the value of ‘n’ is 6 that means the subscript of Carbon and Hydrogen are divided by the whole number 6 and we get an empirical formula for compound 2nd. 6=C6H6EmpiricalformulaEmpiricalformula=C6H66Empiricalformula=CHIn both compounds, the ratio of atoms of Carbon and Hydrogen are 1:1.Thus, the empirical formula of both C2 H2 and C6H6 is CH.learning objectives IntroductionEmpirical measurements are based on a measurable (empirical) quantity like mass. Knowing the mass of each element in a compound we can determine its formula. There are two types of formulas, empirical and molecular. Empirical Formula: Lowest whole number ratio of the elements in a compound Molecular Formula:Actual whole number ratio of the elements in a compound. The Empirical formula is the lowest whole number ratio of the elements in a compound. In (section 2.10), we discovered that benzene and acetylene have the same mass percent composition, and thus it is logical that they have the same ratio of elements to each other, that is, they have the same empirical formula.
Figure \(\PageIndex{1}\): Empirical and molecular formulas of several simple compounds. Empirical FormulaThe Empirical formula is the lowest whole number ratio of the elements in a compound. In (section 2.10), we discovered that benzene and acetylene have the same mass percent composition, and thus it is logical that they have the same ratio of elements to each other, that is, they have the same empirical formula. For salts that do not have homonuclear diatomic ions (like Hg2+2 or O2-2) the empirical formula is the formula we write to describe the salt. Mercury(I)chloride has the empirical formula of HgCl, but the real compound formula is Hg2Cl2 (review table 2.7.3) Multiple molecules can have the same empirical formula. For example, benzene (C6H6) and acetylene (C2H2) both of the empirical formula of CH (see Figure \(\PageIndex{1}\). Calculating Empirical FormulasSteps:
Note The trick is to convert decimals to fractions and then multiply by the lowest common denominator (watch video \(\PageIndex{1}\))
The following video shows how to calculate the empirical formula for aspiring. Calculate the empirical formula for aspirin: Aspirin is made of H, O & C, and was analyzed to contain 60.0% carbon and 35.5% Oxygen. Video \(\PageIndex{1}\): Empirical formula of aspirin Example \(\PageIndex{1}\) A certain compound was found to contain 67.6% C, 22.5% O, and 9.9% H. What is the empirical formula?. Solution \[22.5gO\left ( \frac{1molO}{16.00g} \right )= 1.4\Rightarrow \frac{1.4}{1.4}= 1\] \[67.6gC\left ( \frac{1molC}{12.011g} \right )= 5.63\Rightarrow \frac{5.63}{1.4}= 4\] \[9.9gH\left ( \frac{1molH}{1.007g} \right )= 9.9\Rightarrow \frac{9.9}{1.4}= 7\] C4H7O This Applet comes from the ChemCollective at Carnegie Mellon University. This link will send you to the video and tutorial associated with this applet. Every time you load the page a new problem will load, and there are a series of tiered hints to help you work through the problems. Exercise \(\PageIndex{1}\): empirical formula Calculate the Empirical formula for the following
CuN2O6 Answer bNO2 Answer cN2H4CO Exercise \(\PageIndex{2}\): empirical formula Given the empirical formula of the compound in part (a) of the above:
Cu(NO3)2 Copper(II) nitrate of cupric nitrate Exercise \(\PageIndex{3}\): empirical formula In section 2.10.2 we saw that benzene and acetylene both have the same mass percent composition (92.3% C and and 7.7% H), so calculate their empirical formulas
CH Answer bCH Molecular FormulaTo calculate the molecular formula we need additional information beyond that of the mass or mass percent composition, we need to know the molar mass of the substance. There are many experimental ways that can be determined, and we will learn some as the semester proceeds. The mass spectrometer that we used to determine the isotopic composition in section 2.3 of this Chapter could be used to determine the molar mass of many unknowns. But there are other techniques, and at this point in the semester, the molar mass will be treated as a given. The empirical formula represents the lowest whole number ratio of the elements in a molecule while the molecular formula represents the actual formula of the molecule. So the molecular formula must be an integer multiple of the empircal formula, that is it is n times larger where n=1 or 2 or 3 or.... but it is an integer. This gives the following relationship \[\text{[Molecular Formula = n([Empirical Formula)]}\] then if we have one mole \[\text{[Molecular Weight = n([Empirical Weight)]}\] or \[n=\frac{\text{[Molecular Weight]}}{\text{[Empirical Weight]}}\] So you calculate the Empirical formula as above, then determine the weight of one mole, divide that into the molar mass, and that tells you how many times it is bigger, and then multiple the emprical formula by that number. Let's compare Benzene to acetylene. In video 2.10.2 (section 2.10.2) we saw that benzene and acetylene had the same mass % composition and in exercise 2.11.3 we say they had the same empirical formula
Calculating Molecular FormulasExample \(\PageIndex{2}\) A certain compound was found to contain 67.6% C, 22.5% O, and 9.9% H. If the molecular weight of the compound was found to be approximately 142 g/mol, what is the correct molecular formula for the compound? Solution C4H7O has a empirical weight of (4)(12.01) + (7)(1.007) + (16.00) = 71.09g/mol \[\left ( \frac{142g/mol}{71.09g/mol} \right )= 2\] multiply the coefficients by 2 C8H14O2 Exercise \(\PageIndex{4}\): Molecular formula Calculate the molecular formula for the following
C6H3F3 Answer bC44N4H32 Answer cC21N2H22O2 Contributors and AttributionsRobert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, . You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:
Is C2H2 empirical or molecular?The molecular formula of the gas acetylene is C2H2. What is the empirical formula? C2H2 is divisible by “n ratio factor” of two; thus C1H1 is the empirical formula. An empirical formula is the smallest whole number ratio for a molecular formula.
Which has same empirical and molecular formula?C12H22O11 is the correct answer as it has the same empirical and molecular formula , because in Sucrose is C12H22O11 there is no number that is a devisor to all the three subscripts (12,22,11) that will produce a whole-number. The formula can not be reduced further.
Which compound has the same empirical and molecular formula C2H4?One molecule of ethylene (molecular formula C2H4) contains two atoms of carbon and four atoms of hydrogen. Its empirical formula is CH2. Both have the same empirical formula, yet they are different compounds with different molecular formulas.
Does C2H2 and C6H6 have the same empirical formula?Multiple molecules can have the same empirical formula. For example, benzene (C6H6) and acetylene (C2H2) both of the empirical formula of CH (see Figure 2.11. 1.
|