Molar mass molecular formula from

A differeut method may be used to calculate the molecular formula from perceut compositiou data plus a molar mass First, calculate the mass of each elemeut iu 1.00 mol of compouud. Next, calculate the uumber of moles of each of the elemeuts iu the mole of compouud. Those results yield the molecular formula. Use this method to calculate the molecular formula of a hydro-carbou (a compouud of carbou aud hydrogen only) that contains 87.73% C and has a molar mass of 82.0 g/mol. 

To calculate the properties of each gas in a mixture of gases To calculate molar masses from mass data along with pressure, volume, temperature data, and to use the molar masses thus calculated to enable calculation of molecular formulas from empirical formulas... 

Formulas describe the composition of compounds. Empirical formulas give the mole ratio of the various elements. However, sometimes different compounds have the same ratio of moles of atoms of the same elements. For example, acetylene, C2H2, and benzene, CeHe, each have 1 1 ratios of moles of carbon atoms to moles of hydrogen atoms. That is, each has an empirical formula CH. Such compounds have the same percent compositions. However, they do not have the same number of atoms in each molecule. The molecular formula is a formula that gives all the information that the empirical formula gives (the mole ratios of the various elements) plus the information of how many atoms are in each molecule. In order to deduce molecular formulas from experimental data, the percent composition and the molar mass are usually determined. The molar mass may be determined experimentally in several ways, one of which will be described in Chap. 12. 

Empirical and molecular formulas from experimental data Molar masses from gas density, freezing-point, and boiling-point measurements... 

Determining Molecular Formula from Mass Percent and Molar Mass... 

This is an extension of an ideal gas law calcnlation involving molar mass. If you determine the molar mass of the gas, yon will be able to determine the molecular formula from the empirical formula. 

Check Note that in determining the molecular formula from the empirical formula, we need only know the approximate molar mass of the componnd. The reason is that the true molar mass is an integral multiple (IX, 2X, 3X,. . . ) of the empirical molar mass. Therefore, the ratio (molar mass/empirical molar mass) will always be close to an integer. 

We determine the molecular formula from the empirical formula and the molar mass ... 

Empirical and Molecular Formulas from Laboratory Data We can refer to the relative masses of the elements within a compound to determine the empirical formula of the compound. If the chemist also knows the molar mass of the compound, he or she can also determine its molecular formula. 

Calculating a Molecular Formula from an Empirical Formula and Molar Mass (Section 6.9)... 

The empirical formula gives only the ratio of combination of the atoms in a molecule, so there may be numerous compounds with the same empirical formula. If we know the approximate molar mass of the compound, though, we can determine the molecular formula from the empirical formula. For instance, the molar mass of glucose is about 180 g. The empirical-formula mass of CH2O is about 30 g [12.01 g + 2(1.008 g) + 16.00 g]. To determine the molecular formula, we first divide the molar mass by the empirical-formula mass 180 g/30 g = 6. This tells us that there are six empirical-formula units per molecule in glucose. Multiplying each subscript by 6 (recall that when none is shown, the subscript is understood to be a 1) gives the molecular formula, CeH Of,. 

Convert between moles and mass using the appropriate molar masses, and convert from mol H2O to mol H using the conversion factor derived from the molecular formula. 

EXAMPLE 3.19 Calculating a Molecular Formula from an Empirical Formula and Molar Mass Butanedione—sl main component responsible for the smell and taste of butter and cheese—contains the elements carbon, hydrogen, and oxygen. The anpirical formula of butanedione is C2H3O, and its molar mass is 86.09 g/mol. Find its molecular formula. ... 

Our approach will require several steps (1) Use the combustion data to determine the percent composition of the compound (similar to Example 3-6). (2) Determine the empirical formula from the percent composition (similar to Example 3-5). (3) Obtain the molecular formula from the empirical formula and the molecular mass. (4) Determine how the C, H, and O atoms represented in the molecular formula might be assembled into a dicarboxylic acid. Use molar masses with (at least) one more significant figure than in the measured masses where possible store intermediate results in your calculator without rounding off. 

J 3 Determine the molecular formula of a compound from its empirical formula and its molar mass (Example F.3). 

F.13 Osmium forms a number of molecular compounds with carbon monoxide. One light-vellow compound was analyzed to give the following elemental composition 15.89% C, 21.18% O, and 62.93% Os. (a) What is the empirical formula of this compound (b) From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g-mol 1. What is its molecular formula ... 

Suppose that 10.0 g of an organic compound used as a component of mothballs is dissolved in 80.0 g of benzene. The freezing point of the solution is 1.20°C. (a) What is an approximate molar mass of the organic compound (b) An elemental analysis of that substance indicated that the empirical formula is C3H2C1. What is its molecular formula (c) Using the atomic molar masses from the periodic table, calculate a more accurate molar mass of the compound. 

The flowchart in Figure 3-15 outlines the process. From masses of products, determine masses of elements. Then convert masses of elements to moles of elements. From moles of the elements, find the empirical formula. Finally, use information about the molar mass to obtain the molecular formula. 

The second average is viscosity average molecular weight, M . This expression is obtained by using the exponent from the limiting viscosity number-molar mass relationship, a, as a power for the molecular weight of each molecule in the distribution. The formula for this average is... 

The molar masses of molecular and ionic compounds are calculated from the molar masses of the elements present the molar mass of a compound is the sum of the molar masses of the elements that make up the molecule or the formula unit. We need only note how many times each atom or ion appears in the molecular formula or the formula unit of the ionic compound. For example, the molar mass of the ionic compound Na2S04 is... 

F.14 In 1978, scientists extracted a compound with antitumor and antiviral properties from marine animals in the Caribbean Sea. A 2.52-mg sample of the compound, didemnin-C, was analyzed and found to have the following composition 1.55 mg C, 0.204 mg H, 0.209 mg N, and 0.557 mg O. The molar mass of didemnin-C was found to be 1014 g-mol-1. What is its molecular formula ... 

Take the freezing-point constant from Table 8.9. Use this -molality to calculate the moles of solute in the sample by multiplying it by the mass of solvent in kilograms. At this stage, determine the molar mass of the solute by dividing the given mass of solute by the number of moles present. For the molecular formula, decide how many atoms of sulfur are needed in each molecule to account for the molar mass. 

Each element has a specific symbol that is different from the symbol for any other element. In a chemical formula, the symbol stands for an atom of an element. Molecular substances are composed of two or more atoms that are tightly bound together. The formula for a molecular substance consists of the symbols for the atoms that are found in that molecule. For instance, the formula for carbon dioxide is CO2. Note the use of the subscript to show that each molecule contains two oxygen atoms in addition to the one carbon atom. Also note that the 1 for the one carbon atom is not written. The molecular mass of CO2 is the sum of the atomic mass of carbon plus twice the atomic mass of oxygen and is expressed in u. As was discussed directly above, the molar mass of CO2 is the mass in grams equal to the molecular mass in u. A mole of carbon dioxide is 12.0 u + 2(16.0 u) = 44 u. This result can be expressed as 44 g to indicate one Avogadro s number, Na, of CO2 molecules. Recall that Na is 6.0221 x 1023 things—molecules in this case. 

The empirical formula CH2 is not a stable substance. It is necessary to determine the molar mass to determine the molecular formula. If this hydrocarbon were a gas or an easily volatilized liquid, its molar mass could be determined from the density of the gas, as shown in Chapter 5. Supposing such a determination yields a molar mass of about 55 g/mol, what is the molecular formula ... 

The approximate molar mass, calculated from the gas density data, is 89 g/mol. The empirical formula, calculated from the percentage composition data, is C2H3O with the empirical formula unit mass of 43.0. The exact molar mass must be (2)(43) = 86.0 g/mol since this is the only multiple of 43.0 (whole-number multiple) reasonably close to the approximate molecular formula of 89 g/mol. The molecule must be the equivalent of 2 empirical formulas CqHgO. 

These numbers approximate the numbers of atoms in the molecule with small deviations from values resulting from the approximate nature of the molar mass measurement. The molecular formula, C4H6O2, is obtained without going through the intermediate evaluation of an empirical formula. 

Calculate the molar mass from both Eqs. (12) and (21) using the appropriate constants in Table 1. An extrapolated value of Mean be obtained from the calculated values from either equation by plotting the calculated molar mass against the depression AT). If the elementary analysis or empirical formula of the unknown is given, deduce the molecular formula and the exact molar mass. 

If we calculated the percent compositions of C2H2 and CeHg (Figure 7.3), we would find that both have the same percentages of carbon and the same percentages of hydrogen (compare Problem 7.100 at the end of the chapter). Both have the same empirical formula—CH. This result means that we cannot tell these two compounds apart from percent composition data alone. However, if we also have a molar mass, we can use that information with the percent composition data to determine not only the empirical formula but also the molecular formula. 

---