Finding the empirical formula

The usual method for finding the empirical formula is simply illustrated with ethanol, a compound containing only carbon, hydrogen, and oxygen. A weighed amount of the pure compound is completely burned in oxygen to give carbon dioxide (from the carbon) and water... 

To find the empirical formula of vitamin C from the data in Example F.l we must express the ratios of numbers of atoms as the simplest whole numbers. First, we divide each number by the smallest value (3.41), which gives a ratio of 1.00 1.33 1.00. Molecules contain only whole numbers of atoms, and one of these numbers is still not a whole number. Hence, we must multiply each number by the correct factor so that all numbers can be rounded off to whole numbers. Because 1.33 is Vi (within experimental error), we multiply all three numbers by 3 to obtain 3.00 3.99 3.00, or approximately 3 4 3. Now we know that the empirical formula of vitamin C is C3H403. 

First, follow the four-step process for finding the empirical formula. Then compare the empirical formula with the molar mass information to find the true formula. 

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. 

C03-0086. When galena, an ore of lead that contains only Pb and S, is heated in oxygen, the ore decomposes to produce pure lead. In one such process, 7.85 g of galena gave 6.80 g of lead. Find the empirical formula of galena. 

High resolution is used to determine the exact mass of an ion species in a mixture knowledge of the exact mass of an unknown substance allows its atomic composition to be established. Target analysis exact mass determination proves the presence of a particular ion species (compound) in a mixture. Mass spectrometry is perhaps the only method that can be used to find the empirical formulae of compounds that are not completely pure. 

Na202. First, find the empirical formula mass of NaO, which is 38.99 g/mol. You determine this by adding one Na (22.99) to one O (16.00). Then divide the grcim formula mass of the mystery compound, 78 g/mol, by this empirical formula mass to obtain the quotient, 2. Multiply each of the subscripts within the empirical formula by this number to obtain Na202. You ve just found the molecular formula for sodium peroxide. 

When a new chemical compound is prepared, we do not know its formula. To establish the formula, we find by experiment the weights of the various atoms in the compound, and from these weights we compute the relative number of each kind of atom in the molecule. The formula so computed is the simplest formula, not necessarily the true one. It is therefore called the empirical formula. For example, we would find the empirical formula for benzene to be CH, whereas... 

A sample of chromium weighing 0.1600 g is heated with an excess of sulfur in a covered crucible. After the reaction is complete, the unused sulfur is vaporized by heating and allowed to burn away. The cooled residue remaining in the crucible weighs 0.3082 g. Find the empirical formula of the chromium-sulfur compound that formed. 

A compound contains 90.6% Pb and 9.4% O by weight. Find the empirical formula. 

Using the approximate atomic weight, we can try to find the empirical formula for the metal oxide We can say that 2341 d of M combine with 2743 d - 2341 d = 404 d of O The number of atoms of each is... 

Using the methods of Chapter 10, we find the empirical formula to be C3H2C1. jf this were the true formula, the molecular weight would be 73.5. The freezing point depression gives a molecular weight of approximately 150. This is not an accurate value, for the experimental measurement is subject to some error, but it indicates that the true molecular weight is near 150. 

Using the steps outlined in Figure 3.9, find the empirical formula of caproic acid, calculate a formula mass, and compare with the known molecular mass. 

Find the empirical formula of a compound that is 46.3% lithium and 53.7% oxygen. 

The percentage composition of a fuel is 81.7% carbon and 18.3% hydrogen. Find the empirical formula of the fuel. 

The carbon-hydrogen combustion analyzer can also be used to find the empirical formula of a compound that contains carbon, hydrogen, and one other element, such as oxygen. The difference between the mass of the sample and the mass of the hydrogen and carbon produced is the mass of the third element. 

Step 1 Find the empirical formula of the gas, using the molar masses of carbon and hydrogen and the percent compositions. 

Find the empirical formula of the organic compound of which 3 g contains 0.6 g of hydrogen and 2.4 g carbon. 

We can find the empirical formula from percent composition data. The empirical formula represents a ratio therefore, it does not depend on the size of the sample under consideration. Because the empirical formula reflects a mole ratio, and percent composition data are given in terms of mass, we have to convert the masses to moles. We then convert the mole ratio, which is unlikely to be an integral ratio, to the smallest possible whole-number ratio, from which we write the empirical formula. 

Find the empirical formula for the following compounds containing ... 

Suppose we know that this compound with empirical formula CH5N has a molar mass of 31.06. How do we determine which of the possible choices represents the molecular formula Since the molecular formula is always a whole-number multiple of the empirical formula, we must first find the empirical formula mass for CH5N ... 

But if a chemist does an elemental analysis, she will find the empirical formula to be NH2O, because it shows the simplest ratio of the elements. For some other compounds, the empirical formula and the actual formula are the same. 

A laboratory experiment determines that a compound contains 57.5% sodium, 40.0% oxygen, and 2.5% hydrogen by mass. Find the empirical formula for this compound. 

To find the empirical formula from the percent mass composition, you assume that you have a 100 gram sample. Now the percent translates directly to grams. 

H.21 (a) We can find the empirical formulas from the percent compositions. 

J.9 Since X turns litmus red and conducts electricity poorly, it is a weak acid. We can find the empirical formula from the percent composition. 

To find the empirical formula, assume a sample size of 100 g and find the mole ratios. 

To find the empirical formula of the unknown acid, you need only find the number of moles of each element in the sample. 

Use stoichiometry principles to analyze a mixture or to find the empirical formula of an unknown compound (Section 4.6). 

On examinations, the empirical formula is often not given, but a percentage composition is given instead along with a molecular mass. Then we have a two-step solution first find the empirical formula as in Section 4.4 and then find the molecular formula as just shown. 

Combustion Analysis of Organic Compounds Still another type of compositional data is obtained through combustion analysis, a method used to measure the amounts of carbon and hydrogen in a combustible organic compound. The unknown compound is burned in pure O2 in an apparatus that consists of a combustion chamber and chambers containing compounds that absorb either H2O or CO2 (Figure 3.5). All the H in the unknown is converted to H2O, which is absorbed in the first chamber, and all the C is converted to CO2, which is absorbed in the second. By weighing the contents of the chambers before and after combustion, we find the masses of CO2 and H2O and use them to calculate the masses of C and H in the compound, from which we find the empirical formula. 

What information do we need to find the empirical formula ... 

Dividing each mole value by 2.021 (the smallest number of moles present), we find the empirical formula CICH2. 

Many homes in rural America are heated by propane gas, a compound that contains only carbon and hydrogen. Complete combustion of a sample of propane produced 2.641 g of carbon dioxide and 1.442 g of water as the only products. Find the empirical formula of propane. 

You are given the percent composition of methyl acetate and must find the empirical formula. Because you can assume that each percent by mass represents the mass of the element in a 100.00-g sample, the percent sign can be replaced with the unit grams. Then, convert from grams to moles and find the smallest whole-number ratio of moles of the elements. 

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