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Mass percent determining formula from

Interconvert masses, moles,and formulas Determine percent compositions in compounds Determine formulas from composition data Recognize models from calculated molecular formulas Perform calculations on the purity of substances... [Pg.44]

Suppose that we want to determine the formula of an unknown hydrocarbon. With combustion analysis we can establish the mass percent composition, and from this, we can determine the empirical formula. The method of Example 6-7 gives us a molar mass, in g moP, which is numerically equal to the molecular mass, in u. This is all the information we need to establish the true molecular formula of the hydrocarbon (see Exercise 96). [Pg.211]

If 1.00 g of the unknown contains 0.817 g carbon, the mass percent of carbon is 81.7 percent, leaving the remaining 18.3 percent as hydrogen. Therefore, we need to use the procedures for determining an empirical formula from a percentage composition. The problem will progress as follows (remember with percents, assume a 100 g sample) ... [Pg.287]

To use the basic chemical quantity— the mole—to make calculations convenient To determine the empirical formula from percent composition or other mass-ratio data... [Pg.196]

To determine the molecular formula from percent composition and molecular mass data or from the empirical formula and molecular mass data... [Pg.196]

The formula of a compound gives the relative number of atoms of the different elements present. It also gives the relative number of moles of the different elements present. As was shown in Sec. 7.5, the percent by mass of each element in a compound may be computed from its formula. Conversely, if the formula is not known, it may be deduced from the experimentally determined composition. This procedure is possible because once the relative masses of the elements are found, the relative numbers of moles of each may be determined. Formulas derived in this manner are called empirical formulas or simplest formulas. In solving a problem in which percent composition is given, any size sample may be considered, since the percentage of each element does not depend on the size of the sample. The most convenient size to consider is 100 g, for with that size sample, the percentage of each element is equal to the same number of grams. [Pg.107]

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. [Pg.108]

Determine the empirical and molecular formulas for a compound from mass percent and actual mass data. [Pg.328]

The steps in determining empirical and molecular formulas from percent composition or mass data are outlined below. As in other calculations, the route leads from mass through moles because formulas are based on the relative numbers of moles of elements in each mole of compound. [Pg.337]

Peroxyacylnitrate (PAN) is one of the components of smog. It is a compound of C, H, N, and O. Determine the percent composition of oxygen and the empirical formula from the following percent composition by mass 19.8 percent C, 2.50 percent H, 11.6 percent... [Pg.100]

Determining Molecular Formula from Mass Percent and Molar Mass... [Pg.96]

I Explain what is meant by the percent composition of a compound. I Determine the empirical and molecular formulas for a compound from mass percent and actual mass data. [Pg.341]

Strategy In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form the compound. Therefore, we need to convert from mass percent to moles in order to determine the empirical formula. If we assume an exactly 100 g sample of the compound, do we know the mass of each element in the compound How do we then convert from grams to moles ... [Pg.47]

The mass percent of Cl is given. From the mass of the compound and the number of hydrogen atoms given, we can calculate the mass percent of H. The mass percent of carbon is then obtained by difference. Once the mass percentages of each element are known, the empirical formula can be determined. [Pg.67]

The empirical formula can be calculated from mass percent data. The molar mass can be calculated using the ideal gas equation. The molecular formula can then be determined. [Pg.124]

Strategy First, we can determine the empirical formula from mass percent data. Then, we can determine the molar mass from the freezing-point depressioiL Finally, from the empirical formula and the molar mass, we can find the molecular formula. [Pg.323]

Because any two compounds with the same empirical formula will give the same percent composition, it is impossible to uniquely determine the molecular formula from the percent compositions by mass alone. For example, both acetylene (C2H2) and benzene (CgHg) have the same empirical formula (CH), so any experiment that only gives the percent composition would be unable to distinguish between the two compounds. Therefore, we need to independently determine both the percentage... [Pg.44]

Another way to express how much of an element is in a given compound is to use the element s mass percent composition for that compound. The mass percent composition or simply mass percent of an element is the element s percentage of the total mass of the compound. For example, the mass percent composition of sodium in sodium chloride is 39%. This information tells us that a 100-g sample of sodium chloride contains 39 g of sodium. The mass percent composition for a compound can be determined from experimental data using the formula ... [Pg.180]

In Section 6.7, we learned how to calculate mass percent composition from a chemical formula. But can we go the other way Can we calculate a chemical formula from mass percent composition This is important because laboratory analyses of compounds do not often give chemical formulas directly rather, they give the relative masses of each element present in a compound. For example, if we decompose water into hydrogen and oxygen in the laboratory, we could measure the masses of hydrogen and oxygen produced. Can we determine a chemical formula for water from this kind of data ... [Pg.183]

Determining Mass Percent from a Chemical Formula Each element contributes a fraction of a compound s mass, and that fraction multiplied by 100 gives the element s mass percent. Finding the mass percent is similar on the molecular and molar scales ... [Pg.78]

Given the formula of a compound, calculate the mass percent composition from the mass percent composition, determine the empirical formula of a compound. [Pg.219]

With ratios such as these—which come from the chemical formula—we can directly determine the amounts of the constituent elements present in a given amount of a compound without having to calculate mass percent composition. For example, we calculate the number of moles of Cl in 38.5 mol of CCI2F2 as follows ... [Pg.112]

The mass percent composition of a compound indicates each element s percentage of the total compound s mass. The mass percent composition can be determined from the compound s chemical formula and the molar masses of its elements. [Pg.128]


See other pages where Mass percent determining formula from is mentioned: [Pg.809]    [Pg.809]    [Pg.813]    [Pg.64]    [Pg.51]    [Pg.136]    [Pg.16]    [Pg.342]    [Pg.111]    [Pg.151]    [Pg.734]    [Pg.114]   
See also in sourсe #XX -- [ Pg.96 ]




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