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Molar mass percent composition calculation using

Benzene (1 g, 12.5 mmol) is allowed to react with 1-chloropropane (1 g, 12.5 mmol) and AICI3. The product (1.2 g) is subjected to analysis on a GLC equipped with a thermal conductivity detector. The chromatogram shows two product peaks identified as n-propylbenzene (area = 65 mm = 1.06) and isopropylbenzene (area = 113 mm = 1.09). Calculate the percent yield of each of the two isomeric products obtained in this reaction. Note that since each of the products has the same molar mass of 120, the use of weight factors gives both weight and mole percent composition. [Pg.209]

We can also calculate mass percent composition of a compound by using its molar mass. Then the total mass of each element is divided by the molar mass of the compound and multiphed by 100%. [Pg.220]

Calculate the percent composition of nitroglycerine, C3H5N3O9, using molar masses instead of formula masses. [Pg.204]

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. [Pg.209]

Percent composition from the chemicai formuia If you already know the chemical formula for a compound such as water (H2O), can you calculate its percent composition The answer is yes. You can use the chemical formula to calculate the molar mass of water (18.02 g/mol) and assume you have an 18.02-g sample. Because the percent composition of a compound is always the same, no matter the size of the sample, you can assume that the sample... [Pg.328]

The data used to determine the chemical formula for a compound may be in the form of percent composition or it may be the actual masses of the elements in a given mass of the compound. If percent composition is given, you can assume that the total mass of the compound is 100.00 g and that the percent by mass of each element is equal to the mass of that element in grams. For example, the percent composition of an oxide of sulfiir is 40.05% S and 59.95% O. Thus, as you can see in Figure 11-10, 100.00 g of the oxide contains 40.05 g S and 59.95 g O. The mass of each element can be converted to a number of moles by multiplying by the inverse of the molar mass. Recall that the number of moles of S and O are calculated in this way. [Pg.331]

The formula calculated from percent composition by mass is always the empirical formula because the coefficients in the formula are always reduced to the smallest whole numbers. To calculate the actual, molecular formula we must know the approximate molar mass of the compound in addition to its empirical formula. Knowing that the molar mass of a compound must be an integral multiple of the molar mass of its empirical formula, we can use the molar mass to find the molecular formula, as Example 3.11 demonstrates. [Pg.83]

Calculate the empirical formula of hypo, used in photographic development, consisting of 29.1% Na, 40.5% S, and 30.4% O. Calculate the percent composition of rubbing alcohol, C3H8O. Calculate the molecular formula of a compound with molar mass 104 g/mol composed of 92.3% carbon and 7.7% hydrogen. Consider the formula of hydrazinium nitrate, N2H6(N03)2. [Pg.55]

The sum of the percentages is 5.926% + 94.06% = 99.99%. The small discrepancy from 100 percent is due to the way we rounded off the molar masses of the elements. If we had used the empirical formula HO for the calculation, we would have obtained the same percentages. This is so because both the molecular formula and empirical formula tell us the percent composition by mass of the compound. [Pg.89]

Knowing the chemical formula and the molecular mass of a compound enables us to calculate the percent composition by mass—the percent by mass of each element in a compound. It is useful to know the percent composition by mass if, for example, we needed to verify the purity of a compound for use in a laboratory experiment. From the formula we could calculate what percent of the total mass of the compound is contributed by each element. Then, by comparing the result to the percent composition obtained experimentally for our sample, we could determine the purity of the sample. Mathematically, the percent composition is obtained by dividing the mass of each element in 1 mole of the compound by the molar mass of the compound and multiplying by 100 percent ... [Pg.42]

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. [Pg.101]


See other pages where Molar mass percent composition calculation using is mentioned: [Pg.734]    [Pg.61]    [Pg.113]    [Pg.1251]    [Pg.1666]    [Pg.2216]    [Pg.1662]    [Pg.2200]    [Pg.342]    [Pg.1432]    [Pg.153]   
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