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Mass percentages calculating from formulas

The empirical formula of a compound can be simply related to the mass percentage of its constituent elements using the mole concept. For example, the empirical formula for ethylene (molecular formula C2H4) is CH2. Its composition by mass is calculated from the masses of carbon and hydrogen in 1 mol of CH2 formula units ... [Pg.35]

Calculate the mass percentage of an element in a compound from a formula (Example El). [Pg.74]

Empirical formulae may also be calculated from percentage composition by mass in a similar way. [Pg.18]

In the problems above, the percentage data was calculated from the chemical formula, but the empirical formula can be determined if the percent compositions of the various elements are known. The empirical formula tells us what elements are present in the compound and the simplest whole-number ratio of elements. The data may be in terms of percentage, or mass, or even moles. But the procedure is still the same convert each to moles, divide each by the smallest number, then use an appropriate multiplier if needed. The empirical formula mass can then be calculated. If the actual molecular mass is known, dividing the molecular mass by the empirical formula mass gives an integer (rounded if needed) that is used to multiply each of the subscripts in the empirical formula. This gives the molecular (actual) formula, which tells which elements are in the compound and the actual number of each. [Pg.90]

The OB is calculated from the empirical formula of an explosive as the percentage of oxygen required for complete oxidation of C—>C02 and H—>H20 by the following expression (Equation 1.3) for an explosive with the general formula CaHbNcOd and molecular mass M. The unit of OB is %. [Pg.18]

Calculate the empirical formula of a compound from its mass percentage composition, Self-Test F.3. [Pg.87]

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. [Pg.84]

In the previous Practice Problems, you used mass data to calculate percentage composition. This skill is useful for interpreting experimental data when the chemical formula is unknown. Often, however, the percentage composition is calculated from a known chemical formula. This is useful when you are interested in extracting a certain element from a compound. For example, many metals, such as iron and mercury, exist in mineral form. Mercury is most often found in nature as mercury(II) sulfide, HgS. Knowing the percentage composition of HgS helps a metallurgist predict the mass of mercury that can be extracted from a sample of HgS. [Pg.202]

When determining the percentage composition by mass of a homogeneous sample, the size of the sample does not matter. According to the law of definite proportions, there is a fixed proportion of each element in the compound, no matter how much of the compound you have. This means that you can choose a convenient sample size when calculating percentage composition from a formula. [Pg.202]

The determination of the empirical formula of a compound can be made experimentally, by determining the percentage amounts of elements present in the substance using the methods of quantitative chemical analysis. At the same time the relative molecular mass of the compound has to be measured as well. From these data the empirical formula can be determined by a simple calculation. If, for some reason, it is impossible to determine the relative molecular mass the simplest (assumed) formula only can be calculated from the results of chemical analysis the true formula might contain multiples of the atoms given in the assumed formula. [Pg.2]

If you know the chemical formula of any compound, then you can calculate the percentage composition. From the subscripts, you can determine the mass contributed by each element and add these to get the molar mass. Then, divide the mass of each element by the molar mass. Multiply by 100 to find the percentage composition of that element. [Pg.264]

The soil porosity is the total volume of pores in the soil mass expressed as a percentage of the total intact earth volume. The pores in the soil have different shapes and sizes and they are filled either with water or with air. The determination of the total porosity P is calculated from the ratio of the specific (S) and reduced bulk weight, using formula... [Pg.689]

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]

Just as the percentage composition by mass can be calculated from the formula of a compound, the simplest formula of a compound can be found from the percentage composition by mass of each element. The simplest formula is also known as the empirical formula. Here, empirical means obtained from experimental results . To find the simplest formula ... [Pg.125]

The theoretical chemical formula of a mineral is unique and identifies only one species. Nevertheless, the actual chemical composition is usually variable within a limited range owing to the isomorphic substitutions (i.e., diadochy), or/and low presence of traces of impurities. The relative atomic or molecular mass (based on C = 12.000) of minerals is calculated from the theoretical formula using the last value of atomic masses adopted by the International Union of Pure and Applied Chemistry (lUPAC) in 2001, and the theoretical chemical composition is commonly expressed in percentage by weight (wt.%) of elements and sometimes oxides for oxygenated minerals. [Pg.757]

The empirical formula (simplest formula) of a compound is obtained from the percentage composition of the substance, which is expressed as mass percentages of the elements. To calculate the empirical formula, you convert mass percentages... [Pg.113]

Calculating the percentage composition from the formula Given the formula of a compound, calculate the mass percentages of the elements in it. (EXAMPLE 3.7)... [Pg.113]

Formula mass, molar mass, and percentage composition can be calculated from the chemical formula for a compound. [Pg.239]

Answer Since a molecular formula represents the number of moles of each element per mole of compound, first calculate the mass of each element in 1 mole of compound from its molecular weight and composition, (Recall that mass percentage of an element is its mass in 100 g of the compound.)... [Pg.40]

Sometimes elemental microanalysis results are given as percentages by mass. The process used to calculate the empirical formula from the percentage by mass is similar to that just shown, assuming the total mass of the sample to be 100 g. [Pg.74]

Answer The first step is to calculate the empirical formula. To begin, we need to determine the number of moles of each element. However, the percentage composition does not tell us a mass. To get around this, the commonly used technique is to assume that you have a 100.0-gram sample of the substance. Because 100.0 grams is equal to 100%, you can simply take the percentages and say that the masses are equal to that same amount in grams. From there, the problem proceeds just like the previous example ... [Pg.274]

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]

The sum of the percentages is 5.926 percent + 94.06 percent = 99.99 percent. 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 written... [Pg.80]

When a compound is analyzed to determine the relative amounts of the elements present, the results are usually given in terms of percentages by masses of the various elements. Previously we learned to calculate the percent composition of a compound from its formula. Now we will do the opposite. Given the percent composition, we will calculate the empirical formula. [Pg.204]

The simplest formula is not always the true or molecular formula it is only the simplest ratio of the atoms contained in the substance. If, for example, you calculated the simplest formula of hydrazine (N2H4) from its percentage composition by mass of nitrogen and hydrogen, you would obtain NH2. In order to establish the molecular formula, you need the molecular mass as well as the empirical formula. [Pg.126]

Often, one of the the first steps in the identification of an unknown organic compound is to submit the compound for quantitative elemental analysis. This type of analysis will determine the percentage by mass of the elements present in the compound. From the results of the analysis, the empirical formula of the compound can be calculated. To find out the relative amounts of carbon and hydrogen in a hydrocarbon, a weighed sample of the hydrocarbon is passed through a tube packed with copper(ll) oxide at a temperature of about 700°C. The copper(ll) oxide oxidizes the carbon in the hydrocarbon to carbon dioxide and the hydrogen to steam ... [Pg.322]


See other pages where Mass percentages calculating from formulas is mentioned: [Pg.74]    [Pg.164]    [Pg.547]    [Pg.17]    [Pg.205]    [Pg.547]    [Pg.162]    [Pg.1000]    [Pg.84]    [Pg.95]    [Pg.384]    [Pg.593]    [Pg.160]    [Pg.1169]    [Pg.82]    [Pg.83]    [Pg.59]    [Pg.103]    [Pg.104]   
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