Stoichiometry molar mass

Pressure Units of Pressure The Gas Laws of Boyle, Charles, and Avogadro Boyle s Law Charles s Law Avogadro s Law The Ideal Gas Law Gas Stoichiometry Molar Mass of a Gas Dalton s Law of Partial Pressures... 

As can be seen from equation 8.14, we may improve a method s sensitivity in two ways. The most obvious way is to increase the ratio of the precipitate s molar mass to that of the analyte. In other words, it is desirable to form a precipitate with as large a formula weight as possible. A less obvious way to improve the calibration sensitivity is indicated by the term of 1/2 in equation 8.14, which accounts for the stoichiometry between the analyte and precipitate. Sensitivity also may be improved by forming precipitates containing fewer units of the analyte. 

The general topic of this chapter is stoichiometry (stoy-key-OM-e-tree), the study of mass relations in chemistry. Whether dealing with atomic masses (Section 3.1), molar masses (Section 3.2), chemical formulas (Section 3.3), or chemical reactions (Section 3.4), you will be answering some very practical questions that ask how much or how many—." For example—... 

The most important concept when working stoichiometry problems such as this one is moles. We must have moles to proceed. The mole determination of iodine will involve the molar mass of iodine (2 x 126.9 g/mol), while the mole determination of fluorine will involve Avogadro s number (since we have number of fluorine molecules). We can find the moles of each as follows ... 

The desired quantity is the volume of solution. The available data are the molarity of HCl in solution, the mass of chalk, and the molar mass of CaCOo. In addition, two relations are required. One identifies cnalk as calcium carbonate by stating that the mass of chalk equals the mass of CaC0 . Another gives the stoichiometry by saying that two times the number of moles of CaCO equals the number of moles of HCl. The diagram showing the solution in this case occupies several screens a separate screen is used to show each application of a relation. 

While the concept of atom economy is simple, unlike the E factor it does not take into account the actual yield or stoichiometry (actual masses or molar excesses)... 

Quantification of metal ion binding and the determination of the nature of the binding sites are extremely difficult for humic acid because of its polymeric (molar masses ca. 5 x 104) and heterogeneous nature. The results of such attempts are often contradictory.15,16 Frequently a 1 1 metal humic acid ratio is assumed for the complexes, a very surprising assumption for such large molecules with high concentrations of carboxylic acids. There is evidence for other stoichiometries, including... 

Introduction and Orientation, Matter and Energy, Elements and Atoms, Compounds, The Nomenclature of Compounds, Moles and Molar Masses, Determination of Chemical Formulas, Mixtures and Solutions, Chemical Equations, Aqueous Solutions and Precipitation, Acids and Bases, Redox Reactions, Reaction Stoichiometry, Limiting Reactants... 

Because 3.12 mol H20 is required and 5.55 mol H20 is supplied, all the calcium carbide can react so the calcium carbide is the limiting reactant and water is present in excess, (b) The reaction stoichiometry implies that 1 mol CaC2 — 1 mol C2H2. It follows that the mass of ethyne (of molar mass 26.04 g-mol ) that can be produced is... 

STRATEGY The moles of electrons supplied during the electrolysis is given i by Eq. 9 the Faraday constant is given inside the back cover. Convert moles of electrons to moles of product by using the stoichiometry of the half-reaction. Finally, convert moles of product to mass by using the molar mass. 

We need to calculate the amount of methyl tert-bu tyl ether that could theoretically be produced from 26.3 g of isobutylene and compare that theoretical amount to the actual amount (32.8 g). As always, stoichiometry problems begin by calculating the molar masses of reactants and products. Coefficients of the balanced equation then tell mole ratios, and molar masses act as conversion factors between moles and masses. 

For work in the laboratory, it s necessary to weigh reactants rather than just know numbers of moles. Thus, it s necessary to convert between numbers of moles and numbers of grams by using molar mass as the conversion factor. The molar mass of any substance is the amount in grams numerically equal to the substance s molecular or formula mass. Carrying out chemical calculations using these relationships is called stoichiometry. 

In general, the amount of product formed in an electrode reaction follows directly from the stoichiometry of the reaction and the molar mass of the product. 

Stoichiometry Empirical formula, molar mass (freezing-point depression), molar mass (vapor density)... 

This reaction requires a stoichiometry of FeOOH to Fe2+ of two-to-one. Analysis of the A/in Fig. 12.1, with the Sauerbrey equation (12.1) and the corresponding molar masses of FeOOH and Fe2+ confirms this hypothesis. It is to be noted that this last result could not have been possible should the EQCM had not been used. 

Figure 4 shows the evolution of the polymer content as a function of the excess LjH. The polymer content decreases to 10% when 45% excess of L H was introduced whereas molar masses were maximal at 30% excess. These results show that excess of L H does not really favor the generation of high polymer content, but that stoichiometry does not lead either to high molar masses, albeit in these unpurified conditions [19]. 

Since the previous protocol led to low precursor conversions, we turned our attention to the second system, starting from precursor M and excess L H (Scheme 4). To confirm the validity of working off stoichiometry, we plotted in Fig. 9 the evolution of the molar mass and polymer content of an hybrid silicone prepared from precursor M and by fractionated addition of L H. The molar masses increase smoothly up to the end of the L H addition. In parallel, the polymer content grows more quickly up to a constant level. 

Section 12.1 introduces the concept of pressure and describes a simple way of measuring gas pressures, as well as the customary units used for pressure. Section 12.2 discusses Boyle s law, which describes the effect of the pressure of a gas on its volume. Section 12.3 examines the effect of temperature on volume and introduces a new temperature scale that makes the effect easy to understand. Section 12.4 covers the combined gas law, which describes the effect of changes in both temperature and pressure on the volume of a gas. The ideal gas law, introduced in Section 12.5, describes how to calculate the number of moles in a sample of gas from its temperature, volume, and pressure. Dalton s law, presented in Section 12.6, enables the calculation of the pressure of an individual gas—for example, water vapor— in a mixture of gases. The number of moles present in any gas can be used in related calculations—for example, to obtain the molar mass of the gas (Section 12.7). Section 12.8 extends the concept of the number of moles of a gas to the stoichiometry of reactions in which at least one gas is involved. Section 12.9 enables us to calculate the volume of any gas in a chemical reaction from the volume of any other separate gas (not in a mixture of gases) in the reaction if their temperatures as well as their pressures are the same. Section 12.10 presents the kinetic molecular theory of gases, the accepted explanation of why gases behave as they do, which is based on the behavior of their individual molecules. 

The stoichiometry and molar masses of component subunits in a multi-subunit protein can usually be elucidated from the results of SDS-PAGE and measurements of native molar mass. This does not, however, give any topological information about... 

Solve stoichiometry problems involving mass by using molar mass. 

The conversion factor for converting between mass and amount in moles is the molar mass of the substance. The molar mass is the sum of atomic masses from the periodic table for the atoms in a substance. Skills Toolkit 3 shows how to use the molar mass of each substance involved in a stoichiometry problem. Notice that the problem is a three-step process. The mass in grams of the given substance is converted into moles. Next, the mole ratio is used to convert into moles of the desired substance. Finally, this amount in moles is converted into grams. 

Gases are measured by volume, just as liquids are. In problems with volume, you can use the density to convert to mass and the molar mass to convert to moles. Then use the mole ratio, just as in any other stoichiometry problem. 

Most chemical reactions that occur on the earth s surface, whether in living organisms or among inorganic substances, take place in aqueous solution. Chemical reactions carried out between substances in solution obey the requirements of stoichiometry discussed in Chapter 2, in the sense that the conservation laws embodied in balanced chemical equations are always in force. But here we must apply these requirements in a slightly different way. Instead of a conversion between masses and number of moles, using the molar mass as a conversion factor, the conversion is now between solution volumes and number of moles, with the concentration as the conversion factor. 

The molar mass of glueose is 180.15 g mol". From this, we ean ealculate the number of moles of glucose formed and, using the reaction stoichiometry, determine the number of moles of CO2 needed. With that information and the other information provided in the problem, we ean use the ideal gas law to ealculate the volume of air that is needed ... 

Equation (1.33) does show that high-molar-mass polymer does not form until high extents of reaction at long time and shows the importance of the stoichiometry (and thus purity) of the starting materials, as given in Table 1.3. 

Molar mass conversions and molarity conversions can be combined to solve equation stoichiometry problems, as we see in Example 10.7. 

Because the ideal gas equation applies to all ideal gases, the molar volume at STP applies to all gases that exhibit the characteristics of the ideal gas model. In equation stoichiometry, the molar volume at STP is used in much the way we use molar mass. Molar mass converts between moles and the measurable property of mass molar volume at STP converts between moles and the measurable property of volume of gas. Note that while every substance has a different molar mass, all ideal gases have the same molar volume at STP. Example 13.5 provides a demonstration. 

Between this chapter and Chapter 10, we have now seen three different ways to convert between a measurable property and moles in equation stoichiometry problems. The different paths are summarized in Figure 13.10 in the sample study sheet on the next two pages. For pure liquids and solids, we can convert between mass and moles, using the molar mass as a conversion factor. For gases, we can convert between volume of gas and moles using the methods described above. For solutions, molarity provides a conversion factor that enables us to convert between moles of solute and volume of solution. Equation stoichiometry problems can contain any combination of two of these conversions, such as we see in Example 13.8. 

A basic question raised in the chemical laboratory is, How much product will be formed from specific amounts of starting materials (reactants) Or in some cases we might ask the reverse question How much starting material must be used to obtain a specific amount of product To interpret a reaction quantitatively, we need to apply our knowledge of molar masses and the mole concept. Stoichiometry is the quantitative study of reactants and products in a chemical reaction. 

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