Element moles, converting

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. 

The electron attachment energy or electron affinity is defined as the change in internal energy (i.e. A U) that occurs when one mole of gaseous atoms of an element are converted by electron attachment to give one mole of gaseous negative ions ... 

After we receive the results of a combustion analysis from the laboratory, we need to convert the mass percentage composition to an empirical formula. For this step, we need to determine the relative number of moles of each type of atom. The simplest procedure is to imagine that we have a sample of mass 100 g exactly. That way, the mass percentage composition tells us the mass in grams of each element. Then we can use the molar mass of each element to convert these masses into moles and go on to find the relative numbers of moles of each type of atom. Let s do that for vitamin C, which was once identified in this way, and suppose that the laboratory has reported that the sample you supplied is 40.9% carbon, 4.58% hydrogen, and 54.5% oxygen. 

Use the percents by mass as grams of elements and convert to the number of moles by multiplying by the conversion factor that relates moles to mass based on molar mass. 

Empirical formulas can also be calculated from the mass of each element in a sample of a compound. Analysis of a sample of a compound shows that it contains 1.179 g Na and 0.821 g S. You could calculate the percent of each element from these numbers, but that s not necessary. The mass of each element is converted directly to moles for the empirical formula. From the table of atomic masses, we find the molar masses are 22.99 g for Na and 32.07 g for S. 

First use the molar mass of the element to convert from grams to moles, and then use Avogadro s number to convert moles to number of atoms. 

Use the molar mass of the compound to convert from grams of the compound to moles of the compound. Then use the chemical formula to obtain a conversion factor to convert from moles of the compound to moles of the constituent element. Finally, use the molar mass of the constituent element to convert from moles of the element to grams of the element. 

To calculate the empirical formula, the grams of each element are converted to moles and divided hy the smallest number of moles to obtain the lowest whole-number ratio. 

For yttrium, cerium, neodymium and praseodymium over 90% of the element was converted into phosphate at a phosphate to metal mole ratio close to unity. A threefold excess of phosphate was required in the case of lanthanum and ytterbium. Amongst the elements studied so far, gadolinium and erbium required the highest P04 RE " mole ratio for complete conversion into phosphates, around 4 and 5, respectively. 

Strategy First (1), convert the masses of the three elements to moles. Knowing the number of moles (n) of K, Cr, and O, you can then (2) calculate the mole ratios. Finally (3), equate the mole ratio to the atom ratio, which gives you the simplest formula. 

SOLUTION We consider a sample of exactly 100 g, then convert the masses into I amounts in moles by dividing the mass percentage for each element by the element s molar mass. The mass of carbon in a sample of vitamin C of mass 100 g is 40.9 g. Because the molar mass of carbon is 12.01 g-moD1,... 

Convert each mass to an amount, ttx, in moles by using the molar mass, Mx, of the element, nx = mx/Mx. 

STRATEGY (a) To obtain the amount of iron(II) in the analyte, we use the volume and concentration of the titrant. We follow the first two steps of the procedure in Toolbox L.2. Then we convert moles of Fe2+ ions into mass by using the molar mass of Fe2+ because the mass of electrons is so small, we use the molar mass of elemental iron for the molar mass of iron(II) ions, (b) We divide the mass of iron by the mass of the ore sample and multiply by 100%. 

Avogadro s number and molar mass make it possible to convert readily among the mass of a pure element, the number of moles, and the number of atoms in the sample. These conversions are represented schematically in the flowchart shown in Figure 2-22. 

We are asked to compare an elemental analysis with a chemical formula. To do so, we can either convert mass percentages to mole amounts or convert the formula to mass percentages. It is easier to compute mass percentages and compare them with the measured values. 

The number of moles of each element in a mole of compound is stated in the chemical formula. Hence, the formula can be used to convert the number of moles of the compound to the number of moles of its component elements, and vice versa (Fig. 4-3). 

Ans. The figure is presented as Fig. 4-4. One can convert from mass to moles, moles of component elements, or number of formula units. Additionally, one can convert from number of formula units to moles, to moles of component elements, or to mass. Also from moles of component elements to moles of compound, number of formula units of compound, or mass of compound. Finally, from moles of compound to number of formula units, mass, or number of moles of component elements. 

The volumetric flow rate of the recycle stream is many many times those of the fresh feed and product streams, and the fresh feed and recycle streams are well mixed at the juncture point. If one uses a mole ratio of 3.4 hydrogen to 1 toluene in the fresh feed stream, what fraction of the toluene is converted to benzene under the previously specified conditions The average residence time of a fluid element is 30.1 sec. Explicitly state any assumptions that you make. In order to obtain a numerical answer, a trial and error solution will be necessary. 

SA To answer this question, we start with a 100.00 g sample of the compound. In this way, each elemental mass in grams is numerically equal to its percent. We convert each mass to an amount in moles, and then determine the simplest integer set of molar amounts. This determination begins by dividing all three molar amounts by the smallest. 

Convert each percentage into the mass in 100.00 g, and then to the moles of that element. 

Table 2 gives the compositions of the same four clay samples in atomic percent. The atomic percent is defined as the number of atoms of an element per unit volume divided by the number of atoms per unit volume of the substance containing the element. This is similar to mole fraction when the atomic percent is converted to fractional value. 

In the problem above, we determined the percentage data from the chemical formula. We can determine the empirical formula if we know the percent compositions of the various elements. 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. However, the procedure is still the same—convert each element to moles, divide each by the smallest, and then use an appropriate multiplier if necessary. We can then determine the empirical formula mass. If we know the actual molecular mass, dividing the molecular formula mass by the empirical formula mass, gives an integer (rounded if needed) that we can multiply each of the subscripts in the empirical formula. This gives the molecular (actual) formula, which tells what elements are in the compound and the actual number of each. 

The same mole ratio, a/d, can also be found by simply balancing the common element in the formulas of A and D. Thus, the ratio, a/d, is the same as the ratio QS/QK seen in the equation for a gravimetric factor derived in Section 3.6.3 (Equation (3.12)) and is used to convert the weight of one substance to the weight of another, just as we described in Chapter 3 was the purpose of the gravimetric factor. Thus the concept of a gravimetric factor is based on stoichiometry. 

When conversions are not small, the change in concentration of reactant and product over the length of the catalyst bed must be taken into account. It is therefore convenient to substitute for the concentrations in Eqs. (57) and (58) so that [B] Pb(1 fn /q) [N] - Pb/n [Q] Pb/q Also, in a flow system the number of moles of reactant converted in an element of catalyst bed is equal to the reaction rate per gram times the mass of the catalyst in that element of the bed. That is. 

Dimensionless current representations also exist. If the electrode is placed at the center of the first volume element in the model, this current will be proportional to the material flux into the first element from the second. For electroactive species A, this flux is given by DMA [fA(2) - fA(l)], where fA(J) is the fractional concentration of A in the Jth element. This fractional flux may be converted to moles of flux through multiplication by C (bulk concentration) and A Ax (volume-element volume, assuming a planar electrode of area A). Appropriate electrochemical conversion and the recognition that this material flux occurs during the interval At yield the current expression... 

We saw in Section 3.3 that the coefficients in a balanced equation tell the numbers of moles of substances in a reaction. In actual laboratory work, though, it s necessary to convert between moles and mass to be sure that the correct amounts of reactants are used. In referring to these mole-mass relationships, we use the word stoichiometry (stoy-key-ahm-uh-tree from the Greek stoicheion, "element," and metron, "measure"). Let s look again at the reaction of ethylene with HC1 to see how stoichiometric relationships are used. 

Most of the time that you can write an equation to describe what is happening, it is a good idea to do so. The equation is essentially a set of instructions telling us how to perform a process. Further, the equation tells us how much of each of the reactants to use and how much to expect of the products. These factors are provided by the coefficients and can be considered in moles of substances. If we have moles of substances, we can easily convert to grams of substances using the atomic masses of the atoms involved provided by the Periodic Table of the Elements. The equation for this reaction is... 

Since you know the percentage composition, it is convenient to assume that you have 100 g of the compound. This means that you have 85.6 g of carbon and 14.4 g of hydrogen. Convert each mass to moles. The number of moles can then be converted into a lowest terms ratio of the elements to get the empirical formula. 

---